Chapter 9

Recall that for exponential functions, the rate of growth is proportional to the current population as shown in(1) below. For logistic functions, the rate of growth is proportional to the current population times the differencebetween the carrying capacity c and the current population (2). For further analysis, this relationship can berewritten as in (3), with a technique known as partial fraction decomposition then applied. This technique ispresented in Section 9.4, and further explored in the Connections to Calculus feature for Chapter 9.

(1) (2) (3) ¢P

P1c � P2� k¢t

¢P

¢t� kP1c � P2

¢P

¢t� kP

Connectionsto Calculus

Precalculus—

Systems of Equationsand InequalitiesCHAPTER OUTLINE

9.1 Linear Systems in Two Variables with Applications 838

9.2 Linear Systems in Three Variables with Applications 853

9.3 Systems of Inequalities and Linear Programming 865

9.4 Partial Fraction Decomposition 879

9.5 Solving Linear Systems Using Matrices and Row Operations 893

9.6 The Algebra of Matrices 905

9.7 Solving Linear Systems Using Matrix Equations 917

9.8 Applications of Matrices and Deter minants: Cramer’s Rule,Geometry, and More 933

CHAPTER CONNECTIONSAt the tur n of the centur y, there was anexplosion in the number of handheld electr onicdevices available to consumers. One device inparticular, the popular MP3 player , experienceda phenomenal gr owth in demand. With highdemand and a lar ge market, competitionbetween manufactur ers and suppliers is oftenfie ce, with each fighting to ea n and hold ashare of the market. One significant factor iwho gains the lar gest shar e is the price char gedfor the player , with suppliers willing to supplymore at a gr eater price, and consumers willing tobuy more at a lesser price. Deter mining wherethe price will stabilize is an impor tant componentin the economics of “supply and demand. ” � This application occurs as Exercise 78 in

Section 9.1.

837

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9.1 Linear Systems in Two Variables with Applications

838 9–2

In earlier chapters, we used linear equations in two variables to model a number ofreal-world situations. Graphing these equations gave us a visual image of how thevariables were related, and helped us better understand this relationship. In manyapplications, two different measures of the independent variable must be consideredsimultaneously, leading to a system of two linear equations in two unknowns. Here,a graphical presentation once again supports a better understanding, as we explore sys-tems and their many applications.

A. Solutions to a System of Equations

A system of equations is a set of two or more equations for which a common solutionis sought. Systems are widely used to model and solve applications when the informa-tion given enables the relationship between variables to be stated in different ways. Forexample, consider an amusement park that brought in $3100 in revenue by charging$9.00 for adults and $5.00 for children, while selling 500 tickets. Using a for adult andc for children, we could write one equation modeling the number of tickets sold:

, and a second modeling the amount of revenue brought in:. To show that we’re considering both equations simultaneously, a

large “left brace” is used and the result is called a system of two equations in twovariables:

We note that both equations are linear and will have different slope values, so theirgraphs must intersect at some point. Since every point on a line satisfies the equationof that line, this point of intersection must satisfy both equations simultaneously and isthe solution to the system. The figure that accompanies Example 1 shows the point ofintersecion for this system is (150, 350).

EXAMPLE 1 � Verifying Solutions to a System

Verify that (150, 350) is a solution to

Solution � Substitute the 150 for a and 350 for c in each equation.

first equation second equation

✓ ✓

Since (150, 350) satisfies both equations, it is the solution to the system and wefind the park sold 150 adult tickets and 350 tickets for children.

Now try Exercises 7 through 18 �

To check the solution to Example 1 on a graphing calculator, recall that we can storevalues in any of the characters using the key. The home screen shown inFigure 9.1 shows that we’ve stored a value of 150 in alpha location A, and 350 in loca-tion C. The solution to a system can then be checked using these alpha keys and theoperations indicated, as shown in Figure 9. 2.

STOALPHA

3100 � 3100 500 � 500 911502 � 513502 � 3100 11502 � 13502 � 500

9a � 5c � 3100 a � c � 500

e a � c � 500

9a � 5c � 3100.

number of tickets

amount of revenuee

a � c � 500

9a � 5c � 3100

9a � 5c � 3100a � c � 500

Precalculus—

LEARNING OBJECTIVESIn Section 9.1 you will see

how we can:

A. Verify ordered pairsolutions

B. Solve linear systems bygraphing

C. Solve linear systems bysubstitution

D. Solve linear systems byelimination

E. Recognize inconsistentsystems and dependentsystems

F. Use a system ofequations to model andsolve applications

0

100

100 300 500 700

300

500

700

Chi

ldre

n

Adults

9a � 5c � 3100

a � c � 500

(150, 350)

a

c

cob19537_ch09_837-853.qxd 1/27/11 7:34 PM Page 838

B. Solving Systems Graphically

To solve a system of equations means we apply various methods in an attempt to findordered pair solutions. As Example 1 suggests, one method for finding solutions is tograph the system. Any method for graphing the lines can be employed, but to keep im-portant concepts fresh, the slope-intercept method is used here.

EXAMPLE 2 � Solving a System Graphically

Solve the system by graphing: .

Solution � First write each equation in slope-intercept form (solve for y):

For the first line, with y-intercept The second equation yields with asthe y-intercept. Both are then graphed on the grid asshown. The point of intersection appears to be (3, 1),and checking this point in both equations gives

substitute 3

✓for x and 1 for y

✓

This verifies that (3, 1) is the solution to the system.


Graphical solutions to a system of equations can be found using a graphing calculatorand the intersection-of-graphs method seen earlier. After solving for y and entering theequations on the screen (Figure 9.3), use the keystrokes (CALC)5:Intersect to determine the point of intersection, if it exists (Figure 9.4).

TRACE2ndY=

�5 � �5 9 � 9 �2132 � 112 � �5 4132 � 3112 � 9

�2x � y � �5 4x � 3y � 9

10, �52¢y¢x � 2

1

10, �32.¢y¢x � 4

3

• 4x � 3y � 9

�2x � y � �5S •

y �4

3x � 3

y � 2x � 5

e 4x � 3y � 9

�2x � y � �5

Precalculus—

9–3 Section 9.1 Linear Systems in Two Variables with Applications 839

A. You’ve just seen how

we can verify ordered pair

solutions

Figure 9.1 Figure 9.2

x

y

5�5

�5

5

(3, 1)

(0, �3)

(0, �5)

y � 2x � 5

y � x � 343

B. You’ve just seen how

we can solve linear systems by

graphing

Figure 9.3

�5 5

�5

5Figure 9.4

cob19537_ch09_837-853.qxd 1/27/11 7:34 PM Page 839

C. Solving Systems by Substitution

While a graphical approach best illustrates why the solution must be an ordered pair, itdoes have one obvious drawback—noninteger solutions are difficult to spot. The

ordered pair is the solution to but this would be difficult to

“pinpoint” as a precise location on a hand-drawn graph. To overcome this limitation,we next consider a method known as substitution. The method involves converting asystem of two equations in two variables into a single equation in one variable by

using an appropriate substitution. For the second equation says “y is two

more than x.” We reason that all points on this line are related this way, including thepoint where this line intersects the other. For this reason, we can substitute for yin the first equation, obtaining a single equation in x.

EXAMPLE 3 � Solving a System Using Substitution

Solve using substitution:

Solution � Since we can replace y with in the first equation.

first equation

substitute for y

simplify

result

The x-coordinate is To find the y-coordinate, we substitute for x into either ofthe original equations, a process known as back-substitution. Substituting in thesecond equation gives

second equation

substitute for x

The solution to the system is , whichcan be verified using the intersection-of-graphs method, and noting and

. See Figure 9.5. Recall that thekeystrokes 1:�FRAC can be used toconvert decimal numbers to fractions.


If neither equation allows an immediate substitution, we first solve for one of thevariables, either x or y, and then substitute. The method is summarized here, and canactually be used with either like variables or like variable expressions. See Exer-cises 33 to 36.

ENTERMATH

125 � 2.4

25 � 0.4

125, 125 2

21

�105

, 25

�105

�125

�12

5

25

�2

5� 2

y � x � 2

25

25.

x �2

5

5x � 2 � 4x � 2 4x � 1x � 22 � 4

4x � y � 4

x � 2y � x � 2,

e4x � y � 4

y � x � 2.

x � 2

e4x � y � 4

y � x � 2,

e4x � y � 4

y � x � 2,125, 12

5 2

Precalculus—

840 CHAPTER 9 Systems of Equations and Inequalities 9–4

�5 5

�5

5

Y1 � �4X � 4, Y2 � X � 2

Figure 9.5

cob19537_ch09_837-853.qxd 1/27/11 7:34 PM Page 840

Solving Systems Using Substitution

1. Solve one of the equations for x in terms of y or y in terms of x.2. Substitute for the appropriate variable in the other equation and solve for the

variable that remains.3. Substitute the value from step 2 into either of the original equations and solve

for the other unknown.4. Write the answer as an ordered pair and check the solution in both original

equations.

D. Solving Systems Using Elimination

Now consider the system where solving for any one of the

variables will result in fractional values. The substitution method can still be used, butoften the elimination method is more efficient. The method takes its name from whathappens when you add certain equations in a system (by adding the like terms fromeach). If the coefficients of either x or y are additive inverses—they sum to zero andare eliminated. For the system shown, “adding the equations” produces , giving

, then using back-substitution (verify).If neither of the like-variable terms sum to zero, we can multiply one or both equa-

tions by a nonzero constant to “match up” the coefficients, so an elimination will takeplace. In doing so, we create an equivalent system of equations, meaning one that has

the same solution as the original system. For , multiplying the

second equation by 2 produces , and after adding the equations,

we see that . Note the three systems produced are equivalent, and all have (4, 3)as a solution ( was found using back-substitution).

1. 2. 3.

In summary,

Operations that Produce an Equivalent System

1. Changing the order of the equations.2. Replacing an equation by a nonzero constant multiple of that equation.3. Replacing an equation with the sum of two equations from the system.

Before beginning a solution using elimination, check to make sure the equationsare written in the standard form so that like terms will appearabove/below each other. Throughout this chapter, we will use R1 to represent the equation in row 1 of the system, R2 to represent the equation in row 2, and so on. Thesedesignations are used to help describe and document the steps being used to solve asystem, as in Example 4 where indicates the first equation has been multiplied by two, with the result added to the second equation.

EXAMPLE 4 � Solving a System by Elimination

Solve using elimination: e2x � 3y � 7

6y � 5x � 4

2R1 � R2

Ax � By � C,

e7x � 4y � 16

x � 4e

7x � 4y � 16

�6x � 4y � �12e

7x � 4y � 16

�3x � 2y � �6

y � 3x � 4

e7x � 4y � 16

�6x � 4y � �12

e7x � 4y � 16

�3x � 2y � �6

x � 1y � 32y � 6

e�2x � 5y � 13

2x � 3y � �7,

Precalculus—


C. You’ve just seen how

we can solve linear systems by

substitution

cob19537_ch09_837-853.qxd 1/27/11 7:34 PM Page 841

Precalculus—


{add

solve for y

sum

{add

solve for x

sum

WORTHY OF NOTEAs the elimination method involvesadding two equations, it issometimes referred to as theaddition method for solvingsystems.

Solution � The second equation is not in standard form, so we rewrite the system as

. If we “added the equations” now, we would get

with neither variable eliminated. However, if we multiply both sides of the firstequation by 2, the y-coefficients will become additive inverses. The sum thenresults in an equation with x as the only unknown.

Substituting 2 for x back into either of the original equations yields Theordered pair solution is (2, ). A graphical check is shown in the figure.


The elimination method is summarized here. If either equation has fraction ordecimal coefficients, we can “clear” them using an appropriate constant multiplier.

Solving Systems Using Elimination

1. Write each equation in standard form: 2. Multiply one or both equations by a constant that will create coefficients of x

(or y) that are additive inverses.3. Combine the two equations using vertical addition and solve for the variable

that remains.4. Substitute the value from step 3 into either of the original equations and solve

for the other unknown.5. Write the answer as an ordered pair and check the solution in both original

equations.

EXAMPLE 5 � Solving a System Using Elimination

Solve using elimination:

Solution � Multiplying the first equation by 8 (8R1) and the second equation by 6 (6R2) willclear the fractions from each.

The x-terms can now be eliminated if we use .

3R1

�5R2�

15x � 18y � 6

�15x � 20y � �30

0x � 2y � �24

y � �12

3R1 � 1�5R22

8R1

6R2e

81 1

58 2x � 8

1 134 2y � 8

1 114 2

61 1

12 2x � 6

1 123 2y � 6112

S e5x � 6y � 2

3x � 4y � 6

e58x � 3

4y � 14

12x � 2

3y � 1.

Ax � By � C.

�1y � �1.

2R1�R2

4x � 6y � 14

5x � 6y � 4

9x � 0y � 18

9x � 18

x � 2

7x � 3y � 11,e2x � 3y � 7

5x � 6y � 4

�5 5

�5

5

Y1 �7 � 2X

�3Y2 �

4 � 5X6

,

cob19537_ch09_837-853.qxd 1/27/11 7:34 PM Page 842

sum

add

variables are eliminated

true statement

Substituting in either of the originalequations yields and the solution is

We can check this solution bygraphing, as we did for the solution to Example 4,or by substituting the x- and y-values into theoriginal equations, as we did for Example 1. Thecheck using substitution is shown in the figure.Note that X has scrolled out of view.


CAUTION � Be sure to multiply all terms (on both sides) of the equation when using a constant mul-tiplier. Also, note that for Example 5, we could have eliminated the y-terms using 2R1with �3R2.

E. Inconsistent and Dependent Systems

A system having at least one solution is called a consistent system. As seen in Exam-ple 2, if the lines have different slopes, they intersect at a single point and the systemhas exactly one solution. Here, the lines are independent of each other and the systemis called an independent system. If the lines have equal slopes and the same y-intercept,they are identical or coincident lines. Since one is right atop the other, they intersectat all points, and the system has an infinite number of solutions. Here, one line dependson the other and the system is called a dependent system. Using substitution or elim-ination on a dependent system results in the elimination of all variable terms and leavesa statement that is always true, such as or some other simple identity.

EXAMPLE 6 � Solving a Dependent System

Solve using elimination: .

Solution � Writing the system in standard form gives

. By applying we

can eliminate the variable x:

Although we didn’t expect it, both variables were eliminated and the finalstatement is true This indicates the system is dependent, which the graphverifies (the lines are coincident). Writing both equations in slope-intercept formshows they represent the same line.

μ

y � �3

4x � 3

y � �3

4x � 3

e4y � �3x � 12

8y � �6x � 24e

3x � 4y � 12

6x � 8y � 24

10 � 02.

�6x � 8y � �24

6x � 8y � 24

0x � 0y � 0

0 � 0

e�2R1

�R2

�2R1,e3x � 4y � 12

6x � 8y � 24

e3x � 4y � 12

6x � 24 � 8y

0 � 0

STO�14

1�14, �122.x � �14,

y � �12

Precalculus—


D. You’ve just seen how

we can solve linear systems

by elimination

x

y

5�5

�5

5

(0, 3)

(4, 0)

6x � 24 � 8y

3x � 4y � 12

cob19537_ch09_837-853.qxd 1/27/11 7:34 PM Page 843

Precalculus—


The solutions of a dependent system are often written in set notation as the setof ordered pairs (x, y), where y is a specified function of x. Here the solution wouldbe Using an ordered pair with an arbitrary variable, called a

parameter, is also common:


If we had attempted to solve the system in Example 6by graphing (Figure 9.6), we could be mislead intothinking something is wrong—because only one lineis visible (Figure 9.7). In this case, using the TABLEfeature of the calculator would help verify that the sys-tem is dependent. Since the ordered pair solutions areidentical (try scrolling through positive and negativevalues), the equations must be dependent (Figure 9.8).

Finally, if the lines have equal slopes and differenty-intercepts, they are parallel and the system will have no solution. A system without asolution is called an inconsistent system. An “inconsistent system” produces an “in-consistent answer,” such as or some other false statement when substitution orelimination is applied. In other words, all variable terms are once again eliminated, butthe remaining statement is false. A summary of the three possibilities is shown in Fig-ure 9.9 for arbitrary slope m and y-intercept (0, b).

12 � 0

ap, �3p

4� 3b.

5 1x, y2 0 y � �34x � 36.

Figure 9.6

�5 5

�5

5

Figure 9.7

Figure 9.8

E. You’ve just seen how

we can recognize inconsistent

systems and dependent

systems

y

One point in common

Independentm1 � m2

x

Dependentm1 � m2, b1 � b2

All points in common

x

y

Inconsistentm1 � m2, b1 � b2

x

No points in common

y

Figure 9.9

WORTHY OF NOTEWhen writing the solution to adependent system using aparameter, the solution can bewritten in many different ways. Forinstance, if we let for thefirst coordinate of the solution toExample 6, we have

as the

second coordinate, and thesolution becomes (4b, �3b � 3)for any constant b.

�314b2

4� 3 � �3b � 3

p � 4b

F. Systems and Modeling

In previous chapters, we solved numerous real-world applications by writing all givenrelationships in terms of a single variable. Many situations are easier to model using asystem of equations with each relationship modeled independently using two vari-ables. We begin here with a mixture application. Although they appear in many differ-ent forms (coin problems, metal alloys, investments, merchandising, and so on),mixture problems all have a similar theme. Generally one equation is related to quan-tity (how much of each item is being combined) and one equation is related to value(what is the value of each item being combined).

EXAMPLE 7 � Solving a Mixture ApplicationA jeweler is commissioned to create a piece of artwork that will weigh 14 oz andconsist of 75% gold. She has on hand two alloys that are 60% and 80% gold,respectively. How much of each should she use?

cob19537_ch09_837-853.qxd 1/27/11 7:34 PM 44

Precalculus—


�5 15

�5

15

Y1 � 14 � X, Y2 �105 � 6X

8

Solution � Let x represent ounces of the 60% alloy and y represent ounces of the 80% alloy.The first equation must be since the piece of art must weigh exactly14 oz (this is the quantity equation). The x ounces are 60% gold, the y ounces are80% gold, and the 14 oz will be 75% gold. This gives the value equation:

The system is (after clearing decimals).

Solving for y in the first equation gives Substituting for y inthe second equation gives

second equation

substitute for y

distribute

simplify

solve for x

Substituting for x in the first equation gives She should use 3.5 oz of the60% alloy and 10.5 oz of the 80% alloy. A graphical check is shown in the figure.


A second example involves an application of uniform motion (distance �rate time), and explores concepts of great importance to the navigation of ships andairplanes. As a simple illustration, if you’ve ever walked at your normal rate r on the“moving walkways” at an airport, you likely noticed an increase in your total speed.This is because the resulting speed combines your walking rate r with the speed w ofthe walkway: . If you walk in the opposite direction of the walk-way, your total speed is much slower, as now .

This same phenomenon is observed when an airplane is flying with or against thewind, or a ship is sailing with or against the current.

EXAMPLE 8 � Solving an Application of Systems — Uniform MotionAn airplane flying due south from St. Louis, Missouri, to Baton Rouge, Louisiana,uses a strong, steady tailwind to complete the trip in only 2.5 hr. On the return trip,the same wind slows the flight and it takes 3 hr to get back. If the flight distancebetween these cities is 912 km, what is the cruising speed of the airplane (speedwith no wind)? How fast is the wind blowing?

Solution � Let r represent the rate of the plane and w the rate of the wind. Since , theflight to Baton Rouge can be modeled by , and the return flight

by . This produces the system e912 � 2.5r � 2.5w

912 � 3r � 3w.912 � 1r � w2 132

912 � 1r � w2 12.52D � RT

total speed � r � wtotal speed � r � w

#

y � 212 .7

2

x �7

2

�2x � 112 � 105 6x � 112 � 8x � 105

14 � x 6x � 8114 � x2 � 105 6x � 8y � 105

14 � xy � 14 � x.

ex � y � 14

6x � 8y � 1050.6x � 0.8y � 0.751142.

x � y � 14,WORTHY OF NOTEAs an estimation tool, note that ifequal amounts of the 60% and80% alloys were used (7 oz each),the result would be a 70% alloy(halfway in between). Since a 75% alloy is needed, more of the80% gold will be used.

� Algebraic Solution � Graphical Solution

Using x for w and y for r, we solve each equation for y andobtain:

Y2 �912 � 3X

3 Y1 �

912 � 2.5X

2.5

Dividing R1 by 2.5 and R2 by 3 produces the following sequence:

e364.8 � r � w

304.0 � r � we

912 � 2.5r � 2.5w

912 � 3r � 3wS

R1

2.5

R2

3

cob19537_ch09_837-853.qxd 1/27/11 7:35 PM Page 845


Precalculus—

We then set an appropriate window and graph these equations to find the point of intersection.

The speed of the wind (x) is 30.4 kph, and the speed of the plane (y) is 334.4 kph.


0 50

200

400

Systems of equations also play a significant role in cost-based pricing in the busi-ness world. The costs involved in running a business can broadly be understood aseither a fixed cost k or a variable cost v. Fixed costs might include the monthly rentpaid for facilities, which remains the same regardless of how many items are producedand sold. Variable costs would include the cost of materials needed to produce the item,which depends on the number of items made. The total cost can then be modeled by

for x number of items. Once a selling price p has been determined, therevenue equation is simply (price times number of items sold). We can nowset up and solve a system of equations that will determine how many items must be soldto break even, performing what is called a break-even analysis where C(x) � R(x).

EXAMPLE 9 � Solving an Application of Systems: Break-Even AnalysisIn home businesses that produce items to sellon Ebay®, fixed costs are easily determined byrent and utilities, and variable costs by the priceof materials needed to produce the item.Karen’s home business makes large decorativecandles for all occasions. The cost of materialsis $3.50 per candle, and her rent and utilitiesaverage $900 per month. If her candles sell for$9.50, how many candles must be sold eachmonth to break even?

Solution � Let x represent the number of candles sold. Her total cost is (variable cost plus fixed cost), and projected revenue is . This gives the

system . To break even, Cost Revenue which gives

The analysis shows that Karen must sell 150 candles each month to break even.


x � 150 6x � 900

9.5x � 3.5x � 900

�eC1x2 � 3.5x � 900

R 1x2 � 9.5x

R1x2 � 9.5xC1x2 � 3.5x � 900

R1x2 � pxC1x2 � vx � k

Using gives , showing. The speed of the plane is 334.4 kph.

Substituting 334.4 for r in the second equation,we have:

equation

substitute

multiply

subtract 1003.2

divide by

The speed of the wind is 30.4 kph.

�3 30.4 � w �91.2 � �3w

912 � 1003.2 � 3w 912 � 31334.42 � 3w 912 � 3r � 3w

334.4 � r668.8 � 2rR1 � R2

cob19537_ch09_837-853.qxd 1/27/11 7:35 PM Page 846


Precalculus—

In a “free-market” economy, also referred to as a “supply-and-demand” econ-omy, there are naturally occurring forces that invariably come into play if no out-side forces act on the producers (suppliers) and consumers (demanders). Generallyspeaking, the higher the price of an item, the lower the demand. A good advertisingcampaign can increase the demand, but the increasing demand brings an increase inprice, which moderates the demand—and so it goes until a balance is reached.These free-market forces ebb and flow until market equilibrium occurs, at the spe-cific price where the supply and demand are equal.

In Exercises 75 to 78, the equation models were artificially constructed to yield a“nice” solution. In actual practice, the equations and coefficients are not so “well behaved” and are based on the collection and interpretation of real data. While marketanalysts have sophisticated programs and numerous models to help develop theseequations, here we’ll use our experience with regression to develop the supply anddemand curves.

EXAMPLE 10 � Using Technology to Find Market EquilibriumA manufacturer of MP3 players has hired aconsulting firm to do market research on their“next-generation” player. Over a 10-week period,the firm collected the data shown for the MP3player market (data includes MP3 players soldand expected to sell).

a. Use a graphing calculator to simultaneouslydisplay the demand and supply scatterplots.

b. Calculate a line of best fit for each and graphthem with the scatterplots (identify each curve).

c. Find the equilibrium point.

Solution � a. Begin by clearing all lists. This can be donemanually, or by pressing (MEM)and selecting option 4:ClrAllLists (thecommand appears on the home screen).Pressing will execute the command,and the word DONE will appear.Carefully input price in L1, demand inL2, and supply in L3 (see Figure 9.10).With the window settings given in Figure 9.11, pressing will displaythe price/demand and price/supplyscatterplots shown. If this is not the case,use (STAT PLOT) to be surethat “On” is highlighted in Plot1 andPlot2, and that Plot1 uses L1 and L2, while Plot2 uses L1 and L3 (Figure 9.12).Note we’ve chosen a different mark to indicate the data points in Plot2.

b. Calculate the linear regression equation for L1 and L2 (demand), and paste it inY1: LinReg (ax � b) L1, L2, Y1 . Next, calculate the linear regression for L1 and L3 (supply) and paste it in Y2: LinReg (ax � b) L1, L3, Y2 (recall thatY1 and Y2 are accessed using the key). The resulting equations and graphsare shown in Figures 9.13 and 9.14.

c. Once again we use (CALC) 5:intersect to find the equilibriumpoint, which is approximately (80, 9931). Supply and demand for this MP3player model are approximately equal at a price of about $80, with 9931 MP3players bought and sold.

TRACE2nd

VARS

ENTER

ENTER

Y=2nd

GRAPH

ENTER

+2nd

WORTHY OF NOTEThere are limitations to this model,and this interplay can be affectedby the number of availableconsumers, production limits,“shelf-life” issues, and so on, butat any given moment in the lifecycle of a product, consumerdemand responds to price in thisway. Producers also respond toprice in a very predictable way.

Figure 9.10 16,000

3000

40 115

Figure 9.11

Price Supply(dollars) Demand (Inventory)

107.10 6900 12,200

85.50 7900 9900

64.80 13,200 8000

52.20 13,500 7900

108.00 6700 14,000

91.80 7600 12,000

77.40 9200 9400

46.80 13,800 6100

74.70 10,600 8800

68.40 12,800 8600

Figure 9.12

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Precalculus—

Figure 9.13 16,000

40 115

3000

Figure 9.14


Other interesting applications can be found in the Exercise Set. See Exercises 83through 88.

F. You’ve just seen how

we can use a system of

equations to model and solve

applications

4. If the lines in a system are coincident, the system isreferred to as and .

� CONCEPTS AND VOCABULAR Y

Fill in the blank with the appropriate word or phrase. Carefully reread the section if needed.

1. Systems that have no solution are called systems.

2. Systems having at least one solution are calledsystems.

5. The given systems are equivalent. How do weobtain the second system from the first?

e4x � 3y � 10

2x � 4y � 10•

2

3 x �

1

2 y �

5

3

0.2x � 0.4y � 1

3. If the lines in a system intersect at a single point,the system is said to be and .

6. For

which solution method would be more efficient,substitution or elimination? Discuss/Explain why.

e2x � 5y � 8

3x � 4y � 5,

9.1 EXERCISES

11. C 12. D

13. E 14. F

Substitute the x- and y-values indicated by the orderedpair to determine if it is a solution to the system. Alsocheck using the keys on the home screen of agraphing calculator.

15.

16.

17. e8x � 24y � �17

12x � 30y � 2; a�

7

8,

5

12b

e3x � 7y � �4

7x � 8y � �21; 1�6, 22

e3x � y � 11

�5x � y � �13; 13, 22

ALPHA

� DEVELOPING YOUR SKILLS

Show the lines in each system would intersect in a singlepoint by writing the equations in slope-intercept form.

7. 8.

An ordered pair is a solution to anequation if it makes the equationtrue. Given the graph shown here,determine which equation(s) havethe indicated point as a solution. Ifthe point satisfies more than oneequation, write the system forwhich it is a solution.

9. A 10. B

e0.3x � 0.4y � 2

0.5x � 0.2y � �4e

7x � 4y � 24

4x � 3y � 15

3x � 2y � 6

x � 3y � �3

y � x � 2

B

CD

EF

A

x

y

5�5

�5

5

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Precalculus—

18.

Solve each system by graphing manually. Check resultsby graphing the system on a graphing calculator, andlocating any points of intersection.

19. 20.

21. 22.

Solve each system using substitution. Write solutions asan ordered pair, and verify solutions using a graphingcalculator.

23. 24.

25. 26.

Identify the equation and variable that makes thesubstitution method easiest to use, then solve the system.Verify solutions using a graphing calculator.

27. 28.

29. 30.

31. 32.

The substitution method can be used for like variablesor for like expressions. Solve the following systems, usingthe expression common to both equations (do not solvefor x or y alone).

33. 34.

35. 36. e�6x � 5y � 16

5y � 6x � 4e

5x � 11y � 21

11y � 5 � 8x

e8x � 3y � 24

8x � 5y � 36e

2x � 4y � 6

x � 12 � 4y

e2x � 5y � 5

8x � y � 6e

5x � 6y � 2

x � 2y � 6

e0.8x � y � 7.4

0.6x � 1.5y � 9.3e

0.7x � 2y � 5

x � 1.4y � 11.4

e3x � 2y � 19

x � 4y � �3e

3x � 4y � 24

5x � y � 17

e2x � y � 6

y � 34x � 1

ey � 2

3x � 7

3x � 2y � 19

e4x � 5y � 7

2x � 5 � ye

x � 5y � 9

x � 2y � �6

e3x � y � �2

5x � 3y � 2e

5x � 2y � 5

x � 3y � �16

e5x � 2y � �2

�3x � y � 10e

3x � 2y � 12

x � y � 9

e4x � 15y � 7

8x � 21y � 11; a

1

2,

1

3b

Solve using elimination. In some cases, the system mustfirst be written in standard form. Verify solutions usinga graphing calculator.

37. 38.

39. 40.

41. 42.

43. 44.

45.

46.

47. 48.

Solve using any method and identify the system asconsistent, inconsistent, or dependent. Verify solutionsusing a graphing calculator.

49. 50.

51. 52.

53. 54.

55. 56.

57. 58.

59. 60. e3p � 2q � 4

9p � 4q � �3e

4a � 2 � 3b

6b � 2a � 7

e�2m � 3n � �1

5m � 6n � 4e

7a � b � �25

2a � 5b � 14

e2x � 3y � 4

x � �2.5ye

�10x � 35y � �5

y � 0.25x

e15 � 5y � �9x

�3x � 53y � 5

e6x � 22 � �y

3x � 12y � 11

e1.2x � 0.4y � 5

0.5y � �1.5x � 2e

0.2y � 0.3x � 4

0.6x � 0.4y � �1

e23x � y � 2

2y � 56x � 9

e4x � 3

4y � 14

�9x � 58y � �13

e34x � 1

3y � �232x � 1

5y � 3e

�16u � 1

4v � 4

12u � 2

3v � �11

e0.06g � 0.35h � �0.67

�0.12g � 0.25h � 0.44

e0.32m � 0.12n � �1.44

�0.24m � 0.08n � 1.04

e0.2x � 0.3y � 0.8

0.3x � 0.4y � 1.3e

0.5x � 0.4y � 0.2

0.3y � 1.3 � 0.2x

e2y � 5x � 2

�4x � 17 � 6ye

2x � �3y � 17

4x � 5y � 12

e5y � 3x � �5

3x � 2y � 19e

4x � 3y � 1

3y � �5x � 19

e�x � 5y � 8

x � 2y � 6e

2x � 4y � 10

3x � 4y � 5

� WORKING WITH FORMULAS

61. Uniform motion with current:

The formula shown can be used to solve uniform motion problems involving a current, where D representsdistance traveled, R is the rate of the object with no current, C is the speed of the current, and T is the time.Chan-Li rows 9 mi up river (against the current) in 3 hr. It only took him 1 hr to row 5 mi downstream (with thecurrent). How fast was the current? How fast can he row in still water?

e1R � C2T1 � D1

1R � C2T2 � D2

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62. Fahrenheit and Celsius temperatures:

Many people are familiar with temperature measurement in degrees Celsius and degrees Fahrenheit, but few realizethat the equations are linear and there is one temperature at which the two scales agree. Solve the system using themethod of your choice and find this temperature.

ey � 9

5x � 32 �Fy � 5

9 1x � 322 �C

� APPLICATIONS

Solve each application by modeling the situation with alinear system. Be sure to clearly indicate what eachvariable represents. Check answers using a graphingcalculator and the method of your choice.

Mixture

63. Theater productions: At a recent production of AComedy of Errors, the Community Theater broughtin a total of $30,495 in revenue. If adult ticketswere $9 and children’s tickets were $6.50, howmany tickets of each type were sold if 3800 ticketsin all were sold?

64. Milkfat requirements: A dietician needs to mix 10gal of milk that is milkfat for the day’s rounds.He has some milk that is 4% milkfat and some thatis milkfat. How much of each should be used?

65. Filling the family cars: Cherokee just filled bothof the family vehicles at a service station. The totalcost for 20 gal of regular unleaded and 17 gal ofpremium unleaded was $144.89. The premium gaswas $0.10 more per gallon than the regular gas.Find the price per gallon for each type of gasoline.

66. Household cleaners: As a cleaning agent, a solutionthat is 24% vinegar is often used. How much pure(100%) vinegar and 5% vinegar must be mixed toobtain 50 oz of a 24% solution?

67. Alumni contributions: A wealthy alumnusdonated $10,000 to his alma mater. The collegeused the funds to make a loan to a science major at7% interest and a loan to a nursing student at 6%interest. That year the college earned $635 ininterest. How much was loaned to each student?

68. Investing in bonds: A total of $12,000 is investedin two municipal bonds, one paying 10.5% and theother 12% simple interest. Last year the annualinterest earned on the two investments was $1335.How much was invested at each rate?

69. Saving money: Bryan has been doing odd jobsaround the house, trying to earn enough money to buya new Dirt-Surfer©. He saves all quarters and dimesin his piggy bank, while he places all nickels andpennies in a drawer to spend. So far, he has 225 coins

112%

212%

in the piggy bank, worth a total of $45.00. How manyof the coins are quarters? How many are dimes?

70. Coin investments: In 1990, Molly attended a coinauction and purchased some rare “Flowing Hair”fifty-cent pieces, and a number of very rare two-cent pieces from the Civil War Era. If she bought47 coins with a face value of $10.06, how many ofeach denomination did she buy?

Uniform Motion

71. Canoeing on a stream: On a recent camping trip,it took Molly and Sharon 2 hr to row 4 mi upstreamfrom the drop in point to the campsite. After aleisurely weekend of camping, fishing, andrelaxation, they rowed back downstream to thedrop in point in just 30 min. Use this informationto find (a) the speed of the current and (b) thespeed Sharon and Molly would be rowing in stillwater.

72. Taking a luxury cruise: A luxury ship is taking aCaribbean cruise from Caracas, Venezuela, to just offthe coast of Belize City on the Yucatan Peninsula, adistance of 1435 mi. En route they encounter theCaribbean Current, which flows to the northwest,parallel to the coastline. From Caracas to the Belizecoast, the trip took 70 hr. After a few days of fun inthe sun, the ship leaves for Caracas, with the returntrip taking 82 hr. Use this information to find (a) thespeed of the Caribbean Current and (b) the cruisingspeed of the ship.

73. Airport walkways: As part of an algebra field trip,Jason takes his class to the airport to use theirmoving walkways for a demonstration. The classmeasures the longest walkway, which turns out to be 256 ft long. Using a stop watch, Jason shows ittakes him just 32 sec to complete the walk going inthe same direction as the walkway. Walking in adirection opposite the walkway, it takes him 320 sec—10 times as long! The next day in class,Jason hands out a two-question quiz: (1) What was thespeed of the walkway in feet per second? (2) What ismy (Jason’s) normal walking speed? Create theanswer key for this quiz.

Precalculus—


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74. Racing pigeons: The American Racing PigeonUnion often sponsors opportunities for owners tofly their birds in friendly competitions. During arecent competition, Steve’s birds were liberated inTopeka, Kansas, and headed almost due north totheir loft in Sioux Falls, South Dakota, a distanceof 308 mi. During the flight, they encountered asteady wind from the north and the trip took 4.4 hr.The next month, Steve took his birds to acompetition in Grand Forks, North Dakota, withthe birds heading almost due south to home, also adistance of 308 mi. This time the birds were aidedby the same wind from the north, and the trip tookonly 3.5 hr. Use this information to (a) find theracing speed of Steve’s birds and (b) find the speedof the wind.

75. Lawn service: Dave and his sons run a lawnservice, which includes mowing, edging, trimming,and aerating lawns. His fixed cost includesinsurance, his salary, and monthly payments onequipment, and amounts to $4000/mo. The variablecosts include gas, oil, hourly wages for hisemployees, and miscellaneous expenses, which runabout $75 per lawn. The average charge for full-service lawn care is $115 per visit. Do a break-even analysis to (a) determine how many lawnsDave must service each month to break even and(b) the revenue required to break even.

76. Production of mini-microwave ovens: Due tohigh market demand, a manufacturer decides tointroduce a new line of mini-microwave ovens forpersonal and office use. By using existing factoryspace and retraining some employees, fixed costsare estimated at $8400/mo. The components toassemble and test each microwave are expected torun $45 per unit. If market research showsconsumers are willing to pay at least $69 for thisproduct, find (a) how many units must be made andsold each month to break even and (b) the revenuerequired to break even.

77. Farm commodities: One area where the law ofsupply and demand is clearly at work is farmcommodities. Both growers and consumers watchthis relationship closely, and use data collected bygovernment agencies to track the relationship andmake adjustments, as when a farmer decides toconvert a large portion of her farmland from cornto soybeans to improve profits. Suppose that for x billion bushels of soybeans, supply is modeled by

where y is the current market price(in dollars per bushel). The related demandequation might be (a) Howmany billion bushels will be supplied at a marketprice of $5.40? What will the demand be at this

y � �2.20x � 12.

y � 1.5x � 3,

price? Is supply less than demand? (b) How manybillion bushels will be supplied at a market priceof $7.05? What will the demand be at this price?Is demand less than supply? (c) To the nearestcent, at what price does the market reachequilibrium? How many bushels are beingsupplied/demanded?

78. Digital media: Marketresearch has indicatedthat by 2015, sales ofMP3 players and similarproducts will mushroominto a $70 billion dollarmarket. With a marketthis large, competition isoften fierce—withsuppliers fighting toearn and hold market shares. For x million MP3players sold, supply is modeled by where y is the current market price (in dollars). The related demand equation might be

(a) How many million MP3players will be supplied at a market price of $88?What will the demand be at this price? Is supply lessthan demand? (b) How many million MP3 playerswill be supplied at a market price of $114? What willthe demand be at this price? Is demand less thansupply? (c) To the nearest cent, at what price does themarket reach equilibrium? How many units are beingsupplied/demanded?

79. Pricing wakeboards: A water sports company that manufactures high-end wakeboards has hiredan outside consulting firm to do some marketresearch on their best wakeboard. This consultingfirm collected the following supply and demanddata for this and comparable wakeboards over a10-week period. Find the equilibrium point. Roundyour answer to the nearest integer and dollar.

y � �5.20x � 140.

y � 10.5x � 25,

Average Price Quantity Available (in U.S. dollars) Demanded Inventory

424.85 175 232

445.25 166 247

389.55 291 215

349.98 391 201

402.22 218 226

413.87 200 222

481.73 139 251

419.45 177 235

397.05 220 219

361.90 317 212

Precalculus—


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80. Pricing pet care products: A metal shop thatmanufactures pens for pet rabbits has collectedsome data on sales and production over the past 8 weeks. The following table shows the supply anddemand data for these pens. Find the equilibriumpoint (round to the nearest cent and whole cage).

81. Tracking supply and demand—oil products:The U.S. Bureau of Labor and Statistics tracksimportant data from many different markets. InMay 2008, it collected the following supply-and-demand data for refined gasoline. Data werecollected every Tuesday and Friday. Find theequilibrium point, rounding your answer to thenearest hundred thousand gallons and whole cent.

Average Price Quantity Demanded Available Inventory (in U.S. dollars) ( gal) ( gal)

3.17 8.82 9.10

3.12 8.87 9.05

3.04 9.08 8.97

2.84 9.22 8.91

3.11 8.92 9.02

3.15 8.76 9.08

3.10 9.01 8.99

3.11 8.94 9.01

2.93 9.13 8.93

1 � 1071 � 107

Average Price Quantity Demanded Available Inventory (in U.S. dollars) (in millions) (in millions)

9.40 0.84 1.23

8.51 1.17 0.95

8.78 1.05 1.11

10.82 0.68 1.29

6.77 1.47 0.77

9.33 0.91 1.21

8.34 1.25 0.88

10.37 0.76 1.27

8.62 1.09 1.02

8.44 1.21 0.92

8.58 1.18 0.97

8.96 1.01 1.17

82. Tracking supply and demand—energy efficientlightbulbs: The U.S. Bureau of Labor andStatistics has collected the following supply anddemand data for the energy-efficient fluorescentlightbulbs sold each month for the past year. Findthe equilibrium point, rounding your answer to thenearest ten thousand lightbulbs and whole cent.What is the yearly demand at the equilibrium point?

Descriptive Translation

83. Important dates in U.S. history: If you sum theyear that the Declaration of Independence wassigned and the year that the Civil War ended, youget 3641. There are 89 yr that separate the twoevents. What year was the Declaration signed?What year did the Civil War end?

84. Architecturalwonders: When itwas firstconstructed in1889, the EiffelTower in Paris,France, was thetallest structure inthe world. In1975, the CNTower in Toronto, Canada, became the world’s talleststructure. The CN Tower is 153 ft less than twice theheight of the Eiffel Tower, and the sum of theirheights is 2799 ft. How tall is each tower?

85. Pacific islands land area: In the South Pacific, theisland nations of Tahiti and Tonga have a combinedland area of 692 mi2. Tahiti’s land area is 112 mi2

more than Tonga’s. What is the land area of eachisland group?

86. Card games: On a cold winter night, in the lobbyof a beautiful hotel in Sante Fe, New Mexico, Marcand Klay just barely beat John and Steve in a closegame of Trumps. If the sum of the team scores was990 points, and there was a 12-point margin ofvictory, what was the final score?

Average Price Quantity Sold Production (in U.S. dollars) (Demand) (Supply)

22.99 12 7

21.49 14 6

23.99 11 7

26.99 9 11

25.99 8 10

27.99 8 13

24.49 10 9

26.49 9 11


Precalculus—

cob19537_ch09_837-853.qxd 1/27/11 7:35 PM Page 852

Given any two points, the equation of a line through these points can be found using a system of equations. Whilethere are certainly more efficient methods, using a system here will lead to finding equations for polynomials ofhigher degree. The key is to note that each point will yield an equation of the form . For instance, the

points (3, 6) and yield the system

87. Use a system of equations to find the equation of the line containing the points (2, 7) and .

88. Use a system of equations to find the equation of the line containing the points and .1�3, 7219, �12

1�4, �52

e6 � 3m � b

�4 � �2m � b.1�2, �42

y � mx � b

� EXTENDING THE CONCEPT

89. Federal income tax reform has been a hot politicaltopic for many years. Suppose tax plan A calls fora flat tax of 20% tax on all income (no deductionsor loopholes). Tax plan B requires taxpayers to pay$5000 plus 10% of all income. For what incomelevel do both plans require the same tax?

90. Suppose a certain amount of money was invested at6% per year, and another amount at 8.5% per year,with a total return of $1250. If the amountsinvested at each rate were switched, the yearlyincome would have been $1375. To the nearestwhole dollar, how much was invested at each rate?

� MAINTAINING YOUR SKILLS

91. (2.2) Given the toolbox function sketchthe graph of

92. (6.1) Find two positive and two negative anglesthat are coterminal with � � 112°.

F1x2 � ��x � 3� �2.f 1x2 � �x�, 93. (5.5) Solve for x (rounded to the nearest

thousandth):

94. (7.2) Verify that is an

identity.

sin x � csc xcsc x

� �cos2x

33 � 77.5e�0.0052x � 8.37

9.2 Linear Systems in Three Variables with Applications

The transition to systems of three equations in three variables requires a fair amount of“visual gymnastics” along with good organizational skills. Although the techniquesused are identical and similar results are obtained, the third equation and variable giveus more to track, and we must work more carefully toward the solution.

A. Visualizing Solutions in Three Dimensions

The solution to an equation in one variable is the single number that satisfies the equa-tion. For the solution is and its graph is a single point on the numberline, a one-dimensional graph. The solution to an equation in two variables, such as

is an ordered pair (x, y) that satisfies the equation. When we graph thissolution set, the result is a line on the xy-coordinate grid, a two-dimensional graph.The solutions to an equation in three variables, such as are the orderedtriples (x, y, z) that satisfy the equation. When we graph this solution set, the result is aplane in space, a graph in three dimensions. Recall a plane is a flat surface having infi-nite length and width, but no depth. We can graph this plane using the interceptmethod and the result is shown in Figure 9.15. For graphs in three dimensions, thexy-plane is parallel to the ground (the y-axis points to the right) and z is the verticalaxis. To find an additional point on this plane, we use any three numbers whose sum is 6,such as (2, 3, 1). Move 2 units along the x-axis, 3 units parallel to the y-axis, and 1 unitparallel to the z-axis, as shown in Figure 9.16.

x � y � z � 6,

x � y � 3,

x � 2x � 1 � 3,


how we can:

A. Visualize a solution inthree dimensions

B. Check ordered triplesolutions

C. Solve linear systems inthree variables

D. Recognize inconsistentand dependent systems

E. Use a system of threeequations in threevariables to solveapplications

9–17 Section 9.2 Linear Systems in Three Variables with Applications 853

Precalculus—

cob19537_ch09_837-853.qxd 1/27/11 7:36 PM Page 853


Precalculus—

WORTHY OF NOTEWe can visualize the location of apoint in space by considering alarge rectangular box 2 ft long �3 ft wide � 1 ft tall, placed snuglyin the corner of a room. The floor isthe xy-plane, one wall is the xz-plane, and the other wall is theyz-plane. The z-axis is formedwhere the two walls meet and thecorner of the room is the origin (0, 0, 0). To find the corner of thebox located at (2, 3, 1), first locatethe point (2, 3) in the xy-plane (thefloor), then move up 1 ft.

y

x

z

(0, 0, 6)

(6, 0, 0)

(0, 6, 0)

Figure 9.15

y

x

z

(0, 0, 6)

(6, 0, 0)

(0, 6, 0)2 units along x

(2, 3, 1)3 units parallel y

1 unit parallel z

Figure 9.16

EXAMPLE 1 � Finding Solutions to an Equation in Three V ariablesUse a guess-and-check method to find four additional points on the planedetermined by

Solution � We can begin by letting then use any combination of y and z that sum to 6.Two examples are (0, 2, 4) and (0, 5, 1). We could also select any two values for xand y, then determine a value for z that results in a sum of 6. Two examples are

and


B. Solutions to a System of Three Equations in Three Variables

When solving a system of three equations in three variables, remember each equationrepresents a plane in space. These planes can intersect in various ways, creating differentpossibilities for a solution set (see Figures 9.17 to 9.20). The system could have a uniquesolution (a, b, c), if the planes intersect at a single point (Figure 9.17) (the point satisfiesall three equations simultaneously). If the planes intersect in a line (Figure 9.18), the sys-tem is linearly dependent and there is an infinite number of solutions. Unlike the two-dimensional case, the equation of a line in three dimensions is somewhat complex, and thecoordinates of all points on this line are usually represented by a specialized ordered triple,which we use to state the solution set. If the planes intersect at all points, the system hascoincident dependence (see Figure 9.19). This indicates the equations of the system differ by only a constant multiple—they are all “disguised forms” of the same equation.The solution set is any ordered triple (a, b, c) satisfying this equation. Finally, the systemmay have no solutions. This can happen a number of different ways, most notably ifthe planes either intersect or are parallel, as shown in Figure 9.20 (other possibilitiesare discussed in the exercises). In the case of “no solutions,” an ordered triple maysatisfy none of the equations, only one of the equations, only two of the equations, butnot all three equations.

18, �3, 12.1�2, 9, �12

x � 0,x � y � z � 6.

Figure 9.17 Figure 9.18 Figure 9.19 Figure 9.20

Unique solution Lineardependence

Coincidentdependence

No solutions


we can visualize a solution in

three dimensions

cob19537_ch09_854-864.qxd 1/27/11 7:41 PM Page 854


Precalculus—

EXAMPLE 2 � Determining If an Ordered Triple Is a SolutionDetermine if the ordered triple is a solution to the systems shown.

a. b.

Solution � Substitute 1 for x, for y, and 3 for z in the first system.

a.

No, the ordered triple is not a solution to the first system. Now usethe same substitutions in the second system.

b.

The ordered triple is a solution to the second system only.

Now try Exercises 11 and 12 �

As with systems of two equations in two variables, solutions to larger systems can alsobe checked by storing values using the keys, and entering each equation on thehome screen of a graphing calculator. The screens shown in Figures 9.21 and 9.22illustrate this process for Example 2(b).

ALPHA

11, �2, 32

•

3x � 2y � z � �42x � 3y � 2z � 2

x � y � 2z � 9S •

3112 � 21�22 � 132 � �42112 � 31�22 � 2132 � 2112 � 1�22 � 2132 � 9

S •

�4 � �4 true2 � 2 true9 � 9 true

11, �2, 32

•

x � 4y � z ��102x � 5y � 8z � 4

x � 2y � 3z ��4S •

112 � 41�22 � 132 ��102112 � 51�22 � 8132 � 4112 � 21�22 � 3132 ��4

S •�10 ��10 true 16 � 4 false �4 ��4 true

�2

•

3x � 2y � z � �42x � 3y � 2z � 2

x � y � 2z � 9•

x � 4y � z � �102x � 5y � 8z � 4

x � 2y � 3z � �4

11, �2, 32


we can check ordered triple

solutions

C. Solving Systems of Three Equations in Three Variables

Using Elimination

From Section 9.1, we know that two systems of equations are equivalent if they havethe same solution set. The systems

are equivalent, as both have the unique solution In addition, it is evident thatthe second system can be solved more easily, since R2 and R3 have fewer variablesthan the first system. In the simpler system, mentally substituting 1 for z into R2 im-mediately gives and these values can be back-substituted into the first equationto find that This observation guides us to a general approach for solving largersystems—we would like to eliminate variables in the second and third equations, un-til we obtain an equivalent system that can easily be solved by back-substitution. Tobegin, let’s review the three operations that “transform” a given system, and producean equivalent system.

x � �3.y � 1,

1�3, 1, 12.

and • 2x � y � 2z � �7 y � 4z � 5 z � 1

•

2x � y � 2z � �7 x � y � z � �1 �2y � z � �3


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Precalculus—

Operations That Produce an Equivalent System

1. Changing the order of the equations.2. Replacing an equation by a nonzero constant multiple of that equation.3. Replacing an equation with the sum of two equations from the system.

Building on the ideas from Section 9.1, we develop the following approach forsolving a system of three equations in three variables.

Solving a System of Three Equations in Three V ariables

1. Write each equation in standard form: 2. If the “x” term in any equation has a coefficient of 1, interchange equations

(if necessary) so this equation becomes R1.3. Use the x-term in R1 to eliminate the x-terms from R2 and R3. The original

R1, with the new R2 and R3, form an equivalent system that contains asmaller “subsystem” of two equations in two variables.

4. Solve the subsystem for either x or y and keep the result as the new R3. Theresult is an equivalent system that can be solved using back-substitution.

We’ll begin by solving the system using the elimination

method and the procedure outlined. In Example 3, the notation indicates the equation in row 1 has been multiplied by and added to the equation inrow 2, with the result placed in the system as the new row 2.

�2�2R1 � R2 S R2

•

2x � y � 2z � �7 x � y � z � �1 �2y � z � �3

Ax � By � Cz � D.

EXAMPLE 3 � Solving a System of Three Equations in Three V ariables

Solve using elimination: .

Solution � 1. The system is in standard form.2. If the x-term in any equation has a coefficient of 1, interchange equations so

this equation becomes R1.

3. Use R1 to eliminate the x-term in R2 and R3. Since R3 has no x-term, the onlyelimination needed is the x-term from R2. Using will eliminatethis term:

The new R2 is The original R1 and R3, along with the new R2form an equivalent system that contains a smaller subsystem

•

x � y � z � �1y � 4z � 5

�2y � z � �3S �2R1 � R2 S R2

R3 S R3 • x � y � z � �1 2x � y � 2z � �7

�2y � z � �3

y � 4z � 5.

�2R1 �2x � 2y � 2z � 2 � R2 2x � y � 2z � �7 0x � 1y � 4z � �5 y � 4z � 5

�2R1 � R2

• x � y � z � �1 2x � y � 2z � �7

�2y � z � �3 SR2 4 R1 •

2x � y � 2z � �7 x � y � z � �1 �2y � z � �3

•

2x � y � 2z � �7 x � y � z � �1 �2y � z � �3

sum

simplify

newequivalentsystem

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Precalculus—

4. Solve the subsystem for either y or z, and keep the result as a new R3. Wechoose to eliminate y using

The new R3 is

The new R3, along with the original R1 and R2 from step 3, form anequivalent system that can be solved using back-substitution. Substituting 1 for z in R2 yields y � 1. Substituting 1 for z and 1 for y in R1 yields The solution is Check this result using the keys, as illustratedfollowing Example 2.


While not absolutely needed for the elimination process, there are two reasons forwanting the coefficient of x to be “1” in R1. First, it makes the elimination methodmore efficient since we can more easily see what to use as a multiplier. Second, it laysthe foundation for developing other methods of solving larger systems. If no equationhas an x-coefficient of 1, we simply use the y- or z-variable instead (see Example 7).Since solutions to larger systems generally are worked out in stages, we will some-times track the transformations used by writing them between the original system andthe equivalent system, rather than to the left as we did in Section 9.1.

Here is an additional example illustrating the elimination process, but in abbrevi-ated form. Verify the calculations indicated using a separate sheet.

EXAMPLE 4 � Solving a System of Three Equations in Three V ariables


Solution � 1. Write the equations in standard form:

2.

3. Using will eliminate the x-term from R2, yielding Using eliminates the x-term from R3, yielding

4. Using will eliminate z from the subsystem, leaving

•

x � 3y � z � 55y � 2z � 18

�11y � z � �18 �2R3 � R2 S R3 •

x � 3y � z � 55y � 2z � 18

27y � 54

27y � 54.�2R3 � R2

•

x � 3y � z � 5�x � 2y � 3z � 132x � 5y � z � �8

R1 � R2 S R2

�2R1 � R3 S R3 •x � 3y � z � 5

5y � 2z � 18�11y � z � �18

�11y � z � �18.�2R1 � R35y � 2z � 18.R1 � R2

•

2x � 5y � z � �8�x � 2y � 3z � 13

x � 3y � z � 5 R3 4 R1 •

x � 3y � z � 5�x � 2y � 3z � 13 2x � 5y � z � �8

•

2x � 5y � z � �8�x � 2y � 3z � 13 x � 3y � z � 5

•

�5y � 2x � z � �8�x � 3z � 2y � 13�z � 3y � x � 5

.

ALPHA1�3, 1, 12.x � �3.

•

x � y � z � �1y � 4z � 5

z � 1S 2R2 � R3 S R3 •

x � y � z � �1y � 4z � 5

�2y � z � �3

z � 1.

2R2 2y � 8z � 10 � R3 �2y � z � �3 0y � 7z � 7 z � 1

2R2 � R3:

S

S

S

equivalentsystem

equivalentsystem

equivalentsystem

sum

simplify

newequivalentsystem

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Precalculus—

Solving for y in R3 yields: , showing y � 2. Substituting 2 for y in R2 gives,

substitute 2 for y

simplify

subtract 10

divide by 2

Substituting 2 for y and 4 for z in R1 gives,substitute 2 for y, 4 for z

simplify

subtract 2

The solution is (3, 2, 4).


D. Inconsistent and Dependent Systems

As mentioned, it is possible for larger systems to have no solutions or an infinite num-ber of solutions. As with our work in Section 9.1, an inconsistent system (no solutions)will produce inconsistent results, ending with a statement such as or someother contradiction.

EXAMPLE 5 � Attempting to Solve an Inconsistent System


Solution � 1. This system has no equation where the coefficient of x is 1.2. We can still use R1 to begin the solution process, but this time we’ll use the

variable y since it does have coefficient 1.Using eliminates the y-term from R2, leaving Butusing to eliminate the y-term from R3 results in a contradiction:�2R1 � R3

7x � 2z � �4.2R1 � R2

•

2x � y � 3z � �33x � 2y � 4z � 24x � 2y � 6z � �7

.

0 � �3

x � 3 x � 2 � 5

x � 3122 � 4 � 5

z � 4 2z � 8

10 � 2z � 18 5122 � 2z � 18

27y

27 �5427


we can solve linear systems in

three variables

We conclude the system is inconsistent. The answer is the empty set andwe need work no further.


Unlike our work with systems having only two variables, systems in three variablescan have two forms of dependence— linear dependence (Figure 9.18) or coincidentdependence (Figure 9.19). To help understand linear dependence, consider a system oftwo equations in three variables: Each of these equationsrepresents a plane, and unless the planes are parallel, their intersection will be a line. Asin Section 9.1, we can state solutions to a dependent system using set notation with twoof the variables written in terms of the third, or as an ordered triple using a parameter.The relationships named can then be used to generate specific solutions to the system.

e�2x � 3y � z � 5

x � 3y � 2z � �1.

�,

�4x � 2y � 6z � 6

4x � 2y � 6z � �70x � 0y � 0z � �1

0 � �1

�2R1�R3

2R1

�R2

4x � 2y � 6z � �6

3x � 2y � 4z � 27x � 2z � �4

contradiction

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Precalculus—

Systems with two equations and two variables or three equations and three vari-ables are called square systems, meaning there are exactly as many equations as thereare variables. A system of linear equations cannot have a unique solution unless thereare at least as many equations as there are variables in the system.



Solution � We immediately note that eliminates the y-term from R2, yielding and the new system . This means (x, y, z) will

satisfy both equations only when (the x-coordinate must be 4 less thanthe z-coordinate). Since x is written in terms of z, we substitute for x ineither equation to find how y is related to z. Using R2 in the original system wehave: , which yields (verify). This means they-coordinate of the solution must be more than . In set notation, the solution is

. Randomly choosing 2, and 7,the solutions would be , respectively. Verifythat these satisfy both equations. Using p as a parameter, the solution could bewritten in parameterized form.


The system in Example 6 was nonsquare, and we knew ahead of time the systemwould be dependent. The system in Example 7 is square, but only by applying theelimination process can we determine the nature of its solution(s).



Solution � This system has no equation where the coefficient of x is 1. We will still use R1,but eliminate z in R2 (there is no z-term in R3).

Using eliminates the z-term from R2, yielding

We next solve the subsystem. Using eliminates the y-term in R3, butalso all other terms:

Since R3 is the same as 2R2, the system is linearly dependent and equivalent toWe can solve for y in R2 to write y in terms of x:

Substituting for y in R1 enables us to also write z in terms of x:5x � 4y � 5x � 4.e

3x � 2y � z � �15x � y � 4 .

�10x � 2y ��810x � 2y � 80x � 0y � 0

0 � 0

�2R2�R3

�2R2 � R3

•

3x � 2y � z � �12x � y � z � 5

10x � 2y � 8

R1 � R2 S R2R3 S R3 •

3x � 2y � z � �15x � y � 4

10x � 2y � 8

5x � y � 4.R1 � R2

•

3x � 2y � z � �12x � y � z � 5

10x � 2y � 8.

1 p � 4, 15p � 35, p2

1�7, 0, �32, 1�2, 1, 22, and 13, 2, 72z � �3,5 1x, y, z,2 | x � z � 4, y � 1

5z � 35, z � �6

15z3

5y � 1

5z � 35�1z � 42 � 5y � 2z � 1

z � 4x � z � 4e

2x � 5y � 3z � �5x � z � �4x � z � �4

R1 � R2

e2x � 5y � 3z � �5

�x � 5y � 2z � 1 .

S

sum

result

cob19537_ch09_854-864.qxd 1/27/11 7:42 PM Page 859


Precalculus—

R1

substitute for y

distribute

simplify

solve for z

The solution set is Three of the infinite number of solutions are for for

Verify these triples satisfy all three equations. Againusing the parameter p, the solution could be written as .


For coincident dependence the equations in a system differ by only a constantmultiple. After applying the elimination process—all variables are eliminated from theother equations, leaving statements that are always true (such as or some other).For additional practice solving various kinds of systems, see Exercises 31 through 44.

E. Applications

Applications of larger systems are simply an extension of our work with systems oftwo equations in two variables. Once again, the applications come in a variety of formsand from many fields. In the world of business and finance, systems can be used todiversify investments or spread out liabilities, a financial strategy hinted at in Example 8.

EXAMPLE 8 � Modeling the Finances of a BusinessA small business borrowed $225,000 from three different lenders to expand theirproduct line. The interest rates were 5%, 6%, and 7%. Find how much wasborrowed at each rate if the annual interest came to $13,000 and twice as muchwas borrowed at the 5% rate than was borrowed at the 7% rate.

Solution � Let x, y, and z represent the amounts borrowed at 5%, 6%, and 7%, respectively.This means our first equation is (in thousands). The secondequation is determined by the total interest paid, which was $13,000:

The third is found by carefully reading the problem.“twice as much was borrowed at the 5% rate than was borrowed at the 7% rate,”or

These equations form the system: The x-term of

the first equation has a coefficient of 1. Written in standard form we have:

multiply R2 by 100

Using will eliminate the x term in R2, while willeliminate the x-term in R3.

�R1 �x � y � z � �225�R3 x � 2z � 0

�y � 3z � �225

�5R1 �5x � 5y � 5z � �1125�R2 5x � 6y � 7z � 1300

y � 2z � 175

�R1 � R3�5R1 � R2

•

x � y � z � 2255x � 6y � 7z � 1300

x � 2z � 0

R1R2R3

•

x � y � z � 2250.05x � 0.06y � 0.07z � 13x � 2z

.

x � 2z.

0.05x � 0.06y � 0.07z � 13.

x � y � z � 225

2 � 2

1 p, 5p � 4, 7p � 921�1, �9, �162 for x � �1.

x � 2, andx � 0, 12, 6, 5210, �4, �925 1x, y, z2 |x � �, y � 5x � 4, z � 7x � 96.

z � 7x � 9 �7x � z � �9

3x � 10x � 8 � z � �15x � 4 3x � 215x � 42 � z � �1

3x � 2y � z � �1



and dependent systems

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Precalculus—

The new R2 is and the new R3 (after multiplying by ) is

yielding the equivalent system

Solving the subsystem using yields Back-substitutionshows and yielding the solution (100, 75, 50). This means $50,000was borrowed at the 7% rate, $75,000 was borrowed at 6%, and $100,000 at 5%.

A “substitution check” on a graphing calculator verifies this solution is correct(see Figures 9.23 and 9.24).

x � 100,y � 75z � 50.�R2 � R32 � 2

•

x � y � z � 225y � 2z � 175y � 3z � 225

.y � 3z � 225,

�1y � 2z � 175,



we can use a system of three

equations in three variables to

solve applications


4. If a system is linearly dependent, theordered triple solutions can be written in terms of asingle variable called a(n) .

3 � 3



1. The solution to an equation in three variables is anordered .

2. The graph of the solutions to an equation in threevariables is a(n) .

5. Find a value of z that makes the ordered triplea solution to

Discuss/Explain how this is accomplished.2x � y � z � 4.12, �5, z2

3. Systems that have the same solution set are called.

6. Explain the difference between linear dependenceand coincident dependence, and describe how theequations are related.

9.2 EXERCISES

Use the keys on a graphing calculator to determineif the given ordered triples are solutions to the system. Ifnot a solution, identify which equation(s) are not satisfied.

11.

12. •2x � 3y � z � 95x � 2y � z � �32; 1�4, 5, 22x � y � 2z � �13 15, �4, 112

•

x � y � 2z � �14x � y � 3z � 3 ; 10, 3, 223x � 2y � z � 4 1�3, 4, 12

ALPHA


Find any four ordered triples that satisfy the equationgiven.

7.

8.

9.

10. 2x � y � 3z � �12�x � y � 2z � �63x � y � z � 8x � 2y � z � 9

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Precalculus—

Solve each system using elimination and back-substitution.

13. 14.

15. 16.

17. 18.

19. 20.

Solve using the elimination method. If a system isinconsistent or dependent, so state. For systems withlinear dependence, write solutions in set notation and asan ordered triple in terms of a parameter.

21. 22.

23. 24.

25. 26.

Solve using elimination. If the system is linearlydependent, state the general solution in terms of aparameter (different forms of the solution are possible).Then find four specific ordered triples that satisfy thesystem (answers will vary).

27.

28.

29.

30. •�2x � 3y � 5z � 3

5x � 7y � 12z � �8x � y � 2z � �2

•

x � 2y � 3z � 13x � 5y � 8z � 7

x � y � 2z � 5

•

5x � 3y � 2z � 4�9x � 5y � 4z � �12�3x � y � 2z � �12

•

3x � 4y � 5z � 5�x � 2y � 3z � �3 3x � 2y � z � 1

e2x � 4y � 5z � �23x � 2y � 3z � 7e

6x � 3y � 7z � 23x � 4y � z � 6

e4x � y � 2z � 93x � y � 5z � 5e

4x � y � 3z � 8x � 2y � 3z � 2

•

2x � y � 3z � 83x � 4y � z � 4

�4x � 2y � 6z � 5•

3x � y � 2z � 3x � 2y � 3z � 1

4x � 8y � 12z � 7

•

3x � y � z � 62x � 2y � z � 52x � y � z � 5

•

2x � 3y � 2z � 03x � 4y � z � �20x � 2y � z � 16

•

x � y � 2z ��14x � y � 3z �33x � 2y � z � 4

•

�x � y � 2z � �10x � y � z � 7

2x � y � z � 5

•

�x � y � 5z � 14x � y � 1

�3x � 2y � 8•

x � 3y � 2z �16�2y � 3z � 1

8y � 13z � �7

•

x � y � 2z � �14x � y � 33x � 6

•

x � y � 2z � �10x � z � 1

z � 4

Solve using the elimination method. If a system isinconsistent or dependent, so state. For systems with lineardependence, write the answer in terms of a parameter. Forcoincident dependence, state the solution in set notation.

31.

32.

33. 34.

35. 36.

37. 38.

Some applications of systems lead to systems similar tothose that follow. Solve using elimination.

39.

40.

41.

42.

43.

44. •C � 3

2A � 3C � 103B � 4C � �11

•

C � �25A � 2C � 5

�4B � 9C �16

•

A � 2B � 5 B � 3C � 7

2A � B � C � 1

•

A � 2C � 72A � 3B � 83A � 6B � 8C � �33

•

�A � 3B � 2C � 112B � C � 9B � 2C � 8

•

�2A � B � 3C � 21B � C � 1

A � B � �4

x

2 �y

3 � z2 � 22x

3 � y � z � 8x

6 � 2y �3z

2 � 6

16x �

13y �

12z � 2

34x �

13y �

12z � 9

12x � y �

12z � 2

•

2x � 3y � 5z � 4x � y � 2z � 3x � 3y � 4z � �1

•

x � 5y � 4z � 32x � 9y � 7z � 23x � 14y � 11z � 5

•

4x � 5y � 6z � 52x � 3y � 3z � 0

x � 2y � 3z � 5•

x � 2y � 2z � 62x � 6y � 3z � 133x � 4y � z � �11

•

2x � 5y � 4z � 6x � 2.5y � 2z � 3

�3x � 7.5y � 6z � �9

•

4x � 2y � 8z � 24�x � 0.5y � 2z � �6 2x � y � 4z � 12

u u

cob19537_ch09_854-864.qxd 1/27/11 7:43 PM Page 862


Precalculus—


45. Dimensions of a rectangular solid:

Using the formula shown, the dimensions of arectangular solid can be found if the perimeters ofthe three distinct faces are known. Find thedimensions of the solid shown.

•

2w � 2h � P1

2l � 2w � P2

2l � 2h � P3

46. Distance from a point (x, y, z) to the plane

The perpendicular distance from a given point (x, y, z)to the plane defined by isgiven by the formula shown. Consider the planegiven in Figure 9.15 What is thedistance from this plane to the point (3, 4, 5)?

1x � y � z � 62.

Ax � By � Cz � D

`Ax � By � Cz � D

2A2 � B2 � C2`Ax � By � Cz � D:

P2 � 16 cm (top)P3� 18 cm(large side)

P1� 14 cm(small side)h

lw

� APPLICATIONS

Solve the following applications by setting up andsolving a system of three equations in three variables.Note that some equations may have only two of the threevariables used to create the system. Check answersusing back-substitution or the keys on the homescreen of a graphing calculator.

Investment/Finance and Simple Interest Problems

47. Investing the winnings: After winning $280,000in the lottery, Maurika decided to place the moneyin three different investments: a certificate ofdeposit paying 4%, a money market certificatepaying 5%, and some Aa bonds paying 7%. After 1 yr she earned $15,400 in interest. Find how muchwas invested at each rate if $20,000 more wasinvested at 7% than at 5%.

48. Purchase at auction: At an auction, a wealthycollector paid $7,000,000 for three paintings: aMonet, a Picasso, and a van Gogh. The Monet cost$800,000 more than the Picasso. The price of thevan Gogh was $200,000 more than twice the priceof the Monet. What was the price of each painting?


49. Major wars: The United States has fought threemajor wars in modern times: World War II, theKorean War, and the Vietnam War. If you sum theyears that each conflict ended, the result is 5871.The Vietnam War ended 20 years after the KoreanWar and 28 years after World War II. In what yeardid each end?

50. Animal gestation periods: The average gestationperiod (in days) of an elephant, rhinoceros, andcamel sum to 1520 days. The gestation period of arhino is 58 days longer than that of a camel. Twicethe camel’s gestation period decreased by 162gives the gestation period of an elephant. What isthe gestation period of each?

ALPHA

51. Moments in U.S. history: If you sum the year theDeclaration of Independence was signed, the yearthe 13th Amendment to the Constitution abolishedslavery, and the year the Civil Rights Act wassigned, the total would be 5605. Ninety-nine yearsseparate the 13th Amendment and the Civil RightsAct. The Civil Rights Act was signed 188 yearsafter the Declaration of Independence. What yearwas each signed?

52. Aviary wingspan: Ifyou combine thewingspan of theCalifornia Condor, the WanderingAlbatross (see photo),and the prehistoricQuetzalcoatlus, youget an astonishing 18.6 m (over 60 ft). Ifthe wingspan of the Quetzalcoatlus is equal to fivetimes that of the Wandering Albatross minus twicethat of the California Condor, and six times thewingspan of the Condor is equal to five times thewingspan of the Albatross, what is the wingspanof each?

Mixtures

53. Chemical mixtures: A chemist mixes threedifferent solutions with concentrations of 20%,30%, and 45% glucose to obtain 10 L of a 38%glucose solution. If the amount of 30% solutionused is 1 L more than twice the amount of 20%solution used, find the amount of each solutionused.

54. Value of gold coins: As part of a promotion, alocal bank invites its customers to view a largesack full of $5, $10, and $20 gold pieces,promising to give the sack to the first person ableto state the number of coins for each

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Precalculus—

denomination. Customers are told there are exactly250 coins, with a total face value of $1875. Ifthere are also seven times as many $5 gold piecesas $20 gold pieces, how many of eachdenomination are there?

Nutrition

55. Industrial food production: Acampana Soups iscreating a new sausage and shrimp gumbo thatcontains three different types of fat: saturated,monounsaturated, and polyunsaturated. As a newmember of the “Heart Healthy” menu, this soupmust contain only 2.8 g of total fat per serving.The head chef Yev Kasem demands that the recipeprovides exactly twice as much saturated fat aspolyunsaturated fat. At the same time, the leadnutrition expert Florencia requires the amount ofsaturated fat in a serving to be 0.4 g less than thecombined amount of unsaturated fats. How manygrams of each type of fat will a serving of thissoup contain?

56. Geriatric nutrition: The dietician at McKnightPlace must create a balanced diet for Fred,consisting of a daily total of 1600 calories. For his successful rehabilitation, exact quantitiesof complex carbohydrates, fat, and protein must provide these calories. In this diet,carbohydrates provide 160 more calories than fat and protein together, while fat provides 1.25 times more calories than protein alone.How many calories should each nutrient provideon a daily basis?

In Exercises 87 and 88 from Section 9.1, we found theequation of a line through two points by creating asystem of two equations of the form y � mx � b. Just asany two points determine a unique line, threenoncollinear points determine a unique parabola (aslong as no two have the same first coordinate). Similar tothe method used there, each point gives an equation ofthe form y � ax2 � bx � c, and we create a 3 � 3 systemthat can be solved using elimination. For instance, thepoint (2, 41) gives the equation 41 � a(2)2 � 2b � c.

57. Height of a soccer ball: One second after beingkicked (t � 1), a soccer ball is 26 ft high. After 2 sec(t � 2), the ball is 41 ft high, and after 6 sec the ball is1 ft above the ground. Use the ordered pairs (time,height) to find a function h(t) modeling the height ofthe ball after t sec, then use the equation to find (a) themaximum height of the kick, and (b) the height of theball after 5.4 sec.

58. Height of an arrow: An archer is out in a large fieldtesting a new bow. Pulling back the bow to near itsbreaking point, the archer lets the arrow fly. Supposethat 1 sec after release (t � 1) the arrow was 184 fthigh, four sec later (t � 5) it was 600 ft high, andafter 12 sec, the arrow was 96 ft high. Use the orderedpairs (time, height) to find a function h(t) modelingthe height of the arrow after t sec, then use theequation to (a) find the maximum height of the shotand (b) determine how long the arrow was airborne.


59. The system is dependent if

, and inconsistent otherwise.k �

•

x � 2y � z � 2x � 2y � kz � 5

2x � 4y � 4z � 10

60. One form of the equation of a circle isUse a system to

find the equation of the circle through the pointsand 12, �52.12, �12, 14, �32,

x2 � y2 � Dx � Ey � F � 0.


61. (8.3) Given and , computeand

62. (6.2) Given use a calculator to findthe acute angle A to the nearest tenth of a degree.

63. (5.5) Solve the logarithmic equation:log1x � 22 � log x � log 3

cot A � 1.6831,3u � v.u � 4v

v � H�3, 12Iu � H1, �7I 64. (2.1) Analyze the graph of gshown. Clearly state thedomain and range, the zeroesof g, intervals whereintervals where localmaximums or minimums, andintervals where the function isincreasing or decreasing. Assume each tick mark isone unit and estimate endpoints to the nearest tenths.

g1x2 6 0,g1x2 7 0,

y

x

cob19537_ch09_854-864.qxd 1/27/11 7:43 PM Page 864

9.3 Systems of Inequalities and Linear Programming

In this section, we’ll build on many of the ideas from Section 9.1, with a focus onsystems of linear inequalities. While systems of linear equations have an unlimitednumber of applications, there are many situations that can only be modeled using lin-ear inequalities. For example, decisions in business and industry are often based on alarge number of limitations or constraints, with many different ways these constraintscan be satisfied.

A. Linear Inequalities in Two Variables

A linear equation in two variables is any equation that can be written in the formwhere A and B are real numbers, not simultaneously equal to zero. A

linear inequality in two variables is similarly defined, with the “ ” sign replaced bythe “ ,” “ ,” “ ” or “ ” symbol:

Solving a linear inequality in two variables has many similarities with the one vari-able case. For one variable, we graph a boundary point on a number line, decidewhether the endpoint is included or excluded, and shade the appropriate half line. For

we have the solution with the endpoint (boundary point) includedand the line shaded to the left (Figure 9.25):

x � 2x � 1 � 3,

Ax � By � CAx � By � CAx � By 7 CAx � By 6 C

�� ,76�

Ax � By � C,


how we can:

A. Solve a linear inequalityin two variables

B. Solve a system of linearinequalities

C. Solve applications usinga system of linearinequalities

D. Solve applications usinglinear programming

[02�� 21� 13 3

Interval notation: x � (��, 2]

��Figure 9.25

For linear inequalities in two variables, we graph a boundary line, decide whetherthe boundary line is included or excluded, and shade the appropriate half plane. For

the boundary line is graphed in Figure 9.26. Note it divides thecoordinate plane into two regions called half planes, and it forms the boundarybetween the two regions. If the boundary is included in the solution set, we graph itusing a solid line. If the boundary is excluded, a dashed line is used. Recall that solu-tions to a linear equation are ordered pairs that make the equation true. We use a simi-lar idea to find or verify solutions to linear inequalities. If any one point in a half planemakes the inequality true, all points in that half plane will satisfy the inequality.

x � y � 3x � y � 3,

5�5

�5

5y

x

(0, 3)Upperhalf plane

Lowerhalf plane

x � y � 3

(4, �1)

Figure 9.26

EXAMPLE 1 � Checking Solutions to an Inequality in T wo VariablesDetermine whether the given ordered pairs are solutions to

a. b. c.

Solution � a. Substitute 4 for x and for y: substitute 4 for x, �3 for y

true

is a solution.b. Substitute for x and 1 for y: substitute �2 for x, 1 for y

false

is not a solution.c. Substitute for x and for y: substitute �4 for x, �1 for y

true

is a solution.


1�4, �122 � 2

�1�42 � 21�12 � 2�1�41�2, 12

4 � 2�1�22 � 2112 � 2�2

14, �32 �10 � 2

�142 � 21�32 � 2�3

1�4, �121�2, 1214, �32�x � 2y � 2:

Precalculus—

9–29 865

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Precalculus—

Earlier we graphed linear equations by plotting a small number of ordered pairs orby solving for y and using the slope-intercept method. The line represented all orderedpairs that made the equation true, meaning the left-hand expression was equal to theright-hand expression. To graph linear inequalities, we reason that if the line representsall ordered pairs that make the expressions equal, then any point not on that line mustmake the expressions unequal—either greater than or less than. These ordered pairsolutions must lie in one of the half planes formed by the line, which we shade to indi-cate the solution region. Note this implies the boundary line for any inequality isdetermined by the related equation, created by temporarily replacing the inequalitysymbol with an “ ” sign.

EXAMPLE 2 � Solving an Inequality in Two VariablesSolve the inequality

Solution � The related equation and boundary line is Since the inequality isinclusive (less than or equal to), we graph a solid line. Using the intercepts, wegraph the line through (0, 1) and shown in Figure 9.27. To determine thesolution region and which side to shade, we select (0, 0) as a test point, whichresults in a true statement: . Since (0, 0) is in the “lower” halfplane, we shade this side of the boundary (see Figure 9.28).

�102 � 2102 � 2 ✓

1�2, 02

�x � 2y � 2.

�x � 2y � 2.

�

WORTHY OF NOTEThis relationship is often called thetrichotomy axiom or the “three-parttruth.” Given any two quantities,they are either equal to each other,or the first is less than the second,or the first is greater than thesecond.

5�5

�5

5

(0, 0)Test point

(4, 3)

(�2, 0)

Upperhalf plane

Lowerhalf plane

(0, 1)

y

x

Figure 9.27

5�5

�5

5

(0, 0)Test point

(4, 3)

(�2, 0)(0, 1)

y

x

Figure 9.28


Solving a Linear Inequality

1. Graph the boundary line by solving for y and using the slope-intercept form.• Use a solid line if the boundary is included in the solution set.• Use a dashed line if the boundary is excluded from the solution set.

2. For “greater than” inequalities shade the upper half plane. For “less than”inequalities shade the lower half plane.

The same solution would be obtained if we first solved for y and graphed theboundary line using the slope-intercept method. However, using the slope-interceptmethod offers a distinct advantage—test points are no longer necessary since solu-tions to “less than” inequalities will always appear below the boundary line and solu-tions to “greater than” inequalities appear above the line. Written in slope-interceptform, the inequality from Example 2 is Note that (0, 0) still results in atrue statement, but the “less than or equal to” symbol now indicates directly that solu-tions will be found in the lower half plane. These observations lead to our generalapproach for solving linear inequalities:

y � 12 x � 1.

cob19537_ch09_865-878.qxd 1/27/11 8:21 PM Page 866

EXAMPLE 3 � Solving linear Inequalities in Two Variables Using TechnologySolve the inequality using a graphing calculator.

Solution � We begin by solving the inequality for y, so we can enter the equation on the screen.

given inequality

subtract 3x (isolate y-term)

divide by 5

Entering on the screen, and graphing the boundary line using

6:ZStandard produces the graph shown in Figure 9.29. The “less than”inequality indicates the region below the line should be shaded, so we return to the

screen and move the cursor to the far left. From the default “connected line”setting, pressing three times brings the “shade below the graph” marker “�” intoview (Figure 9.30), and pressing gives the solution shown in Figure 9.31. As acheck, note that (0, 0) is in the solution region, and is a solution to .3x � 5y � 10

GRAPH

ENTER

Y=

ZOOM

Y=Y1 ��3

5X � 2

y ��3

5 x � 2

5y � �3x � 103x � 5y � 10

Y=

3x � 5y � 10

�10 10

�10

10

Figure 9.29Figure 9.30

�10 10

�10

10

Figure 9.31


A. You’ve just seen how we

can solve a linear inequality in

two variables

9–31 Section 9.3 Systems of Inequalities and Linear Programming 867

Precalculus—

B. Solving Systems of Linear Inequalities

To solve a system of inequalities, we apply the procedure outlined above to all in-equalities in the system, and note the ordered pairs that satisfy all inequalities simulta-neously. In other words, we find the intersection of all solution regions (where theyoverlap), which then represents the solution for the system. In the case of verticalboundary lines, the designations “above” or “below” the line cannot be applied, andinstead we simply note that for any vertical line points with x-coordinates largerthan k will occur to the right.

x � k,

EXAMPLE 4 � Solving a System of Linear Inequalities

Solve the system of inequalities:

Solution � Solving for y, we obtain and The line willbe a solid boundary line (included), while will be dashed (not included).Both inequalities are “greater than” and so we shade the upper half plane for each.The regions overlap and form the solution region (the lavender region shown). Thissequence of events is illustrated here:

y � x � 2y � �2x � 4y 7 x � 2.y � �2x � 4

e2x � y � 4

x � y 6 2.

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Precalculus—

5�5

�5

5

2x � y � 4

Shade above y � �2x � 4 (in blue)y

x 5�5

�5

5

2x � y � 4

x � y � 2

y

x

Shade above y � x � 2 (in pink)

5�5

�5

5

2x � y � 4

x � y � 2

y

x

Solutionregion

Cornerpoint

Overlapping region

The solutions are all ordered pairs found in this region and its includedboundaries. To verify the result, test the point (2, 3) from inside the region, and

from outside the region [the point (2, 0) is not a solution since it does notsatisfy ].


For future reference, the point of intersection (2, 0) is called a corner point orvertex of the solution region. If the point of intersection is not easily found from thegraph, we can find it by solving a linear system using the two lines. For Example 4,the system is

and solving by elimination gives and (2, 0) as the point of intersection.A graphing calculator can also be used to solve a system of linear inequalities. One

method (there are several) involves these three steps, which are performed on eachequation:

1. Solve for y and enter the results on the screen to create the boundary lines.2. Graph each line and test the related half plane.3. Shade the appropriate half plane.

Y=

3x � 6, x � 2,

e2x � y � 4

x � y � 2

x � y 6 215, �22

This process is illustrated in Example 5. Since many real-world applications of lin-ear inequalities do not use negative numbers, we often include and aspart of the system, and set Xmin � 0 and Ymin � 0 as part of the size.The value of Xmax and Ymax will depend on equations or context given.

EXAMPLE 5 � Solving a System of Inequalities Using T echnology

Use a graphing calculator to solve the system:

Solution � Following the steps outlined above, we have1. Enter the related equations. For we have

For , we have Enter these as Y1 and Y2 on thescreen.

2. Graph the boundary lines. Note the x- and y-intercepts of both lines are lessthan 10, so we can graph them using a “friendly window” where x [0, 9.4]and y [0, 6.2]. After setting the window, press to graph the lines.GRAPH�

�

Y=

y � �0.5x � 4.x � 2y � 8y � �1.5x � 7.3x � 2y � 14,

μ

3x � 2y 6 14

x � 2y 6 8

x � 0

y � 0

WINDOW

y � 0x � 0

cob19537_ch09_865-878.qxd 1/27/11 8:22 PM Page 868

3. Shade the appropriate half plane. Since both equations are in slope-interceptform and solving for y resulted in two “less than” inequalities, we shade belowboth lines, using the “�” feature located to the far left of Y1 and Y2. Simplyoverlay the diagonal line and press repeatedly until the symbol appears(Figure 9.32). After pressing the GRAPH key, the calculator draws both linesand shades the appropriate regions (Figure 9.33). Note the calculator usestwo different kinds of shading. This makes it easy to identify the solutionregion—it will be the “checker-board area” where the horizontal and verticallines cross.

ENTER


Precalculus—


we can solve a system of

linear inequalities

Figure 9.32

9.40

0

6.2

Figure 9.33

As a final check, we can navigate the position marker into the solution regionand test a few points in both inequalities. Using the test point (2, 2) from within the region yields

first inequality second inequality

substitute 2 for x, 2 for y substitute 2 for x, 2 for y

true true


C. Applications of Systems of Linear Inequalities

Systems of inequalities give us a way to model the decision-making process when cer-tain constraints must be satisfied. A constraint is a fact or consideration that somehowlimits or governs possible solutions, like the number of acres a farmer plants—whichmay be limited by time, size of land, government regulations, and so on.

6 � 810 � 14122 � 2122 � 83122 � 2122 � 14

x � 2y 6 83x � 2y 6 14

EXAMPLE 6 � Solving Applications of Linear InequalitiesAs part of their retirement planning, James and Lily decide to invest up to $30,000in two separate investment vehicles. The first is a bond issue paying 9% and thesecond is a money market certificate paying 5%. A financial adviser suggests theyinvest at least $10,000 in the certificate and not more than $15,000 in bonds. Whatvarious amounts can be invested in each?

Solution � Consider the ordered pairs (B, C) where B represents the money invested in bondsand C the money invested in the certificate. Since they plan to invest no more than$30,000, the investment constraint would be (in thousands).Following the adviser’s recommendations, the constraints on each investmentwould be and Since they cannot invest less than zero dollars, thelast two constraints are and

μ

B � C � 30

B � 15

C � 10

B � 0

C � 0

C � 0.B � 0C � 10.B � 15

B � C � 30

Solutionregion

(15, 15)

(0, 30)

(0, 10)

(15, 10)

QI

30 402010

10

20

30

40

B

C B � 15

C � 10

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Precalculus—

The resulting system is shown in the figure, and indicates solutions will be in thefirst quadrant.

There is a vertical boundary line at with shading to the left (less than)and a horizontal boundary line at with shading above (greater than). Aftergraphing we see the solution region is a quadrilateral with vertices at(0, 10), (0, 30), (15, 10), and (15, 15), as shown.


From Example 6, any ordered pair in this region or on its boundaries would repre-sent an investment of the form (money in bonds, money in CDs) S (B, C ), and wouldsatisfy all constraints in the system. A natural follow-up question would be—Whatcombination of (money in bonds, money in CDs) would offer the greatest return? Thiswould depend on the interest being paid on each investment, and introduces us to astudy of linear programming, which follows soon.

D. Linear Programming

To become as profitable as possible, corporations look for ways to maximize their rev-enue and minimize their costs, while keeping up with delivery schedules and productdemand. To operate at peak efficiency, plant managers must find ways to maximizeproductivity, while minimizing related costs and considering employee welfare, unionagreements, and other factors. Problems where the goal is to maximize or minimizethe value of a given quantity under certain constraints or restrictions are calledprogramming problems. The quantity we seek to maximize or minimize is called theobjective function. For situations where linear programming is used, the objectivefunction is given as a linear function in two variables and is denoted f (x, y). A functionin two variables is evaluated in much the same way as a single variable function. Toevaluate at the point (4, 5), we substitute 4 for x and 5 for y:f 14, 52 � 2142 � 3152 � 23.

f 1x, y2 � 2x � 3y

C � 30 � B,C � 10

B � 15


we can solve applications

using a system of linear

inequalities

EXAMPLE 7 � Determining Maximum ValuesDetermine which of the following ordered pairs maximizes the value of

(0, 6), (5, 0), (0, 0), or (4, 2).

Solution � Organizing our work in table form gives

5x � 4y:f 1x, y2 �

Given EvaluatePoint

(0, 6)

(5, 0)

(0, 0)

(4, 2) f 14, 22 � 5142 � 4122 � 28

f 10, 02 � 5102 � 4102 � 0

f 15, 02 � 5152 � 4102 � 25

f 10, 62 � 5102 � 4162 � 24

f 1x, y2 � 5x � 4y

The function is maximized at (4, 2).


When the objective is stated as a linear function in two variables and the con-straints are expressed as a system of linear inequalities, we have what is called a linearprogramming problem. The systems of inequalities solved earlier produced solutionregions that were either bounded (as in Example 6) or unbounded (as in Example 4).

f 1x, y2 � 5x � 4y

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Precalculus—

EXAMPLE 8 � Finding the Maximum of an Objective FunctionFind the maximum value of the objective function given the

constraints shown: .

Solution � Begin by noting that the solutions must be in QI,since and Graph the boundary lines

and shading the lowerhalf plane in each case since they are “less than”inequalities. This produces the feasible regionshown in lavender. There are four corner points tothis region: (0, 0), (0, 4), (2, 0), and (1, 3). Three ofthese points are intercepts and can be found quickly.The point (1, 3) was found by solving the

system Knowing that the objective

function will be maximized at one of the corner points, we test them in theobjective function, using a table to organize our work.

e x � y � 4

3x � y � 6.

y � �3x � 6,y � �x � 4y � 0.x � 0

μ

x � y � 4

3x � y � 6

x � 0

y � 0

f 1x, y2 � 2x � y

x

y

54321�5 �4 �3 �2 �1�1

�2

5

6

7

8

4(1, 3)

Feasibleregion

3

2

1

Convex Not convex

Figure 9.34 We interpret the word bounded to mean we can enclose the solution region within acircle of appropriate size. If we cannot draw a circle around the region because it ex-tends indefinitely in some direction, the region is said to be unbounded. In this study,we will consider only situations that produce bounded solution regions, meaning theregions will have three or more vertices. The regions we study will also be convex,meaning that for any two points in the enclosed region, the line segment between themis also in the region (Figure 9.34). Under these conditions, it can be shown that themaximum or minimum values must occur at one of the corner points of the solutionregion, also called the feasible region.

Corner Objective FunctionPoint

(0, 0)

(0, 4)

(2, 0)

(1, 3) f 11, 32 � 2112 � 132 � 5

f 12, 02 � 2122 � 102 � 4

f 10, 42 � 2102 � 142 � 4

f 10, 02 � 2102 � 102 � 0

f 1x, y2 � 2x � y

The objective function is maximized at (1, 3).


To help understand why solutions must occur at a vertex, note the objective func-tion f (x, y) is maximized using only (x, y) ordered pairs from the feasible region. If welet K represent this maximum value, the function from Example 8 becomes

or which is a line with slope and y-intercept K. Thetable in Example 8 suggests that K should range from 0 to 5 and graphing

for and produces the family of parallel linesshown in Figure 9.35. Note that values of K larger than 5 will cause the line to miss thesolution region, and the maximum value of 5 occurs where the line intersects the fea-sible region at the vertex (1, 3). These observations lead to the following principles,which we offer without a formal proof.

K � 5K � 1, K � 3,y � �2x � K

�2y � �2x � K,K � 2x � y

f 1x, y2 � 2x � y

x

y

(1, 3)

K � 1K � 3

K � 554321�5 �4 �3 �2 �1

�1

5

6

7

8

4

3

2

1

Figure 9.35

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Precalculus—

Linear Programming Solutions

1. If the feasible region is convex and bounded, a maximum and a minimumvalue exist.

2. If a unique solution exists, it will occur at a vertex of the feasible region.3. If more than one solution exists, at least one of them occurs at a vertex of the

feasible region with others on a boundary line.4. If the feasible region is unbounded, a linear programming problem may have

no solutions.

Solving linear programming problems depends in large part on two things: (1) identifying the objective and the decision variables (what each variable representsin context), and (2) using the decision variables to write the objective function andconstraint inequalities. This brings us to our five-step approach for solving linear pro-gramming applications.

Solving Linear Programming Applications

1. Identify the main objective and the decision variables (descriptive variablesmay help) and write the objective function in terms of these variables.

2. Organize all information in a table, with the decision variables andconstraints heading up the columns, and their components leading each row.

3. Complete the table using the information given, and write the constraintinequalities using the decision variables, constraints, and the domain.

4. Graph the constraint inequalities, determine the feasible region, and identifyall corner points.

5. Test these points in the objective function to determine the optimal solution(s).

EXAMPLE 9 � Solving an Application of Linear ProgrammingThe owner of a snack food business wants to create two nut mixes for the holidayseason. The regular mix will have 14 oz of peanuts and 4 oz of cashews, while thedeluxe mix will have 12 oz of peanuts and 6 oz of cashews. The owner estimateshe will make a profit of $3 on the regular mixes and $4 on the deluxe mixes. Howmany of each should be made in order to maximize profit, if only 840 oz ofpeanuts and 348 oz of cashews are available?

Solution � Our objective is to maximize profit, and the decision variables could be r torepresent the regular mixes sold, and d for the number of deluxe mixes. This gives

as our objective function. The information is organized inTable 9.1, using the variables r, d, and the constraints to head each column. Sincethe mixes are composed of peanuts and cashews, these lead the rows in the table.

P1r, d2 � $3r � $4d

Table 9.1

Regular Deluxe Constraints: Total r d Ounces Available

Peanuts 14 12 840

Cashews 4 6 348

TTP1r, d2 � $3r � $4 d

After filling in the appropriate values, reading the table from left to right along the“peanut” row and the “cashew” row, gives the constraint inequalities 14r � 12d � 840

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Precalculus—

Table 9.2

Profit will be maximized if 24 boxes of the regular mix and 42 boxes of the deluxemix are made and sold.


Linear programming can also be used to minimize an objective function, as inExample 10.

and Realizing we won’t be making negative numbers of mixes, theremaining constraints are and The complete system is

Note once again that the solutions must be in QI, sinceand Graphing the first two inequalities

using slope-intercept form gives andproducing the feasible region shown

in lavender. The four corner points are (0, 0), (60, 0),(0, 58), and (24, 42). Three of these points areintercepts and can be read from a table of values or thegraph itself. The point (24, 42) was found by solving

the system Knowing the solution

must occur at one of these points, we test them inthe objective function (Table 9.2).

e14r � 12d � 840

4r � 6d � 348.

d � �23r � 58

d � �76r � 70

d � 0.r � 0

μ

14r � 12d � 840

4r � 6d � 348

r � 0

d � 0

d � 0.r � 04r � 6d � 348.

10080 9070605010 20 30 40

40

50

30

20

10

70

80

90

100

60

r

d

Feasibleregion

Corner Objective FunctionPoint P(r, d )

(0, 0)

(60, 0)

(0, 58)

(24, 42) P124, 422 � $31242 � $41422 � $240

P10, 582 � $3102 � $41582 � $232

P160, 02 � $31602 � $4102 � $180

P10, 02 � $3102 � $4102 � $0

� $3r � $4d

EXAMPLE 10 � Minimizing Costs Using Linear ProgrammingA beverage producer needs to minimize shipping costs from its two primary plantsin Kansas City (KC) and St. Louis (STL). All wholesale orders within the state areshipped from one of these plants. An outlet in Macon orders 200 cases of softdrinks on the same day an order for 240 cases comes from Springfield. The plant inKC has 300 cases ready to ship and the plant in STL has 200 cases. The cost ofshipping each case to Macon is $0.50 from KC, and $0.70 from STL. The cost ofshipping each case to Springfield is $0.60 from KC, and $0.65 from STL. Howmany cases should be shipped from each warehouse to minimize costs?

Solution � Our objective is to minimize costs, which depends on the number of cases shippedfrom each plant. To begin we use the following assignments:

D S cases shipped from STL to SpringfieldC S cases shipped from STL to MaconB S cases shipped from KC to SpringfieldA S cases shipped from KC to Macon

cob19537_ch09_865-878.qxd 1/27/11 8:22 PM Page 873


Precalculus—

From this information, the equation for total cost T is

an equation in four variables. To make the cost equation more manageable, notesince Macon ordered 200 cases, . Similarly, Springfield ordered 240 cases, so . After solving for C and D, respectively, these equationsenable us to substitute for C and D, resulting in an equation with just two variables.For and we have

The constraints involving the KC plant are with . Theconstraints for the STL plant are with . Since we wanta system in terms of A and B only, we again substitute and

in all the STL inequalities:

STL inequalities

substitute for C, for D

simplify

result240 � A � B240 � B200 � A

240 � B

440 � A � B � 200

240 � B � 0200 � A � 0200 � A1200 � A2 � 1240 � B2 � 200

D � 0C � 0C � D � 200

D � 240 � BC � 200 � A

C � 0, D � 0C � D � 200A � 0, B � 0A � B � 300

� 296 � 0.2A � 0.05B � 0.5A � 0.6B � 140 � 0.7A � 156 � 0.65B

T1A, B2 � 0.5A � 0.6B � 0.71200 � A2 � 0.651240 � B2

D � 240 � BC � 200 � A

B � D � 240A � C � 200

0.65D,T � 0.5A � 0.6B � 0.7C �

u



using linear programming

To find the minimum cost, we check each vertex in the objective function.

The minimum cost occurs when and meaning the producershould ship the following quantities:


D S cases shipped from STL to Springfield � 140C S cases shipped from STL to Macon � 0B S cases shipped from KC to Springfield � 100A S cases shipped from KC to Macon � 200

B � 100,A � 200

Objective FunctionVertices T(A, B)

(0, 240)

(60, 240)

(200, 100)

(200, 40) P1200, 402 � 296 � 0.212002 � 0.051402 � $254

P1200, 1002 � 296 � 0.212002 � 0.0511002 � $251

P160, 2402 � 296 � 0.21602 � 0.0512402 � $272

P10, 2402 � 296 � 0.2102 � 0.0512402 � $284

� 296 � 0.2A � 0.05B

Combining the new STL constraints with those from KC produces the followingsystem and solution. All points of intersection were read from the graph or locatedusing the related system of equations.

A � B � 300

A � B � 240

A � 200

B � 240

A � 0

B � 0

(60, 240)

A400100 200 300

100

200

300

400

B � 240

A � 200B

(200, 40)

(200, 100)Feasibleregion

cob19537_ch09_865-878.qxd 1/28/11 7:56 PM Page 874


Precalculus—

2. For the line drawn in the coordinateplane, solutions to are found in theregion the line.

y 7 mx � by � mx � b



9.3 EXERCISES

1. Any line drawn in the coordinateplane divides the plane into two regions called

.

y � mx � b

3. The overlapping region of two or more linearinequalities in a system is called the region.

4. If a linear programming problem has a uniquesolution (x, y), it must be a of the feasibleregion.

6. Describe the conditions necessary for a linearprogramming problem to have multiple solutions.(Hint: Consider the diagram in Figure 9.35, and theslope of the line from the objective function.)

5. Suppose two boundary lines in a system of linearinequalities intersect, but the point of intersectionis not a vertex of the feasible region. Describe howthis is possible.


Determine whether the ordered pairs given are solutions.

7. (0, 0),

8. (0, 0), ,

9. (0, 0), , ,

10. (0, 0),

Solve the linear inequalities by shading the appropriatehalf plane. Verify your answer using a graphingcalculator.

11. 12.

13. 14.

Solve the following inequalities using a graphingcalculator. Your answer should include a screen shot orfacsimile, and comments regarding a test point.

15. 3x � 2y � 8 16. 2x � 5y � 10

17. 4x � 5y �20 18. 6x � 3y 18

Determine whether the ordered pairs given are solutionsto the accompanying system.

19.

1�2, 12, 1�5, �42, 1�6, 22, 1�8, 2.22

e5y � x � 10

5y � 2x � �5;

4x � 5y � 152x � 3y � 9

x � 3y 7 6x � 2y 6 8

17, �321�1, 62,13, 52,3x � 5y � 15;

1�1, 121�3, �221�3, 524x � 2y � �8;

11, �221�1, �52,14, �123x � y 7 5;

1�3, 921�3, �42,13, �52,2x � y 7 3; 20.

Solve each system of inequalities by graphing thesolution region. Verify the solution using a test point.

21. 22.

23. 24.

25. 26.

27. 28.

29. 30.

31. 32.

33. 34. •3x � 4y 7 12

y 62

3 x

• y �

3

2x

4y � 6x � 12

ex 7 �0.4y � 2.2

x � 0.9y � �1.2e

0.2x 7 �0.3y � 1

0.3x � 0.5y � 0.6

e10x � 4y � 20

5x � 2y 7 �1e

5x � 4y � 20

x � 1 � y

e2x � 5y 6 15

3x � 2y 7 6e

x 7 �3y � 2

x � 3y � 6

ex � 2y 6 �7

2x � y 7 5e

2x � y 6 4

2y 7 3x � 6

e3x � 2y

y � 4x � 3e

3x � y 7 4

x 7 2y

e�x � 5y 6 5

x � 2y � 1e

x � 2y � 1

2x � y � �2

•

8y � 7x � 56

3y � 4x � �12

y � 4; 11, 52, 14, 62,

18, 52, 15, 32

cob19537_ch09_865-878.qxd 1/27/11 8:24 PM Page 875


Precalculus—

35. 36.

37. 38.

39. 40.

41. 42.

Use a graphing calculator to find the solution region foreach system of linear inequalities. Your answer shouldinclude a screen shot or facsimile, and the location ofany points of intersection.

43. 44.

45. 46.

Use the equations given to write the system of linearinequalities represented by each graph.

47. 48.

x

y

y � x � 1

x � y � 354321�5�4�3�2�1�1

�2�3�4�5

2345

1x

y

y � x � 1

x � y � 354321�5�4�3�2�1�1

�2�3�4�5

2345

1

μ

y � 2x � 10

2y � x � 11

x � 0

y � 0

μ

�2x � y 7 �8

�x � 2y 6 �7

x � 0

y � 0

μ

x � 2y 6 10

x � y 6 7

x � 0

y � 0

μ

y � 2x 6 8

y � x 6 6

x � 0

y � 0

•

8x � 5y � 40

x � y � 7

x � 0, y � 0

•

2x � 3y � 18

2x � y � 10

x � 0, y � 0

•

4y 6 3x � 12

y � x � 1

x � 0, y � 0

•

y � x � 3

x � 2y � 4

x � 0, y � 0

•

2x � y � 5

x � 3y � 6

x � 0, y � 0

•

x � y � �4

2x � y � 4

x � 0, y � 0

μ

1

2x �

2

5y � 5

5

6x � 2y � �5

μ

�2

3 x �

3

4 y � 1

1

2 x � 2y � 3

49. 50.

Determine which of the ordered pairs given producesthe maximum value of f (x, y).

51. (0, 0), (0, 8.5), (7, 0), (5, 3)

52. (0, 0), (0, 21), (15, 0), (7.5, 12.5)

Determine which of the ordered pairs given producesthe minimum value of f (x, y).

53. (0, 20), (35, 0), (5, 15), (12, 11)

54. (0, 9), (10, 0), (4, 5), (5, 4)

For Exercises 55 and 56, find the maximum value of theobjective function f (x, y) and where thisvalue occurs, given the constraints shown.

55. 56.

For Exercises 57 and 58, find the minimum value of theobjective function f (x, y) and where thisvalue occurs, given the constraints shown.

57. 58. μ

2x � y � 10

x � 4y � 3

x � 2

y � 0

μ

3x � 2y � 18

3x � 4y � 24

x � 0

y � 0

� 36x � 40y

μ

2x � y � 7

x � 2y � 5

x � 0

y � 0

μ

x � 2y � 6

3x � y � 8

x � 0

y � 0

� 8x � 5y

f 1x, y2 � 75x � 80y;

f 1x, y2 � 8x � 15y;

f 1x, y2 � 50x � 45y;

f 1x, y2 � 12x � 10y;

x

y

y � x � 1

x � y � 354321�5�4�3�2�1�1

�2�3�4�5

2345

1x

y

y � x � 1

x � y � 354321�5�4�3�2�1�1

�2�3�4�5

2345

1


Area Formulas

59. The area of a triangle is usually given aswhere B and H represent the base and

height, respectively. The area of a rectangle can bestated as If the base of both a triangle andrectangle is equal to 20 in., what are the possiblevalues for H if the triangle must have an areagreater than 50 in2 and the rectangle must have anarea less than 200 in2?

A � BH.

A � 12 BH,

Volume Formulas

60. The volume of a cone is where r is theradius of the base and h is the height. The volumeof a cylinder is If the radius of both acone and cylinder is equal to 10 cm, what are thepossible values for h if the cone must have avolume greater than 200 cm3 and the volume ofthe cylinder must be less than 850 cm3?

V � �r2h.

V � 13�r2h,

cob19537_ch09_865-878.qxd 1/27/11 8:24 PM Page 876


Precalculus—

� APPLICATIONS

65. Manufacturing screws: A machine shopmanufactures two types of screws—sheet metalscrews and wood screws, using three differentmachines. Machine Moe can make a sheet metalscrew in 20 sec and a wood screw in 5 sec.Machine Larry can make a sheet metal screw in 5 sec and a wood screw in 20 sec. Machine Curly,the newest machine (nyuk, nyuk) can make a sheetmetal screw in 15 sec and a wood screw in 15 sec.(Shemp couldn’t get a job because he failed themath portion of the employment exam.) Eachmachine can operate for only 3 hr each day beforeshutting down for maintenance. If sheet metalscrews sell for 10 cents and wood screws sell for12 cents, how many of each type should themachines be programmed to make in order tomaximize revenue? (Hint: Standardize time units.)

66. Hauling hazardous waste: A waste disposalcompany is contracted to haul away somehazardous waste material. A full container of liquidwaste weighs 800 lb and has a volume of Afull container of solid waste weighs 600 lb and hasa volume of The trucks used can carry atmost 10 tons (20,000 lb) and have a carryingvolume of If the trucking company makes$300 for disposing of liquid waste and $400 fordisposing of solid waste, what is the maximumrevenue per truck that can be generated?

67. Maximizing profit—food service: P. Barrett &Justin, Inc., is starting up a fast-food restaurantspecializing in peanut butter and jelly sandwiches.Some of the peanut butter varieties are smooth,crunchy, reduced fat, and reduced sugar. The jellieswill include those expected and common, as wellas some exotic varieties such as kiwi and mango.Independent research has determined the two mostpopular sandwiches will be the traditional P&J(smooth peanut butter and grape jelly), and theDouble-T (three slices of bread). A traditional P&Juses 2 oz of peanut butter and 3 oz of jelly. TheDouble-T uses 4 oz of peanut butter and 5 oz ofjelly. The traditional sandwich will be priced at$2.00, and a Double-T at $3.50. If the restauranthas 250 oz of smooth peanut butter and 345 oz ofgrape jelly on hand for opening day, how many ofeach should they make and sell to maximizerevenue?

800 ft3.

30 ft3.

20 ft3.

Write a system of linear inequalities that models theinformation given, then solve. Verify the solution regionusing a graphing calculator.

61. Gifts to grandchildren: Grandpa Augustus isconsidering how to divide a $50,000 gift betweenhis two grandchildren, Julius and Anthony. Afterweighing their respective positions in life andfamily responsibilities, he decides he mustbequeath at least $20,000 to Julius, but no morethan $25,000 to Anthony. Determine the possibleways that Grandpa can divide the $50,000.

62. Guns versus butter: Every year, governmentsaround the world have to make the decision as tohow much of their revenue must be spent onnational defense and domestic improvements (gunsversus butter). Suppose total revenue for these twoneeds was $120 billion, and a government decidesthey need to spend at least $42 billion on butter andno more than $80 billion on defense. Determine thepossible amounts that can go toward each need.

Solve the following applications of linear programming.

63. Land/crop allocation: A farmer has 500 acres ofland to plant corn and soybeans. During the lastfew years, market prices have been stable and thefarmer anticipates a profit of $900 per acre on thecorn harvest and $800 per acre on the soybeans.The farmer must take into account the time it takesto plant and harvest each crop, which is 3 hr/acrefor corn and 2 hr/acre for soybeans. If the farmerhas at most 1300 hr to plant, care for, and harvesteach crop, how many acres of each crop should beplanted in order to maximize profits?

64. Coffee blends: The owner of a coffee shop hasdecided to introduce two new blends of coffee inorder to attract new customers—a Deluxe Blendand a Savory Blend. Each pound of the deluxeblend contains 30% Colombian and 20% Arabiancoffee, while each pound of the savory blendcontains 35% Colombian and 15% Arabian coffee(the remainder of each is made up of cheap andplentiful domestic varieties). The profit on thedeluxe blend will be $1.25 per pound, while theprofit on the savory blend will be $1.40 per pound.How many pounds of each should the owner makein order to maximize profit, if only 455 lb ofColombian coffee and 250 lb of Arabian coffee arecurrently available?

cob19537_ch09_865-878.qxd 1/27/11 8:24 PM Page 877


Precalculus—

68. Maximizing profit—construction materials:Mooney and Sons produces and sells two varietiesof concrete mixes. The mixes are packaged in 50-lbbags. Type A is appropriate for finish work, andcontains 20 lb of cement and 30 lb of sand. Type Bis appropriate for foundation and footing work, andcontains 10 lb of cement and 20 lb of sand. Theremaining weight comes from gravel aggregate.The profit on type A is $1.20/bag, while the profiton type B is $0.90/bag. How many bags of eachshould the company make to maximize profit, if2750 lb of cement and 4500 lb of sand arecurrently available?

69. Minimizing transportation costs: Robert’s LasVegas Tours needs to drive 375 people and 19,450 lbof luggage from Salt Lake City, Utah, to Las Vegas,Nevada, and can charter buses from two companies.The buses from company X carry 45 passengers and2750 lb of luggage at a cost of $1250 per trip.Company Y offers buses that carry 60 passengers

and 2800 lb of luggage at a cost of $1350 per trip.How many buses should be chartered from eachcompany in order for Robert to minimize the cost?

70. Minimizing shipping costs: An oil company istrying to minimize shipping costs from its twoprimary refineries in Tulsa, Oklahoma, andHouston, Texas. All orders within the region areshipped from one of these two refineries. An orderfor 220,000 gal comes in from a location inColorado, and another for 250,000 gal from alocation in Mississippi. The Tulsa refinery has320,000 gal ready to ship, while the Houstonrefinery has 240,000 gal. The cost of transportingeach gallon to Colorado is $0.05 from Tulsa and$0.075 from Houston. The cost of transportingeach gallon to Mississippi is $0.06 from Tulsa and$0.065 from Houston. How many gallons shouldbe distributed from each refinery to minimize thecost of filling both orders?


71. Graph the feasible region formed by the system

(a) How would you describe this region?

(b) Select random points within the region or on anyboundary line and evaluate the objective function

At what point (x, y) will thisfunction be maximized? (c) How does this relate tooptimal solutions to a linear programing problem?

f 1x, y2 � 4.5x � 7.2y.

μ

x � 0

y � 0

y � 3

x � 3

.

72. Find the maximum value of the objective functiongiven the constraints

μ

2x � 5y � 24

3x � 4y � 29

x � 6y � 26

x � 0

y � 0

.

f 1x, y2 � 22x � 15y


73. (6.7) Given the point is on the terminalside of angle with in standard position, find

a.b.c.

74. (4.6) Solve the rational inequality. Write your

answer in interval notation. x � 2

x2 � 97 0

cot �

csc �

cos �

��1�3, 42 75. (2.6) The resistance to current flow in copper wire

varies directly as its length and inversely as thesquare of its diameter. A wire 8 m long with a0.004-m diameter has a resistance of Findthe resistance in a wire of like material that is 2.7 m long with a 0.005-m diameter.

76. (7.4) Use a half-angle identity to find an exact

value for cos a7�

12b.

1500 �.

cob19537_ch09_865-878.qxd 1/27/11 8:24 PM Page 878

9.4 Partial Fraction Decomposition

One application of linear systems often seen in higher mathematics involves rewrit-ing a rational expression as a sum of its partial fractions. While most often used asa prelude to the application of other mathematical techniques, practical applicationsof the process range from a study of chemical reactions to the analysis of thermo-dynamic experiments.

A. Setting Up a Decomposition Template

Recall that a rational expression is one of the form where P and Q are polynomials

and The addition of rational expressions is widely taught in courses prior tocollege algebra, and involves combining two rational expressions into a single termusing a common denominator. For the decomposition of rational expressions, we seekto reverse this process. To begin, we make the following observations:

1. Consider the sum noting both terms are proper fractions (the degree

of the numerator is less than the degree of the denominator) and have distinctlinear denominators.

common denominator

combine numerators

result

Assuming we didn’t have the original sum to look at, reversing the processwould require us to begin with the template

and solve for the constants A and B. We know the numerators must be constant,otherwise the fraction(s) would be improper while the original expression is not.

2. Consider the sum again noting both terms are proper

fractions.

factor denominators

common denominator

combine numerators

result

Note that while the new denominator is the repeated factor bothand were denominators in the original sum. Assuming we didn’t know theoriginal sum, reversing the process would require us to begin with the template

3x � 2

1x � 122�

A

x � 1�

B

1x � 122

1x � 1221x � 121x � 122,

�3x � 2

1x � 122

�13x � 32 � 5

1x � 12 1x � 12

�31x � 12

1x � 12 1x � 12�

5

1x � 12 1x � 12

3

x � 1�

5

x2 � 2x � 1�

3

x � 1�

5

1x � 12 1x � 12

3

x � 1�

5

x2 � 2x � 1,

12x � 11

1x � 22 1x � 32�

A

x � 2�

B

x � 3

�12x � 11

1x � 22 1x � 32

�71x � 32 � 51x � 22

1x � 22 1x � 32

7

x � 2�

5

x � 3�

71x � 32

1x � 22 1x � 32�

51x � 22

1x � 32 1x � 22

7

x � 2�

5

x � 3,

Q1x2 � 0.

P1x2

Q1x2,


how we can:

A. Set up a decompositiontemplate to help rewrite arational expression as thesum of its partial fractions

B. Decompose a rationalexpression usingconvenient values

C. Decompose a rationalexpression by equatingcoefficients and using asystem of equations

D. Apply partial fractiondecomposition to atelescoping sum

Precalculus—

9–43 879

cob19537_ch09_879-892.qxd 1/27/11 9:04 PM Page 879

and solve for the constants A and B. As before, the numerator of the first termmust be constant. While the second term would still be a proper fraction if thenumerator were linear (degree 1), the denominator is a repeated linear factorand using a single constant in the numerator of all such fractions will ensure weobtain unique values for A and B (see Exercise 58). Note that for any repeatedlinear factor in the original denominator, terms of the form

must appear in

the decomposition template, although some of these numerators may turn out tobe zero.

EXAMPLE 1 � Setting Up the Decomposition Template for Linear FactorsWrite the decomposition template for

a. b.

Solution � a. Factoring the denominator gives Since the denominator

consists of two distinct linear factors, the decomposition template is:

decomposition template

b. After factoring the denominator we have , where the denominator

consists of a repeated linear factor. Using our previous observations, thedecomposition template would be:



When both distinct and repeated linear factors are present in the denominator, thedecomposition template maintains the elements illustrated above. Each distinct linearfactor appearing in the denominator, and all powers of a repeated linear factor, willhave a constant numerator.

EXAMPLE 2 � Setting Up the Decomposition Template for Repeated Linear Factors

Write the decomposition template for

Solution � Factoring the denominator gives or after factoring

completely. With a distinct linear factor of x, and the repeated linear factor the decomposition template becomes:



Returning to our observations:

x2 � 4x � 15

x1x � 122�

Ax

�B

x � 1�

C

1x � 122

1x � 122,

x2 � 4x � 15

x1x � 122x2 � 4x � 15

x1x2 � 2x � 12

x2 � 4x � 15

x3 � 2x2 � x.

x � 1

1x � 322�

A

x � 3�

B

1x � 322

x � 1

1x � 322

x � 8

12x � 32 1x � 12�

A

2x � 3�

B

x � 1

x � 8

12x � 32 1x � 12.

x � 1

x2 � 6x � 9

x � 8

2x2 � 5x � 3

A1

x � a�

A2

1x � a22�

A3

1x � a23� p �

An�1

1x � a2n�1 �An

1x � a2n

1x � a2n


Precalculus—

cob19537_ch09_879-892.qxd 1/27/11 9:04 PM Page 880

3. Consider the sum noting the denominator of the first term is linear,

while the denominator of the second is an irreducible quadratic.

find common denominator

combine numerators

result

Here, reversing the process would require us to begin with the template

allowing that the numerator of the second term might be linear since it is quadraticbut not a repeated linear factor, and noting the fraction would still be proper incases where

4. Finally, consider the sum where we note the denominator of

the first term is an irreducible quadratic, with the denominator of the second termbeing the same factor with multiplicity two.

common denominator

combine numerators

result after simplifying

Reversing the process would require us to begin with the template

allowing that the numerator of either term might be nonconstant for the reasons inobservation 3. Similar to our reasoning in observation 2, all powers of a repeatedquadratic factor must be present in the decomposition template.

EXAMPLE 3 � Setting Up the Decomposition Template for Quadratic FactorsWrite the decomposition template for

a. b.

Solution � a. With the denominator having one distinct linear factor and one irreduciblequadratic factor, the decomposition template would be:


b. The denominator consists of a repeated quadratic factor. Using our previousobservations, the decomposition template would be:



x2

1x2 � 223�

Ax � B

x2 � 2�

Cx � D

1x2 � 222�

Ex � F

1x2 � 223

x2 � 10x � 1

1x � 12 1x2 � 3x � 42�

A

x � 1�

Bx � C

x2 � 3x � 4

x2

1x2 � 223x2 � 10x � 1

1x � 12 1x2 � 3x � 42

x2 � x � 1

1x2 � 322�

Ax � B

x2 � 3�

Cx � D

1x2 � 322

�x2 � x � 1

1x2 � 322

�1x2 � 32 � 1x � 22

1x2 � 32 1x2 � 32

1

x2 � 3�

x � 2

1x2 � 322�

11x2 � 32

1x2 � 32 1x2 � 32�

x � 2

1x2 � 32 1x2 � 32

1

x2 � 3�

x � 2

1x2 � 322,

B � 0.

6x2 � 3x � 4

x1x2 � 12�

Ax

�Bx � C

x2 � 1

�6x2 � 3x � 4

x1x2 � 12

�14x2 � 42 � 12x2 � 3x2

x1x2 � 12

4x

�2x � 3

x2 � 1�

41x2 � 12

x1x2 � 12�12x � 32x

1x2 � 12x

4x

�2x � 3

x2 � 1,

9–45 Section 9.4 Partial Fraction Decomposition 881

Precalculus—

WORTHY OF NOTENote that the second term in thedecomposition template would stillbe a proper fraction if the numeratorwere quadratic or cubic, but sincethe denominator is a repeatedquadratic factor, using only a linearform ensures we obtain uniquevalues for all coefficients.

cob19537_ch09_879-892.qxd 1/27/11 9:04 PM Page 881

When both distinct and repeated factors are present in the denominator, thedecomposition template maintains the essential elements determined by observations1 through 4. Using these observations, we can formulate a general approach to thedecomposition template.

The Decomposition Template

For the rational expression and constants A, B, C, D, ... ,

1. If the degree of P is greater than or equal to the degree of Q, find the quotientand remainder using polynomial division. Only the remainder portion needbe decomposed into partial fractions.

2. Factor Q completely into linear factors and irreducible quadratic factors.3. For the linear factors, each distinct linear factor and each power of a repeated

linear factor must appear in the decomposition template and have a constantnumerator.

4. For the irreducible quadratic factors, each distinct quadratic factor and eachpower of a repeated quadratic factor must appear in the decompositiontemplate and have a linear numerator.

B. Decomposition Using Convenient Values

Once the decomposition template is obtained, we multiply both sides of the equationby the factored form of the original denominator and simplify. The resulting equa-tion is an identity (a true statement for all real numbers x), and in many cases, allthat’s required is a choice of convenient values for x to identify the constants A, B,C, and so on.

EXAMPLE 4 � Decomposing a Rational Expression Using Con venient Values

Decompose the expression into partial fractions. Use the TABLE

feature of a calculator to check your answer.

Solution � Factoring the denominator gives and we note there are two distinct

linear factors in the denominator. The decomposition template will be


Multiplying both sides by clears all denominators and yields

clear denominators

Since the equation must be true for all x, using will conveniently eliminatethe term with B, and enable us to solve for A directly:

substitute for x

simplify

term with B is eliminated

solve for A 3 � A �9 � �3A

�20 � 11 � �3A � B102�5 41�52 � 11 � A1�5 � 22 � B1�5 � 52

x � �5

4x � 11 � A1x � 22 � B1x � 52

1x � 52 1x � 22

4x � 11

1x � 52 1x � 22�

A

x � 5�

B

x � 2

4x � 11

1x � 52 1x � 22,

4x � 11

x2 � 7x � 10

P1x2

Q1x2


Precalculus—


we can set up a decomposition

template

cob19537_ch09_879-892.qxd 1/27/11 9:04 PM Page 882

To find B, we repeat this procedure, using a value that conveniently eliminates theterm with A, namely

original equation

substitute �2 for x

simplify

term with A is eliminated

solve for B

With and the complete decomposition is:

Check � In the screen, enter the original expression as and the

decomposition as (see Figure 9.36). Next, set the TABLE to

start at –3 with a step size of 1 and press to produce the table shown inFigure 9.37. Scrolling through the table, note the Y1 and Y2 outputs for any given Xare the same, providing strong evidence that Y1 � Y2. For a rigorous check that

, simply add the rational expressions on the right.4x � 11

x2 � 7x � 10�

3

x � 5�

1

x � 2

GRAPH2nd

Y2 �3

X � 5�

1

X � 2

Y1 �4X � 11

X2 � 7X � 10Y=

4x � 11

1x � 52 1x � 22�

3

x � 5�

1

x � 2

B � 1,A � 3

1 � B 3 � 3B

�8 � 11 � A102 � 3B 41�22 � 11 � A1�2 � 22 � B1�2 � 52

4x � 11 � A1x � 22 � B1x � 52

x � �2.


Precalculus—

WORTHY OF NOTEKeep in mind the zeroes of anydenominator in rational functionsare excluded from the domain. The calculator reminds us of this in Example 4 by indicating that X ��2 and X ��5 result in an“ERROR,” for both Y1 and Y2.




Decompose the expression into partial fractions.

Solution � Factoring the denominator gives or in

simplified form. With one distinct linear factor and a repeated linear factor, the

decomposition template will be .

Multiplying both sides by clears all denominators and yields

Using will conveniently eliminate the terms with A and B, enabling us tosolve for C directly:


simplify

terms with A and B are eliminated

solve for C �3 � C 9 � �3C 9 � A102 � B1�32 102 � 3C 9 � A1�5 � 522 � B1�5 � 22 1�5 � 52 � C1�5 � 22

x � �5

9 � A1x � 522 � B1x � 22 1x � 52 � C1x � 22

1x � 22 1x � 522

9

1x � 22 1x � 522�

A

x � 2�

B

x � 5�

C

1x � 522

9

1x � 22 1x � 5229

1x � 52 1x � 22 1x � 52

9

1x � 52 1x2 � 7x � 102

cob19537_ch09_879-892.qxd 1/27/11 9:04 PM Page 883

Using will conveniently eliminate the terms with B and C, enabling us tosolve for A:

original equation


simplify

terms with B and C are eliminated

solve for A

To find the value of B, we can substitute and into the previousequation, and any value of x that does not eliminate B. For efficiency’s sake, weoften elect to use or for this purpose (if possible).

original equation

substitute 1 for A, �3 for C, 0 for x

simplify

solve for B �1 � B �10 � 10B

9 � 25 � 10B � 6 9 � 110 � 522 � B10 � 22 10 � 52 � 310 � 22 9 � A1x � 522 � B1x � 22 1x � 52 � C1x � 22

x � 1x � 0

C � �3A � 1

1 � A 9 � 9A 9 � A1322 � B102 132 � C102 9 � A1�2 � 522 � B1�2 � 22 1�2 � 52 � C1�2 � 22 9 � A1x � 522 � B1x � 22 1x � 52 � C1x � 22

x � �2

With and , the complete decomposition is:

which can also be written as Figure 9.38 shows a

TABLE check, where the original expression and decomposition have been entered into Y1 and Y2, respectively.



Decompose the expression into partial fractions.

Solution � After inspection, we note the denominator can be factored by grouping, resulting

in the expression . With one distinct linear factor and one

irreducible quadratic factor, the decomposition template will be

Multiplying both sides by yields

clear denominators

Using will conveniently eliminate the term with B and C, giving

substitute 3 for x

simplify

term with B and C is eliminated

solve for A

Noting will conveniently eliminate the term with B, we substitute 0 for x and2 for A in order to solve for C:

original equation

substitute 2 for A and 0 for x 3102 � 11 � 2102 � 12 � 1B 30 4 � C2 10 � 32 3x � 11 � A1x2 � 12 � 1Bx � C2 1x � 32

x � 0

2 � A 20 � 10A 20 � 10A � 1B 33 4 � C2 102

3132 � 11 � A132 � 12 � 1B 33 4 � C2 13 � 32

x � 3

3x � 11 � A1x2 � 12 � 1Bx � C2 1x � 32

1x � 32 1x2 � 12

3x � 11

1x � 32 1x2 � 12�

A

x � 3�

Bx � C

x2 � 1.

3x � 11

1x � 32 1x2 � 12

3x � 11

x3 � 3x2 � x � 3

1

x � 2�

1

x � 5�

3

1x � 522.

9

1x � 22 1x � 522�

1

x � 2�

�1

x � 5�

�3

1x � 522,

C � �3A � 1, B � �1,


Precalculus—

Figure 9.38

cob19537_ch09_879-892.qxd 1/27/11 9:04 PM Page 884

simplify

solve for C

result

To find the value of B, we substitute 2 for A, for C, and any value of x that doesnot eliminate B. Here we chose

original equation

substitute 2 for A, �3 for C, and 1 for x

simplify

distribute

solve for B

result

With and , the complete decomposition is:

.

The result can actually be written with fewer negative signs, as

Check the result by combining these fractions.


C. Decomposition Using a System of Equations

As an alternative to using convenient values, a system of equations can be set up bymultiplying out the right-hand side (after clearing fractions) and equating coefficientsof the terms with like degrees. Here we’ll re-solve Example 6 using this method.

EXAMPLE 7 � Decomposing a Rational Expression Using a System of Equations

Decompose the expression into partial fractions by equating

coefficients of like degree terms and using a system.

Solution � From Example 6 we obtain


Multiplying both sides by yields

clear denominators

distribute/FOIL

collect like terms

By comparing the like terms on the left and right, we find

, and resulting in the system .

Using (from R1) in R3 results in the subsystem

and of this system yields giving as before. Back-substitution again verifies and


A � 2.C � �3B � �2�10B � 20,3R1 � R2

e�3B � C � 3

�B � 3C � 11,2 � 2A � �B

•

A � B � 0

�3B � C � 3

A �3C � 11

A � 3C � 11,�3B � C � 3

A � B � 0,

� 1A � B2x2 � 1�3B � C2x � 1A � 3C2 � Ax2 � A � Bx2 � 3Bx � Cx � 3C

3x � 11 � A1x2 � 12 � 1Bx � C2 1x � 32

1x � 32 1x2 � 12

3x � 11

1x � 32 1x2 � 12�

A

x � 3�

Bx � C

x2 � 1

3x � 11

x3 � 3x2 � x � 3

2

x � 3�

2x � 3

x2 � 1.

3x � 11

1x � 32 1x2 � 12�

2

x � 3�

�2x � 3

x2 � 1

C � �3A � 2, B � �2,

�2 � B 4 � �2B

14 � 4 � 2B � 6 14 � 2122 � 1B � 32 1�22

3112 � 11 � 2112 � 12 � 1B 31 4 � 3�3 4 2 11 � 32 3x � 11 � A1x2 � 12 � 1Bx � C2 1x � 32

x � 1.�3

�3 � C 9 � �3C

11 � 2 � C1�32


Precalculus—


we can decompose a rational

expression using convenient

values

cob19537_ch09_879-892.qxd 1/27/11 9:04 PM Page 885

In some cases, there are no “convenient values” to use and a system of equationsis our best and only approach. More often than not, this happens when one or more ofthe denominators are irreducible quadratic factors.

EXAMPLE 8 � Decomposing a Rational Expression Using a System of Equations

Decompose the expression into partial fractions by equating

coefficients of like degree terms and using a system.

Solution � By inspection or using a u-substitution, we find the denominator factors into aperfect square: The decomposition template is then

After clearing fractions we obtain The only “candidate” for a convenient value is but this still leaves theunknowns B and D. Instead, we multiply out the right-hand side and equatecoefficients of like terms as before.

By equating coefficients we find since the left-hand side has no cubic term,and by direct comparison. This yields the system

With the third equation shows and by substituting 5 for B in thefourth equation we find that The final form of the decomposition is


D. Partial Fractions and Telescoping Sums

From movies or the popular media (Pirates of the Caribbean,Dances With Wolves, etc.), you might be aware that the tele-scopes of old were retractable. Since they were constructed asa series of nested tubes, you could compress the length withthe lengths of the interior tubes being “negated” (see figure).In a similar fashion, some very extensive sums, called tele-scoping sums, can be rewritten in a more “compressed” form, where the interior termsare likewise negated and the resulting sum easily computed.

EXAMPLE 9 � Using Decomposition to Evaluate a Telescoping Sum

For

a. Evaluate the function for x � 1, 2, 3, and 4, then find the sum of these four terms.b. Decompose the expression into partial fractions.

t1x2 �3

x2 � x

5x2 � 2x � 12

1x2 � 322�

5

x2 � 3�

2x � 3

1x2 � 322

D � �3.C � 2,A � 0

μ

A � 0

B � 5

3A � C � 2

3B � D � 12

B � 5A � 0,

� Ax3 � Bx2 � 13A � C2x � 13B � D2 � Ax3 � 3Ax � Bx2 � 3B � Cx � D

5x2 � 2x � 12 � 1Ax � B2 1x2 � 32 � Cx � D

x � 0,� Cx � D.5x2 � 2x � 12 � 1Ax � B2 1x2 � 32

5x2 � 2x � 12

1x2 � 322�

Ax � B

x2 � 3�

Cx � D

1x2 � 322

x4 � 6x2 � 9 � 1x2 � 322.

5x2 � 2x � 12

x4 � 6x2 � 9


Precalculus—


we can decompose a rational

expression by equating

coefficients and using a

system of equations

cob19537_ch09_879-892.qxd 1/27/11 9:05 PM Page 886

c. Evaluate the decomposed form for 2, 3, and 4, leaving each result inunsimplified form.

d. Rewrite the sum from part (a) using the decomposed form of each term, andsee if you can detect a pattern that enables you to compute the original summore efficiently.

e. Note and use the pattern noted in part (d) to find the

following sum: .

Use a calculator to check your result.

Solution � a.

b. The denominator has two distinct linear factors and the decompositiontemplate will be


Multiplying both sides by clears all denominators and gives

clear denominators

Since the equation must be true for all x, using will eliminate the termwith A, and enable us to solve for B directly:


simplify, term with A is eliminated

solve for B

Repeat this procedure using to solve for A.

from template

substitute �3 for B, 1 for x

add 3 to both sides

result

With and the decomposition is

decomposed form

c. For we have

�3

4�

3

5 �

3

3�

3

4 �

3

2�

3

3 � 3 �

3

2

t142 �3

4�

3

4 � 1 t132 �

3

3�

3

3 � 1 t122 �

3

2�

3

2 � 1t112 �

3

1�

3

1 � 1

t1x2 �3x

�3

x � 1,

3

x1x � 12�

3x

�3

x � 1

B � �3,A � 3

3 � A 6 � 2A 3 � A11 � 12 � 3112 3 � A1x � 12 � Bx

x � 1

�3 � B 3 � �B 3 � A1�1 � 12 � B1�12

x � �1

3 � A1x � 12 � Bx

x1x � 12

3

x1x � 12�

Ax

�B

x � 1

�48

20�

12

5

�30

20�

10

20�

5

20�

3

20

3

1 � 1�

3

4 � 2�

3

9 � 3�

3

16 � 4�

3

2�

1

2�

1

4�

3

20

3

1 # 2�

3

2 # 3�

3

3 # 4� p �

3

24 # 25

3

x2 � x�

3

1x2 1x � 12

x � 1,


Precalculus—

cob19537_ch09_879-892.qxd 1/27/11 9:05 PM Page 887

d. Replacing each term in the given sum by the equivalent term in decomposedform yields

and we note that all interior terms will cancel (add to zero) leaving only the

first and last terms of the sum. The result is

e. From the pattern in part (d) we note that regardless of the number of terms inthe sum, all interior terms will cancel (add to zero) leaving only the first andlast terms.

Check � To check this result using a calculator, we begin by clearing out lists L1, L2, andL3 and entering the integers 1 through 24 into L1. These values will serve as thefirst factors of each denominator in the given sum. Next, move the cursor onto the L2 column header and use the keystrokes (L1�1)to fill L2 with the second factors of each denominator. Now we can use L3 togenerate each individual term in the sum, as shown in Figure 9.39. At this point,pressing will cause these terms appear and we QUIT to the home screen. From here, use the calculator to sum the L3 terms with the keystrokes (LIST) 5:sum( . As Figure 9.40 shows, this sum is equal

to and the result from part (e) checks.72

25

ENTER)32nd

STAT2nd

ENTER

ENTER)1+12nd

�72

25

�3

1�

3

25�

75

25�

3

25

� a3

1�

3

2b � a

3

2�

3

3b � a

3

3�

3

4b � p � a

3

24�

3

25b

3

1 # 2 �

3

2 # 3�

3

3 # 4� p �

3

24 # 25

3

1�

3

5�

15

5�

3

5�

12

5.

� a3

1�

3

2b � a

3

2�

3

3b � a

3

3�

3

4b � a

3

4�

3

5b

3

1 � 1�

3

4 � 2�

3

9 � 3 �

3

16 � 4


Precalculus—



As a final note, if the degree of the numerator is greater than the degree of thedenominator in the original expression, divide using long division and apply themethods above to the rational remainder. For instance, you can check that

and decomposing the remainder expression

gives a final result of 3x �2

x � 1�

9

1x � 122.

3x3 � 6x2 � 5x � 7

x2 � 2x � 1� 3x �

2x � 7

1x � 122,D. You’ve just seen

how we can apply partial

fraction decomposition to

a telescoping sum

cob19537_ch09_879-892.qxd 1/27/11 9:05 PM Page 888

2. Before beginning the process of partial fractiondecomposition, the rational expression must be

. If not, use polynomial division.



9.4 EXERCISES

1. In order to rewrite a rational expression as the sumof its partial fractions, we must set up adecomposition .

3. If the denominator of a rational expression containsa factor, each power of the factormust appear in the decomposition template andhave a constant numerator.

4. If the denominator of a rational expression containsa distinct irreducible quadratic factor, it mustappear in the decomposition template with a

numerator.

6. Discuss/Explain the first steps of the process of

writing as a sum of partial fractions.x2

x2 � 7x � 12

5. Discuss/Explain the process of writing as a sum of partial fractions.

8x � 3

x2 � x


The exercises below are designed solely to reinforce thevarious possibilities for decomposing a rationalexpression. All are proper fractions whose denominatorsare completely factored. Write the decompositiontemplate, but do not solve.

7. 8.

9. 10.

11. 12.

13. 14.

15. 16.

17. 18.

19. 20.

21. 22.

23. 24.2x3 � 3x2 � 4x � 1

x1x2 � 322x3 � 3x � 2

1x � 12 1x2 � 222

7x2 � 3x � 1

1x � 12 1x2 � 2x � 52

2x2 � 3

1x � 32 1x2 � 5x � 72

3x2 � 5

x212x � 122x3 � 2x � 5

x21x � 522

x2 � 3x � 5

1x � 32 1x � 222x2 � x � 1

x21x � 22

x2 � 2x � 3

1x � 423x2 � 2x � 4

1x � 523

x2 � 7

1x � 42 1x � 22x

x2 � 5

x1x � 32 1x � 12

�2x2 � 3x � 4

1x � 32 1x � 12 1x � 22

3x2 � 2x � 5

1x � 12 1x � 22 1x � 32

x � 7

1x � 3222x � 5

1x � 122

�4x � 1

1x � 22 1x � 52

3x � 2

1x � 32 1x � 22

Decompose each rational expression into partialfractions using convenient values. Use the TABLEfeature of a calculator to verify your results.

25. 26.

27. 28.

29. 30.

31. 32.

33. 34.

35. 36.

37. 38.

39. 40.2x2 � 14x � 7

x3 � 2x2 � 5x � 10

3x2 � 10x � 4

8 � x3

2x3 � x � 6

x4 � 2x3 � 3x2

3x3 � 2x2 � 7x � 3

x4 � x3 � 3x2

�3x4 � 11x2 � x � 12

x5 � 4x3 � 4x

x4 � 3x2 � 2x � 1

x5 � 2x3 � x

3x2 � 4x � 1

x3 � 1

6x2 � x � 13

x3 � 2x2 � 3x � 6

2x3 � 2x2 � 6x

x4 � 4x2 � 3

3x3 � 3x2 � 3x � 5

x4 � 3x2 � 2

�2x2 � 7x � 28

x3 � 4x2 � 4x

3x2 � 7x � 1

x3 � 2x2 � x

x2 � 24x � 12

x3 � 4x

8x2 � 3x � 7

x3 � x

�11x � 6

5x2 � 4x � 12

2x � 27

2x2 � x � 15


Precalculus—

cob19537_ch09_879-892.qxd 1/27/11 9:05 PM Page 889

When the denominator of the rational expression contains repeated factors, using a system of equations sometimesoffers a more efficient way to find the needed coefficients. Decompose each rational expression into partial fractionsby equating coefficients and using a system of equations.

41. 42. 43.

44. 45. 46.


Logistic equations and population size: Logistic equations are often used to model the growth of various populations. Initially growth is very near to exponential, but then due to certain limitingfactors (food, limited resources, space, etc.), growth slows and reaches a limit called the carryingcapacity c. In the process of solving the logistic equation for P(t), we often need to decompose the expression shown into partial fractions, where P(0) represents the initial population. Decompose the expression for the following values given.

47. 48. 49. 50.

� APPLICATIONS

Telescoping sums: For each function given, use the method illustrated in Example 9 to find a pattern that enables you tocompute the sum shown quickly.

51.

52.

2

1 # 2�

2

2 # 3�

2

3 # 4� p �

2

19 # 20

t1x2 �2

x2 � x;

1

1 # 2�

1

2 # 3�

1

3 # 4� p �

1

49 # 50

t1x2 �1

x2 � x;

P102 � 20, c � 80P102 � 10, c � 100P102 � 80, c � 20P102 � 100, c � 10

5x2 � 20x � 21

x3 � 6x2 � 12x � 8

2x2 � 4x � 5

x3 � 3x2 � 3x � 1

x3 � 5x2 � 6x � 37

x4 � 14x2 � 49

2x3 � x2 � 5x � 1

x4 � 2x2 � 1

14 � 3x

x2 � 8x � 16

5x � 13

x2 � 6x � 9


Precalculus—

1

P aP102 �P 102

cPb

53.

54.

1

1 # 2�

3

2 # 5�

5

5 # 10� p �

21

101 # 122

t1x2 �2x � 1

1x2 � 12 1x2 � 2x � 22;

1

1 # 3�

1

3 # 5�

1

5 # 7� p �

1

123 # 125

t1x2 �1

12x � 12 12x � 12;


Decompose the following expressions into partial fractions.

55.

56.

57. As written, the rational expressiongiven cannot be decomposed usingthe standard template (two distinctlinear factors). Try to apply this template and see what happens. What can be done to decomposethe expression?

1 � ex

1ex � 32 1e2x � 2ex � 12

ln x � 2

1ln x � 22 1ln x � 12258. Try to decompose the rational expression

using the decomposition template

(note the second rational expression

in this template is a proper fraction). (a) How doesthis affect the decomposition process? Is thedecomposition unique? (b) Next, use the standardtemplate as outlined in this section. What do younotice?

Ax

�Bx � C

x2

3x � 1

x2

x � 2

1x � 12 11 � x2

cob19537_ch09_879-892.qxd 1/27/11 9:05 PM Page 890


59. (4.1) Use polynomial division to rewrite

in the form .

60. (4.1) Use the remainder theorem to find f (3) for.

61. (7.2) Verify that is an

identity.

cos3�

sin �� cot � � cos � sin �

f 1x2 � 2x3 � x2 � 10

q1x2 �r1x2

d1x2

2x3 � 3x2 � 13x � 5

x2 � x � 6

62. (7.5) Evaluate the following expressions:

a.

b. cos�1 c cosa��

6b d

cos c cos�1a�23

2b d

9–55 Mid-Chapter Check 891

Precalculus—

MID-CHAPTER CHECK

1. Solve by graphing the system on a graphingcalculator. State whether the system is consistent,inconsistent, or dependent.

2. Solve the system using elimination. State whetherthe system is consistent, inconsistent, ordependent:

3. Solve using a system of linear equations and anymethod you choose. How many ounces of a 40%acid should be mixed with 10 oz of a 64% acid toobtain a 48% acid solution?

4. Determine whether the ordered triple is a solution tothe system. If not, identify which equation(s) arenot satisfied.

5. The system given is a dependent system. Withoutsolving, state why.

6. Solve the system of equations:

•

x � 2y � 3z � �4

2y � z � 7

5y � 2z � 4

•

x � 2y � 3z � 3

2x � 4y � 6z � 6

x � 2y � 5z � �1

•

5x � 2y � 4z � 22

2x � 3y � z � �1

3x � 6y � z � 2

12, 0, �32

ex � 3y � �4

2x � y � 13

ex � 3y � �2

2x � y � 3

7. Solve using elimination:

8. Decompose the expression into

partial fractions.

9. Child prodigies: If you add Mozart’s age when hewrote his first symphony, with the age ofAmerican chess player Paul Morphy when hebegan dominating the international chess scene,and the age of Blaise Pascal when he formulatedhis well-known Essai pour les coniques (Essayon Conics), the sum is 37. At the time of eachevent, Paul Morphy’s age was 3 yr less than twiceMozart’s, and Pascal was 3 yr older than Morphy.Set up a system of equations and find the age of each.

10. Candle manufacturing: David and Karen maketable candles and holiday candles and sell themout of their home. Dave works 10 min and Karenworks 30 min on each table candle, while Karenworks 40 min and David works 20 min on eachholiday candle. Dave can work at most 3 hr and 20 min (200 min) per day on the home business,while Karen can work at most 7 hr. If table candlessell for $4 and holiday candles sell for $6, howmany of each should made to maximize theirrevenue?

�x2 � 4x � 15

1x � 12 1x � 222

•

2x � 3y � 4z � �4

x � 2y � z � 0

�3x � 2y � 2z � �1

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Precalculus—

REINFORCING BASIC CONCEPTS

Window Size and Graphing T echnologySince most substantial applications involve noninteger values, technology can play an important role in applyingmathematical models. However, with its use comes a heavy responsibility to use it carefully. A very real effort must bemade to determine the best approach and to secure a reasonable estimate. This is the only way to guard against (theinevitable) keystroke errors, or ensure a window size that properly displays the results.

RationaleOn October 1, 1999, the newspaper USA TODAY ran an article titled, “Bad Math added up to Doomed Mars Craft.” Thearticle told of how a $125,000,000.00 spacecraft was lost, apparently because the team of scientists that plotted thecourse for the craft used U.S. units of measurement, while the team of scientists guiding the craft were using metricunits. NASA’s space chief was later quoted, “The problem here was not the error, it was the failure of . . . the checksand balances in our process to detect the error.”

No matter how powerful the technology, always try to begin your problem-solving efforts with an estimate. Beginby exploring the context of the problem, asking questions about the range of possibilities: How fast can a human run?How much does a new car cost? What is a reasonable price for a ticket? What is the total available to invest? There isno calculating involved in these estimates, they simply rely on “horse sense” and human experience. In many appliedproblems, the input and output values must be positive—which means the solution will appear in the first quadrant,narrowing the possibilities considerably. This information will be used to set the viewing window of your graphingcalculator, in preparation for solving the problem using a system and graphing technology.

Illustration 1 � Erin just filled both her boat and Blazer with gas, at a total cost of $211.14. She purchased 35.7 gallons of premium for her boat and 15.3 gal of regular for her Blazer. Premium gasoline cost $0.10 per gallonmore than regular. What was the cost per gallon of each grade of gasoline?

Solution � Asking how much you paid for gas the last time you filled up should serveas a fair estimate. Certainly (in 2011) a cost of $6.00 or more per gallon in the UnitedStates is too high, and a cost of $1.50 per gallon or less would be too low. Also, we canestimate a solution by assuming that both kinds of gasoline cost the same. This wouldmean 51 gal were purchased for about $211, and a quick division would place theestimate at near per gallon. A good viewing window would be restricted tothe first quadrant with maximum values of and .

Exercise 1: Solve Illustration 1 using graphing technology.

Exercise 2: Re-solve Exercises 63 and 64 from Section 9.1 using graphing technology. Verify results are identical.

Ymax � 6Xmax � 61since cost 7 02

21151 � $4.14

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Precalculus—

9.5 Solving Linear Systems Using Matrices and Row Operations

Just as synthetic division streamlines the process of polynomial division, matrices androw operations streamline the process of solving systems using elimination. With theequations of the system in standard form, the location and order of the variable termsand constant terms are predetermined, and we simply apply the elimination process onthe coefficients and constants alone.

A. Introduction to Matrices

In general terms, a matrix is simply a rectangular arrangement of numbers, calledthe entries of the matrix. Matrices (plural of matrix) are denoted by enclosing theentries between a left and right bracket, and named using a capital letter, such as

and They occur in many different sizes

as defined by the number of rows and columns each has, with the number of rowsalways given first. Matrix A is said to be a 2 � 3 (two by three) matrix, since it has tworows and three columns. Matrix B is a 3 � 3 (three by three) matrix.

EXAMPLE 1A � Identifying the Size and Entries of a MatrixDetermine the size of each matrix and identify the entry located in the second rowand first column.

a. b.

Solution � a. Matrix C is 3 � 2. The row 2, column 1 entry is 1.b. Matrix D is 3 � 4. The row 2, column 1 entry is �0.4.

If a matrix has the same number of rows and columns, it’s called a square matrix.Matrix B above is a square matrix, while matrix A is not. For square matrices, thevalues on a diagonal line from the upper left to the lower right are called the diagonalentries and are said to be on the diagonal of the matrix. When solving systems usingmatrices, much of our focus is on these diagonal entries.

EXAMPLE 1B � Identifying the Diagonal Entries of a Square MatrixName the diagonal entries of each matrix.

a. b.

Solution � a. The diagonal entries of matrix E are 1 and �3.b. For matrix F, the diagonal entries are 0.2, 0.3, and 0.6.


B. The Augmented Matrix of a System of Equations

A system of equations can be written in matrix form by augmenting or joining the coefficient matrix, formed by the variable coefficients, with the matrix of constants.

F � £

0.2 �0.5 0.7

�0.4 0.3 1

2.1 �0.1 0.6

§E � c1 4

�2 �3d

D � £

0.2 �0.5 0.7 3.3

�0.4 0.3 1 2

2.1 �0.1 0.6 4.1

§C � £

3 �2

1 5

�4 3

§

B � £

2 �1 3

4 6 �2

1 0 �1

§ .A � c1 �3 2

5 1 �1d


how we can:

A. State the size of a matrixand identify its entries

B. Form the augmentedmatrix of a system ofequations

C. Solve a system ofequations using rowoperations

D. Solve a system ofequations usingtechnology

E. Recognize inconsistentand dependent systems

F. Solve applications usingmatrix methods


we can state the size of a

matrix and identify its entries

9–57 893

cob19537_ch09_893-905.qxd 1/27/11 9:11 PM Page 893


Precalculus—

The coefficient matrix for the system is with

column 1 for the coefficients of x, column 2 for the coefficients of y, and so on. The

matrix of constants is These two are joined to form the augmented matrix,

with a dotted line often used to separate the two as shown here: .

It’s important to note the use of a zero placeholder for the y-variable in the second rowof the matrix, signifying there is no y-variable in the corresponding equation.

EXAMPLE 2 � Forming Augmented MatricesForm the augmented matrix for each system, and name the diagonal entries of eachcoefficient matrix.

a. b.

Solution � a.

Diagonal entries: 2 and 3.

b.

Diagonal entries: and 1.Notice that in the third row of part (b), the fact that is indicated by zeroesin the x- and y-columns, 1 in the z-column, and a 6 in the column of constants.


This process can easily be reversed to write a system of equations from a givenaugmented matrix.

EXAMPLE 3 � Writing the System Cor responding to an Augmented MatrixWrite the system of equations corresponding to each matrix, then solve the systemusing back-substitution.

a. b.

Solution � a.

With , we have

substitute 4 for y

multiply

add 20

divide by 3

The solution is (2, 4) x � 2

3x � 6 3x � 20 � �14

3x � 5142 � �14

y � 4

e3x � 5y � �14

1y � 4¡c

3 �5 �14

0 1 4d

£

1 4 �1 �10

0 �3 10 7

0 0 1 1

§c3 �5 �14

0 1 4d

z � 6

12, 23,

£

12 1 0 5

1 23

56 11

0 0 1 6

§¡•

12 x � y � 5

x � 23 y � 56z � 11

z � 6

c2 1 11

�1 3 �2d¡e

2x � y � 11

�x � 3y � �2

•

12x � y � 5

x � 23y � 5

6z � 11

z � 6

e 2x � y � 11

�x � 3y � �2

£

2 3 �1 1

1 0 1 2

1 �3 4 5

§

£

1

2

5

§ .

£

2 3 �1

1 0 1

1 �3 4

§•

2x � 3y � z � 1

x � z � 2

x � 3y � 4z � 5

cob19537_ch09_893-905.qxd 1/27/11 9:11 PM Page 894

b.

With , we first substitute 1 for z in the second equation.

substitute 1 for z

multiply

subtract 10

divide by �3

We then substitute 1 for z and 1 for y in the first equation.

substitute 1 for y and 1 for z

simplify

subtract 3

The solution is (�13, 1, 1).


C. Solving a System Using Matrices

When a system of equations is written in augmented matrix form, we can solve the sys-tem by applying the same operations to each row of the matrix, that would be appliedto the equations in the system. In this context, the operations are referred to as elemen-tary row operations.

Elementary Row Operations

1. Any two rows in a matrix can be interchanged.2. The elements of any row can be multiplied by a nonzero constant.3. Any two rows can be added together, and the sum used to replace

one of the rows.

In this section, we’ll use these operations to triangularize the augmented matrix,employing a solution method known as Gaussian elimination. A square matrix is saidto be in triangular form when all of the entries below the diagonal are zero. For example,

the matrix is in triangular form:

In system form we have meaning a matrix written in triangular

form can be used to solve the system using back-substitution. We’ll illustrate by

solving , using elimination to the left, and row operations on the

augmented matrix to the right. As before, R1 represents the first equation in the systemand the first row of the matrix, R2 represents equation 2 and row 2, and so on. The cal-culations involved are shown for the first stage only and are designed to offer a carefulcomparison. In actual practice, the format shown in Example 4 is used.

•

1x � 4y � 1z � 4

2x � 5y � 8z � 15

1x � 3y � 3z � 1

•

x � 4y � z � �10

� 3y � 10z � 7

z � 1

,

£

1 4 �1 �10

0 �3 10 7

0 0 1 1

§ .£

1 4 �1 �10

0 �3 10 7

0 0 1 1

§

x � �13 x � 3 � �10

x � 4112 � 1112 � �10

y � 1 �3y � �3

�3y � 10 � 7 �3y � 10112 � 7

z � 1

•

1x � 4y � 1z � �10

� 3y � 10z � 7

1z � 1

¡£

1 4 �1 �10

0 �3 10 7

0 0 1 1

§

9–59 Section 9.5 Solving Linear Systems Using Matrices and Row Operations 895

Precalculus—


we can form the augmented

matrix of a system of

equations

cob19537_ch09_893-905.qxd 1/27/11 9:11 PM Page 895


Precalculus—

Elimination Row Operations (System of Equations) (Augmented Matrix)

To eliminate the x-term in R2, we use Identical operations areperformed on the matrix, which begins the process of triangularizing the matrix.

System Form Matrix Form

For R3, the operations are

As always, we should look for opportunities to simplify any equation in the system(and any row in the matrix). Note that �1R3 will make the coefficients and relatedmatrix entries positive. Here is the new system and matrix.

New System New Matrix

On the left, we would finish by solving the subsystem using. In matrix form, we eliminate the corresponding entry (third row,

second column) to triangularize the matrix.

Dividing R3 by 16 gives z � 1 in the system, and entries of 0 0 1 1 in the aug-mented matrix. Completing the solution by back-substitution in the system gives theordered triple (1, 1, 1). For practice using these row operations, see Exercises 19through 27.

The general solution process is summarized here.

Solving Systems by Triangularizing the Augmented Matrix

1. Write the system as an augmented matrix.2. Use row operations to obtain zeroes below the first diagonal entry.3. Use row operations to obtain zeroes below the second diagonal entry.4. Continue until the matrix is triangularized (entries below diagonal are zero).5. Divide to obtain a “1” in the last diagonal entry (if it is nonzero), then convert

to equation form and solve using back-substitution.

Note: At each stage, look for opportunities to simplify row entries using multiplica-tion or division. Also, to begin the process any equation with an x-coefficientof 1 can be made R1 by interchanging the equations.

£

1 4 �1 4

0 �3 10 7

0 0 16 16

§ £

1 4 �1 4

0 �3 10 7

0 1 2 3

§

R2 � 3R3 S R32 � 2

£

1 4 �1 4

0 �3 10 7

0 1 2 3

§•

1x � 4y � 1z � 4

�3y � 10z � 7

1y � 2z � 3

�1 �4 1 �4

1 3 �3 10 �1 �2 �3

�1R1�R3

New R3

�1x � 4y � 1z � �4

1x � 3y � 3z � 1� 1y � 2z � �3

�1R1�R3

New R3

�1R1 � R3 S R3.

�2 �8 2 �8

2 5 8 150 �3 10 7

�2R1�R2

New R2

�2x � 8y � 2z � �8

2x � 5y � 8z � 15� 3y � 10z � 7

�2R1�R2

New R2

�2R1 � R2 S R2.

£

1 4 �1 4

2 5 8 15

1 3 �3 1

§•

1x � 4y � 1z � 4

2x � 5y � 8z � 15

1x � 3y � 3z � 1

R2 � 3R3 R3¬¬¬¬¡S

WORTHY OF NOTEThe procedure outlined for solvingsystems using matrices is virtuallyidentical to that for solving systemsby elimination. Using a 3 � 3system for illustration, the “zeroesbelow the first diagonal entry”indicates we’ve eliminated the x-term from R2 and R3, the “zeroesbelow the second entry” indicateswe’ve eliminated the y-term fromthe subsystem, and the division “to obtain a ‘1’ in the final entry”indicates we have just solved for z.

cob19537_ch09_893-905.qxd 1/27/11 9:11 PM Page 896

EXAMPLE 4 � Solving Systems Using the Augmented Matrix

Solve by triangularizing the augmented matrix:

Solution � matrix form S R1 4 R2

�2R1 � R2 S R2 �1R2 S R2

2R2 � R3 S R3

Converting the augmented matrix back into equation form yields Back-substituting 1 for zin the second equation gives , with showing . Back-substituting 1for z and 1 for y in the first equation gives , or . Subtracting 2 fromboth sides shows . The solution is .


As mentioned, the process used in Example 4 is called Gaussian elimination(Carl Friedrich Gauss, 1777–1855), with the last matrix written in row-echelon form.It’s also possible to solve a system entirely using only the augmented matrix, by contin-uing to use row operations until the diagonal entries are 1’s, with 0’s for all other entries

in the coefficient matrix: . The process is then called Gauss-Jordan

elimination (Wilhelm Jordan, 1842–1899), with the final matrix written in reducedrow-echelon form (see Appendix D).

Note that with Gauss-Jordan elimination, our initial focus is less on getting 1’salong the diagonal, and more on obtaining zeroes for all entries other than the diagonalentries. This will enable us to work with integer values in the solution process, asshown in Example 5.

EXAMPLE 5 � Solving a System Using Gauss-Jordan Elimination

Solve using Gauss-Jordan elimination

standard form

Solution � matrix form S �R1 � R2 S R2

£

10 0 13 43

0 5 �6 �16

0 0 1 1

§R329

S R3£

10 0 13 43

0 5 �6 �16

0 0 29 29

§

2R2 � 5R1 S R1

�4R2 � 5R3 S R3

£

2 �2 5 15

0 5 �6 �16

0 4 1 �7

§

£

2 �2 5 15

0 5 �6 �16

0 4 1 �7

§£

2 �2 5 15

2 3 �1 �1

0 4 1 �7

§•

2x � 2y � 5z � 15

2x � 3y � 1z � �1

0x � 4y � 1z � �7

•

2x � 5z � 15 � 2y

2x � 3y � �1 � z

4y � z � �7

.

£

1 0 0 a

0 1 0 b

0 0 1 c

§

1�3, 1, 12x � �3x � 2 � �1x � 1 � 1 � �1

y � 1y � 4 � 5y � 4112 � 5z � 1.

£

1 1 1 �1

0 1 4 5

0 0 1 1

§R37

S R3£

1 1 1 �1

0 1 4 5

0 0 7 7

§£

1 1 1 �1

0 1 4 5

0 �2 �1 �3

§

£

1 1 1 �1

0 1 4 5

0 �2 �1 �3

§£

1 1 1 �1

0 �1 �4 �5

0 �2 �1 �3

§£

1 1 1 �1

2 1 �2 �7

0 �2 �1 �3

§

£

1 1 1 �1

2 1 �2 �7

0 �2 �1 �3

§£

2 1 �2 �7

1 1 1 �1

0 �2 �1 �3

§•

2x � y � 2z � �7

x � y � z � �1

�2y � z � �3

•

2x � y � 2z � �7

x � y � z � �1

�2y � z � �3


Precalculus—

cob19537_ch09_893-905.qxd 1/27/11 9:11 PM Page 897


Precalculus—

The final matrix shows the solution is or .


D. Solving Systems of Equations Using Technology

Graphing calculators offer a very efficient way to solve systems of equations usingmatrices. Once the system has been written in matrix form, it can easily be entered intoa calculator and solved by having the calculator instantly perform the row operationsneeded to produce the diagonal of 1’s (with zeroes for all other coefficient matrix entries).On many calculators, pressing (MATRIX) gives a screen similar to the oneshown in Figure 9.41, where we begin by selecting the EDIT option (push the rightarrow twice). Pressing places you on a screen where you can EDIT matrix A,changing the size as needed, then input the entries of the matrix. To use the 3 � 4 matrixfrom Example 4, we press 3 and then 4 , giving the screen shown in Figure 9.42.The dash marks to the right indicate that there is a fourth column that cannot be seen,but that comes into view as you enter the elements of the matrix. Begin entering thefirst row of the matrix, which is . Press after each entry and the cur-sor automatically goes to the next position in the matrix (note the calculator automati-cally shifts left and right to allow all four columns to be entered). After entering thesecond row and the third row the completed matrixshould look like the one shown in Figure 9.43 (the matrix is currently shifted to theright, showing the fourth column).

50, �2, �1, �3651, 1, 1, �16

ENTER52, 1, �2, �76

ENTERENTER

ENTER

x-12nd

13, �2, 12x � 3, y � �2, and z � 1,

£

1 0 0 3

0 1 0 �2

0 0 1 1

§£

10 0 0 30

0 5 0 �10

0 0 1 1

§

�13R3 � R1 S R1

6R3 � R2 S R2

£

10 0 13 43

0 5 �6 �16

0 0 1 1

§ R25

S R2

R110

S R1

Figure 9.41 Figure 9.42 Figure 9.43


we can solve a system

of equations using row

operations

Most graphing calculators have an rref function, which is short for reduced rowechelon form. To access this function, press (MATRIX) and select theMATH option, then scroll upward (or downward) until you get to the B:rref option.Pressing places this function on the home screen, where we must tell it to performthe rref operation on matrix [A]. Pressing the once again returns us to thematrix options, enabling us to select matrix NAMES(notice that NAMES is automatically highlighted).With the cursor over 1:[A] 3�4 we press andmatrix [A] is placed on the home screen as the objectof the rref function. After pressing and thecalculator quickly computes the reduced row echelonform and displays it on the screen, as in Figure 9.44.The solution is easily read as and

, just as we found in Example 4.z � 1x � �3, y � 1,

ENTER)

ENTER

x-12nd

ENTER

x-12nd

Figure 9.44

cob19537_ch09_893-905.qxd 2/11/11 5:12 PM Page 898



we can solve a system of

equations using technology

EXAMPLE 6 � Solving a System of Equations Using T echnologySolve the following system using a graphing calculator.

Solution � Entering the system as matrix [A] produces the screen shown in Figure 9.45. Usingthe rref([A]) command as shown above produces the screen in Figure 9.46. Thesolution is the ordered triple (�2, 1, 3).

Check by substituting �2 for x, 1 for y, and 3 for z in the original equations.

•

2x � 3y � z � �4

x � 3y � 2z � 1

�3x � y � 4z � �5


Precalculus—


E. Inconsistent and Dependent Systems

Due to the strong link between a linear system and its augmented matrix, inconsistentand dependent systems can be recognized just as in Sections 9.1 and 9.2. An inconsis-tent system will yield an inconsistent or contradictory statement such as meaning all entries in a row of the matrix of coefficients are zero, but the constant isnot. A linearly dependent system will yield an identity statement such as meaning all entries in one row of the matrix are zero. If the system has coincidentdependence, all rows will be zero except one.


Solve the system using Gauss-Jordan elimination:

Solution �

standard form S matrix form S

Since all entries in the last row are zeroes and it’s the only row of zeroes, we conclude the system

is linearly dependent and equivalent to (a calculator solution is shown in the

figure). As in Section 9.2, we demonstrate this dependence by writing the (x, y, z) solution in terms of

ex � 2z � 1

y � 3z � 2

£

1 0 �2 1

0 1 �3 2

0 0 0 0

§

�1R2 � R1 S R1

3R2 � R3 S R3

£

1 1 �5 3

0 1 �3 2

0 �3 9 �6

§

R1 � R2 S R2

�2R1 � R3 S R3

£

1 1 �5 3

�1 0 2 �1

2 �1 �1 0

§

£

1 1 �5 3

�1 0 2 �1

2 �1 �1 0

§•

x � y � 5z � 3

�x � 0y � 2z � �1

2x � y � z � 0

•

x � y � 5z � 3

�x � 2z � �1

2x � y � z � 0

•

x � y � 5z � 3

�x � 2z � �1

2x � y � z � 0

0 � 0,

0 � �12,

cob19537_ch09_893-905.qxd 1/27/11 9:11 PM Page 899


Precalculus—

a parameter. Solving for y in R2 gives y in terms of z: Solving for x in R1 gives x in terms of z: . As written, thesolutions all depend on z: and Selecting p as the parameter (or some other “neutral” variable), wewrite the solution as Two of the infinite numberof solutions would be (1, 2, 0) for and for

Test these triples in the original equations.


Since there was only one row of zeroes in the system from Example 7, we knew thesystem was linearly dependent. As mentioned previously, for coincident dependencethe system will have only one nonzero row, with all other rows consisting entirely ofzeroes. Note that if we change the constant in the third equation of the original system(the entry for row 3, column 4 was changed to a 5), an inconsistent system results asone equation is no longer dependent on the others (see Figures 9.47 and 9.48).

p � �1. 1�1, �1, �12p � 0,

12p � 1, 3p � 2, p2.

z � z.x � 2z � 1, y � 3z � 2,x � 2z � 1

y � 3z � 2.




and dependent systems

When the number of variables and equations in a system increases, solutions usingtechnology can be particularly effective. In addition, some applications generate sys-tems where the number of equations and number of variables are unequal, which thetechnology still handles efficiently.

EXAMPLE 8 � Solving a System of Equations Using T echnologySolve the following system using a graphing calculator.

Solution � After entering the system as matrix [A], using therref([A]) command produces the screen shown inthe figure. Note that in this case, while the solutioncontains a row of zeroes, the system is not linearlydependent as there are more equations thanvariables, and unique values have been found for x,y, and z. The solution is (�1, 2, 1).


F. Solving Applications Using Matrices

As in other areas, solving applications using systems relies heavily on the ability tomathematically model information given verbally or in context. As you work throughthe exercises, read each problem carefully. Look for relationships that yield a systemof two equations in two variables, three equations in three variables, and so on.

μ

x � 3y � z � 4

�2x � y � 3z � 7

�2y � 4z � 0

�x � 4y � 2z � 11

cob19537_ch09_893-905.qxd 1/27/11 9:11 PM Page 900


Precalculus—

F. You’ve just seen how


using matrix methods

EXAMPLE 9 � Determining the Original Value of Collector’s ItemsA museum purchases a famous painting, a ruby tiara, and a rare coin for its collection, spendinga total of $30,000. One year later, the painting has tripled in value, while the tiara and the coinhave doubled in value. The items now have a total value of $75,000. Find the purchase price ofeach if the original price of the painting was $1000 more than twice the coin.

Solution � Let P represent the price of the painting, T the tiara, and C the coin.

Total spent was $30,000:One year later:Value of painting versus coin:

standard form S matrix form S

�1R2 � R3 S R3

From R3 of the triangularized form, directly. SinceR2 represents we find the tiara was purchasedfor Substituting these values into the first equationshows the painting was purchased for $15,000. The solution is(15,000, 8000, 7000). The solution found using a graphingcalculator is also shown.


T � $8000.T � C � 15,000,

C � $7000

£

1 1 1 30000

0 1 1 15000

0 0 1 7000

§R32

S R3£

1 1 1 30000

0 1 1 15000

0 0 2 14000

§£

1 1 1 30000

0 1 1 15000

0 1 3 29000

§

£

1 1 1 30000

0 1 1 15000

0 1 3 29000

§�1R2 S R2

�1R3 S R3

£

1 1 1 30000

0 �1 �1 �15000

0 �1 �3 �29000

§

�3R1 � R2 S R2

�1R1 � R3 S R3

£

1 1 1 30000

3 2 2 75000

1 0 �2 1000

§

£

1 1 1 30000

3 2 2 75000

1 0 �2 1000

§•

1P � 1T � 1C � 30,000

3P � 2T � 2C � 75,000

1P � 0T � 2C � 1000

•

P � T � C � 30,000

3P � 2T � 2C � 75,000

P � 2C � 1000

S P � 2C � 1000S 3P � 2T � 2C � 75,000S P � T � C � 30,000

9.5 EXERCISES

2. When the coefficient matrix is used with the matrixof constants, the result is a(n) matrix.



1. A matrix with the same number of rows andcolumns is called a(n) matrix.

3. Matrix is a by

matrix. The entry in the second row and thirdcolumn is .

A � c2 4 �3

1 �2 1d 4. Given matrix B shown

here, the diagonal entries are , , and . B � £

1 4 3

�1 5 2

3 �2 1

§

5. The notation indicates that anequivalent matrix is formed by performing whatoperations/replacements?

�2R1 � R2 S R2 6. Describe how to tell an inconsistent system apartfrom a dependent system when solving usingmatrix methods (row reduction).

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Precalculus—


Determine the size of each matrix and identify the thirdrow and second column entry. If the matrix given is asquare matrix, identify the diagonal entries.

7. 8.

9.

Form the augmented matrix, then name the diagonalentries of the coefficient matrix.

10.

11.

12.

Write the system of equations for each matrix. Then useback-substitution to find its solution.

13.

14.

15.

16.

17.

18. £1 2 �1 3

0 1 16

23

0 0 1 227

§

£

1 3 �4 29

0 1 �32

212

0 0 1 3

§

£

1 0 7 �5

0 1 �5 15

0 0 1 �26

§

£

1 2 �1 0

0 1 2 2

0 0 1 3

§

c1 �5 �15

0 �1 �2d

c1 4 5

0 1 12

d

•

2x � 3y � z � 5

2y � z � 7

x � y � 2z � 5

•

x � 2y � z � 1

x � z � 3

2x � y � z � 3

•

2x � 3y � 2z � 7

x � y � 2z � �5

3x � 2y � z � 11

≥

1 0 4

1 3 �7

5 �1 2

2 �3 9

¥

£

1 0 4

1 3 �7

5 �1 2

§£

1 0

2.1 1

�3 5.8

§

Perform the indicated row operation(s) in the ordergiven and write the new matrix.

19.

20.

21.

22.

23.

24.

What row operations would produce zeroes beneath thefirst entry in the diagonal?

25.

26.

27.

Solve each system by triangularizing the augmentedmatrix and using back-substitution. Simplify byclearing fractions or decimals before beginning.

28.

29.

30. 31.

32. •2x � 3y � 2z � 7

x � y � 2z � �5

3x � 2y � z � 11

•

x � 2y � 2z � 7

2x � 2y � z � 5

3x � y � z � 6

e�1

5u � 14v � 1

110u � 1

2v � 7

e 0.15g � 0.35h � �0.5

�0.12g � 0.25h � 0.1

e2y � 5x � 4

�5x � 2 � 4y

£

1 2 0 10

5 1 2 6

�4 3 �3 2

§

£

1 1 �4 �3

3 0 1 5

�5 3 2 3

§

£

1 3 0 2

�2 4 1 1

3 �1 �2 9

§

£

2 1 �1 �3

3 1 1 0

4 3 2 3

§

�3R1 � 2R2 S R2,

�2R1 � R3 S R3

£

3 1 1 8

6 �1 �1 10

4 �2 �3 22

§

�2R1 � R2 S R2,

�4R1 � 3R3 S R3

£

�3 2 0 0

1 1 2 6

4 1 �3 2

§

R1 4 R2,

�4R1 � R3 S R3

£

�2 1 0 4

5 8 3 �5

1 �3 3 2

§

R1 4 R3,

�5R1 � R2 S R2

c7 4 3

4 �8 12d

14R2 S R2,

R1 4 R2

c12 �3 �1

�5 2 4d

2R1 S R1,

5R1 � R2 S R2

cob19537_ch09_893-905.qxd 1/27/11 9:12 PM Page 902


Precalculus—

Solve using Gauss-Jordan elimination.

33. 34.

35. 36.

Solve each system of equations using a graphingcalculator. Verify each solution on the home screen usingthe keys, as in Section 9.2.

37.

38.

39.

40.

41.

42.

Solve each system by triangularizing the augmentedmatrix and using back-substitution. If the system islinearly dependent, give the solution in terms of aparameter. If the system has coincident dependence,answer in set notation as in Section 9.2.

43.

44.

45. •x � 3y � 5z � 20

2x � 3y � 4z � 16

x � 2y � 3z � 12

•

3x � y � z � �2

x � 2y � 3z � 1

2x � 3y � 5z � 3

•

4x � 8y � 8z � 24

2x � 6y � 3z � 13

3x � 4y � z � �11

•

54x � 1

2y � z � �14

�x � 32y � 7

8z � �35

14x � y � 1

4z � 25

•

12x � 3

4y � z � 34

�x � 14y � 3

2z � �58

54x � y � 1

2z � �12

•

0.1x � 2y � 0.6z � �36.4

�3x � 0.8y � 0.2z � �35

0.5x � y � 0.4z � 25

•

0.2x � 0.5y � z � 13

�x � 0.7y � 0.4z � �9

�0.5x � y � 0.8z � �4.6

•

3x � 5y � z � 6

�4x � 7y � 2z � 24

x � 10y � 3z � �40

•

x � 2y � 3z � 9

�2x � y � 5z � �20

3x � 5y � z � 14

ALPHA

•

x � y � 2z � �1

4x � y � 3z � 3

3x � 2y � z � 4

•

�x � y � 2z � 2

x � y � z � 1

2x � y � z � 4

•

2x � 3y � z � 5

2y � z � 7

x � y � 2z � 5

•

x � 2y � z � 1

x � z � 3

2x � y � z � 3

46.

47.

48.

49.

50.

51.

52.

Solve each system of equations using a graphingcalculator. Verify each solution on the home screen usingthe keys, as in Section 9.2.

53.

54.

55.

56. μ

�2x � 3y � 4

� 2y � 5z � �19

x � 4z � 13

�x � 3y � 4z � 17

μ

x � 3y � z � 1

x � 2z � 7

�2y � 3z � �6

2x � 3y � z � 8

μ

�2x � 5y � �4

3x � 4y � z � 0

x � y � 3z � �8

5x � 9y � z � 4

μ

2x � 3y � z � 0

x � 2y � z � 5

3x � 2z � 4

x � 3y � z � 4

ALPHA

•

�2x � 4y � 3z � 4

5x � 6y � 7z � �12

x � 2y � z � �4

•

x � 2y � z � 4

3x � 4y � z � 4

6x � 8y � 2z � 8

•

x � 2y � 3z � 2

3x � 4y � z � 6

4x � 2y � 2z � 7

•

2x � y � 3z � 1

2y � 6z � 2

x � 12y � 3

2z � 5

•

2x � y � 3z � 1

4x � 2y � 6z � 2

10x � 5y � 15z � 5

•

3x � 4y � 2z � �2

32x � 2y � z � �1

�6x � 8y � 4z � 4

•

�x � 2y � 3z � �6

x � y � 2z � �4

3x � 6y � 9z � 18

cob19537_ch09_893-905.qxd 1/27/11 9:13 PM Page 903


Precalculus—

57. μ

x � 2y � 3z � w � �13

�2x � 3y � z � 2w � �3

�x � y � 2z � 3w � 4

3x � 2y � z � w � 5

58. μ

2x � 3y � 4z � w � �10

x � 2y � 3w � 6

3x � 5z � 2w � 1

�x � y � 2z � 5w � 21


Area of a triangle in the plane: (x1 y2 � x2 y1 � x2 y3 � x3 y2 � x3 y1 � x1 y3)

The area of a triangle in the plane is given by the formula shown, where the vertices of the triangle are located at thepoints (x1, y1), (x2, y2), and (x3, y3), and the sign is chosen to ensure a positive value.

A � �12

� APPLICATIONS

Model each problem using a system of linear equations. Then solve using the augmented matrix.


61. The distance (via air travel) from Los Angeles (LA), California, toSaint Louis (STL), Missouri, to Cincinnati (CIN), Ohio, to New YorkCity (NYC), New York, is approximately 2480 mi. Find the length ofeach flight if the distance from LA to STL is 50 mi more than fivetimes the distance between STL and CIN, and 110 mi less than threetimes the distance from CIN to NYC.

62. In the 2006 NBA Championship Series, Dwayne Wade of the MiamiHeat carried his team to the title after the first two games were lost to the Dallas Mavericks. If 187 points were scored in the title game and the Heat won by 3 points, what was the final score?

63. Moe is lecturing Larry and Curly once again (Moe, Larry, and Curly of The Three Stooges fame) claiming he istwice as smart as Larry and three times as smart as Curly. If he is correct and the sum of their IQs is 165, what isthe IQ of each stooge?

New York City

St. Louis Cincinnati

Los Angeles

59. Find the area of a triangle whose vertices are(5, 2), and (1, 8).1�1, �32,

60. Find the area of a triangle whose vertices are, and 1�1, 72.16, �22, 1�5, 42

64. A collector of rare books buys a handwritten,autographed copy of Edgar Allan Poe’s AnnabelLee, an original advance copy of L. Frank Baum’sThe Wonderful Wizard of Oz, and a first print copyof The Caine Mutiny by Herman Wouk, paying atotal of $100,000. Find the cost of each one, giventhat the cost of Annabel Lee and twice the cost ofThe Caine Mutiny sum to the price paid for TheWonderful Wizard of Oz, and The Caine Mutinycost twice as much as Annabel Lee.

Geometry

65. A right triangle has a hypotenuse of 39 m. If theperimeter is 90 m, and the longer leg is 6 m longerthan twice the shorter leg, find the dimensions ofthe triangle.

66. In triangle ABC, the sum of angles A and C is equalto three times angle B. Angle C is 10 degrees morethan twice angle B. Find the measure of each angle.

Investment and Finance

67. Suppose $10,000 is invested in three differentinvestment vehicles paying 5%, 7%, and 9%annual interest. Find the amount invested at eachrate if the interest earned after 1 yr is $760 and theamount invested at 9% is equal to the sum of theamounts invested at 5% and 7%.

68. The trustee of a union’s pension fund hasinvested funds in three ways: a savings fundpaying 4% annual interest, a money market fundpaying 7%, and government bonds paying 8%.Find the amount invested in each if the interestearned after one year is $0.178 million and theamount in government bonds is $0.3 millionmore than twice the amount in money marketfunds. The total amount invested is $2.5 milliondollars.

cob19537_ch09_893-905.qxd 1/27/11 9:13 PM Page 904

9–69 Section 9.6 The Algebra of Matrices 905

Precalculus—

71. (8.5) a. Convert to trigonometric form.

b. Convert to rectangular

form.

72. (6.2) State the exact value of the following trigfunctions:

a. b.

c. d.

73. (5.6) Since 2005, cable installations for an Internetcompany have been modeled by the function

, where C(t) represents cableinstallations in thousands, t yr after 2005. In whatC1t2 � 15 ln1t � 12

csca3�

2btana

2�

3b

cosa5�

4bsina�

�

6b

z2 � 5 cisa2�

3b

z1 � �1� 3i year will the number of installations be greaterthan 30,000?

74. (5.6) If a set amount of money p is depositedregularly (daily, weekly, monthly, etc.) n times peryear at a fixed interest rate r, the amount of money Aaccumulated in t years is given by the formulashown. If a parent deposits $250 per month for 18 yr at 4.6% beginning when her first child wasborn, how much has been accumulated to help payfor college expenses?

A �

p c a1 �rnb

nt

� 1 d

rn


69. Given the drawing shown, use a system ofequations and the matrix method to find themeasure of the angles labeled as x and y. Recallthat vertical angles are equal and that the sum ofthe angles in a triangle is 180°.

70. The system given here has a solution of Find the value of a and b.

£

1 a b 1

2b 2a 5 13

2a 7 3b �8

§

11, �2, 32.

(x � 59)�xy

71�


Matrices serve a much wider purpose than just a convenient method for solving sys-tems. To understand their broader application, we need to know more about matrixtheory, the various ways matrices can be combined, and some of their more practicaluses. The common operations of addition, subtraction, multiplication, and division areall defined for matrices, as are other operations. Practical applications of matrix theorycan be found in the social sciences, inventory management, genetics, operationsresearch, engineering, and many other fields.

A. Equality of Matrices

To effectively study matrix algebra, we first give matrices a more general definition.For the general matrix A, all entries will be denoted using the lowercase letter “a,” withtheir position in the matrix designated by the dual subscript aij. The letter “i” gives therow and the letter “j ” gives the column of the entry’s location. The general matrix A is written

m � n


how we can:

A. Determine if two matricesare equal

B. Add and subtractmatrices

C. Compute the product of two matrices

9.6 The Algebra of Matrices

cob19537_ch09_893-905.qxd 1/27/11 9:13 PM Page 905

Precalculus—



we can determine if two

matrices are equal

The size of a matrix is also referred to as its order, and we say the order of generalmatrix A is Note that diagonal entries have the same row and column number,aij, where Also, where the general entry of matrix A is aij, the general entry ofmatrix B is bij, of matrix C is cij, and so on.

EXAMPLE 1 � Identifying the Order and Entries of a MatrixState the order of each matrix and name the entries corresponding to a22, a31; b22,b31; and c22, c31.

a. b. c.

Solution � a. matrix A: order Entry (the row 2, column 2 entry is ).There is no a31 entry (A is only ).

b. matrix B: order Entry , entry .c. matrix C: order Entry , entry .


Equality of Matrices

Two matrices are equal if they have the same order and their corresponding entries areequal. In symbols, this means that if for all i and j.

EXAMPLE 2 � Determining If Two Matrices Are EqualDetermine whether the following statements are true, false, or conditional. If false,explain why. If conditional, find values that will make the statement true.

a. b.

c.

Solution � a. is false. The matrices have the same order and

entries, but corresponding entries are not equal.

b. is false. Their orders are not equal.

c. is conditional. The statement is true when

and is false otherwise.


2b � 4 1b � 22, c � �2,a � 2 � 1 1a � 32,

� ca � 2 2b

c �3dc

1 4

�2 �3d

� c3 �2 1

5 �4 3d£

3 �2

1 5

�4 3

§

� c�3 �2

4 1dc

1 4

�2 �3d

c1 4

�2 �3d � c

a � 2 2b

c �3d

� c3 �2 1

5 �4 3d£

3 �2

1 5

�4 3

§c1 4

�2 �3d � c

�3 �2

4 1d

aij � bij A � B

c31 � 2.1c22 � 0.33 � 3.b31 � �4b22 � 53 � 2.

2 � 2�3a22 � �32 � 2.

C � £

0.2 �0.5 0.7

�1 0.3 1

2.1 �0.1 0.6

§B � £

3 �2

1 5

�4 3

§A � c1 4

�2 �3d

i � j.m � n.

row m S

row i S

row 3 Srow 2 Srow 1 S

col 1 col 2 col 3 col j col n

a11 a12 a13p a1j

p a1n

a21 a22 a23p a2j

p a2n

a31 a32 a33p a3j

p a3n

o o o o oai1 ai2 ai3

p aijp ain

o o o o oam1 am2 am3

p amjp amn

is a general matrix elementaij

p

z

l

p

{

}

cob19537_ch09_906-917.qxd 1/27/11 9:26 PM Page 906


Precalculus—

B. Addition and Subtraction of Matrices

A sum or difference of matrices is found by combining the corresponding entries. Thislimits the operations to matrices of like orders, so that every entry in one matrix has a“corresponding entry” in the other. This also means the result is a new matrix of likeorder, whose entries are the corresponding sums or differences. If you attempt to addmatrices of unlike size on a graphing calculator, an error message is displayed: ERR:DIM MISMATCH. Note that since aij represents a general entry of matrix A, [aij] rep-resents the entire matrix.

Addition and Subtraction of Matrices

Given matrices A, B, and C having like orders.

The sum The difference where where

EXAMPLE 3 � Adding and Subtracting MatricesCompute the sum or difference of the matrices indicated.

a. b. c.

Solution � a. sum of A and C

add corresponding entries

b.

c. difference of C and A

subtract corresponding entries


Operations on matrices can be very laborious for larger matrices and for matrices withnoninteger or large entries. For these, we can turn to available technology forassistance. This shifts our focus from a meticulous computation of entries, to carefullyentering each matrix into the calculator, double-checking each entry, and appraisingresults to see if they’re reasonable.

� £

1 �8

0 5

�5 6

§ � £

3 � 2 �2 � 6

1 � 1 5 � 0

�4 � 1 3 � 1�32

§

� £

2 6

1 0

1 �3

§C � A � £

3 �2

1 5

�4 3

§

c�3 2 �1

�5 4 3dA � B � £

2 6

1 0

1 �3

§ �

� £

5 4

2 5

�3 0

§ � £

2 � 3 6 � 1�22

1 � 1 0 � 5

1 � 1�42 �3 � 3

§

£

3 �2

1 5

�4 3

§ A � C � £

2 6

1 0

1 �3

§ �

C � AA � BA � C

C � £

3 �2

1 5

�4 3

§B � c�3 2 �1

�5 4 3dA � £

2 6

1 0

1 �3

§

3aij � bij 4 � 3cij 4 .3aij � bij 4 � 3cij 4 .A � B � C,A � B � C,

Addition and subtraction are notdefined for matrices of unlike order.

cob19537_ch09_906-917.qxd 1/27/11 9:26 PM Page 907


Precalculus—


we can add and subtract

matrices

EXAMPLE 4 � Using Technology for Matrix OperationsUse a calculator to compute the difference for the matrices given.

Solution � The entries for matrix A are shown in Figure 9.49. After entering matrix B, exit tothe home screen [ (QUIT)], call up matrix A, press the (subtract)key, then call up matrix B and press . The calculator quickly finds the differenceand displays the results shown in Figure 9.50. The last line on the screen shows theresult can be stored for future use in a new matrix C by pressing the key,calling up matrix C, and pressing .


In Figure 9.50 the dots to the right on the calculator screen indicate there are addi-tional digits or matrix columns that can’t fit on the display, as often happens with largermatrices or decimal numbers. Sometimes, converting entries to fraction form will pro-vide a display that’s easier to read. Here, this is done by calling up the matrix C, andusing the 1: � Frac option. After pressing , all entries are converted to frac-tions in simplest form (where possible), as in Figure 9.51. The third column can beviewed by pressing the right arrow.

Since the addition of two matrices is defined as the sum of corresponding entries,we find the properties of matrix addition closely resemble those of real number addi-tion. Similar to standard algebraic properties, represents the product andany subtraction can be rewritten as an algebraic sum: As noted inthe properties box, for any matrix A, the sum will yield the zero matrix Z,a matrix of like size whose entries are all zeroes. Also note that matrix is theadditive inverse for A, while Z is the additive identity.

Properties of Matrix Addition

Given matrices A, B, C, and Z are matrices, with Z the zero matrix. Then,

I. matrix addition is commutativeII. matrix addition is associativeIII. Z is the additive identityIV. is the additive inverse of A

C. Matrices and Multiplication

The algebraic terms 2a and ab have counterparts in matrix algebra. The product 2Arepresents a constant times a matrix and is called scalar multiplication. The productAB represents the product of two matrices.

�AA � 1�A2 � 1�A2 � A � ZA � Z � Z � A � A1A � B2 � C � A � 1B � C2A � B � B � A

m � n

�AA � 1�A2

A � B � A � 1�B2.�1 # A�A

ENTERMATH


ENTER

STO

ENTER

�MODE2nd

B � £

16

�710 0.75

1125 0 �5

�4 �59

�512

§A � £

211 �0.5 6

5

0.9 34 �4

0 6 � 512

§

A � B

Figure 9.51

cob19537_ch09_906-917.qxd 1/27/11 9:26 PM Page 908

Scalar Multiplication

Scalar multiplication is defined by taking the product of the constant with each entryin the matrix, forming a new matrix of like size. In symbols, for any real number k andmatrix A,

EXAMPLE 5 � Computing Operations on Matrices

Given and compute the following:

a. b.

Solution � a.

b. rewrite using algebraic addition

simplify

result


Matrix Multiplication

Consider a cable company offering three different levels of Internet service: Bronze—fast, Silver—very fast, and Gold—lightning fast. Table 9.3 shows the number andtypes of programs sold to households and businesses for the week. Each program hasan incentive package consisting of a rebate and a certain number of free weeks, asshown in Table 9.4.

£

�352 �11

�2 �7

2 11.8

§� £

�16 � 1�32 2 �12 � 1

�2 � 0 �4 � 1�32

0 � 2 12 � 1�0.22

§ �

£

�32 1

0 �3

2 �0.2

§� £

�16 �12

�2 �4

0 12

§ �

£

1�12 2 132 1�1

2 2 1�22

1�12 2 102 1�1

2 2 162

1�12 2 1�42 1�1

2 2 10.42

§� £

1�42 142 1�42 132

1�42 112 2 1�42 112

1�42 102 1�42 1�32

§ �

�4A �1

2 B � �4A � a�

1

2b B

£

32 �1

0 3

�2 0.2

§� £

112 2 132 112 2 1�22

112 2 102 112 2 162

112 2 1�42 112 2 10.42

§ �

1

2 B � a

1

2b £

3 �2

0 6

�4 0.4

§

�4A � 12 B1

2 B

B � £

3 �2

0 6

�4 0.4

§ ,A � £

4 312 1

0 �3

§

kA � 3kaij 4 .


Precalculus—

Table 9.3 Matrix A Table 9.4 Matrix B

Bronze Silver Gold

Homes 40 20 25

Businesses 10 15 45

Rebate Free Weeks

Bronze $15 2

Silver $25 4

Gold $35 6

To compute the amount of rebate money the cable company paid to households for the week, we would take the first row (R1) in Table 9.3 and multiply by the

cob19537_ch09_906-917.qxd 1/27/11 9:26 PM Page 909


Precalculus—

corresponding entries (bronze with bronze, silver with silver, and so on) in the first column (C1) of Table 9.4 and add these products. In matrix form, we have

Using R1 of Table 9.3

with C2 from Table 9.4 gives the number of free weeks awarded to homes:

Using the second row (R2) of

Table 9.3 with the two columns from Table 9.4 will give the amount of rebate moneyand the number of free weeks, respectively, awarded to business customers. When allcomputations are complete, the result is a product matrix P with order This isbecause the product of R1 from matrix A, with C1 from matrix B, gives the entry in

position P11 of the product matrix: .

Likewise, the product will give entry P12 (310), the product of R2 with C1 willgive P21 (2100), and so on. This “row column” multiplication can be generalized,and leads to the following. Given matrix A and matrix B,s � tm � n

�R1 # C2

� c1975 310

2100 350d£

15 2

25 4

35 6

§c40 20 2510 15 45

d #

2 � 2.

340 20 25 4 # £2

4

6§ � 40 # 2 � 20 # 4 � 25 # 6 � 310.

25 # 35 � $1975.340 20 25 4 # £15

25

35§ � 40 # 15 � 20 # 25 �

A B A B

c c c c1s � t 21m � n21s � t 21m � n2

result will be anmatrixm � t

matrix multiplication ispossible only when

n � s

In less formal terms, matrix multiplication involves multiplying the row entries ofthe first matrix with the corresponding column entries of the second, and adding themtogether. In Example 6, two of the matrix products [parts (a) and (b)] are shown in fulldetail, with the first entry of the product matrix color-coded.

EXAMPLE 6 � Multiplying MatricesGiven the matrices A through E shown here, compute the following products:

a. AB b. CD c. DC d. AE e. EA

E � £

�2 �1

3 0

1 2

§D � £

2 5 1

4 �1 1

0 3 �2

§C � £

�2 1 3

1 0 2

4 1 �1

§B � c4 3

6 1dA � c

�2 1

3 4d

In more formal terms, we have the following definition of matrix multiplication.

Matrix Multiplication

Given the matrix and the matrix If thenmatrix multiplication is possible and the product AB is an matrix where pij is product of the ith row of A with the jth column of B.

P � 3pij 4 ,m � tn � s,B � 3bij 4 .s � tA � 3aij 4m � n

cob19537_ch09_906-917.qxd 1/27/11 9:27 PM Page 910


Precalculus—

C D C D(3 � 3) (3 � 3) (3 � 3) (3 � 3)

c c c cresult will be a 3 � 3 matrix

multiplication is possiblesince 3 � 3

D C D C(3 � 3) (3 � 3) (3 � 3) (3 � 3)



E A E A(3 � 2) (2 � 2) (3 � 2) (2 � 2)



A E(2 � 2) (3 � 2)

c cmultiplication is not possible since 2 � 3

A B A B(4 � 3) (3 � 3) (4 � 3) (3 � 3)



Solution � a.

Computation:

b.

Computation:

c.

d.

e.


Example 6 shows that in general, matrix multiplication is not commutative. Parts(b) and (c) show since we get different results, and parts (d) and (e) show

since AE is not defined while EA is.As with the addition and subtraction of matrices, matrix multiplication becomes

cumbersome and time consuming for larger matrices, and we will often turn to thetechnology available in such cases.

EXAMPLE 7 � Using Technology for Matrix Operations

Use a calculator to compute the product AB.

Solution � Carefully enter matrices A and B into the calculator, then press (QUIT) to get to the home screen. Use [A][B] , and the calculator finds the product shown in the figure. Just for “fun,” we’ll

ENTER

MODE2nd

B � £

12 �0.7 1

0.5 3.2 �3

�2 34 4

§A � ≥

2 �3 0

�1 5 4

6 0 2

3 2 �1

¥

AE � EA,CD � DC

EA � £

�2 �1

3 0

1 2

§ c�2 1

3 4d � £

1 �6

�6 3

4 9

§

AE � c�2 1

3 4d £

�2 �1

3 0

1 2

§

DC � £

2 5 1

4 �1 1

0 3 �2

§ £

�2 1 3

1 0 2

4 1 �1

§ � £

5 3 15

�5 5 9

�5 �2 8

§

£

1�22 122 � 112 142 � (3)(0) 1�22 152 � 112 1�12 � (3)132 1�22 112 � 112 112 � (3)1�22

112 122 � 102 142 � 122(0) 112 152 � 102 1�12 � 122 132 112 112 � 102 112 � 122 1�22

142 122 � 112 142 � 1�12(0) 142 152 � 112 1�12 � 1�12 132 142 112 � 112 112 � 1�12 1�22

§

CD � £

�2 1 31 0 2

4 1 �1

§ £

2 5 1

4 �1 1

0 3 �2

§ � £

0 �2 �7

2 11 �3

12 16 7

§

c1�22 142 � 112 162 1�22 132 � 112 112

132 142 � 142 162 132 132 � 142 112d

AB � c�2 1

3 4d c

4 3

6 1d � c

�2 �5

36 13d

A B A B(2 � 2) (2 � 2) (2 � 2) (2 � 2)



cob19537_ch09_906-917.qxd 2/11/11 5:10 PM Page 911


Precalculus—

Table 9.5 Matrix A

A Age Group

Sex 18–19 20–21 22–23

Female 1000 1500 500

Male 2500 3000 2000

Table 9.6 Matrix B

B Likelihood of Joining

Age Group Army Navy Air Force Marines

18–19 0.42 0.28 0.17 0.13

20–21 0.38 0.26 0.27 0.09

22–23 0.33 0.25 0.35 0.07

double check the first entry of the product matrix: (2) � (�3)(0.5) � (0)(�2) � �0.5. ✓


Properties of Matrix Multiplication

Earlier, Example 6 demonstrated that matrix multiplication is not commutative. Hereis a group of properties that do hold for matrices. You are asked to check these proper-ties in the Exercise Set using various matrices. See Exercises 53 through 56.

Properties of Matrix Multiplication

Given matrices A, B, and C for which the products are defined:

I. matrix multiplication is associativeII. matrix multiplication is distributive from the leftIII. matrix multiplication is distributive from the rightIV. a constant k can be distributed over addition

We close this section with an application of matrix multiplication. There are manyother interesting applications in the Exercise Set.

EXAMPLE 8 � Using Matrix Multiplication to Track Volunteer EnlistmentsIn a certain country, the number of males and females that will join the military depends on their age.This information is stored in matrix A (Table 9.5). The likelihood a volunteer will join a particularbranch of the military also depends on their age, with this information stored in matrix B (Table 9.6).(a) Compute the product P � AB and discuss/interpret what is indicated by the entries P11, P13, andP24 of the product matrix. (b) How many males are expected to join the Navy this year?

k1A � B2 � kA � kB1B � C2A � BA � CAA1B � C2 � AB � ACA1BC2 � 1AB2C

£

12 �0.7 1

0.5 3.2 �3

�2 34 4

§AB � ≥

2 �3 0

�1 5 4

6 0 2

3 2 �1

¥

112 2

Solution � a. Matrix A has order and matrix B has order The product matrix P can be foundand is a matrix. Carefully enter the matrices in your calculator. Figure 9.52 shows theentries of matrix B. Using , the calculator finds the product matrix shown inFigure 9.53. Pressing the right arrow shows the complete product matrix is

The entry P11 is the product of R1 from A and C1 from B, and indicates that for the year, 1155females are projected to join the Army. In like manner, entry P13 shows that 750 females areprojected to join the Air Force. Entry P24 indicates that 735 males are projected to join theMarines.

P � c1155 795 750 300

2850 1980 1935 735d .

ENTER3A 4 3B 42 � 4

3 � 4.2 � 3

cob19537_ch09_906-917.qxd 1/27/11 9:27 PM Page 912


Precalculus—



we can compute the product

of two matrices

b. The product R2 (males) (Navy) gives meaning 1980 males areexpected to join the Navy.

Now try Exercise 59 through 66 �

P22 � 1980,# C2



9.6 EXERCISES

2. The sum of two matrices (of like size) is found byadding the corresponding entries. In symbols,

.A � B �

1. Two matrices are equal if they are like size and thecorresponding entries are equal. In symbols, if .�

A � B

3. The product of a constant times a matrix is calledmultiplication.

4. The size of a matrix is also referred to as its .

The order of is .A � c1 2 3

4 5 6 d

5. Give two reasons why matrix multiplication isgenerally not commutative. Include severalexamples using matrices of various sizes.

6. Discuss the conditions under which matrixmultiplication is defined. Include several examplesusing matrices of various sizes.


State the order of each matrix and name the entries inpositions a12 and a23 if they exist. Then name theposition aij of the “5” in each.

7. 8.

9. 10.

11. 12. £89 55 34 21

13 8 5 3

2 1 1 0

§£

�2 1 �7

0 8 1

5 �1 4

§

£

2 0.4

�0.1 50.3 �3

§c2 �3 0.5

0 5 6 d

£

19

�11

5§c

1 �3

5 �7d

Determine if the following statements are true, false, orconditional. If false, explain why. If conditional, find valuesof a, b, c, p, q, and r that will make the statement true.

13.

14.

15.

16. £2p � 1 �5 9

1 12 0

q � 5 9 �2r

§ � £

7 �5 2 � q

1 3r 0

�2 3p �8

§

£

�2 3 a

2b �5 4

0 �9 3c

§ � £

c 3 �4

6 �5 �a

0 �3b �6

§

≥

3

2

�7

5

13

10

�1

2

�2

5

1

3

¥ � c1.5 �1.4 1.3

�0.5 �0.4 0.3d

c11 14 18

116 132 164d � c

1 2 212

4 412 8d

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Precalculus—

For matrices A through J as given, perform theindicated operation(s), if possible. If an operationcannot be completed, state why. Use a calculator onlyfor those exercises designated by an icon.

17. 18.

19. 20.

21. 22.

23. 24.

25. 26.

27. 28.

29. ED 30. DE

31. AH 32. HA

33. CB 34. FH

35. HF 36. EB

37. H2 38. F2

39. FE 40. EF

F �2

3F

1

2E � 3D

2E � 3G3H � 2A

I � GG � I

A � JH � J

G � DF � H

E � GA � H

J � ≥

7

32

5

8

�5

16

3

16

¥I �

1

2�

3

8�

1

4

1

4

3

2�

5

8

1

8

3

4�

5

2

H � c8 �3

�5 2dG � £

�1 2 0

0 1 �2

�4 �3 6

§

F � c6 �3 9

12 0 �6dE � £

1 �2 0

0 �1 2

4 3 �6

§

D � £

1 0 0

0 1 0

0 0 1

§C � £

2 0.5

0.2 5

�1 3

§

B � £

2

1

�3

§A � c2 3

5 8d

For matrices A through H as given, use a calculator toperform the indicated operation(s), if possible. If anoperation cannot be completed, state why.

41. AH 42. HA

43. EG 44. GE

45. HB 46. BH

47. DG 48. GD

49. C2 50. E2

51. FG 52. AF

For Exercises 53 through 56, use a calculator andmatrices A, B, and C to verify each statement.

53. Matrix multiplication is not generally commutative:(a) (b) and (c)

54. Matrix multiplication is distributive from the left:

55. Matrix multiplication is distributive from the right:

56. Matrix multiplication is associative:1AB2C � A1BC2.

1B � C2A � BA � CA.

A1B � C2 � AB � AC.

BC � CB.AC � CA,AB � BA,

C � £

45 �1 3

�6 10 �15

21 �28 36

§

B � £

0.3 �0.4 1.2

�2.5 2 0.9

1 �0.5 0.2

§A � £

�1 3 5

2 7 �1

4 0 6

§

H � ≥

�3

19

4

57

1

19

5

57

¥G �

03

4

1

4

�1

2

3

8

1

8

�1

4

11

16

1

16

F � c12 �8 32

4 8 16dE � £

1 �2 0

0 �1 2

4 3 �6

§

D � £

1 0 0

0 1 0

0 0 1

§C � £

13

2

13

3

13 213§

B � c1 0

0 1dA � c

�5 4

3 9d

cob19537_ch09_906-917.qxd 1/27/11 9:28 PM Page 914


Precalculus—


The perimeter and area of a rectangle can be simultaneously calculated using the matrix formula shown, where Lrepresents the length and W represents the width of the rectangle. Use the matrix formula and your calculator to findthe perimeter and area of the rectangles shown, then check the results using and

57. 58.

A � LW.P � 2L � 2W

c2 2W 0

d # c LWd � c

PerimeterArea

d

6.374 cm

4.35 cm

5.02 cm

3.75 cm

� APPLICATIONS

59. Custom T’s designs and sells specialty T-shirts andsweatshirts, with factories in Verdi and Minsk. Thecompany offers this apparel in three quality levels:standard, deluxe, and premium. Last fall the Verdiplant produced 3820 standard, 2460 deluxe, and1540 premium T-shirts, along with 1960 standard,1240 deluxe, and 920 premium sweatshirts. TheMinsk plant produced 4220 standard, 2960 deluxe,and 1640 premium T-shirts, along with 2960standard, 3240 deluxe, and 820 premiumsweatshirts in the same time period.

a. Write a “production matrix” for eachplant Verdi, Minsk], with a T-shirtcolumn, a sweatshirt column, and three rowsshowing how many of the different types ofapparel were manufactured.

b. Use the matrices from part (a) to determinehow many more or fewer articles of clothingwere produced by Minsk than Verdi.

c. Use scalar multiplication to find how manyshirts of each type will be made at Verdi andMinsk next fall, if each is expecting a 4%increase in business.

d. Write a matrix that shows Custom T’s totalproduction next fall (from both plants), foreach type of apparel.

60. Terry’s Tire Store sells automobile and truck tiresthrough three retail outlets. Sales at the Cahokia storefor the months of January, February, and Marchbroke down as follows: 350, 420, and 530 auto tiresand 220, 180, and 140 truck tires. The Shady Oakbranch sold 430, 560, and 690 auto tires and 280,320, and 220 truck tires during the same 3 months.Sales figures for the downtown store were 864, 980,and 1236 auto tires and 535, 542, and 332 truck tires.

a. Write a “sales matrix” for each storeCahokia, Shady Oak,

Downtown], with January, February, andD SS S3C S

2 � 3

M S3V S3 � 2

March columns, and two rows showing thesales of auto and truck tires respectively.

b. Use the matrices from part (a) to determinehow many more or fewer tires of each type thedowntown store sold (each month) over theother two stores combined.

c. Market trends indicate that for the same threemonths in the following year, the Cahokiastore will likely experience a 10% increase insales, the Shady Oak store a 3% decrease, withsales at the downtown store remaining level(no change). Write a matrix that shows thecombined monthly sales from all three storesnext year, for each type of tire.

61. Home improvements: Dream-Makers HomeImprovements specializes in replacement windows,replacement doors, and new siding. During thepeak season, the number of contracts that camefrom various parts of the city (North, South, East,and West) are shown in matrix C. The averageprofit per contract is shown in matrix P. Computethe product PC and discuss what each entry of theproduct matrix represents.

N S E W

Windows Doors Siding

62. Classical music: Station 90.7—The Home ofClassical Music—is having their annual funddrive. Being a loyal listener, Mitchell decides thatfor the next 3 days he will donate money accordingto his favorite composers, by the number of timestheir music comes on the air: $3 for every piece byMozart (M), $2.50 for every piece by Beethoven(B), and $2 for every piece by Vivaldi (V).

31500 500 2500 4 � P

Windows

Doors

Siding

£

9 6 5 4

7 5 7 6

2 3 5 2

§ � C

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Precalculus—

This information is displayed in matrix D. Thenumber of pieces he heard from each composer isdisplayed in matrix C. Compute the product DCand discuss what each entry of the product matrixrepresents.

Mon. Tue. Wed.

M B V

63. Pizza and salad: The science department and mathdepartment of a local college are at a pre-semesterretreat, and decide to have pizza, salads, and softdrinks for lunch. The quantity of food ordered byeach department is shown in matrix Q. The cost ofthe food item at each restaurant is shown in matrixC using the published prices from three popularrestaurants: Pizza Home (PH), Papa Jeff’s (PJ), andDynamos (D).

a. What is the total cost to the math department ifthe food is ordered from Pizza Home?

b. What is the total cost to the science departmentif the food is ordered from Papa Jeff’s?

c. Compute the product QC and discuss themeaning of each entry in the product matrix.

Pizza Salad Drink

PH PJ D

64. Manufacturing pool tables: Cue BallIncorporated makes three types of pool tables, forhomes, commercial use, and professional use. Theamount of time required to pack, load, and installeach is summarized in matrix T, with all times inhours. The cost of these components in dollars perhour, is summarized in matrix C for two of itswarehouses, one on the west coast and the other inthe midwest.

a. What is the cost to package, load, and install acommercial pool table from the coastalwarehouse?

b. What is the cost to package, load, and install acommercial pool table from the warehouse inthe midwest?

c. Compute the product TC and discuss themeaning of each entry in the product matrix.

Pizza

Salad

Drink

£

8 7.5 10

1.5 1.75 2

0.90 1 0.75

§ � C

Science

Math c

8 12 20

10 8 18d � Q

33 2.5 2 4 � D

M

B

V

£

4 3 5

3 2 4

2 3 3

§ � C

Pack Load Install

Coast Midwest

65. Joining a club: Each school year, among thestudents planning to join a club, the likelihood astudent joins a particular club depends on theirclass standing. This information is stored in matrixC. The number of males and females from eachclass that are projected to join a club each year isstored in matrix J. Compute the product JC and usethe result to answer the following:

a. Approximately how many females joined thechess club?

b. Approximately how many males joined thewriting club?

c. What does the entry p13 of the product matrixtells us?

Fresh Soph Junior

Spanish Chess Writing

66. Designer shirts: The SweatShirt Shoppe sellsthree types of designs on its products: stenciled(S), embossed (E), and applique (A). The quantityof each size sold is shown in matrix Q. The retailprice of each sweatshirt depends on its size andwhether it was finished by hand or machine. Retailprices are shown in matrix C. Assuming all stock issold, compute the product QC and use the result toanswer the following.

a. How much revenue was generated by the largesweatshirts?

b. How much revenue was generated by theextra-large sweatshirts?

c. What does the entry p11 of the product matrixQC tell us?

S E A Hand MachineS

E

A

£

40 25

60 40

90 60

§ � C

med

large

x-large

£

30 30 15

60 50 20

50 40 30

§ � Q

Fresh

Soph

Junior

£

0.6 0.1 0.3

0.5 0.2 0.3

0.4 0.2 0.4

§ � C

Female

Male c

25 18 21

22 19 18d � J

Pack

Load

Install

£

10 8

12 10.5

13.5 12.5

§ � C

Home

Comm

Prof

£

1 0.2 1.5

1.5 0.5 2.2

1.75 0.75 2.5

§ � T

cob19537_ch09_906-917.qxd 1/27/11 9:28 PM Page 916

9–81 Section 9.7 Solving Linear Systems Using Matrix Equations 917

Precalculus—


67. For the matrix A shown, use your calculator tocompute A2, A3, A4, and A5. Do you notice apattern? Try to write a “matrix formula” for An,where n is a positive integer, then use your formulato find A6. Check results using a calculator.

A � £

1 0 1

1 1 1

1 0 1

§

68. The matrix has some very

interesting properties. Compute the powers M 2, M3,M4, and M5, then discuss what you find. Try tofind/create another matrix that has similarproperties.

2 � 2

M � c2 1

�3 �2 d


69. (9.2) Solve the system using elimination.

70. (7.5) Evaluate .cos1cos�10.32112

•

x � 2y � z � 3

�2x � y � 3z � �5

5x � 3y � 2z � 2

71. (8.6) Solve using the nth rootstheorem. Leave your answer in trigonometric form.

72. (4.1) Find the quotient using synthetic division,then check using multiplication.

x3 � 9x � 10

x � 2

z4 � 81i � 0

9.7 Solving Linear Systems Using Matrix Equations

While using matrices and row operations offers a degree of efficiency in solving sys-tems, we are still required to solve for each variable individually. Using matrix mul-tiplication we can actually rewrite a given system as a single matrix equation, inwhich the solutions are computed simultaneously. As with other kinds of equations,the use of identities and inverses are involved, which we now develop in the contextof matrices.

A. Multiplication and Identity Matrices

From the properties of real numbers, 1 is the identity for multiplication sinceA similar identity exists for matrix multiplication. Consider the

matrix While matrix multiplication is not generally commutative,

if we can find a matrix B where then B is a prime candidate for theidentity matrix, which is denoted I. For the products AB and BA to be possible and havethe same order as A, we note B must also be a matrix. Using the arbitrary matrix

we have the following.B � ca b

c dd ,

2 � 2

AB � BA � A,

A � c1 4

�2 3d .

2 � 2n # 1 � 1 # n � n.


how we can:

A. Recognize the identitymatrix for multiplication

B. Find the inverse of asquare matrix

C. Solve systems usingmatrix equations

D. Use determinants to findwhether a matrix isinvertible

cob19537_ch09_906-917.qxd 1/27/11 9:28 PM Page 917

Precalculus—

EXAMPLE 1A � Solving AB � A to Find the Identity Matrix

For use matrix multiplication, the equality of

matrices, and systems of equations to find the value of a, b, c, and d.

Solution � The product on the left gives

Since corresponding entries must be equal (shown by matching colors), we can find

a, b, c, and d by solving the systems and . For

the first system, shows and Using for the second

shows and It appears is a candidate for the identity matrix.

Before we name B as the identity matrix, we must show that

EXAMPLE 1B � Verifying AB � BA � A

Given and determine if and

Solution �

✓ ✓

Since B is the identity matrix I.


By replacing the entries of with those of the general matrix

we can show that is the identity for all matrices. In

considering the identity for larger matrices, we find that only square matrices have in-verses, since is the primary requirement (the multiplication must be possiblein both directions). This is commonly referred to as multiplication from the right andmultiplication from the left. Using the same procedure as before we can show

is the identity for matrices (denoted I3). The identity matrix

In consists of 1’s down the main diagonal and 0’s for all other entries. Also, the iden-tity In for a square matrix is unique.

n � n3 � 3£

1 0 0

0 1 0

0 0 1

§

AI � IA

2 � 2I � c1 0

0 1dc

a11 a12

a21 a22d ,

A � c1 4

�2 �3d

AB � A � BA,

� c1 4

�2 3d � A � c

1 4

�2 3d � A

� c1112 � 01�22 1142 � 0132

0112 � 11�22 0142 � 1132d � c

1112 � 4102 1102 � 4112

�2112 � 3102 �2102 � 3112d

BA � c1 0

0 1d c

1 4

�2 3d AB � c

1 4

�2 3d c

1 0

0 1d

BA � A.AB � AB � c1 0

0 1d ,A � c

1 4

�2 3d

AB � BA � A.

c1 0

0 1dd � 1.b � 0

2R1 � R2c � 0.a � 12R1 � R2

eb � 4d � 4

�2b � 3d � 3e

a � 4c � 1�2a � 3c � �2

ca � 4c b � 4d

�2a � 3c �2b � 3dd � c

1 4�2 3

d .

c1 4

�2 3d c

a b

c dd � c

1 4

�2 3d ,


cob19537_ch09_918-933.qxd 1/27/11 9:49 PM Page 918

As in Section 9.6, a graphing calculator canbe used to investigate operations on matricesand matrix properties. For the matrix

and a

calculator will confirm that Carefully enter A into your calculator asmatrix A, and I3 as matrix B. Figure 9.54shows and . See Exercises 11through 14.

B. The Inverse of a Matrix

Again from the properties of real numbers, we know the multiplicative inverse for a is

since the products and yield the identity 1. To show

that a similar inverse exists for matrices, consider the square matrix and

an arbitrary matrix If we can find a matrix B, where then B

is a prime candidate for the inverse matrix of A, which is denoted In an attempt tofind such a matrix B, we proceed as in Examples 1A and 1B.

EXAMPLE 2A � Solving AB � I to find A�1

For use matrix multiplication, the equality of matrices,

and systems of equations to find the entries of B.

Solution � The product on the left gives Since corresponding

entries must be equal (shown by matching colors), we find the values of a, b, c, and

d by solving the systems and Using

for the first system shows and while for the second

system shows and Matrix is the

prime candidate for

It may have occurred to you that the systems used in Example 2A have exactly

the same coefficients. In matrix form, these systems are and .

Instead of solving these two systems separately, we can solve them simultaneously by sim-

ply using the coefficient matrix augmented with the identity matrix: .

Row operations are then used to transform the given matrix into the identity:

, in a sense “preserving” the inverse operations needed. The inverse

matrix is then .ce f

g hd

c1 0 e f

0 1 g hd

c6 5 1 0

2 2 0 1d

c6 5 0

2 2 1dc

6 5 1

2 2 0d

A�1.

B � ca b

c dd � c

1 �2.5

�1 3dd � 3.b � �2.5

�3R2 � R1c � �1,a � 1

�3R2 � R1e6b � 5d � 0

2b � 2d � 1.e

6a � 5c � 12a � 2c � 0

c6a � 5c 6b � 5d

2a � 2c 2b � 2dd � c

1 0

0 1d .

c6 5

2 2d c

a b

c dd � c

1 0

0 1d ,

A�1.

AB � BA � I,B � ca b

c dd .

A � c6 5

2 2d

a�1 # aa # a�1a�1 �1a

1a � 02,

BA � AAB � A

AI3 � A � I3 A.

I3 � £

1 0 0

0 1 0

0 0 1

§ ,A � £

2 5 1

4 �1 1

0 3 �2

§

3 � 3


Precalculus—


we can recognize the identity

matrix for multiplication

Figure 9.54

cob19537_ch09_918-933.qxd 1/27/11 9:49 PM Page 919


EXAMPLE 2B � Finding a 2 � 2 Inverse Using the Augmented Matrix

Use the augmented matrix method to find for . Then use a

calculator to verify .

Solution � First we augment the given matrix with the identity: . Then we use

row operations to transform A into the identity matrix.

As in Example 2A, a prime candidate for . Using a calculator

with and confirms that (see figure).


These observations guide us to the following definition of an inverse matrix.

The Inverse of a Matrix

Given an matrix A, if there exists an matrix such that , then is the inverse of matrix A.

We will soon discover that while only square matrices have inverses, not everysquare matrix has an inverse. If an inverse exists, the matrix is said to be invertible. For

matrices that are invertible, a simple formula exists for computing the inverse.The formula is derived in the Strengthening Core Skills feature at the end of Chapter 9.

The Inverse of a 2 � 2 Matrix

If then provided

To “test” the formula, again consider the matrix where

and

✓

See Exercises 63 through 66 for more practice with this formula.

�1

2c

2 �5

�2 6d � c

1 �2.5

�1 3d

A�1 �1

162 122 � 152 122c

2 �5

�2 6d

d � 2:c � 2,

a � 6, b � 5,A � c6 5

2 2d ,

ad � bc � 0A�1 �1

ad � bcc

d �b

�c adA � c

a b

c dd ,

2 � 2

A�1AA�1 � A�1A � In

A�1n � nn � n

A�1A � AA�1 � IA�1 � 3B 4A � 3A 4

A�1 is c1 �2.5

�1 3d

�1R2 S R2c1 0 1 �2.5

0 1 �1 3d

R1

6S R1

c6 0 6 �15

0 �1 1 �3d

c6 0 6 �15

0 �1 1 �3d5R2 � R1 S R1c

6 5 1 0

0 �1 1 �3d

c6 5 1 0

0 �1 1 �3d�3R2 � R1 S R2c

6 5 1 0

2 2 0 1d

c6 5 1 0

2 2 0 1d

A�1A � AA�1 � I

A � c6 5

2 2dA�1

Precalculus—

cob19537_ch09_918-933.qxd 1/27/11 9:49 PM Page 920

Almost without exception, real-world applications involve much larger matrices,with entries that are not integer-valued. Although the augmented matrix method fromExample 2 can be extended to find the inverse of larger matrices, the process becomesvery tedious and too time consuming to be useful. The process for a 3 � 3 matrix isdiscussed in the Strengthening Core Skills feature at the end of Chapter 9. But forpractical reasons, we will rely on a calculator to produce these larger inverse matrices.This is done by (1) carefully entering a square matrix A into the calculator, (2) return-ing to the home screen, and (3) calling up matrix A and pressing the key and to find . In the context of matrices, calculators are programmed to compute an in-verse matrix using this key, rather than to somehow find a reciprocal. See Exercises23 through 26.

C. Solving Systems Using Matrix Equations

One reason matrix multiplication has its definition is to assist in writ-ing a linear system of equations as a single matrix equation. The equation consists ofthe matrix of constants B on the right, and a product of the coefficient matrix A with the

matrix of variables X on the left: For the matrix

equation is Note that computing the product on the left

will yield the original system.Once written as a matrix equation, the system can be solved using an inverse ma-

trix and the following sequence. If A represents the matrix of coefficients, X the matrixof variables, B the matrix of constants, and I the appropriate identity, the sequence is

matrix equation

multiply from the left by the inverse of A

associative property

A �1 A � I

IX � X

Lines 1 through 5 illustrate the steps that make the method work. In actual prac-tice, after carefully entering the matrices, only step 5 is used when solving matrixequations using technology. Once matrix A is entered, the calculator will automaticallyfind and use as we enter

EXAMPLE 3 � Using Technology to Solve a Matrix EquationUse a calculator and a matrix equation to solve the system

Solution � As before, the matrix equation is £

1 4 �1

2 5 �3

8 1 �2

§ £

x

y

z

§ � £

10

7

11

§ .

•

x � 4y � z � 10

2x � 5y � 3z � 7.

8x � y � 2z � 11

A�1B.A�1

152 X � A�1B142 IX � A�1B

132 1A�1A2X � A�1B

122 A�11AX2 � A�1B

112 AX � B

£

1 4 �1

2 5 �3

8 1 �2

§ £

x

y

z

§ � £

10

7

11

§ .

•

x � 4y � z � 10

2x � 5y � 3z � 7

8x � y � 2z � 11

,AX � B.

row � column

A�1

ENTERx

-1


Precalculus—


we can find the inverse of a

square matrix

cob19537_ch09_918-933.qxd 1/27/11 9:49 PM Page 921

Carefully enter (and double-check) the matrix of coefficients as matrix A in yourcalculator, and the matrix of constants as matrix B (see Figure 9.55). The product

shows the solution is (see Figure 9.56). Verify bysubstitution.


The matrix equation method does have a few shortcomings. Consider the system

whose corresponding matrix equation is After entering

the matrix of coefficients A and matrix of constants B, attempting to compute results in the error message shown in Figure 9.57. The calculator is unable to return asolution due to something called a “singular matrix.” To investigate further, we attempt

to find for using the formula for a matrix. With

and we have

Since division by zero is undefined, we conclude that matrix A has no inverse. Amatrix having no inverse is said to be singular or noninvertible. Solving systemsusing matrix equations is only possible when the matrix of coefficients is nonsingular.

D. Determinants and Singular Matrices

As a practical matter, it becomes important to know ahead of time whether a particularmatrix has an inverse. To help with this, we introduce one additional operation on asquare matrix, that of calculating its determinant. For a matrix the determinant

is the entry itself. For a matrix the determinant of A, written

as det(A) or denoted with vertical bars as is computed as a difference of diagonalproducts beginning with the upper-left entry:

2nd diagonalproduct

1st diagonalproduct

The Determinant of a 2 � 2 Matrix

Given any matrix ,

det1A2 � �A� � a11a22 � a21a12

A � ca11 a12

a21 a22d2 � 2

det1A2 � `a11 a12

a21 a22` � a11a22 � a21a12

�A�,

A � ca11 a12

a21 a22d ,2 � 2

1 � 1

�1

0c5 10

2 4d A�1 �

1

ad � bcc

d �b

�c ad �

1

142 152 � 1�102 1�22c5 10

2 4d

d � 5,b � �10, c � �2,

a � 4,2 � 2c4 �10

�2 5dA�1

A�1B

c4 �10

�2 5d c

x

yd � c

�8

13d .


x � 2, y � 3, z � 4A �1B


Precalculus—


we can solve systems using

matrix equations

Figure 9.57

cob19537_ch09_918-933.qxd 1/27/11 9:50 PM Page 922

EXAMPLE 4 � Calculating DeterminantsCompute the determinant of each matrix.

a. b. c.

Solution � a.

b. Determinants are only defined for square matrices (see figure).

c.


Notice from Example 4(c), the determinant of is zero, and this is the

same matrix we earlier found had no inverse. This observation can be extended tolarger matrices and offers the connection we seek between a given matrix, its inverse,and matrix equations.

Singular Matrices

If A is a square matrix and the inverse matrix does not existand A is said to be singular or noninvertible.

In summary, inverses exist only for square matrices, but not every square matrixhas an inverse. If the determinant of a square matrix is zero, an inverse does not existand the method of matrix equations cannot be used to solve the system.

To use the determinant test for a system, we need to compute a deter-minant. At first glance, our experience with determinants appears to be of littlehelp. However, every entry in a matrix is associated with a smaller matrix,formed by deleting the row and column of that entry and using the entries that remain.These ’s are called the associated minor matrices or simply the minors. Usinga general matrix of coefficients, we’ll identify the minors associated with the entries inthe first row.

Entry: a11 Entry: a12 Entry: a13

associated minor associated minor associated minor

To illustrate, consider the system shown, and (1) form the matrix of coefficients,(2) identify the minor matrices associated with the entries in the first row, and(3) compute the determinant of each minor.

ca21 a22

a31 a32dc

a21 a23

a31 a33dc

a22 a23

a32 a33d

£

a11 a12 a13

a21 a22 a23

a31 a32 a33

§£

a11 a12 a13

a21 a22 a23

a31 a32 a33

§£

a11 a12 a13

a21 a22 a23

a31 a32 a33

§

2 � 2

2 � 23 � 32 � 2

3 � 33 � 3

det1A2 � 0,

c4 �10

�2 5d

det1D2 � `4 �10

�2 5` � 142 152 � 1�22 1�102 � 20 � 20 � 0

det1B2 � `3 2

1 �6` � 132 1�62 � 112 122 � �20

D � c4 �10

�2 5dC � c

5 2 1

�1 �3 4dB � c

3 2

1 �6d


Precalculus—

WORTHY OF NOTEFor the determinant of a general

matrix using cofactors, seeAppendix D.n � n

cob19537_ch09_918-933.qxd 1/27/11 9:50 PM Page 923

(1) Matrix of coefficients

Entry a11: 2 Entry a12: 3 Entry a13: �1associated minor associated minor associated minor

(3) Determinant Determinant of Determinant of of minor minor minor

For computing a determinant, we illustrate a technique called expansion by minors.

The Determinant of a 3 � 3 Matrix — Expansion by Minors

For the matrix M shown, det(M) is the unique number computedas follows:1. Select any row or column and form the product of each

entry with its minor matrix. The illustration here uses theentries in row 1:

2. The signs used between terms of the expansion dependson the row or column chosen, according to the sign chartshown.

The determinant of a matrix is unique and any row or column can be used. For thisreason, it’s helpful to select the row or column having the most zero, positive, and/orsmaller entries.

EXAMPLE 5 � Calculating a 3 � 3 Determinant

Compute the determinant of .

Solution � Since the second row has the “smallest” entries as well as a zero entry, we computethe determinant using this row. According to the sign chart, the signs of the termswill be negative–positive–negative, giving

� �7 � (�2) � 0


The value of det1M2 is �9. � �9

� �114 � 32 � 1�12 18 � 62 � 102 12 � 22

det1M2 � �112 `1 �3

1 4` � 1�12 `

2 �3

�2 4` � 102 `

2 1

�2 1`

M � £

2 1 �3

1 �1 0

�2 1 4

§

det1M2 � �a11 `a22 a23

a32 a33` � a12 `

a21 a23

a31 a33` � a13 `

a21 a22

a31 a32`

3 � 3

112 112 � 132 1�42 � 13112 102 � 132 122 � �61�42 102 � 112 122 � �2

c1 �4

3 1dc

1 2

3 0dc

�4 2

1 0d

£

2 3 �1

1 �4 2

3 1 0

§£

2 3 �1

1 �4 2

3 1 0

§122 £

2 3 �1

1 �4 2

3 1 0

§

£

2 3 �1

1 �4 2

3 1 0

§•

2x � 3y � z � 1

x � 4y � 2z � �3

3x � y � �1


Precalculus—

matrix M

£

a11 a12 a13

a21 a22 a23

a31 a32 a33

§

Sign Chart

£

� � �

� � �

� � �

§

cob19537_ch09_918-933.qxd 1/27/11 9:50 PM Page 924

Try computing the determinant of M two more times, using a different row or col-umn each time. Since the determinant is unique, you should obtain the same result.

There are actually other alternatives for computing a determinant. The firstis called determinants by column rotation, and takes advantage of patterns generatedfrom the expansion of minors. This method is applied to the matrix shown, which usesalphabetical entries for simplicity.

Although history is unsure of who should be credited, notice that if you repeat thefirst two columns to the right of the given matrix (“rotation of columns”), identicalproducts are obtained using the six diagonals formed—three in the downward direc-tion using addition, three in the upward direction using subtraction.

gec hfa idb

aei bfg cdh

Adding the products in blue (regardless of sign) and subtracting the products inred (regardless of sign) gives the determinant. This method is more efficient thanexpansion by minors, but can only be used for matrices!

EXAMPLE 6 � Calculating det(A) Using Column Rotation

Use the column rotation method to find the determinant of .

Solution � Rotate columns 1 and 2 to the right, and compute the diagonal products.

72 0 �10

�8 0 66

Adding the products in blue (regardless of sign) and subtracting the products inred (regardless of sign) shows


The final method is presented in the Extending the Concept feature of the ExerciseSet, and shows that if certain conditions are met, the determinant of a matrix can befound using its triangularized form.

As with the operations studied in Section 9.6, the process of computing a determi-nant becomes very cumbersome for larger matrices, or those with rational or radicalentries. Most graphing calculators are programmed to handle these computations easily.After accessing the matrix menu ( ), calculating a determinant is the firstoption under the MATH submenu (Figure 9.58). The calculator results for det([A]) anddet([B]) as defined are shown in Figures 9.59 and 9.60. See Exercises 57 and 58.

x-12nd

�8 � 0 � 66 � 72 � 0 � 1�102 � �4.

det1A2 � �4:

£

1 5 3

�2 �8 0

�3 �11 1

§ 1 5

�2 �8

�3 �11

A � £

1 5 3

�2 �8 0

�3 �11 1

§

3 � 3

£

a b c

d e f

g h i

§ a b

d e

g h

expansion using R1

distribute

rewrite result

� a1ei � fh2 � b1di � fg2 � c1dh � eg2

� aei � afh � bdi � bfg � cdh � ceg

� aei � bfg � cdh � afh � bdi � ceg

det £

a b c

d e f

g h i

§

3 � 3


Precalculus—

cob19537_ch09_918-933.qxd 1/27/11 9:50 PM Page 925

EXAMPLE 7 � Solving a System after Verifying A is InvertibleGiven the system shown here, (1) form the matrix equation (2) computethe determinant of the coefficient matrix and determine if you can proceed; and (3) if so, solve the system using a matrix equation.

Solution � 1. Form the matrix equation :

2. Since det(A) is nonzero (from Example 5 and Figure 9.61), we can proceed.3. For , input on your calculator and press (Figure 9.62).

Notice that matrix name [B] has scrolled out of view.

ENTERA�1BX � A�1B

£

2 1 �3

1 �1 0

�2 1 4

§ £

x

y

z

§ � £

11

1

�8

§

AX � B

•

2x � 1y � 3z � 11

1x � 1y � 1

�2x � 1y � 4z � �8

AX � B;


Precalculus—

Figure 9.58 Figure 9.59 Figure 9.60



We close this section with an application involving a system. There is alarge variety of additional applications in the Exercise Set.

EXAMPLE 8 � Solving an Application Using Technology and Matrix EquationsA local theater sells four sizes of soft drinks: 32 oz @ $2.25; 24 oz @ $1.90; 16 oz@ $1.50; and 12 oz @ $1.20/each. As part of a “free guest pass” promotion, themanager asks employees to try and determine the number of each size sold, giventhe following information: (1) the total revenue from soft drinks was $719.80; (2) there were 9096 oz of soft drink sold; (3) there was a total of 394 soft drinkssold; and (4) the number of 24-oz and 12-oz drinks sold was 12 more than thenumber of 32-oz and 16-oz drinks sold. Write a system of equations that modelsthis information, then solve the system using a matrix equation.

4 � 4

The solution is the ordered triple .13, 2, �12

cob19537_ch09_918-933.qxd 1/27/11 9:51 PM Page 926

Solution � If we let x, l, m, and s represent the number of 32-oz, 24-oz, 16-oz, and 12-oz softdrinks sold, the following system is produced:

When written as a matrix equation the system becomes:

To solve, carefully enter the matrix of coefficients as matrix A (see Figure 9.63), andthe matrix of constants as matrix B, then compute [since ].This gives a solution of 1x, l, m, s2 � 1112, 151, 79, 522 1Figure 9.642.

det1A2 � 0A�1B � X

≥

2.25 1.9 1.5 1.2

32 24 16 12

1 1 1 1

�1 1 �1 1

¥ ≥

x

l

m

s

¥ � ≥

719.8

9096

394

12

¥

μ

2.25x � 1.90l � 1.50m � 1.20s � 719.8

32x � 24l � 16m � 12s � 9096

x � l � m � s � 394

l � s � x � m � 12

revenue:

ounces sold:

quantity sold:

amounts sold:


Precalculus—

Figure 9.63


we can use determinants to

find whether a matrix is

invertible

2. The product of a square matrix A and its inverseyields the matrix.A�1



1. The identity matrix In consists of 1’s downthe and for all other entries.

n � n

9.7 EXERCISES

3. Given square matrices A and B of like size, B is theinverse of A if � � . Notationally wewrite B � .

4. If the determinant of a matrix is zero, the matrix issaid to be or , meaning noinverse exists.

5. Explain why inverses exist only for squarematrices, then discuss why some square matricesdo not have an inverse. Illustrate each point with anexample.

6. What is the connection between the determinant ofa matrix and the formula for finding itsinverse? Use the connection to create a matrix that is invertible, and another that is not.

2 � 22 � 2

Figure 9.64


cob19537_ch09_918-933.qxd 1/27/11 9:51 PM Page 927


Precalculus—


Use matrix multiplication, equality of matrices, and the

arbitrary matrix given to show that

7.

8.

9.

10.

For and

show for the

matrices of like size. Use a calculator for Exercise 14.

11. 12.

13. 14.

Find the inverse of each matrix using matrixmultiplication, equality of matrices, and a system ofequations.

15. 16.

Find the inverse of each matrix by augmenting of thethe identity matrix and using row operations.

17. 18.

Demonstrate that by showing .Do not use a calculator.

19. 20.

B � c5.5 3

�2 �1dB � c

�9 �5

2 1d

A � c�2 �6

4 11dA � c

1 5

�2 �9d

AB � BA � IB � A�1,

c�2 0.4

1 0.8dc

1 �3

4 �10d

c1 �5

0 �4dc

5 �4

2 2d

2 � 2

≥

9 1 3 �1

2 0 �5 3

4 6 1 0

0 �2 4 1

¥£

�4 1 6

9 5 3

0 �2 1

§

c0.5 �0.2

�0.7 0.3dc

�3 8

�4 10d

AI � IA � AI4 � ≥

1 0 0 00 1 0 00 0 1 00 0 0 1

¥ ,

I2 � c1 00 1

d , I3 � £

1 0 00 1 00 0 1

§ ,

A � c12

14

13

18

d ca b

c dd � c

12

14

13

18

d

A � c0.4 0.6

0.3 0.2d c

a b

c dd � c

0.4 0.6

0.3 0.2d

A � c9 �7

�5 4d c

a b

c dd � c

9 �7

�5 4d

A � c2 5

�3 �7d c

a b

c dd � c

2 5

�3 �7d

ca bc d

d � c1 00 1

d .21. 22.

Use a calculator to find , then confirm theinverse by showing

23.

24.

25.

26.

Write each system in the form of a matrix equation. Donot solve.

27.

28.

29.

30.

31.

32. μ

1.5w � 2.1x � 0.4y � z � 1

0.2w � 2.6x � y � 5.8

3.2x � z � 2.7

1.6w � 4x � 5y � 2.6z � �1.8

μ

�2w � x � 4y � 5z � �3

2w � 5x � y � 3z � 4

�3w � x � 6y � z � 1

w � 4x � 5y � z � �9

•

2x � 3y � 2z � 4

14x �

25y �

34z � �1

3

�2x � 1.3y � 3z � 5

•

x � 2y � z � 1

x � z � 3

2x � y � z � 3

e 0.5x � 0.6y � 0.6

�0.7x � 0.4y � �0.375

e 2x � 3y � 9

�5x � 7y � 8

A �1

12≥

12 �6 3 0

0 �4 8 �12

12 �12 0 0

0 12 0 �12

¥

A � £

�7 5 �3

1 9 0

2 �2 �5

§

A � £

0.5 0.2 0.1

0 0.3 0.6

1 0.4 �0.3

§

A � £

�2 3 1

5 2 4

2 0 �1

§

AB � BA � I. A�1 � B

B � c47

57

37

27

dB � c14

58

0 12

d

A � c�2 5

3 �4dA � c

4 �5

0 2d

cob19537_ch09_918-933.qxd 1/27/11 9:52 PM 8

Write each system as a matrix equation and solve (ifpossible) using inverse matrices and your calculator. Ifthe coefficient matrix is singular, write no solution.

33.

34.

35.

36.

37.

38.

39.

40.

41.

42.

43.

44.

Compute the determinant of each matrix and statewhether an inverse matrix exists. Do not use acalculator.

45. 46.

47. 48. c�2 6

�3 9dc

1.2 �0.8

0.3 �0.2d

c0.6 0.3

0.4 0.5dc

4 �7

3 �5d

μ

2w � 5x � 3y � 4z � 7

1.6w � 4.2y � 1.8z � 5.4

3w � 6.7x � 9y � 4z � �8.5

0.7x � 0.9z � 0.9

μ

�2w � 3x � 4y � 5z � �3

0.2w � 2.6x � y � 0.4z � 2.4

�3w � 3.2x � 2.8y � z � 6.1

1.6w � 4x � 5y � 2.6z � �9.8

•

4x � 5y � 6z � 3318x � 3

5y � 54z � 9

�0.5x � 2.4y � 4z � �32

•

x � 2y � 2z � 9

2x � 1.5y � 1.8z � 12�23 x �

12y �

35z � �4

•

1.7x � 2.3y � 2z � 41.5

1.4x � 0.9y � 1.6z � �10

�0.8x � 1.8y � 0.5z � 16.5

•

0.2x � 1.6y � 2z � �1.9

�0.4x � y � 0.6z � �1

0.8x � 3.2y � 0.4z � 0.2

e312a � 213b � 12

512a � 313b � 1

e�52 a � 3

5b � � 310

516a � 3

2b � � 716

e12a � 13b � 216

16a � b � 412

e�16 u � 1

4v � 112u � 2

3v � �2

e 0.3x � 1.1y � 3.5

�0.5x � 2.9y � �10.1

e0.05x � 3.2y � �15.8

0.02x � 2.4y � 12.08

Compute the determinant of each matrix without usinga calculator. If the determinant is zero, write singularmatrix.

49. 50.

51.

52.

Compute the determinant of each matrix using thecolumn rotation method.

53. 54.

55. 56.

Use a calculator to compute the determinant of eachmatrix. If the determinant is zero, write singular matrix.If the determinant is nonzero, find and store theresult as matrix B ( 2: [B] ). Thenverify the inverse by showing

57.

58.

For each system shown, form the matrix equation; compute the determinant of the coefficient

matrix and determine if you can proceed; and ifpossible, solve the system using the matrix equation.

59. 60.

61. 62. •5x � 2y � z � 1

3x � 4y � 9z � �2

4x � 3y � 5z � 6

•

x � 3y � 4z � �1

4x � y � 5z � 7

3x � 2y � z � �3

•

2x � 3y � 2z � 7

x � y � 2z � �5

3x � 2y � z � 11

•

x � 2y � 2z � 7

2x � 2y � z � 5

3x � y � z � 6

AX � B

M � ≥

1 2 1 1

0 1 �3 2

�1 0 2 �3

2 �1 1 4

¥

A � ≥

1 0 3 �4

2 5 0 1

8 15 6 �5

0 8 �4 1

¥

AB � BA � I.

ENTERx

-12ndSTO

A�1

£

5 6 2

�2 1 �2

3 4 �1

§£

1 �1 2

3 �2 4

4 3 1

§

£

�3 2 4

�1 �2 0

3 1 5

§£

2 �3 1

4 �1 5

1 0 �2

§

D � £

1 2 �0.8

2.5 5 �2

3 0 �2.5

§

C � £

�2 3 4

0 6 2

1 �1.5 �2

§

B � £

�2 2 1

0 �1 2

4 �4 0

§A � £

1 0 �2

0 �1 �1

2 1 �4

§


Precalculus—

cob19537_ch09_918-933.qxd 1/27/11 9:53 PM Page 929



Precalculus—

The inverse of a matrix:

The inverse of a matrix can be found using the formula shown, as long as . Use the formula to findinverses for the matrices here (if possible), then verify by showing

63. 64. 65. 66. c0.2 0.3

�0.4 �0.6dC � c

0.3 �0.4

�0.6 0.8dB � c

2 3

�5 �4dA � c

3 �5

2 1d

A # A�1 � A # A�1 � I.ad � bc � 02 � 2

A � ca bc d

d S A�1 �1

ad � bc# c d �b

�c ad2 � 2

Solve each application using a matrix equation.


67. Convenience store sales: The local Moto-Martsells four different sizes of Slushies—behemoth,60 oz @ $2.59; gargantuan, 48 oz @ $2.29;mammoth, 36 oz @ $1.99; and jumbo, 24 oz @ $1.59. As part of a promotion, the owner offersfree gas to any customer who can tell how many ofeach size were sold last week, given the followinginformation: (1) The total revenue for the Slushieswas $402.29; (2) 7884 ounces were sold; (3) a totalof 191 Slushies were sold; and (4) the number ofbehemoth Slushies sold was one more than thenumber of jumbo. How many of each size were sold?

68. Cartoon characters: In America, four of the mostbeloved cartoon characters are Foghorn Leghorn,Elmer Fudd, Bugs Bunny, and Tweety Bird.Suppose that Bugs Bunny is four times as tall asTweety Bird. Elmer Fudd is as tall as the combinedheight of Bugs Bunny and Tweety Bird. FoghornLeghorn is 20 cm taller than the combined height ofElmer Fudd and Tweety Bird. The combined heightof all four characters is 500 cm. How tall is each one?

69. Rolling Stones music: One of the most prolificand popular rock-and-roll bands of all time is theRolling Stones. Four of their many great hitsinclude: Jumpin’ Jack Flash, Tumbling Dice, YouCan’t Always Get What You Want, and WildHorses. The total playing time of all four songs is20.75 min. The combined playing time of Jumpin’Jack Flash and Tumbling Dice equals that of YouCan’t Always Get What You Want. Wild Horses is2 min longer than Jumpin’ Jack Flash, and YouCan’t Always Get What You Want is twice as long asTumbling Dice. Find the playing time of each song.

70. Mozart’s arias: Mozart wrote some of vocalmusic’s most memorable arias in his operas,

including Tamino’s Aria, Papageno’s Aria, theChampagne Aria, and the Catalogue Aria. Thetotal playing time of all four arias is 14.3 min.Papageno’s Aria is 3 min shorter than theCatalogue Aria. The Champagne Aria is 2.7 minshorter than Tamino’s Aria. The combined time ofTamino’s Aria and Papageno’s Aria is five timesthat of the Champagne Aria. Find the playing timeof all four arias.

Manufacturing

71. Resource allocation: Time Pieces Inc.manufactures four different types of grandfatherclocks. Each clock requires these four stages: (1) assembly, (2) installing the clockworks, (3) inspection and testing, and (4) packaging fordelivery. The time required for each stage is shownin the table, for each of the four clock types. At theend of a busy week, the owner determines thatpersonnel on the assembly line worked for 262 hr,the installation crews for 160 hr, the testingdepartment for 29 hr, and the packagingdepartment for 68 hr. How many clocks of eachtype were made?

� APPLICATIONS

72. Resource allocation: Figurines Inc. makes andsells four sizes of metal figurines, mostly historicalfigures and celebrities. Each figurine goes throughfour stages of development: (1) casting, (2) trimming,(3) polishing, and (4) painting. The time requiredfor each stage is shown in the table, for each of thefour sizes. At the end of a busy week, the managerfinds that the casting department put in 62 hr, and

Dept. Clock A Clock B Clock C Clock D

Assemble 2.2 2.5 2.75 3

Install 1.2 1.4 1.8 2

Test 0.2 0.25 0.3 0.5

Pack 0.5 0.55 0.75 1.0

cob19537_ch09_918-933.qxd 1/27/11 9:53 PM Page 930

the trimming department worked for 93.5 hr, withthe polishing and painting departments logging 138 hr and 358 hr, respectively. How manyfigurines of each type were made?

73. Thermal conductivity:In lab experimentsdesigned to measurethe heat conductivityof a square metal plateof uniform density, theedges are held at fourdifferent (constant)temperatures. Themean-value principle from physics tells us that thetemperature at a given point pi on the plate is equalto the average temperature of nearby points. Usethis information to form a system of four equationsin four variables, and determine the temperature atinterior points p1, p2, p3, and p4 on the plate shown.(Hint: Use the temperature of the four points closestto each.)

74. Thermal conductivity: Repeat Exercise 73 if (a) the temperatures at the top and bottom of theplate were increased by , with the temperaturesat the left and right edges decreased by (whatdo you notice?); (b) the temperature at the top andthe temperature to the left were decreased by ,with the temperatures at the bottom and right heldat their original temperature.

Curve Fitting

75. Quadratic fit: Use a matrix equation to find aquadratic function of the form such that , , and (2, 7) are on thegraph of the function.

76. Quadratic fit: Use a matrix equation to find aquadratic function of the form such that , (1, 5), and, are on thegraph of the function.

77. Cubic fit: Use a matrix equation to find a cubicfunction of the form suchthat and (3, 8) are onthe graph of the function.

78. Cubic fit: Use a matrix equation to find a cubicfunction of the form suchthat and (3, 25) are on thegraph of the function.1�2, 52, 10, 12, 12, �32,

y � ax3 � bx2 � cx � d

1�4, �62, 1�1, 02, 11, �162,y � ax3 � bx2 � cx � d

12, �621�4, �02y � ax2 � bx � c

10, �521�4, �52y � ax2 � bx � c

10°

10°10°

Investing

79. Wise investing: Morgan received an $800 giftfrom her grandfather, and showing wisdom beyondher years, decided to place the money in acertificate of deposit (CD) and a money marketfund (MM). At the time, CDs were earning 3.5%and MMs were earning 2.5%. At the end of 1 yr,she cashed both in and received a total of $824.50.How much was deposited in each?

80. Baseball cards: Gary has a passion for baseball,which includes a collection of rare baseball cards. His most prized cards feature Willie Mays(1953 Topps) and Mickey Mantle (1959 Topps).The Willie Mays card has appreciated 28% and theMickey Mantle card 25% since he purchased them,and together they are now worth $17,100. If hepaid a total of $13,507.50 at auction for both cards,what was the original price of each?


Precalculus—

64°C

96°C

70°C 80°Cp1 p2

p3 p4

Willie Mays (1953 Topps) $7187.50

Dept. Small Medium Large X-Large

Casting 0.5 0.6 0.75 1

Trimming 0.8 0.9 1.1 1.5

Polishing 1.2 1.4 1.7 2

Painting 2.5 3.5 4.5 6

81. Retirement planning: Using payroll deduction,Jeanette was able to put aside $4800 per month lastyear for her impending retirement. Last year, hercompany retirement fund paid 4.2% and her mutualfunds returned 5.75%, but her stock fund actuallydecreased 2.5% in value. If her net gain for theyear was $104.50 and $300 more was placed instocks than in mutual funds, how much was placedin each investment vehicle?

82. Charitable giving: Thehyperbolic funnels seenat many shopping mallsare primarily used bynonprofit organizations toraise funds for worthycauses. A coin islaunched down a ramp into the funnel andseemingly makes endless circuits before finallydisappearing down a “black hole” (the collectionbin). During one such collection, the bin was foundto hold $112.89, and 1450 coins consisting ofpennies, nickels, dimes, and quarters. How many of each denominator were there, if the number ofquarters and dimes was equal to the number ofnickels, and the number of pennies was twice the number of quarters.

cob19537_ch09_918-933.qxd 1/28/11 8:18 PM Page 931

Nutrition

83. Animal diets: A zoo dietician needs to create a specialized diet thatregulates an animal’s intake of fat, carbohydrates, and protein duringa meal. The table given shows three different foods and the amount ofthese nutrients (in grams) that each ounce of food provides. Howmany ounces of each should the dietician recommend to supply 20 gof fat, 30 g of carbohydrates, and 44 g of protein?

84. Training diet: A physical trainer is designing a workout diet for oneof her clients, and wants to supply him with 24 g of fat, 244 g ofcarbohydrates, and 40 g of protein for the noontime meal. The tablegiven shows three different foods and the amount of these nutrients (ingrams) that each ounce of food provides. How many ounces of eachshould the trainer recommend?

85. Some matrix applications require that you solve a matrix equation of the form where A, B, and Care matrices with the appropriate number of rows and columns and exists. Investigate the solution process

for such equations using and then solve for X

symbolically (using I, and so on).

86. Another alternative for finding determinants uses the triangularized form of a matrix and is offered withoutproof: If nonsingular matrix A is written in triangularized form using standard row operations but withoutexchanging any rows and without using the operation kRi to replace any row (k a constant), then det(A) isequal to the product of resulting diagonal entries. Compute the determinant of each matrix using this method.Be careful not to interchange rows and do not replace any row by a multiple of that row in the process.

a. b. c. d. £3 �1 4

0 �2 6

�2 1 �3

§£

�2 4 1

5 7 �2

3 �8 �1

§£

2 5 �1

�2 �3 4

4 6 5

§£

1 �2 3

�4 5 �6

2 5 3

§

A�1,

AX � B � CX � cx

yd ,A � c

2 3

�5 �4d , B � c

4

9d , C � c

12

�4d ,

A�1AX � B � C,


Precalculus—


Nutrient Food I Food II Food III

Fat 2 4 3

Carb. 4 2 5

Protein 5 6 7

Nutrient Food I Food II Food III

Fat 2 5 0

Carb. 10 15 18

Protein 2 10 0.75


87. (6.3) Find the amplitude and period of

88. (2.2/5.3) Match each equation to its related graph.Justify your answers.

a. b.y � log2 x � 2y � log2 1x � 22

y � �125 cos13t2.89. (2.3) Solve the absolute value inequality:

90. (7.6) Find all solutions of in30, 2�2.

�7 tan2x � �21

�3�2x � 5� � 7 � �19.

1098765432�1 1

�5�4

�6

�3�2�1

54321

y

x

x � 2

1098765432�1 1

�5�4

�6

�3�2�1

54321

y

x

cob19537_ch09_918-933.qxd 1/27/11 9:53 PM Page 932

Precalculus—

9.8 Applications of Matrices and Determinants: Cramer’s Rule, Geometry, and More

In addition to solving systems, matrices can be used to accomplish such diverse things asfinding the volume of a three-dimensional solid or establishing certain geometrical rela-tionships in the coordinate plane. Numerous uses are also found in higher mathematics,such as checking whether solutions to a differential equation are linearly independent.

A. Solving Systems Using Determinants and Cramer’s Rule

In addition to identifying singular matrices, determinants can actually be used todevelop a formula for the solution of a system. Consider the following solution to ageneral system, which parallels the solution to a specific system. With aview toward a solution involving determinants, the coefficients of x are written as a11

and a21 in the general system, and the coefficients of y are a12 and a22.

Specific System General System

eliminate the x-term eliminate the x-term

�3R1 � 2R2 �a21R1 � a11R2

sums to zero sums to zero

Notice the x-terms sum to zero in both systems. We are deliberately leaving the so-lution on the left unsimplified to show the pattern developing on the right. Next wesolve for y.

Factor Out y Factor Out y

On the left we find and back-substitution shows . But moreimportant, on the right we obtain a formula for the y-value of any system:

. If we had chosen to solve for x, the solution would be

. Note these formulas are defined only if

You may have already noticed, but this denominator is the determinant of the matrix of

coefficients from the previous section! Since the numerator is also a difference

of two products, we investigate the possibility that it too can be expressed as a determi-nant. Working backward, we’re able to reconstruct the numerator for x in determinant

form as where it is apparent this matrix was formed by replacing the coef-

ficients of the x-variables with the constant terms.

`c1 a12

c2 a22``

a12

a22``

a11 a12

a21 a22`

cc1 a12

c2 a22d ,

ca11 a12

a21 a22d

a11a22 � a21a12 � 0.x �a22c1 � a12c2

a11a22 � a21a12

y �a11c2 � a21c1

a11a22 � a21a12

2 � 2x � 2y � �7

�7 � 1

y �a11c2 � a21c1

a11a22 � a21a12 y �

2 # 10 � 3 # 9

2 # 4 � 3 # 5

1a11a22 � a21a122y � a11c2 � a21c1 12 # 4 � 3 # 52y � 2 # 10 � 3 # 9

a11a22y � a21a12y � a11c2 � a21c12 # 4y � 3 # 5y � 2 # 10 � 3 # 9

e �a21a11x � a21a12y � �a21c1

a11a21x � a11a22y � a11c2e

�3 # 2x � 3 # 5y � �3 # 9

2 # 3x � 2 # 4y � 2 # 10

ea11x � a12y � c1

a21x � a22y � c2e

2x � 5y � 9

3x � 4y � 10

2 � 22 � 2


how we can:

A. Solve a system usingdeterminants andCramer’s rule

B. Use determinants inapplications involvinggeometry in thecoordinate plane

C. Decompose a rationalexpression into partialfractions using matricesand technology

D. Use matrices to solvestatic systems

E. Use matrices forencryption/decryption

remove coefficients of x

(removed)

replace with constants

divide divide

9–97 933

cob19537_ch09_918-933.qxd 1/27/11 9:54 PM Page 933


Precalculus—

It is also apparent the numerator for y can be also written in determinant form as

, or the determinant formed by replacing the coefficients of the y-variables

with the constant terms:

If we use the notation Dy for this determinant, Dx for the determinant where x co-efficients were replaced by the constants, and D as the determinant for the matrix ofcoefficients—the solution of the system can be written as shown next, with the resultknown as Cramer’s rule.

Cramer’s Rule for 2 � 2 Systems

Given a system of linear equations

the solution is the ordered pair (x, y), where

and

provided

EXAMPLE 1 � Solving a System Using Cramer’s Rule

Use Cramer’s rule to solve the system

Solution � Begin by finding the value of D, Dx, and Dy.

� 7

This gives and The solution is

Check by substituting these values into the original equations.


Regardless of the method used to solve a system, always be aware that a consistent,

inconsistent, or dependent system is possible. The system yields

in standard form, with

Since Cramer’s rule cannot be applied, and the system is either inconsistent ordependent. To find out which, we write the equations in function form (solve

D � 0,

D � `�2 1

4 �2` � 1�22 1�22 � 142 112 � 0.e

�2x � y � �3

4x � 2y � �6

ey � 2x � �3

4x � 6 � 2y

12, �12.

y �Dy

D�

7

�7� �1.x �

Dx

D�

�14

�7� 2

� �14� �7

� 122 1�102 � 1�32 192� 192 142 � 1�102 1�52� 122 142 � 1�32 1�52

Dy � `2 9

�3 �10`Dx � `

9 �5

�10 4`D � `

2 �5

�3 4`

e2x � 5y � 9

�3x � 4y � �10.

D � 0.

y �Dy

D�

à11 c1

a21 c2`

à11 a12

a21 a22`

x �Dx

D�

`c1 a12

c2 a22`

à11 a12

a21 a22`

ea11x � a12y � c1

a21x � a22y � c2

2 � 2

à11 c1

a21 c2``

a11

a21 `à11 a12

a21 a22`

à11 c1

a21 c2`

remove coefficients of y

replace with constants

(removed)

cob19537_ch09_934-947.qxd 1/27/11 10:19 PM Page 934

for y). The result is , showing the system consists of two parallel lines

and has no solutions.

Cramer’s Rule for 3 � 3 Systems

Cramer’s rule can be extended to a system of linear equations, using the samepattern as for systems. Given the general system

the solutions are and where Dx, Dy, and Dz are again formed

by replacing the coefficients of the indicated variable with the constants, and D is thedeterminant of the matrix of coefficients .

Cramer’s Rule Applied to 3 � 3 Systems

Given a system of linear equations

The solution is an ordered triple (x, y, z), where

provided

EXAMPLE 2 � Solving a 3 � 3 System Using Cramer’s Rule

Solve using Cramer’s rule:

Solution � Begin by computing the determinant of the matrix of coefficients, to ensure that Cramer’s rulecan be applied. Using the third row, we have

Since we continue, electing to compute the remaining determinants using a calculator(see figures).

Dz � †

1 �2 �1

�2 1 1

3 3 2

† � �6Dy � †

1 �1 3

�2 1 �5

3 2 4

† � 0Dx � †

�1 �2 3

1 1 �5

2 3 4

† � 12

D � 0

� 3172 � 3112 � 41�32 � 6

D � †

1 �2 3

�2 1 �5

3 3 4

† � �3 `�2 3

1 �5` � 3 `

1 3

�2 �5` � 4 `

1 �2

�2 1`

•

x � 2y � 3z � �1

�2x � y � 5z � 1

3x � 3y � 4z � 2

D � 0.

z �

†

a11 a12 c1

a21 a22 c2

a31 a32 c3

†

†

a11 a12 a13

a21 a22 a23

a31 a32 a33

†

,y �

†

a11 c1 a13

a21 c2 a23

a31 c3 a33

†

†

a11 a12 a13

a21 a22 a23

a31 a32 a33

†

x �

†

c1 a12 a13

c2 a22 a23

c3 a32 a33

†

†

a11 a12 a13

a21 a22 a23

a31 a32 a33

†

•

a11x � a12y � a13z � c1

a21x � a22y � a23z � c2

a31x � a32y � a33z � c3

3 � 3

1D � 02

z �Dz

D,y �

Dy

D,x �

Dx

D,

•

a11x � a12y � a13z � c1

a21x � a22y � a23z � c2

a31x � a32y � a33z � c3

3 � 32 � 23 � 3

ey � 2x � 3

y � 2x � 3

7–99 Section 9.8 Applications of Matrices and Determinants: Cramer’s Rule, Geometry, and More 935

Precalculus—

cob19537_ch09_934-947.qxd 1/27/11 10:20 PM Page 935

As the final screen shows, the

solution is , and

or in triple form. Check this

solution in the original equations.


12, 0, �12z �Dz

D�

�6

6� �1,

y �Dy

D�

0

6� 0,x �

Dx

D�

12

6� 2


Precalculus—


we can solve a system using

determinants and Cramer’s

rule B. Determinants, Geometry, and the Coordinate Plane

As mentioned in the introduction, the use of determinants extends far beyond solvingsystems of equations. Here, we’ll demonstrate how determinants can be used to findthe area of a triangle whose vertices are given as three points in the coordinate plane.

The Area of a Triangle in the xy-Plane

Given a triangle with vertices at (x1, y1), (x2, y2), and (x3, y3),

EXAMPLE 3 � Finding the Area of a T riangle Using DeterminantsFind the area of a triangle with vertices at (3, 1), (�2, 3), and (1, 7) (see Figure 9.65).

Solution � Begin by forming matrix T and computing det(T) (see Figure 9.66):

The area of this triangle is 13 units2.


As an extension of this formula, what if the three points were collinear? After amoment, it may occur to you that the formula would give an area of 0 units2, since notriangle could be formed. This gives rise to a test for collinear points.

� 13

Compute the area: A � `det1T 2

2` � `

�26

2`

� �12 � 3 � 1�172 � �26 � 313 � 72�11�2 � 12 � 11�14 � 32

det1T2 � †

x1 y1 1

x2 y2 1

x3 y3 1

† � †

3 1 1

�2 3 1

1 7 1

†

Area � `det1T2

2` where T � £

x1 y1 1

x2 y2 1

x3 y3 1

§

Figure 9.66

T � [A]Figure 9.65

x

y

54321�5 �4 �3 �2 �1

1

�1�2�3

4567

3

(3, 1)(�2, 3)

(1, 7)

2

cob19537_ch09_934-947.qxd 1/27/11 10:20 PM Page 936


Precalculus—

B. You’ve just seen how we

can use determinants in

applications involving geometry

in the coordinate plane

Test for Collinear Points

Three points (x1, y1), (x2, y2), and (x3, y3) are collinear if

.

See Exercises 33 through 38.

C. More on Partial Fraction Decomposition

Occasionally, the decomposition template for a rational expression becomes rather ex-tensive. Instead of attempting a solution using convenient values or row operations as inSection 9.4, solutions can more readily be found using a matrix equation and technology.

EXAMPLE 4 � Decomposing a Rational Expression Using Matrices and T echnologyUse matrix equations and a graphing calculator to decompose the given expression

into partial fractions:

Solution � The degree of the numerator is less than that of the denominator, so we begin byfactoring the denominator. After removing the common factor of x and applyingthe rational roots theorem with synthetic division the completely

factored form is . The decomposed form will be

Clearing denominators and simplifying yields

clear denominators

After expanding the powers on the right, grouping like terms, and factoring, we obtain

By equating the coefficients of like terms, the following system and matrixequation are obtained:

After carefully entering the matrices F (coefficients)and G (constants) (see Figure 9.67), we obtain thesolution , and as shown in Figure 9.68. The decomposed form is


12x3 � 62x2 �102x �56

x1x � 223�

7x

�5

1x � 221�

2

1x � 223.

D � �2A � 7, B � 5, C � 0

μ

A � B � 12

6A � 4B � C � 62

12A � 4B � 2C � D � 102

8A � 56

S ≥

1 1 0 0

6 4 1 0

12 4 2 1

8 0 0 0

¥ ≥

A

B

C

D

¥ � ≥

12

62

102

56

¥ .

� 112A � 4B � 2C � D2x � 8A 12x3 � 62x2 � 102x � 56 � 1A � B2x3 � 16A � 4B � C2x2

� Cx1x � 22 � Dx 12x3 � 62x2 � 102x � 56 � A1x � 223 � Bx1x � 222

12x3 � 62x2 � 102x � 56

x1x � 223�

Ax

�B

1x � 221�

C

1x � 222�

D

1x � 223.

12x3 � 62x2 � 102x � 56

x1x � 223

1c � �22,

12x3 � 62x2 � 102x � 56

x4 � 6x3 � 12x2 � 8x.

det1A2 � †

x1 y1 1

x2 y2 1

x3 y3 1

† � 0

Figure 9.68

Figure 9.67

C. You’ve just seen how we

can decompose a rational

expression into partial fractions

using matrices and technology

cob19537_ch09_934-947.qxd 1/27/11 10:21 PM Page 937

D. Solving Static Systems with Varying Constraints

When the considerations of a business or industry involve more than two variables,solutions using matrix methods have a distinct advantage over other methods. Compa-nies often have to perform calculations using basic systems weekly, daily, or evenhourly, to keep up with trends, market changes, changes in cost of raw materials, andso on. In many situations, the basic requirements remain the same, but the frequentlychanging inputs require a recalculation each time they change.


Precalculus—

EXAMPLE 5 � Determining Supply Inventories Using MatricesBNN Soft Drinks receives new orders daily for its most popular drink, SaratogaCola. It can deliver the carbonated beverage in a twelve-pack of 12-ounce (oz)cans, a six-pack of 20-oz bottles, or in a 2-L bottle. The ingredients required toproduce a twelve-pack include 1 gallon (gal) of carbonated water, 1.25 pounds (lb)of sugar, 2 cups (c) of flavoring, and 0.5 grams (g) of caffeine. For the six-pack, 0.8 gal of carbonated water, 1 lb of sugar, 1.6 c of flavoring, and 0.4 g of caffeineare needed. The 2-L bottle contains 0.47 gal of carbonated water, 0.59 lb of sugar,0.94 c of flavoring, and 0.24 g of caffeine. How much of each ingredient must beon hand for Monday’s order of 300 twelve-packs, 200 six-packs, and 500 2-Lbottles? What quantities must be on hand for Tuesday’s order: 410 twelve-packs,320 six-packs, and 275 2-L bottles?

Solution � Begin by setting up a general system of equations, letting x represent the numberof twelve-packs, y the number of six-packs, and z the number of 2-L bottles:

As a matrix equation we have

Enter the matrix as matrix A, and the size ofthe order as matrix as B. Using a calculator, we find

,

and BNN Soft Drinks will need 695 gal ofcarbonated water, 870 lb of sugar, 1390 c of flavoring, and 350 g of caffeine for Monday’s order. After entering

for Tuesday’s orders, computing the product AC shows 795.25 gal of carbonatedwater, 994.75 lb of sugar, 1590.5 c of flavoring, and 399 g of caffeine are neededfor Tuesday.


Example 5 showed how the creation of a static matrix can help track and controlinventory requirements. In Example 6, we use a static matrix to solve a system that willidentify the amount of data traffic used by a company during various hours of the day.

C � £

410

320

275

§

AB � ≥

1 0.8 0.47

1.25 1 0.59

2 1.6 0.94

0.5 0.4 0.24

¥ £

300

200

500

§ � ≥

695

870

1390

350

¥

4 � 3

≥

1 0.8 0.47

1.25 1 0.59

2 1.6 0.94

0.5 0.4 0.24

¥ £

x

y

z

§ � ≥

w

s

f

c

¥

0.5x � 0.4y � 0.24z � grams of caffeine 2x � 1.6y � 0.94z � cups of flavoring

1.25x � 1y � 0.59z � pounds of sugar 1x � 0.8y � 0.47z � gallons of carbonated water

μ

cob19537_ch09_934-947.qxd 1/27/11 10:21 PM Page 938

EXAMPLE 6 � Identifying the Source of Data T raffic Using MatriceMariño Imports is a medium-size company that is considering upgrading from a 1.544 megabytes per sec (Mbps) T1 Internet line to a fractional T3 line with abandwidth of 7.72 Mbps. They currently use their bandwidth for phone traffic, officedata, and Internet commerce. The IT (Information Technology) director devises a planto monitor how much data traffic each resource uses on an hourly basis. Because ofthe physical arrangement of the hardware, she cannot monitor each resourceindividually. The table shows the information she collected for the first 3 hr. Determinehow many gigabytes (GB) each resource used individually during these 3 hr.

Solution � Using p to represent the phone traffic, d for office data, and cfor Internet commerce, we create the system shown, whichmodels data use for the 9:00 o’clock hour. Since we actuallyneed to solve two more systems whose only difference is theconstant terms (for the 10 and 11 o’clock hours), using a matrix equation to solvethe system (Section 9.7) will be most convenient. Begin by writing the relatedmatrix equation for this system:

Using to solve the system (see figure),we find there was 1.2 GB of phone traffic, 2.8 GBof office data, and 1.4 GB of Internet commerceduring this hour. Note that the IT director maymake this calculation 10 or more times a day(once for every hour of business). While we couldsolve for the 10 and 11 o’clock hours

using , then calculate and , these

calculations can all be performed simultaneously by combining the matrices B, C,and D into one matrix and multiplying by . Due to the properties ofmatrix multiplication, each column of the product will represent the datainformation for a given hour, as shown here:

In the second hour, there was 1.1 GB of phone traffic, 2.7 GB of office data, and1.5 GB of Internet commerce. In the third hour, 1.5 GB of phone traffic, 2.0 GB ofoffice data, and 1.6 GB of Internet commerce bandwidth was used.


3A�1 4 £

5.4 5.3 5.14.0 3.8 3.54.2 4.2 3.6

§ � £

1.2 1.1 1.5

2.8 2.7 2.0

1.4 1.5 1.6

§

A�13 � 3

A�1DA�1CC � £

5.3

3.8

4.2

§ and D � £

5.13.53.6§

X � A�1B

AX � B: £

1 1 1

1 1 0 §

0 1 1

£

p

d

c

§ � £

5.4

4.0

4.2

§


Precalculus—

Phone, Data, Phone and Data andand Commerce Data Commerce

9–10:00 A.M. 5.4 GB 4.0 GB 4.2 GB

10–11:00 A.M. 5.3 GB 3.8 GB 4.2 GB

11–12:00 P.M. 5.1 GB 3.5 GB 3.6 GB

•

p � d � c � 5.4

p � d � 4.0

d � c � 4.2


we can use matrix equations

to solve static systems

cob19537_ch09_934-947.qxd 1/27/11 10:21 PM Page 939

E. Using Matrices to Encrypt Messages

In The Gold-Bug, by Edgar Allan Poe, the narrator deciphers the secret message on atreasure map by accounting for the frequencies of specific letters and words. The codingused was a simple substitution cipher, which today can be broken with relative ease. Inmodern times where information wields great power, a simple cipher like the one usedthere will not suffice. When you pay your tuition, register for classes, make on-line pur-chases, and so on, you are publicly transmitting very private data, which needs to be pro-tected. While there are many complex encryption methods available (including the nowfamous symmetric-key and public-key techniques), we will use a matrix-based tech-nique. The nature of matrix multiplication makes it very difficult to determine the exactmatrices that yield a given product, and we’ll use this fact to our advantage. Beginningwith a fixed, invertible matrix A, we will develop a matrix B such that the product AB ispossible, and our secret message is encrypted in AB. At the receiving end, they will needto know to decipher the message, since , which is theoriginal message. Note that in case an intruder were to find matrix A (perhaps purchas-ing the information from a disgruntled employee), we must be able to change it easily.This means we should develop a method for generating a matrix A, with integer entries,where A is invertible and also has integer entries.

EXAMPLE 7 � Finding an Invertible Matrix A Where Both A and A�1 Have Integer EntriesFind an invertible matrix A as just described, and its inverse .

Solution � Begin with any matrix that has only 1s or on its main diagonal, and 0sbelow the diagonal. The upper triangle can consist of any integer values youchoose, as in

.

Now, use any of the elementary row operations tomake the matrix more complex. For instance, we’ll use a calculator to create a new matrix by (1) using to create matrix [C], then(2) using to create matrix [D], then(3) using to create matrix [E], andfinally (4) to obtain our finalmatrix [A]. To begin, enter the initial matrix asmatrix [B]. For (1) , go to the(MATRIX) MATH submenu, select option

( and press to bring this option to thehome screen. This feature requires us to name thematrix we’re using, and to indicate what rows toadd, so we enter ([B], 1, 2). The screenshown in Figure 9.69 indicates we’ve placed theresult in matrix [C]. For , recall

([B], 1, 2) using and change it to([C], 1, 3) [D] (Figure 9.70). Repeat

this process for (3) to create matrix[E]: ([D], 3, 2) [E] (Figure 9.71).Finally, we compute (4) usingthe (new) option F:*row�(�2, [E], 1, 3) [A]STO

�2R1 � R3 S R3STOD:row�

R3 � R2 S R2STOD:row�

ENTER2ndD:row�R1 � R3 S R3

D:row�

ENTERD:row�

R1 � R2 S R2

�2 R1 � R3 S R3R3 � R2 S R2R1 � R3 S R3R1 � R2 S R2

£

�1 5 �1

0 1 8

0 0 1

§

�1s3 � 3

A�13 � 3

A�1

A�11AB2 � 1A�1A2B � BA�1


Precalculus—

Figure 9.69

Figure 9.70

Figure 9.71

WORTHY OF NOTEPerforming row operations isexplained in more detail in thegraphing calculator manualaccompanying this text.

cob19537_ch09_934-947.qxd 1/27/11 10:21 PM Page 940

and the process is complete (Figure 9.72). The entries of matrix A are all integers,and exists and also has integer entries (Figure 9.73). This will always be thecase for matrices created in this way.


EXAMPLE 8 � Using Matrices to Encr ypt MessagesSet up a substitution cipher to encode the message MATH IS SWEET, and then usethe matrix A from Example 7 to encrypt it.

Solution � For the cipher, we will associate a unique number to every letter in the alphabet. Thiscan be done randomly or using a systematic approach. Here we choose to associate 0with a blank space, and assign 1 to A, to B, 2 to C, to D, and so on.�2�1

A�1


Precalculus—

Figure 9.73Figure 9.72

M A T H I S S W E E T

7 1 0 5 10 0 10 12 3 3 �10�4�10

N O P Q R S T U V W X Y Z

8 9 10 11 12 13 �13�12�11�10�9�8�7

Blank A B C D E F G H I J K L M

0 1 2 3 4 5 6 7�6�5�4�3�2�1

Now encode the secret message as shown:

We next enter the coded message into a new matrix B, by entering it letter by letterinto the columns of B. Note that since the encrypting matrix A is , B musthave 3 rows for multiplication to be possible. The result is

Since the message is too short to fill matrix B, we use blank spaces to complete thefinal column. Computing the product AB encrypts the message, and only someonewith access to will be able to read it:

The encrypted message is 8, , , , 43, 6, , 50, 30, 0, 30, 3, 10, 20, .


�10�20�1�18�73

� £

8 �1 �20 0 10

�73 43 50 30 20

�18 6 30 3 �10

§

AB � £

�1 5 �1

�2 11 7

1 �5 2

§ £

7 �4 10 12 �10

1 0 0 3 0

�10 5 10 3 0

§

A�1

B � £

M H S W T

A * * E *

T I S E *

§ � £

7 �4 10 12 �10

1 0 0 3 0

�10 5 10 3 0

§

3 � 3

cob19537_ch09_934-947.qxd 1/27/11 10:22 PM Page 941

EXAMPLE 9 � Deciphering Encrypted Messages Using an Inverse MatrixDecipher the encrypted message from Example 8 using from Example 7.

Solution � The received message is 8, , , , 43, 6, , 50, 30, 0, 30, 3, 10, 20, , and is the result of the product AB. To find matrix B, we apply since

. Writing the received message in matrix form we have

Next multiply AB by on the left, to determine matrix B:

Writing matrix B in sentence form gives 7, 1, , , 0, 5, 10, 0, 10, 12, 3, 3,, 0, 0, and using the substitution cipher to replace numbers with letters, reveals

the message MATH IS SWEET.


�10�4�10

� £

7 �4 10 12 �10

1 0 0 3 0

�10 5 10 3 0

§ � B

A�11AB2 � £

�57 5 �46

�11 1 �9

1 0 1

§ £

8 �1 �20 0 10

�73 43 50 30 20

�18 6 30 3 �10

§

A�1

AB � £

8 �1 �20 0 10

�73 43 50 30 20

�18 6 30 3 �10

§

A�1 1AB2 � 1A�1 A2B � BA�1�10

�20�1�18�73

A�1


Precalculus—


we can use matrices for

encryption/decryption

1. The determinant is evaluated

as: .

`a11 a12

a21 a22` 2. rule uses a ratio of determinants to

solve for the unknowns in a system.

3. Given the matrix of coefficients D, the matrix Dx isformed by replacing the coefficients of x with the

terms.

4. The three points (x1, y1), (x2, y2), and (x3, y3) are

collinear if has a value of .0T 0 � †

x1 y1 1

x2 y2 1

x3 y3 1

†

5. Discuss/Explain the process of writing as a

sum of partial fractions.

8x � 3

x2 � x6. Discuss/Explain why Cramer’s rule cannot be

applied if Use an example to illustrate.D � 0.

9.8 EXERCISES




Write the determinants D, Dx, and Dy for the systemsgiven. Do not solve.

7. 8. e�x � 5y � 12

3x � 2y � �8e

2x � 5y � 7

�3x � 4y � 1

Solve each system of equations using Cramer’s rule, ifpossible. Do not use a calculator.

9. 10. ex � �2y � 11

y � 2x � 13e

4x � y � �11

3x � 5y � �60

cob19537_ch09_934-947.qxd 1/27/11 10:22 PM Page 942


Precalculus—


Area of a Norman window: The determinant shown can be used to find the area of a Norman

window (rectangle � half-circle) with length L, width W, and radius Find the area of the following windows.

23. 24.

32 cm

58 cm20 in.

16 in.

r �W2

.

A � †L r2

��

2W† .

� APPLICATIONS

Area of a triangle: Find the area of the triangle with thevertices given. Assume units are in centimeters.

25. (2, 1), (3, 7), and (5, 3)

26. and

Area of a parallelogram: Find the area of the parallelogramwith vertices given (Hint: Use two triangles.) Assume unitsare in feet.

27. and

28. , (5, 0), (5, 4), and 1�5, �221�5, �62

15, 221�4, 22, 1�6, �12, 13, �12,

1�6, 121�2, 32, 1�3, �42,

Volume of a pyramid: The volume of a triangular pyramidis given by the formula where B represents thearea of the triangular base and h is the height of thepyramid. Find the volume of a triangular pyramid whoseheight is given and whose base has the coordinates shown.Assume units are in meters.

29. vertices (3, 5),

30. vertices and 1�6, 121�2, 32, 1�3, �42,h � 7.5 m;

1�4, 22, and 1�1, 62h � 6 m;

V � 13Bh,

11. 12.

13. 14.

The two systems given in Exercises 15 and 16 areidentical except for the third equation. For the firstsystem given, (a) write the determinants D, Dx, Dy, andDz then (b) determine if a solution using Cramer’s ruleis possible by computing |D| without the use of a calculator(do not solve the system). Then (c) compute |D| for thesecond system and try to determine how the equationsin the second system are related.

15. 16.

•

2x � 3z � �2

�x � 5y � z � 12

3x � 5y � 4z � 10

•

4x � y � 2z � �5

�3x � 2y � z � 8

x � y � z � 3

•

2x � 3z � �2

�x � 5y � z � 12

3x � 2y � z � �8

•

4x � y � 2z � �5

�3x � 2y � z � 8

x � 5y � 3z � �3

e�2.5x � 6y � �1.5

0.5x � 1.2y � 3.6e

0.6x � 0.3y � 8

0.8x � 0.4y � �3

μ

2

3x �

3

8y �

7

5

5

6x �

3

4y �

11

10

μ

x

8�

y

4� 1

y

5�

x

2� 6

Use Cramer’s rule to solve each system of equations.Verify computations using a graphing calculator.

17. 18.

19.

20.

21.

22. μ

w � 2x � 3y � z � 11

3w � 2y � 6z � �13

2x � 4y � 5z � 16

3x � 4z � 5

μ

w � 2x � 3y � �8

x � 3y � 5z � �22

4w � 5x � 5

�y � 3z � �11

•

x � 2y � 5z � 10

3x � z � 8

�y � z � �3

•

y � 2z � 1

4x � 5y � 8z � �8

8x � 9z � 9

•

x � 3y � 5z � 6

2x � 4y � 6z � 14

9x � 6y � 3z � 3

•

x � 2y � 5z � 10

3x � 4y � z � 10

x � y � z � �2

cob19537_ch09_934-947.qxd 1/28/11 7:30 PM Page 943


Precalculus—

Volume of a prism: The volume ofa right prism is given by theformula , where Brepresents the area of the base and his the height of the prism. Find thevolume of a triangular prism whoseheight is given and whose base hasthe coordinates shown. Assumeunits are in inches.

31. (3, 0), (9, 0), (6, 4),

32. (2, 3), (9, 5), (0, 6),

Determine if the following sets of points are collinear.

33. (1, 5), , and (4, 11)

34. (1, 1), and

35. , and

36. and (3, 3.95)

For each linear equation given, substitute the first twopoints to verify they are solutions. Then use the test forcollinear points to determine if the third point is also asolution.

37.

38.

Decompose each rational expressing using a matrixequation and a graphing calculator.

39.

40.

41.

42.

43. Slammin’ Drums manufactures several differenttypes of drums. Its most popular drums are the 22�bass drum, the 12� tom, and the 14� snare drum.The 22� bass drum requires 7 ft2 of skin, 8.5 ft2 ofwood veneer, 8 tension rods, and 11.5 ft of hoop.The 12� tom requires 2 ft2 of skin, 3 ft2 of woodveneer, 6 tension rods, and 6.5 ft of hoop. The 14�snare requires 2.5 ft2 of skin, 1.5 ft2 of woodveneer, 10 tension rods, and 7 ft of hoop. InFebruary, Slammin’ Drums received orders for 15 bass drums, 21 toms, and 27 snares. Use yourcalculator and a matrix equation to determine how

16x3 � 66x2 � 98x � 54

12x2 � 3x2 14x2 � 12x � 92

x3 � 17x2 � 76x � 98

1x2 � 6x � 92 1x2 � 2x � 32

�3x4 � 13x2 � x � 12

x5 � 4x3 � 4x

x4 � x2 � 2x � 1

x5 � 2x3 � x

5x � 2y � 4; 12, �32, 13.5, �6.752, 1�2.7, 8.752

2x � 3y � 7; 12, �12, 1�1.3, �3.22, 1�3.1, �4.42

1�0.5, 2.552, 1�2.8, 1.632,

12.2, �8.521�2.5, 5.22, 11.2, �5.62

1�2, 9213, �52,

1�2, �12

h � 5

h � 8

V � Bh

much of each raw material they need to have onhand to fill these orders.

44. In March, Slammin’ Drums’ orders consisted of 19 bass drums, 19 toms, and 25 snares. Use yourcalculator and a matrix equation to determine howmuch of each raw material they need to have onhand to fill their orders. (See Exercise 43.)

45. The following table represents Slammin’s ordersfor the months of April through July. Use yourcalculator and a matrix equation to determine howmuch of each raw material they need to have onhand to fill these orders. (See Exercise 43.) (Hint:Using a clever and matrix can reducethis problem to a single step.)

46. The following table represents Slammin’s ordersfor the months of August through November. Useyour calculator and a matrix equation to determinehow much of each raw material they need to haveon hand to fill their orders (see Exercise 43). (Hint: Using a clever and matrix canreduce this problem to a single step.)

47. Roll-X Watches makes some of the finestwristwatches in the world. Their most popularmodel is the Clam. It comes in three versions:Silver, Gold, and Platinum. Management thinksthere might be a thief in the production line, so theydecide to closely monitor the precious metalconsumption. A Silver Clam contains 1.2 oz ofsilver and 0.2 oz of gold. A Gold Clam contains 0.5 oz of silver, 0.8 oz of gold, and 0.1 oz ofplatinum. A Platinum Clam contains 0.2 oz ofsilver, 0.5 oz of gold, and 0.7 oz of platinum.During the first week of monitoring, the productionteam used 10.9 oz of silver, 9.2 oz of gold, and 2.3 oz of platinum. Use your calculator and amatrix equation to determine the number of eachtype of watch that should have been produced.

48. The following table contains the precious metalconsumption of the Roll-X Watch production line

3 � 14 � 3

3 � 14 � 3

April May June July

Bass drum 23 21 17 14

Tom 20 18 15 17

Snare drum 29 35 27 25

August September October November

Bass drum 17 22 16 12

Tom 15 14 13 11

Snare drum 32 28 27 21

cob19537_ch09_934-947.qxd 1/27/11 10:23 PM Page 944


Precalculus—

during the next five weeks (see Exercise 47). Useyour graphing calculator to determine the numberof each type of watch that should have beenproduced each week. For which week does the dataseem to indicate a possible theft of precious metal?

49. There are three classes of grain, of which threebundles from the first class, two from the second,and one from the third make 39 measures. Two ofthe first, three of the second, and one of the thirdmake 34 measures. And one of the first, two of thesecond, and three of the third make 26 measures.How many measures of grain are contained in onebundle of each class? (This is the historic problemfrom the Chiu chang suan shu.)

50. During a given week, the measures of grain thatmake up the bundles in Exercise 49 can varyslightly. Three local Chinese bakeries always buythe same numbers of bundles, as outlined inExercise 49. That is to say, bakery 1 buys threebundles of the first class, two of the second, and oneof the third. Bakery 2 buys two of the first, three ofthe second, and one of the third. And finally, bakery3 buys one of the first, two of the second, and threeof the third. The following table outlines how manymeasures of grain each bakery received each day.How many measures of grain were contained in onebundle of each class, on each day?

For Exercises 51–56, use the criteria indicated to findmatrices A and , where the entries of both are

all integers.

51. The lower triangle is all zeroes.

52. The upper triangle is all zeroes.

53.

54.

55. and

56. and a1,3 � 1a2,1 � �3

a2,3 � 2a3,1 � 1

a3,2 � �2

a2,1 � 5

A�13 � 3

57. Use the matrix A you created in Exercise 51 andthe substitution cipher from Example 8 to encryptyour full name.

58. Use the matrix A you created in Exercise 52 andthe substitution cipher from Example 8 to encryptyour school’s name.

59. Design your own substitution cipher. Then use itand the matrix A you created in Exercise 53 toencrypt the title of your favorite movie.

60. Design your own substitution cipher. Then use itand the matrix A you created in Exercise 54 toencrypt the title of your favorite snack food.

61. Design your own substitution cipher. Then use itand the matrix A you created in Exercise 55 toencrypt the White House switchboard phonenumber, 202-456-1414.

62. Design your own substitution cipher. Then use itand the matrix A you created in Exercise 56 toencrypt the Casa Rosada switchboard phonenumber 54-11-4344-3600. The Casa Rosada, orPink House, consists of the offices of the presidentof Argentina.

63. Use the matrix from Exercise 51, and theappropriate substitution cipher to decrypt themessage from Exercise 57.






Use a linear system to model each application. Thensolve using Cramer’s rule.

69. Return on investments: If $15,000 is invested at acertain interest rate and $25,000 is invested atanother interest rate, the total return was $2900. Ifthe investments were reversed the return would be$2700. What was the interest rate paid on eachinvestment?

A�1

A�1

A�1

A�1

A�1

A�1

Mon Tues Wed Thurs Fri

Bakery 1 39 38 38 37.75 39.75(measures)

Bakery 2 34 33 33.5 32.5 35(measures)

Bakery 3 26 26 27 26.25 27.25(measures)

Ounces Week 1 Week 2 Week 3 Week 4 Week 5

Silver 13.1 9 12.9 11.9 11.2

Gold 11 7.7 8.6 8.4 9.5

Platinum 2.5 1.5 0.9 2.8 1.7

cob19537_ch09_934-947.qxd 1/27/11 10:23 PM Page 945

Precalculus—

70. CD and DVD clearance sale: To generate interest ina music store clearance sale, the manger sets out alarge box full of $2 CDs and $7 DVDs, with anadvertised price of $800 for the lot. When asked howmany of each are in the box, the manager will onlysay the box holds a total of 200 disks. How manyCDs and DVDs are in the box?

71. Cost of fruit: Many years ago, two pounds ofapples, 2 lb of kiwi, and 10 lb of pears cost $3.26.Three pounds of apples, 2 lb of kiwi, and 7 lb ofpears cost $2.98. Two pounds of apples, 3 lb ofkiwi, and 6 lb of pears cost $2.89. Find the cost ofa pound of each fruit.

72. Vending machine receipts: The vending machinesat an amusement park are stocked with variouscandies that sell for 5¢, 10¢, and 25¢. At week’send, the park collected $54.30 from one of themachines. How many of each type of candy weresold, if 484 total sales were made and the machinevended twice as many 5¢ candies as 25¢ candies?

73. Coffee blends: To make its morning coffees, acoffee shop uses three kinds of beans costing$1.90/lb, $2.25/lb, and $3.50/lb, respectively. By theend of the week, the shop went through 24 lb ofcoffee beans, having a total value of $58. Find howmany pounds of each type of bean were used, giventhat the number of pounds used of the cheapestbeans was four more than the most expensive beans.

74. Manufacturing ball bearings: A ball bearingproducer makes three sizes of bearings, half-inch-diameter size weighing 30 grams (g), a three-quarter-inch-diameter size weighing 105 g, and a 1-in.-diameter size weighing 250 g. A large storagecontainer holds ball bearings that have been rejecteddue to small defects. If the net weight of thecontainer’s contents is 95.6 kilograms (95,600 g),and automated tallies show 920 bearings have beenrejected, how many of each size is in the reject bin,given that statistical studies show there are twice asmany rejects of the smallest bearing, as compared tothe largest?


75. Solve the given system four different ways: (1) elimination, (2) row reduction, (3) Cramer’s rule,and (4) using a matrix equation. Which methodseems to be the least error-prone? Which methodseems most efficient (takes the least time)? Discussthe advantages and drawbacks of each method.

76. Find the area of the pentagon whose vertices are:(8, 6), and (0, 12.5).

77. The polynomial form for the equation of a circle isFind the equation of

the circle that contains the points (2, 8),and

78. For square matrix A, calculating A2 is a simplematter of matrix multiplication. But what of those applications that require higher powers?

For any matrix we define the

characteristic polynomial of A to be

p1�2 � `� � a �b

�c � � d` � 1� � a2 1� � d2 � cb.

A � ca b

c dd ,

15, �12.1�1, 72,

x2 � y2 � Dx � Ey � F � 0.

1�8, 62,1�5, �52, 15, �52,

•

x � 3y � 5z � 6

2x � 4y � 6z � 14

9x � 6y � 3z � 3

The Cayley-Hamilton theorem states that if theoriginal matrix A is substituted for the resultmust be zero, or in other words, every squarematrix satisfies its own characteristic equation.This fact is often used to compute higher powers ofa square matrix, since all higher powers can thenbe expressed in terms of A. Specifically, for

with the theorem stating that (note I2 is used to ensure we remain in the contextof matrices). For we havethe follow sequence for any power of A:

For find A2 in terms of A and I2,

then use the Cayley-Hamilton theorem to find A3,A4, and A5. Verify A3 using scalar multiplicationand matrix addition.

A � c2 �1

4 1d ,

� 145A � 54I2 � 27A � 10I2

� 2715A � 2I22 � 10A � 515A � 2I22 � 2A � 27A2 � 10A � 5A2 � 2A � 127A � 10I22A � 15A � 2I22A

A4 � A3A A3 � A2A

A2 � 5A � 2I2,2 � 2

A2 � 5A � 2I2 � 01� � 12 1� � 42 � 1�32 1�22 � �2 � 5� � 2,

A � c1 2

3 4d , p1�2 � `

� � 1 �2

�3 � � 4` �

�,


79. (4.3) Graph the polynomial using information about end-behavior, y-intercept, x-intercept(s), and midintervalpoints: .f 1x2 � x3 � 2x2 � 7x � 6


cob19537_ch09_934-947.qxd 1/27/11 10:24 PM Page 946

Precalculus—

80. (2.2) Which is the graph (left or right) ofJustify your answer.g1x2 � ��x � 1� � 3?

81. (8.1) Solve the triangle:side in.side in.

82. (6.4) Graph over the interval Note the period, asymptotes, zeroes and

value of A.3�1

2, 12 4 .y � 3 tan12�t2

�A � 49.0°b � 11.2a � 8.7

54321�5�4�3�2�1�1�2�3�4�5

234

1

y5

x 54321�5�4�3�2�1�1�2�3�4�5

234

1

5y

x

MAKING CONNECTIONS

Making Connections: Graphically , Symbolically , Numerically, and V erballyEight graphs (a) through (h) are given. Match the characteristics shown in 1 through 16 to one of the eight graphs.

5�5

�5

5

x

y

5�5

�5

5

x

y

5�5

�5

5

x

y

1. ____

2. ____

3. ____

4. ____ coincident dependence

5. ____ inconsistent

6. ____ linear dependence

7. ____ vertices: (1, 2), (5, 0), (0, 0), (0, 1)

8. ____ independent and consistent

μ

y x � 1

y 0

y �1

2x �

5

2

ex y � 1

x � 2y 5

`1 �1 �1

1 2 5`

(a) (b) (c) (d)

5�5

�5

5

x

y

5�5

�5

5

x

y

5�5

�5

5

x

y(e) (f) (g) (h)

9. ____ three equations in three variables

10. ____ (3, 2) does not satisfy any inequality

11. ____

12. ____

13. ____ (1, 2) is the solution

14. ____ (3, 2) is a solution

15. ____

16. ____ A �1

2†

�1 0 1

1 2 1

5 0 1

†

ey x � 1

x � 2y 5

`�1 1 1

1 �1 3`

•

y x � 1

x � 2y 5

x 0, y 0

9–111 Making Connections 947

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Precalculus—

SECTION 9.1 Linear Systems in T wo Variables with Applications

KEY CONCEPTS• A solution to a linear system in two variables is any ordered pair (x, y) that makes all equations in the system true.

• Since every point on the graph of a line satisfies the equation of that line, a point where two lines intersect mustsatisfy both equations and is a solution of the system.

• A system with at least one solution is called a consistent system.

• If the lines have different slopes, there is a unique solution to the system (they intersect at a single point). Thesystem is called a consistent and independent system.

• If the lines have equal slopes and the same y-intercept, they form identical or coincident lines. Since one line isright atop the other, they intersect at all points with an infinite number of solutions. The system is called aconsistent and dependent system.

• If the lines have equal slopes but different y-intercepts, they will never intersect. The system has no solution and iscalled an inconsistent system.

EXERCISESSolve each system by graphing manually. Verify your answer by graphing on a graphing calculator. If the system isinconsistent or dependent, so state.

1. 2. 3.

Solve using substitution. Indicate whether each system is consistent, inconsistent, or dependent. Write uniquesolutions as an ordered pair. Check your answer using a graphing calculator.

4. 5. 6.

Solve using elimination. Indicate whether each system is consistent, inconsistent, or dependent. Write uniquesolutions as an ordered pair. Check your answer using a graphing calculator.

7. 8.

9. When it was first constructed in 1968, the John Hancock building in Chicago, Ilinois, was the tallest structure inthe world. In 1974, the Willis Tower in Chicago (formerly known as the Sears Tower) became the world’s talleststructure. The Willis Tower is 323 ft taller than the John Hancock Building, and the sum of their heights is 2577 ft. How tall is each structure?

10. The manufacturer of a revolutionary automobile spark plug finds that demand for the plugcan be modeled by the function D(p) � �0.8p � 110, where D(p) represents the number ofplugs bought (demanded, in tens of thousands) at price p in cents. The supply of these sparkplugs is modeled by S(p) � 0.24p � 14.8, where S(p) represents the number of plugsmanufactured/supplied (in tens of thousands) at price p. Find the price for marketequilibrium using a graphing calculator.

SECTION 9.2 Linear Systems in Thr ee Variables with Applications

KEY CONCEPTS• The graph of a linear equation in three variables is a plane.

• A linear system in three variables has the following possible solution sets:

• If the planes intersect at a point, the system could have one unique solution (x, y, z).

• If the planes intersect at a line, the system has linear dependence and the solution (x, y, z) can be written aslinear combinations of a single variable (a parameter).

e2x � 3y � 6

2.4x � 3.6y � 6e

2x � 4y � 10

3x � 4y � 5

ex � 2y � 3

x � 4y � �1e

x � y � 4

0.4x � 0.3y � 1.7e

y � 5 � x

2x � 2y � 13

e2x � y � 2

x � 2y � 4e

0.2x � 0.5y � �1.4

x � 0.3y � 1.4e

3x � 2y � 4

�x � 3y � 8

SUMMAR Y AND CONCEPT REVIEW

cob19537_ch09_948-957.qxd 1/27/11 10:28 PM Page 948

• If the planes are coincident, the equations in the system differ by a constant multiple, meaning they are all“disguised forms” of the same equation. The solutions have coincident dependence, and the solution set can berepresented by any one of the equations.

• In all other cases, the system has no solutions and is an inconsistent system.

EXERCISESSolve using elimination. If a system is inconsistent or dependent, so state. For systems with linear dependence, givethe answer as an ordered triple using a parameter. Verify solutions on the home screen using the keys.

11. 12. 13.

Solve using a system of three equations in three variables.14. A large coin jar is full of nickels, dimes, and quarters. There are 217 coins in all. The number of nickels is 12

more than the number of quarters. The value of the dimes is $4.90 more than the value of the nickels. How manycoins of each type are in the bank?

15. If the point (x, y) is on the graph of a parabola, it must satisfy the equation Use a system ofthree equations in three variables to find the equation of the parabola containing the points (2, 7), and (6, 15).

SECTION 9.3 Systems of Inequalities and Linear Pr ogramming

KEY CONCEPTS• To solve a system of inequalities, we find the intersecting or overlapping regions of the solution regions from the

individual inequalities. The common area is called the solution region for the system.

• The process known as linear programming seeks to maximize or minimize the value of a given quantity undercertain constraints or restrictions.

• The quantity we attempt to maximize or minimize is called the objective function.

• The solution(s) to a linear programming problem occur at one of the corner points of the feasible (solution) region.

• The process of solving a linear programming application contains these six steps:

• Identify the main objective and the decision variables.

• Write the objective function in terms of these variables.

• Organize all information in a table, using the decision variables and constraints.

• Fill in the table with the information given and write the constraint inequalities.

• Graph the constraint inequalities and determine the feasible region.

• Identify all corner points of the feasible region and test these points in the objective function.

EXERCISESGraph the solution region for each system of linear inequalities by first solving equation for y. Verify each solutionusing a test point. Solve Exercise 18 using a graphing calculator.

16. 17. 18.

19. Carefully graph the feasible region for the system of inequalities shown, then maximize the

objective function:

20. After retiring, Oliver and Lisa Douglas buy and work a small farm (near Hooterville) that consists mostly of milkcows and egg-laying chickens. Although the price of a commodity is rarely stable, suppose that milk sales bringin an average of $85 per cow and egg sales an average of $50 per chicken over a period of time. During this timeperiod, the new ranchers estimate that care and feeding of the animals took about 3 hr per cow and 2 hr per

•

2x � y � 10

2x � 3y � 18

x � 0, y � 0

f 1x, y2 � 30x � 45y

ex � 2y � 1

2x � y � �2e

x � 4y � 5

�x � 2y � 0e

�x � y 7 �2

�x � y 6 �4

10, �92,y � ax2 � bx � c.

•

3x � y � 2z � 3

x � 2y � 3z � 1

4x � 8y � 12z � 7

•

�x � y � 2z � 2

x � y � z � 1

2x � y � z � 4

•

x � y � 2z � �1

4x � y � 3z � 3

3x � 2y � z � 4

ALPHA

9–113 Summary and Concept Review 949

Precalculus—

cob19537_ch09_948-957.qxd 1/27/11 10:28 PM Page 949

chicken, while maintaining the related equipment took 2 hr per cow and 1 hr per chicken. How many animals ofeach type should be maintained in order to maximize profits, if at most 1000 hr can be spent on care and feeding,and at most 525 hr on equipment maintenance?

SECTION 9.4 Partial Fraction Decomposition

KEY CONCEPTS• Partial fraction decomposition is an attempt to reverse the process of adding rational expressions.

• The process begins by factoring the denominator and setting up a “decomposition template.”

• For each linear factor of the denominator, the template will have a term of the form a constant.

• For each irreducible quadratic factor of the denominator, the template will have a term of the form

for constants A1 and B1.

• If the factor is repeated n times the template will have n terms of the form for to n.

• If the irreducible quadratic factor is repeated n times, the template will have n terms of the form

for to n.

• After the template has been set multiply both sides by the LCDand (1) use convenient values or (2) simplify the right-hand side and equate coefficients to find the constants Ai

and Bi needed. Note that some of the coefficients found may be zero.

• If the degree of the numerator is greater than the degree of the denominator, divide using long division anddecompose the remainder portion.

EXERCISESDecompose each of the following expressions into partial fractions.

21. 22. 23.

24. 25. 26.

27. 28. 29.

SECTION 9.5 Solving Linear Systems Using Matrices and Row Operations

KEY CONCEPTS• A matrix is a rectangular arrangement of numbers. An matrix has m rows and n columns.

• The matrix derived from a system of linear equations is called the augmented matrix. It is created by augmentingthe coefficient matrix (formed by the variable coefficients) with the matrix of constants.

• One matrix method for solving systems involves the augmented matrix and row-reduction.

• If possible, interchange equations so that the coefficient of x is a “1” in R1.

• Write the system in augmented matrix form (coefficient matrix with matrix of constants).

• Use row operations to obtain zeroes below the first entry of the diagonal.

• Use row operations to obtain zeroes below the second entry of the diagonal.

• Continue until the matrix is triangularized (all entries below the diagonal are zero).

• Convert the augmented matrix back into equation form and solve for z.

• Graphing calculators can quickly and accurately perform row-reduction.

m � n

2x2 � x � 1

x4 � 6x2 � 9

�x2 � 15x � 22

x3 � 3x2 � 9x � 5

2x2 � 8x � 33

x3 � x2 � 8x � 12

6x2 � 2x � 7

x3 � 1

�2x2 � 13x � 24

x3 � 27

�2x2 � 3x � 19

x3 � 5x2 � 3x � 15

x2 � 2x � 9

x3 � 4x2 � x � 4

16x � 1

2x2 � 5x � 3

6x � 14

x2 � 2x � 15

1original expression � decomposition template2,

i � 1Aix � Bi

1ax2 � bx � c2i,

i � 1Ai

1x � c2i1x � c2

A1x � B1

ax2 � bx � c

1ax2 � bx � c2

A1

x � c, A11x � c2


Precalculus—

cob19537_ch09_948-957.qxd 1/27/11 10:28 PM Page 950

EXERCISESSolve by triangularizing the augmented matrix. Use a calculator for Exercise 32.

30. 31. 32.

SECTION 9.6 The Algebra of Matrices

KEY CONCEPTS• The entries of a matrix are denoted aij, where i gives the row and j gives the column of its location.

• The size of a matrix is also referred to as its order.

• Two matrices A and B are equal if corresponding entries are equal: if

• The sum or difference of two matrices is found by combining corresponding entries:

• The identity matrix for addition is an matrix whose entries are all zeroes.

• The product of a constant times a matrix is called scalar multiplication, and is found by taking the product of thescalar with each entry, forming a new matrix of like size. For matrix

• Matrix multiplication is performed as according to the following procedure: For anmatrix and a matrix matrix multiplication is possible if and the result will

be an matrix P. In symbols where pij is product of the ith row of A with the jth column of B.

• When technology is used to perform operations on matrices, the focus shifts from a meticulous computation ofnew entries, to carefully entering each matrix into the calculator, double checking that each entry is correct, andappraising the results to see if they are reasonable.

EXERCISESCompute the operations indicated below (if possible), using the following matrices.

33. 34. 35. 36. 8A 37. BA

38. 39. 40. BC 41. 42. CD

SECTION 9.7 Solving Linear Systems Using Matrix Equations

KEY CONCEPTS• The identity matrix I for multiplication has 1’s on the main diagonal and 0’s for all other entries. For any

matrix A, the identity matrix is also an matrix , where

• For an (square) matrix A, the inverse matrix for multiplication is a matrix B such that Onlysquare matrices have inverses. For matrix A the inverse is denoted

• Any system of equations can be written as a matrix equation and solved (if solutions exist) using an inverse

matrix. For the matrix equation is

• Every square matrix has a real number associated with it called its determinant. For a matrix

det(A)

• If the determinant of a matrix is zero, the matrix is said to be singular or noninvertible.

• For matrix equations, if the coefficient matrix is noninvertible, the system has no unique solution.

� a11a22 � a21a12.A � `a11 a12

a21 a22` ,

2 � 2

£

2 3 �1

�3 2 4

1 �5 2

§ # £x

y

z

§ � £

5

13

�3

§ .•

2x � 3y � z � 5

�3x � 2y � 4z � 13

x � 5y � 2z � �3

,

n � nA�1.

AB � BA � I.n � nAIn � InA � A.Inn � n

n � n

�4DD � CC � D

C � BB � AA � B

D � £

2 �3 0

0.5 1 �1

4 0.1 5

§C � £

�1 3 4

5 �2 0

6 �3 2

§B � c�7 6

1 �2dA � c

�14

�34

�18

�78

d

A # B � 3pij 4 ,m � qn � p,B � 3bij 4 ,p � qA � 3aij 4m � n

row entry � column entryA: kA � kaij.

m � nA � B � 3aij � bij 4 .

aij � bij.A � Bm � n

μ

2w � x � 2y � 3z � �19

w � 2x � y � 4z � 15

x � 2y � z � 1

3w � 2x � 5z � �60

•

x � 2y � 2z � 7

2x � 2y � z � 5

3x � y � z � 6

ex � 2y � 6

4x � 3y � 4

9–115 Summary and Concept Review 951

Precalculus—

cob19537_ch09_948-957.qxd 1/27/11 10:29 PM Page 951

EXERCISESComplete Exercises 43 through 45 using the following matrices:

43. Exactly one of the matrices given is singular. Compute each determinant to identify it.

44. Show that What can you conclude about matrix A?

45. Show that What can you conclude about matrix C?

Use a graphing calculator to complete Exercises 46 through 49, using these matrices:

46. Exactly one of the matrices above is singular. Determine which one.

47. Show that What can you conclude about the matrix G?

48. Show that What can you conclude about the matrix H?

49. Verify that and and comment.

Solve manually using a matrix equation.

50.

Solve using a matrix equation and your calculator.

51.

SECTION 9.8 Applications of Matrices and Deter minants: Cramer’s Rule, Geometr y, and Mor e

KEY CONCEPTS• Cramer’s rule uses a ratio of determinants to solve systems of equations (if they exist).

• To compute the value of and larger determinants, a calculator is generally used.

• Determinants can be used to find the area of a triangle in the plane if the vertices are known, and as a test to see ifthree points are collinear.

• A system of equations can be used to write a rational expression as a sum of its partial fractions.

EXERCISESSolve using Cramer’s rule. Use a graphing calculator for Exercise 54.

52. 53. 54.

55. Find the area of a triangle whose vertices have the coordinates (6, 1), , and

56. Use a matrix equation and a graphing calculator to find the partial fraction decomposition for 7x2 � 5x � 17

x3 � 2x2 � 3x � 6.

1�6, 22.1�1, �62

•

2x � y � �2

�x � y � 5z � 12

3x � 2y � z � �8

•

2x � y � z � �1

x � 2y � z � 5

3x � y � 2z � 8

e5x � 6y � 8

10x � 2y � �9

3 � 3

•

0.5x � 2.2y � 3z � �8

�0.6x � y � 2z � �7.2

x � 1.5y � 0.2z � 2.6

e2x � 5y � 14

�3y � 4x � �14

EF � FEEH � HE

FH � HF � I.

GF � FG � F.

H � £

�1 0 �1

0 1 0

2 1 1

§G � £

1 0 0

0 1 0

0 0 1

§F � £

1 �1 1

0 1 0

�2 1 �1

§E � £

1 �2 3

�2 1 �5

�1 �1 �2

§

BC � CB � I.

AB � BA � B.

D � c10 �6

�15 9dC � c

2 �1

3 1dB � c

0.2 0.2

�0.6 0.4dA � c

1 0

0 1d


Precalculus—

cob19537_ch09_948-957.qxd 1/27/11 10:29 PM Page 952

57. Midwest Petroleum (MP) produces three types of combustibles using common refined gasoline and vegetableproducts. The first is E10 (also known as gasohol), the second is E85, and the third is biodiesel. One gallon ofE10 requires 0.90 gal of gasoline, 2 lb of corn, 1 oz of yeast, and 0.5 gal of water. One gallon of E85 requires0.15 gal of gasoline, 17 lb of corn, 8.5 oz of yeast, and 4.25 gal of water. One gallon of biodiesel requires 20 lb ofcorn and 3 gal of water. One week’s production at MP consisted of 100,000 gal of E10, 15,000 gal of E85, and7000 gal of biodiesel. Use your calculator and a matrix equation to determine how much of each raw materialthey used to fill their orders.

58. The following table represents Midwest Petroleum’s production for thenext 3 weeks. Use your calculator and a matrix equation to determinethe total amount of raw material they used to fill their orders. SeeExercise 57.

59. In addition to being one of the world’s first celebrity chefs, Marie-Antoine Carême (1784–1833) was a master pastry chef. Had heused matrices to encode his favorite culinary math secret, perhaps his message would have been:

If Carême used the cipher on page 941 (Example 8) and the matrix for the

encoding process, decode the message to find the unconventional culinary secret that is actually familiar to students of mathematics.

A � £

1 2 3

1 3 1

2 4 7

§

11, 10, 25, �25, �6, �59, 33, 13, 76, 34, 43, 69, �3, 0, �6

9–117 Practice Test 953

Precalculus—

Week 2 Week 3 Week 4

E10 110,000 95,000 105,000

E85 17,000 18,000 20,000

Biodiesel 6000 8000 10,000

PRACTICE TEST

Solve each system and state whether the system isconsistent, inconsistent, or dependent.Verify solutionsusing a graphing calculator.

1. Solve graphically:

2. Solve using substitution:



5. Given matrices A and B, compute:

a. b. c. AB

d. e.

B � c3 3

�3 �5dA � c

�3 �2

5 4d

�A�A�1

2

5BA � B

•

x � 2y � z � �4

2x � 3y � 5z � 27

�5x � y � 4z � �27

e5x � 8y � 1

3x � 7y � 5

e3x � y � 2

�7x � 4y � �6

e3x � 2y � 12

�x � 4y � 10

6. Given matrices C and D, use a calculator to find:a. b. c. DCd. e.

7. Solve using matrices and row reduction:

8. Solve using Cramer’s rule:

9. Solve using a matrix equation and your calculator:

10. Solve using a matrix equation and your calculator:

•

x � 2y � 2z � 7

2x � 2y � z � 5

3x � y � z � 6

e2x � 5y � 11

4x � 7y � 4

•

2x � 3y � z � 3

x � 2y � z � 4

x � y � 2z � �1

•

4x � 5y � 6z � 5

2x � 3y � 3z � 0

x � 2y � 3z � 5

D � £

0.5 0.1 0.2

�0.1 0.1 0

0.3 0.4 0.8

§C � £

0.5 0 0.2

0.4 �0.5 0

0.1 �0.4 �0.1

§

�D�D�1�0.6DC � D

cob19537_ch09_948-957.qxd 1/27/11 10:29 PM Page 953

Cramer’s RuleIn Section 9.8, we saw that one interesting application of matrices is Cramer’s rule. You may have noticed that whentechnology is used with Cramer’s rule, the chances of making an error are fairly high, as we need to input the entriesfor numerous matrices. However, as we mentioned in the chapter introduction, one of the advantages of matrices isthat they are easily programmable, and we can actually write a very simple program that will make Cramer’s rule amore efficient method.

To begin, press the key, and then the right arrow twice and to create a name for our program. At theprompt, we’ll enter CRAMER2. As we write the program, note that the needed commands (ClrHome, Disp, Pause,Prompt, Stop) are all located in the submenus of the key, and the sign is found under the TEST menu,accessed using the keys (the arrows “S” are used to indicate the key.

The following program takes the coefficients and constants of a 2 � 2 linear system, and returns the ordered pairsolution in the form of x � h and y � k (for constants h and k). Even with minimal programming experience, readingthrough the program will help you identify that Cramer’s rule is being used.

STOMATH2nd

=PRGM

ENTERPRGM


Precalculus—

CALCULATOR EXPLORA TION AND DISCOVER Y

Use a system of equations to model each exercise, thensolve using the method of your choice.

11. The perimeter of a “legal-size” paper is 114.3 cm.The length of the paper is 7.62 cm less than twicethe width. Find the dimensions of a legal-size sheetof paper.

12. The island nations of Tahiti and Tonga have acombined land area of 692 mi2. Tahiti’s land area is112 mi2 more than Tonga’s. What is the land area ofeach island group?

13. After inheriting $30,000 from a rich aunt, Daviddecides to place the money in three differentinvestments: a savings account paying 5%, a bondaccount paying 7%, and a stock account paying 9%.After 1 yr he earned $2080 in interest. Find howmuch was invested at each rate if $8000 less wasinvested at 9% than at 7%.

14. The company from Exercise 10 of the Summary and Concept Review, hired a market research firm to collect data about the next-generation spark plug that will increase fuel efficiency by over 20%. The data they collected areshown in the table. Enter the data into a graphingcalculator, calculate the regression line for each setof data, then (a) display the lines and scatterplot on asingle screen, and (b) use the lines to find theestimated price and quantity at which marketequilibrium is achieved.

Using a radar gun placed in a protective cage beneath afalling baseball, the following data points are taken: (1, 128), (2, 80), (2.5, 44). Assume the data are of theform (time in seconds, distance from radar gun).

15. Use the data to set up a system of three equations inthree variables to find the quadratic equation thatmodels the ball’s height at time t.

16. From what height is the baseball being released? Inhow many seconds will it strike the cage protectingthe radar gun?

17. Solve the system of inequalities by graphing.

18. Maximize the objective function:

Solve the linear programming problem.

19. A company manufactures two types of T-shirts, a plainT-shirt and a deluxe monogrammed T-shirt. Toproduce a plain shirt requires 1 hr of working time onmachine A and 2 hr on machine B. To produce adeluxe shirt requires 1 hr on machine A and 3 hron machine B. Machine A is available for at most50 hr/week, while machine B is available for at most 120 hr/week. If a plain shirt can be sold at a profit of $4.25 each and a deluxe shirt can be sold at a profit of $5.00 each, how many of each should bemanufactured to maximize the profit?

20. Decompose the expression into partial fractions:4x2 � 4x � 3

x3 � 27.

•

x � 2y � 8

8x � 5y � 40

x � 0, y � 0

P � 50x � 12y

ex � y � 2

x � 2y � 8

Price Demand Supply(cents) (10,000s) (10,000s)

135 2.9 19.7

90 37.2 8.0

62 59.6 0.45

112 19.7 12.7

75 46.0 3.69

121 12.4 15.9

cob19537_ch09_948-957.qxd 1/27/11 10:29 PM Page 954

9–119 Strengthening Core Skills 955

Precalculus—

STRENGTHENING CORE SKILLS

Augmented Matrices and Matrix InversesThe formula for finding the inverse of a matrix has its roots in the more general method of computing the inverseof an matrix. This involves augmenting a square matrix M with its corresponding identity In on the right (formingan matrix), and using row operations to transform M into the identity. In some sense, as the original matrix istransformed, the “identity part” keeps track of the operations we used to convert M and we can use the results to “getback home,” so to speak. We’ll illustrate with the matrix from Section 9.7, Example 2B, where we found that

was the inverse matrix for We begin by augmenting with the identity matrix.

As you can see, the identity is automatically transformed into the inverse matrix when this method is applied.

Performing similar row operations on the general matrix results in the formula given earlier. As you might

imagine, attempting this on a general matrix is problematic at best, and instead we simply apply the augmentedmatrix method to find for the matrix shown in blue.

£

1 0 0 1 �0.5 �0.5

0 1 0 �1 1 1

0 0 1 �2 1.25 1.75

§ .

R1

�14S R1

R2

7S R2

£

�14 0 0 �14 7 7

0 7 0 �7 7 7

0 0 1 �2 1.25 1.75

§4R3 � R2 S R2

4R3 � R1 S R1

£

�14 0 �4 �6 2 0

0 7 �4 1 2 0

0 0 1 �2 1.25 1.75

§R3

8S R3£

�14 0 �4 �6 2 0

0 7 �4 1 2 0

0 0 8 �16 10 14

§R2 � 7R1 S R1

5R2 � 7R3 S R3

£

2 1 0 1 0 0

0 7 �4 1 2 0

0 �5 4 �3 0 2

§R1 � 2R2 S R2

�3R1 � 2R3 S R3£

2 1 0 1 0 0

�1 3 �2 0 1 0

3 �1 2 0 0 1

§

3 � 3A�1 3 � 3

ca b

c dd

c6 5 1 0

0 1 �1 3d �5R2 � R1 S R1 c

6 0 6 �15

0 1 �1 3d

R1

6S R1 c

1 0 1 �2.5

0 1 �1 3d

c6 5 1 0

2 2 0 1d �3R2 � R1 S R2 c

6 5 1 0

0 �1 1 �3d �1R2 S R2 c

6 5 1 0

0 1 �1 3d

2 � 2c6 5

2 2dc

6 5

2 2d .c

1 �2.5

�1 3d

2 � 2

n � 2nn � n

2 � 2

ClrHomeDisp ''2�2 SYSTEMS''PauseDisp ''AX�BY � C''Disp ''DX�EY � F''Disp ''''Disp ''ENTER THE VALUES''Disp ''FOR A,B,C,D,E,F''Disp ''''Prompt A,B,C,D,E,F

Exercise 1: Use the program to check the answers to Exercises 1, 2, and 3 of the Practice Test.Exercise 2: Create systems of your own that are (a) consistent, (b) inconsistent, and (c) dependent. Then verifyresults using the program.Exercise 3: Use the box on page 935 of Section 9.8 to write a similar program for systems. Call the programCRAMER3, and repeat parts (a), (b), and (c) from Exercise 2.

3 � 3

2 � 2

(CE–BF)/(AE–BD)SX(AF–DC)/(AE–BD)SYClrHomeDisp ''THE SOLUTION IS''Disp ''''Disp ''X�''Disp XDisp ''Y�''Disp YStop

→

→

→

cob19537_ch09_948-957.qxd 1/27/11 10:29 PM Page 955


Precalculus—

To verify, we show ✓ ( A � I also checks).

Exercise 1: Use the preceding inverse and a matrix equation to solve the system

•

2x � y � �2

�x � 3y � 2z � �15

3x � y � 2z � 9

.

A�1£

2 1 0

�1 3 �2

3 �1 2

§ £

1 �0.5 �0.5

�1 1 1

�2 1.25 1.75

§ � £

1 0 0

0 1 0

0 0 1

§AA�1 � I:

1. Solve each equation by factoring.a.b.c.d.

2. Solve for x.

3. A torus is a donut-shaped solid figure. Its surfacearea is given by the formula where R is the outer radius of the donut, and r is theinner radius. Solve the formula for R in terms of rand A.

4. State the value of all six trig functions given (21, ) is a point on the terminal side of .

5. Sketch the graph of using

transformations of

6. A jai alai player in an open-air court becomesfrustrated and flings the pelota out and beyond thecourt. If the initial velocity of the ball is 136 mphand it is released at height of 5 ft, (a) how high is theball after 3 sec? (b) What is the maximum height ofthe ball? (c) How long until the ball hits the ground(hopefully in an unpopulated area)? Recall theprojectile equation is .

7. For a complex number (a) verify the sum ofa complex number and its conjugate is a realnumber. (b) Verify the product of a complex numberand its conjuate is a real number.

8. Solve using the quadratic formula:

9. Solve by completing the square:

10. Given and approximate the value of and without using a calculator.

cos 125°cos 19°cos 72° � 0.3,cos 53° � 0.6

3x2 � 72x � 427 � 0.

5x2 � 8x � 2 � 0.

a � bi,

h1t2 � �16t2 � v0t � k

y � cos x.

y � 4 cos a�

6x �

�

3b

��28

A � �21R2 � r22,

�3

x � 3�

3x

x2 � x � 6�

1

x � 2

x3 � 4x � 3x23x3 � 15x2 � 6x � 30x2 � 7x � 09x2 � 12x � �4

11. Given is a point on the unit circle, find thevalue of , and if .

12. State the domain of each function:a.b.

c.

13. Write the following formulas from memory:a. slope formulab. midpoint formulac. quadratic formulad. distance formulae. interest formula (compounded continuously)

14. Find the equation of the line perpendicular to theline , that contains the point (0, 1).

15. For the force vectors F1 and F2 given, find a thirdforce vector F3 that will bring these vectors intoequilibrium:

16. A commercial fishery stocks a lake with 250 fish.Based on previous experience, the population offish is expected to grow according to the model

where t is the time in months.

From on this model, (a) how many months arerequired for the population to grow to 7500 fish?(b) If the fishery expects to harvest three-fourths ofthe fish population in 2 yr, approximately how manyfish will be taken?

17. Evaluate each expression by drawing a right triangleand labeling the sides.

a.

b. sin c csc�1a29 � x2

xb d

sec c sin�1ax

2121 � x2b d

P 1t2 �12,000

1 � 25e�0.2t,

F1 H�5, 12I, F2 H8, 6I

4x � 5y � �20

h 1x2 �x � 3

x2 � 5

g 1x2 � logb1x � 32f 1x2 � 12x � 3

y 7 0tan �sin �, cos �A13

4 , yB

CUMULATIVE REVIEW CHAPTERS 1–9

cob19537_ch09_948-957.qxd 1/27/11 10:29 PM Page 956

18. An luxury ship is traveling at 15 mph on aheading of . There is a strong, 12 mph oceancurrent flowing from the southeast, at a heading of . What is the true course and speed of the ship?

19. Use the Guidelines for Graphing to sketch the graphof function f given, then use it to solve

20. Use the Guidelines for Graphing to sketch the graphof function g given, then use it to name the intervals

where and

21. Find using De Moivre’s theorem.

22. Solve

23. If I deposited $200 each month into an annuityprogram that paid 8% annual interest compoundedmonthly, how long would it take to save $10,000?

24. Mount Tortolas lies on the Argentine-Chilean border.When viewed from a distance of 5 mi, the angle ofelevation to the top of the peak is How tall isMount Tortolas? State the answer in feet.

25. The graph given is of the form Find the values of A, B, and C.

y � A sin1Bx � C2.

38°.

ln1x � 22 � ln1x � 32 � ln14x2.

11 � i1328

g1x2 �x2 � 4

x2 � 1g 1x2c:g 1x2T

f 1x2 � x3 � 4x2 � x � 6f 1x2 6 0:

330°

10°Exercises 26 through 30 require the use of a graphingcalculator.

26. At Leroy’s Butcher Shop, last week’s specials wereboneless chuck steaks, chicken breasts, and chorizosausages. Pamela spent a total of $17.81 on 3.21 lbof chuck, 2.45 lb of chicken, and 1.15 lb of chorizo.Shannon purchased 1.57 lb of chuck, 4.21 lb ofchicken, and 0.89 lb of chorizo with $15.24, whilethe 2.41 lb of chuck, 1.39 lb of chicken, and 1.46 lbof chorizo Dusty took home cost a total of $14.05.To the nearest cent, how much did each type of meatcost per pound?

27. Apply long division to find the quotient and remainder

for . Determine where the graph

of the function crosses the horizontal asymptote.

28. Find all real solutions of the equation: tan t � cos t � cos t.

29. Find the local minimum value of y � x3 � 3x2.

30. In New Zealand’s Abel Tasman National Park, thedepth of the water in one particular tidal pool can be

modeled by the function

where d(t) is the depth in feet and t is the time of theday (using a 24-hr clock). Determine the times of theday when there is at most 2 ft of water in this pool.Round your answers to the nearest minute.

d1t2 � 1.9 sin a�

6 tb � 2.2,

v1x2 ��2x2 � �x

x2 � �

9–121 Cumulative Review Chapters 1–9 957

Precalculus—

x

y

�2

�1

2

1

3�2

�2

� 2��2�

cob19537_ch09_948-957.qxd 1/27/11 10:29 PM Page 957

Precalculus—

958 9–122

CONNECTIONS TO CALCULUS

As mentioned in the chapter opener, in calculus as well as algebra we often look for ex-pressions that fit a special form, then apply a general formula to all expressions of thatform. Here this general idea is applied to partial fraction decomposition, a techniquewe studied in Section 9.4 and an important part of integral calculus.

More on Partial Fraction Decomposition

To rewrite the expression as a sum of its partial fractions, we would use the

template and proceed as in Section 9.4. Using the

ideas developed there, we’ll attempt to develop a decomposition formula that works

for all rational expressions of the form , a form that certainly applies to theexpression at hand.

EXAMPLE 1 � Decomposing Special FormsUse the techniques from Section 9.4 to rewrite the expression as the sumof its partial fractions.

Solution � After factoring the denominator, we obtain the following template:


clear denominators

distribute

collect like terms

This gives the system , yielding the solutions and As it

turns out, any rational expression of the form can be written in decomposed

form as


EXAMPLE 2 � Decomposing Special FormsUse the “formula” from Example 1 to decompose the expression as thesum of its partial fractions, then verify the result.

Solution � By comparing with , we note and so must

be equal to . To check, we combine the fractions using the LCD

giving or ✓.


10

x2 � 25

�11x � 52 � 11x � 52

1x � 52 1x � 52�

�x � 5 � x � 5

1x � 52 1x � 521x � 52 1x � 52,

�1

x � 5�

1

x � 5

10

x2 � 25k � 5,2k � 10

10

x2 � 25

2k

x2 � k2

10

x2 � 25

�1

x � k�

1

x � k.

2k

x2 � k2

A � �1.B � 1eA � B � 0

B � A � 2

� 1A � B2x � 1B � A2k � Ax � Ak � Bx � Bk

2k � A1x � k2 � B1x � k2

2k

1x � k2 1x � k2�

A

x � k�

B

x � k

2k

x2 � k2

2k

x2 � k2

10

1x � 52 1x � 52�

A

x � 5�

B

x � 5

10

x2 � 25

cob19537_ch09_958-960.qxd 1/27/11 10:32 PM Page 958

The Geometry of Vectors and Determinants

Although vectors (Section 8.3) and determinants (Section 9.8)are studied in two separate chapters in Precalculus, in acalculus course you’ll see there is an intriguing connectionbetween the two. In Section 8.3 we noted the tip-to-tail ad-dition of vectors u and v form a parallelogram, and used theparallelogram to illustrate a vector sum. In a calculuscourse, we say the vectors span the parallelogram, and incertain applications we need to know the area of this

figure. Using the vectors and and relation for the

angle between them, it can be shown that the area of the parallelogram is the absolute

value of the determinant (see Figure 9.74). For a complete proof, see Appendix B.

EXAMPLE 3 � Using Determinants to Find AreaFor the vectors and ,

a. Compute the vector sum using the parallelogram (tail-to-tip)method.

b. Compute and find the angle between u and w.

c. Use the formula from Section 8.2 to find the area of the

triangle formed.

d. Use the determinant to find the area of the

parallelogram spanned by these vectors, then verify the area is twice as largeas the triangle found in part (c).

Solution � a. The vector sum is

b. . For

and (verify this), we have and .

This indicates

c. area formula

substitute

result

d. For and

the determinant

and the area of the parallelogram is indeed twice the area of the triangle.


1122 1152 � 182 152 � 140 units2✓,

à b

c d` � `

12 5

8 15` �

v � H8, 15I,u � H12, 5I

� 70 units2

�1

21132 120122sin 22.38°

A �1

2�u��w�sin �

� � 22.38°.

� � cos�1a17

1312bcos � �

340

1132 120122�u� � 13

u # w � 121202 � 51202 � 340,�w� � 2202 � 202 � 1800 � 2012

H12, 5I � H8, 15I � H20, 20I � w.

à b

c d` � `

12 5

8 15`

A �1

2�u��w�sin �

��w�

w � u � vv � H8, 15Iu � H12, 5I

à b

c d`

�

cos � �u # v�u��v�

v � Hc, dIu � Ha, bI

Ha, bI

Hc, dI

Figure 9.74

H8, 15I

H12, 5I

H20, 20I

2018161412

8

14161820

1210

642

108642

9–123 Connections to Calculus 959

Precalculus—

cob19537_ch09_958-960.qxd 1/27/11 10:32 PM Page 959

960 Connections to Calculus 9–124

Precalculus—

Connections to Calculus Exercises

Use the techniques from Section 9.4 to complete Exercises 1 and 2.

1. A common form of partial fraction decompositioninvolves a constant over a factorable quadraticexpression. (a) Rewrite the expression

as the sum of partial fractions to

show the system obtained is (b) Use

the fact that to show the system can be

rewritten as . (c) Solve for A in the

second equation and use the result to quickly write

as the sum of its partial fractions.16

1x � 32 1x � 52

eA � B � 0

A1b � a2 � k

A � B � 0

eA � B � 0

Ab � Ba � k.

k

1x � a2 1x � b2

2. Another common form involves a linear expressionover a factorable quadratic expression.

(a) Rewrite the expression as the

sum of partial fractions to show the system obtained

is . (b) Use the fact that

to show that system can be written as

(c) Solve for A in the

second equation, and use the result to write

as the sum of its partial fractions.x � 10

1x � 32 1x � 42

eA � B � 1

A1b � a2 � a � k.

A � B � 1eA � B � 1

Ab � Ba � k

x � k

1x � a2 1x � b2

x � k

3. 4. 5.

6. 7. 8.

Find the area of the parallelogram spanned by the vectors given, then verify your answer using the method shown inExample 3.

x � 17

x2 � 9x � 14

x � 11

x2 � 13x � 40

15

x2 � 3x � 4

�22

x2 � 3x � 28

18

x2 � 81

6

x2 � 9

9. and

11. and 6i � 3j6i � 3j

v � H15, 3Iu � H2, 8I 10. and

12. 9i and 7j

v � H4, 4Iu � H�4, 1I

13. Find the volume of the parallelepiped spanned byand using the

determinant formula, then verify your answer usinga more familiar formula.

H0, 0, 8IH5, 0, 0I, H0, 6, 0I,14. Find the volume of the parallelepiped spanned by

and H0, 10, �7I.u � H2, 5, �8I, H3, 0, 4I,

Use the “formulas” developed in Example 1 and Exercises 1 and 2 to decompose each expression into a sum ofpartial fractions.

Similar to how the absolute value of gives the area of a parallelogram spanned by and the absolute

value of the determinant gives the volume of a parallelepiped spanned by and in

three dimensions.

Hg, h, iIHa, b, c,I, Hd, e, f I,†

a b cd e fg h i

†

Hc, dI,Ha, bI`a bc d`

cob19537_ch09_958-960.qxd 1/27/11 10:33 PM Page 960

Chapter 9

Documents

Transcript of Chapter 9