CHAPTER 1 Systems of Linear Equations - Cengage...CHAPTER 1 Systems of Linear Equations Section 1.1...

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C H A P T E R 1 Systems of Linear Equations

Section 1.1 Introduction to Systems of Linear Equations........................................2

Section 1.2 Gaussian Elimination and Gauss-Jordan Elimination ..........................9

Section 1.3 Applications of Systems of Linear Equations .....................................15

Review Exercises ..........................................................................................................21

2 Copyright © Houghton Mifflin Harcourt Publishing Company. All rights reserved.

C H A P T E R 1 Systems of Linear Equations

Section 1.1 Introduction to Systems of Linear Equations

1. Because the equation is in the form 1 2 ,a x a y b+ = it is linear in the variables x and y.

3. Because the equation cannot be written in the form 1 2 ,a x a y b+ = it is not linear in the variables x and y.


7. Choosing y as the free variable, let y t= and obtain

2 4 02 4

2 .

x tx tx t

− =

=

=

So, you can describe the solution set as 2x t= and ,y t= where t is any real number.

9. Choosing y and z as the free variables, let y s= and ,z t= and obtain 1x s t+ + = or 1 .x s t= − − So,

you can describe the solution set as 1 ,x s t= − − y s= and ,z t= where s and t are any real numbers.

11. From Equation 2 you have 2 3.x = Substituting this value into Equation 1 produces 1 3 2x − = or

1 5.x = So, the system has exactly one solution:

1 5x = and 2 3.x =

13. From Equation 3 you can conclude that 0.z = Substituting this value into Equation 2 produces

32

2 0 3

.

y

y

+ =

=

Finally, by substituting 32y = and 0z = into Equation

1, you obtain

32

32

0 0

.

x

x

− + − =

=

So, the system has exactly one solution: 3 32 2, ,x y= =

and 0.z =

15. Begin by rewriting the system in row-echelon form. The equations are interchanged.

1 2

1 2 3

2 05 2 0

x xx x x

+ =

+ + =

The first equation is multiplied by 12.

1 2

1 2 3

12

0

5 2 0

x x

x x x

+ =

+ + =

Adding 5− times the first equation to the second equation produces a new second equation.

1 2

32

1212

0

0

x x

xx

+ =

=− +

The second equation is multiplied by 2.−

1 2

2 3

12 0

2 0x x

x x+ =

− =

To represent the solutions, choose 3x to be the free variable and represent it by the parameter t. Because

2 32x x= and 1 212 ,x x= − you can describe the solution

set as 1 2, 2 ,x t x t= − = 3 ,x t= where t is any real number.

17. 2 42

x yx y+ =

− =

Adding the first equation to the second equation produces a new second equation, 3 6,x = or 2.x = So,

0,y = and the solution is 2,x = 0.y = This is the point where the two lines intersect.

x123

3

4

−2−2

−4

−4 4

x − y = 2

2x + y = 4

y

Section 1.1 Introduction to Systems of Linear Equations 3


19. 12 2 5

x yx y

− =

− + =

Adding 2 times the first equation to the second equation produces a new second equation.

10 7

x y− =

=

Because the second equation is a false statement, you can conclude that the original system of equations has no solution. Geometrically, the two lines are parallel.

21. 3 5 72 9

x yx y− =

+ =

Adding the first equation to 5 times the second equation produces a new second equation, 13 52,x = or 4.x = So, ( )2 4 9,y+ = or 1,y = and the solution is: 4,x =

1.y = This is the point where the two lines intersect.

23. 2 55 11

x yx y− =

− =

Subtracting the first equation from the second equation produces a new second equation, 3 6,x = or 2.x = So, ( )2 2 5,y− = or 1,y = − and the solution is: 2,x =

1.y = − This is the point where the two lines intersect.

25. 3 1 14 3

2 12

x y

x y

+ −+ =

− =

Multiplying the first equation by 12 produces a new first equation.

3 4 72 12

x yx y+ =

− =

Adding the first equation to 4 times the second equation produces a new second equation, 11 55,x = or

5.x = So, ( )2 5 12,y− = or 2,y = − and the solution

is: 5,x = 2.y = − This is the point where the two lines intersect.

27. 0.05 0.03 0.070.07 0.02 0.16

x yx y− =

+ =

Multiplying the first equation by 200 and the second equation by 300 produces new equations.

10 6 1421 6 48

x yx y− =

+ =

Adding the first equation to the second equation produces a new second equation, 31 62,x = or 2.x = So, ( )10 2 6 14,y− = or 1,y = and the solution is:

2,x = 1.y = This is the point where the two lines intersect.

−2 2 4 6 8 10x

2

4

6

3x − 5y = 7

2x + y = 98

10

y

1 2 4 5 6 7 8−2−4−6

2468

0.05x − 0.03y = 0.07

0.07x + 0.02y = 0.16

x

y

x

y

−2−4−6 4 6

2x − y = 55x − y = 11

−2

−4

−12

1 2 3 6 7−2−4−6−8

−10−12

46

x + 34

y − 13

= 1+

2x − y = 12x

y

x1

34

−2−4

−3−4

2

−2x + 2y = 5

x − y = 1

y

1 3 4

4 Chapter 1 Systems of Linear Equations


29. 14 6

3

x y

x y

+ =

− =

Adding 6 times the first equation to the second equation produces a new second equation, 5

2 9,x = or 185 .x =

So, 185 3,y− = or 3

5,y = and the solution is: 185 ,x =

35.y = This is the point where the two lines intersect.

31. (a)

(b) The system is inconsistent.

33. (a)

(b) The system is consistent

(c) The solution is approximately 1 12 4, .x y= = −

(d) Adding 14− times the first equation to the second

equation produces the new second equation 343 ,y = − or 1

4.y = − So, 12,x = and the solution

is 12,x = 1

4.y = −

(e) The solutions in (c) and (d) are the same.

35. (a)

(b) The system is consistent. (c) There are infinite solutions. (d) The second equation is the result of multiplying both

sides of the first equation by 0.2. A parametric representation of the solution set is given by

94 2 , ,x t y t= + = where t is any real number.

(e) The solutions in (c) and (d) are consistent.

37. Adding 3− times the first equation to the second equation produces a new second equation.

1 2

2

01

x xx

− == −

Now, using back-substitution you can conclude that the system has exactly one solution: 1 1x = − and 2 1.x = −

39. Interchanging the two equations produces the system

2 120

.2 120u vu v+ =

+ =


2 1203 120

u vv

+ =− = −

Solving the second equation you have 40.v = Substituting this value into the first equation gives

80 120u + = or 40.u = So, the system has exactly one solution: 40u = and 40.v =

41. Dividing the first equation by 9 produces a new first equation.

1 13 9

1 2 15 5 3

x y

x y

− = −

+ = −

Adding 15− times the first equation to the second

equation produces a new second equation. 1 1

3 97 14

15 45

x y

y

− = −

= −

Multiplying the second equation by 157 produces a new

second equation. 1 1

3 923

x y

y

− = −

= −

Now, using back-substitution, you can substitute 23y = −

into the first equation to obtain 2 19 9x + = − or 1

3.x = −

So, you can conclude that the system has exactly one solution: 1

3x = − and 23.y = −

−1 1 3 4−2

2

4

6

8

x − y = 3

x4

= 1 +y6

x

y

−4

−3

4

3−3x − y = 3

6 + 2 = 1x y

−4

−3

4

32x − 8y = 3

12x + y = 0

−4

−3

4

34x − 8y = 9

0.8x − 1.6y = 1.8



43. To begin, change the form of the first equation.

231 12 3 6

2 5

x y

x y

+ =

− =

Multiplying the first equation by 2 yields a new first equation.

2323 3

2 5

x y

x y

+ =

− =

Subtracting the first equation from the second equation yields a new second equation.

2323 38 83 3

x y

y

+ =

− = −

Dividing the second equation by 83− yields a new second

equation.

2323 3

1

x y

y

+ =

=

Now, using back-substitution you can conclude that the system has exactly one solution:

7x = and 1.y =

45. Multiplying the first equation by 50 and the second equation by 100 produces a new system.

1 2

1 2

2.5 9.53 4 52

x xx x

− = −

+ =


1 2

2

2.5 9.511.5 80.5

x xx

− = −

=

Now, using back-substitution, you can conclude that the system has exactly one solution:

1 8x = and 2 7.x =

47. Adding 2− times the first equation to the second equation yields a new second equation.

63 9

3 0

x y zy z

x z

+ + =

− − = −

− =

Adding 3− times the first equation to the third equation yields a new third equation.

63 93 4 18

x y zy zy z

+ + =

− − = −

− − = −

Dividing the second equation by 3− yields a new second equation.

13

63

3 4 18

x y zy z

y z

+ + =

+ =

− − = −

Adding 3 times the second equation to the third equation yields a new third equation.

13

6

3

3 9

x y z

y z

z

+ + =

+ =

− = −

Dividing the third equation by 3− yields a new third equation.

13

6

3

3

x y z

y z

z

+ + =

+ =

=

Now, using back-substitution you can conclude that the system has exactly one solution:

1,x = 2,y = and 3.z =

49. Dividing the first equation by 3 yields a new first equation.

1 2 3

1 2 3

1 2 3

2 4 13 3 3

2 32 3 6 8

x x x

x x xx x x

− + =

+ − =

− + =

Subtracting the first equation from the second equation yields a new second equation.

1 2 3

2 3

1 2 3

2 4 13 3 35 10 83 3 3

2 3 6 8

x x x

x x

x x x

− + =

− =

− + =


1 2 3

2 3

2 3

2 4 13 3 35 10 83 3 35 10 22

33 3

x x x

x x

x x

− + =

− =

− + =

At this point you should recognize that Equations 2 and 3 cannot both be satisfied. So, the original system of equations has no solution.



51. Dividing the first equation by 2 yields a new first equation.

1 2 3

1 3

1 2 3

1 32 2

2

4 2 102 3 13 8

x x x

x xx x x

+ − =

+ =

− + − = −


1 2 3

2 3

1 2 3

1 32 2

2

2 8 22 3 13 8

x x x

x xx x x

+ − =

− + =

− + − = −

Adding 2 times the first equation to the third equation produces a new third equation.

1 2 3

2 3

2 3

1 32 2

2

2 8 24 16 4

x x x

x xx x

+ − =

− + =

− = −


1 2 3

2 3

2 3

1 32 2

2

4 14 16 4

x x x

x xx x

+ − =

− = −

− = −

Adding 4− times the second equation to the third equation produces a new third equation.

1 2 3

2 3

1 32 2

2

4 10 0

x x x

x x

+ − =

− = −

=

Adding 12− times the second equation to the first

equation produces a new first equation. 1 3

2 3

1 52 2

4 1

x x

x x

+ =

− = −

Choosing 3x t= as the free variable, you can describe

the solution as 15 12 2 ,x t= − 2 4 1,x t= − and

3 ,x t= where t is any real number.

53. Adding 5− times the first equation to the second equation yields a new second equation.

3 2 180 72

x y z− + =

= −

Because the second equation is a false statement, you can conclude that the original system of equations has no solution.

55. Adding 2− times the first equation to the second, 3 times the first equation to the third, and 1− times the first equation to the fourth, produces

62 3 12

7 4 5 22.2 6

x y z wy z wy z wy z

+ + + =

− − = −

+ + =

− = −

Adding 7− times the second equation to the third, and 1− times the second equation to the fourth, produces

62 3 12

18 26 1063 6.

x y z wy z w

z ww

+ + + =

− − = −

+ =

=

Using back-substitution, you find the original system has exactly one solution: 1,x = 0,y = 3,z = and 2.w =

Answers may vary slightly for Exercises 57–63.

57. Using a computer software program or graphing utility, you obtain

1 2 3 415, 40, 45, 75x x x x= − = = = −


1.2, 0.6, 2.4.x y z= − = − =


1 2 31 4 15 5 2, , .x x x= = − =


6.8813, 163.3111,x y= = − 210.2915,z = − 59.2913.w = −



65. 0x y z= = = is clearly a solution.

Dividing the first equation by 4 yields a new first equation.

3 174 4

0

5 4 22 04 2 19 0

x y z

x y zx y z

+ + =

+ + =

+ + =

Adding 5− times the first equation to the second equation yields a new second equation.

3 174 41 34 4

0

0

4 2 19 0

x y z

y z

x y z

+ + =

+ =

+ + =


3 174 41 34 4

0

0

2 0

x y z

y z

y z

+ + =

+ =

− + =

Multiplying the second equation by 4 yields a new second equation.

3 174 4

0

3 02 0

x y z

y zy z

+ + =

+ =

− + =

Adding the second equation to the third equation yields a new third equation.

3 174 4

0

3 05 0

x y z

y zz

+ + =

+ =

=

Dividing the third equation by 5 yields a new third equation.

3 174 4

0

3 00

x y z

y zz

+ + =

+ =

=

Now, using back-substitution, you can conclude that the system has exactly one solution: 0,x = 0,y = and

0.z =

67. 0x y z= = = is clearly a solution.

Dividing the first equation by 5 yields a new first equation.

15

0

10 5 2 05 15 9 0

x y z

x y zx y z

+ − =

+ + =

+ − =


15

0

5 4 05 15 9 0

x y z

y zx y z

+ − =

− + =

+ − =


15

0

5 4 010 8 0

x y z

y zy z

+ − =

− + =

− =


1545

00

10 8 0

x y zy zy z

+ − =

− =

− =

Adding 10− times the second equation to the third equation yields a new third equation.

1545

0

0

0 0

x y z

y z

+ − =

− =

=

Adding 1− times the second equation to the first equation yields a new first equation.

3545

0

0

x z

y z

+ =

− =

Choosing z t= as the free variable you find the solution to be 3

5 ,x t= − 45 ,y t= and ,z t= where t is any real

number.

69. (a) True. You can describe the entire solution set using parametric representation. ax by c+ = Choosing y t= as the free variable, the solution is

,c bx ta a

= − ,y t= where t is any real number.

(b) False. For example, consider the system 1 2 3

1 2 3

12

x x xx x x

+ + =

+ + =

which is an inconsistent system. (c) False. A consistent system may have only one

solution.



71. Because 1x t= and 2 13 4 3 4,x t x= − = − one answer is the system 1 2

1 2

3 4.3 4

x xx x

− =

− + = −

Letting 2 ,x t= you get 14 4 .

3 3 3t tx +

= = +

73. Substituting 1A x= and 1B y= into the original system yields

12 12 7.3 4 0

A BA B− =

+ =

Reduce this system to row-echelon form. Dividing the first equation by 12 yields a new first equation.

712

3 4 0

A B

A B

− =

+ =


71274

7

A B

B

− =

= −

Dividing the second equation by 7 yields a new second equation.

71214

A B

B

− =

= −

So, 1 4A = − and 1 3.B = Because 1A x= and 1 ,B y= the solution of the original system of equations

is: 3x = and 4.y = −

75. Substituting 1 ,A x= 1 ,B y= and 1 ,C z= into the original system yields

2 3 44 2 102 3 13 8.

A B CA CA B C

+ − =

+ =

− + − = −

Reduce this system to row-echelon form. 1 3

2 22

2 8 24 16 4

A B C

B CB C

+ − =

− + =

− = −

312 2 2

4 1A B C

B C+ − =

− = −

Letting t C= be the free variable, you have

,C t= 4 1B t= − and ( )5.

2t

A− +

= So, the solution

to the original problem is

2 ,5

xt

=−

1 ,4 1

yt

=−

1,zt

= where 15, , 0.4

t ≠

77. Reduce the system to row-echelon form. Dividing the first equation by cos θ yields a new second equation.

( ) ( )

sin 1cos cos

sin cos 0

x y

x y

θθ θ

θ θ

⎛ ⎞+ =⎜ ⎟⎝ ⎠

− + =

Multiplying the first equation by sin θ and adding to the second equation yields a new second equation.

sin 1cos cos

1 sincos cos

x y

y

θθ θ

θθ θ

⎛ ⎞+ =⎜ ⎟⎝ ⎠

⎛ ⎞=⎜ ⎟

⎝ ⎠

2 2 2sin sin cos 1Because cos

cos cos cosθ θ θθθ θ θ

⎛ ⎞++ = =⎜ ⎟

⎝ ⎠

Multiplying the second equation by cosθ yields a new second equation.

sin 1cos cos

sin

x y

y

θθ θ

θ

⎛ ⎞+ =⎜ ⎟⎝ ⎠

=

Substituting siny θ= into the first equation yields

2 2

sin 1sincos cos

1 sin cos cos .cos cos

x

x

θ θθ θ

θ θ θθ θ

⎛ ⎞+ =⎜ ⎟⎝ ⎠

−= = =

So, the solution of the original system of equations is: cosx θ= and sin .y θ=

79. For this system to have an infinite number of solutions both equations need to be equivalent.

Multiply the second equation by 2.−

4 62 2 6

x kykx y

+ =

− − =

So, when 2k = − the system will have an infinite number of solutions.

81. Reduce the system to row-echelon form.

( )2

0

1 0

x ky

k y

+ =

− =

2

0

, 1 00

x ky

ky

+ =

− ≠=

2

01 00,

xky

=

− ≠=

If 21 0,k− ≠ that is if 1,k ≠ ± the system will have exactly one solution.

Section 1.2 Gaussian Elimination and Gauss-Jordan Elimination 9


83. To begin, reduce the system to row-echelon form.

( )2 6

8 3 14

x y kz

k z

+ + =

− = −

This system will have no solution if 8 3 0,k− = that is, 83.k =

85. Reducing the system to row-echelon form, you have

( ) ( )( ) ( )2

3

1 1 1

1 1 1 3

x y kz

k y k z

k y k z k

+ + =

− + − = −

− + − = −

( ) ( )( )2

3

1 1 1

.2 3

x y kz

k y k z

k k z k

+ + =

− + − = −

− − + = −

If 2 2 0,k k− − + = then there is no solution. So, if 1k = or 2,k = − there is not a unique solution.

87. (a) All three of the lines will intersect in exactly one point (corresponding to the solution point). (b) All three of the lines will coincide (every point on

these lines is a solution point). (c) The three lines have no common point.

89. Answers vary. ( :Hint Choose three different values for x

and solve the resulting system of linear equations in the variables a, b, and ).c

91. 4 35 6 13

x yx y− = −

− =

4 314 28

x yy

− = −

=

4 32

x yy

− = −

=

52

xy=

=

At each step, the lines always intersect at ( )5, 2 , which is

the solution to the system of equations.

93. Solve each equation for y.

1100199

2

2

y x

y x

= +

= −

The graphs are misleading because, while they appear parallel, when the equations are solved for y they have slightly different slopes.

Section 1.2 Gaussian Elimination and Gauss-Jordan Elimination

1. Because the matrix has 3 rows and 3 columns, it has size 3 3.×


5. Because the matrix has 1 row and 5 columns, it has size 1 5.×


9. The matrix satisfies all three conditions in the definition of row-echelon form. Moreover, because each column that has a leading 1 (columns one and two) has zeros elsewhere, the matrix is in reduced row-echelon form.

11. Because the matrix has two non-zero rows without leading 1’s, it is not in row-echelon form.

13. Because the matrix has a non-zero row without a leading 1, it is not in row-echelon form.

15. Because the matrix is in reduced row-echelon form, convert back to a system of linear equations

1

2

02

xx

=

=

So, the solution is: 1 0x = and 2 2.x =

x

x − 4y = −35−2−4 43

−4−3−2

−5

345

14y = 28

y

x

x − 4y = −35−2−4 43

−4−3−2

−5

345

y = 2

y

x−2−4 321

−4−3−2

−5

3

1

45

y = 2

x = 5y

4

x

x − 4y = −3

5x − 6y = 13

5−2−4 3 4

2

−4−5

345

y



17. Because the matrix is in row-echelon form, convert back to a system of linear equations.

1 2

2 3

3

32 1

1

x xx x

x

− =

− =

= −

Solve this system by back-substitution.

( )2

2

12 11

xx

− − == −

Substituting 2 1x = − into equation 1,

( )1

1

31.2

xx

− − ==

So, the solution is: 1 22, 1,x x= = − and 3 1.x = −

19. Interchange the first and second rows.

1 1 1 02 1 1 30 1 2 1

−⎡ ⎤⎢ ⎥−⎢ ⎥⎢ ⎥⎣ ⎦

Interchange the second and third rows.

1 1 1 00 1 2 12 1 1 3

−⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥−⎣ ⎦

Add 2− times the first row to the third row to produce a new third row.

1 1 1 00 1 2 10 3 3 3

−⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥−⎣ ⎦

Add 3− times the second row to the third row to produce a new third row.

1 1 1 00 1 2 10 0 9 0

−⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥−⎣ ⎦

Divide the third row by 9− to produce a new third row.

1 1 1 00 1 2 10 0 1 0

−⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

Convert back to a system of linear equations. 1 2 3

2 3

3

02 1

0

x x xx x

x

− + =

+ =

=


( )2 3

1 2 3

1 2 1 12 0

1 0 1

x x

x x x

= − = − =

= − = − =

So, the solution is: 1 1,x = 2 1,x = and 3 0.x =

21. Because the matrix is in row-echelon form, convert back to a system of linear equations.

1 2 4

2 3 4

3 4

4

2 42 3

2 14

x x xx x x

x xx

+ + =

+ + =

+ =

=


( )( )( )

3 4

2 3 4

1 2 4

1 2 1 2 4 7

3 2 3 2 7 4 13

4 2 4 2 13 4 26

x x

x x x

x x x

= − = − = −

= − − = − − − =

= − − = − − = −

So, the solution is: 1 26,x = − 2 13,x = 3 7,x = − and

4 4.x =

23. The augmented matrix for this system is

1 2 7

.2 1 8⎡ ⎤⎢ ⎥⎣ ⎦

Adding 2− times the first row to the second row yields a new second row.

1 2 70 3 6⎡ ⎤⎢ ⎥− −⎣ ⎦

Dividing the second row by 3− yields a new second row.

1 2 70 1 2⎡ ⎤⎢ ⎥⎣ ⎦

Converting back to a system of linear equations produces 2 7

2.x y

y+ =

=

Finally, using back-substitution you find that 3x = and 2.y =


1 2 1.5

.2 4 3−⎡ ⎤⎢ ⎥−⎣ ⎦

Gaussian elimination produces the following.

1 2 1.52 4 3−⎡ ⎤

⇒⎢ ⎥−⎣ ⎦

32

1 2

2 4 3

−⎡ ⎤−⇒⎢ ⎥

−⎢ ⎥⎣ ⎦

32

1 2

0 0 6

−⎡ ⎤−⎢ ⎥⎢ ⎥⎣ ⎦

Because the second row of this matrix corresponds to the equation 0 6,= you can conclude that the original system has no solution.




3 5 223 4 4 .4 8 32

− −⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥−⎣ ⎦

Dividing the first row by 3− yields a new first row.

5 223 31

3 4 44 8 32

⎡ ⎤−⎢ ⎥⎢ ⎥⎢ ⎥−⎣ ⎦


5 223 31

0 9 184 8 32

⎡ ⎤−⎢ ⎥

−⎢ ⎥⎢ ⎥−⎣ ⎦

Adding 4− times the first row to the third row yields a new third row.

5 223 3

843 3

10 9 180

⎡ ⎤−⎢ ⎥

−⎢ ⎥⎢ ⎥−⎣ ⎦

Dividing the second row by 9 yields a new second row.

5 223 3

843 3

10 1 20

⎡ ⎤−⎢ ⎥

−⎢ ⎥⎢ ⎥−⎣ ⎦

Adding 43 times the second row to the third row

yields a new third row.

5 223 31

0 1 20 0 0

⎡ ⎤−⎢ ⎥

−⎢ ⎥⎢ ⎥⎣ ⎦

Converting back to a system of linear equations produces

5 223 3

2.

x y

y

− =

= −

Finally, using back-substitution you find that the solution is: 4x = and 2.y = −


1 0 3 23 1 2 5 .2 2 1 4

− −⎡ ⎤⎢ ⎥−⎢ ⎥⎢ ⎥⎣ ⎦


1 0 3 2 1 0 3 2 1 0 3 23 1 2 5 0 1 7 11 0 1 7 112 2 1 4 0 2 7 8 0 0 7 14

− − − − − −⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥− ⇒ ⇒⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥− −⎣ ⎦ ⎣ ⎦ ⎣ ⎦

Back substitution now yields

( )( )

3

2 3

1 3

211 7 11 7 2 3

2 3 2 3 2 4.

xx x

x x

=

= − = − = −

= − + = − + =

So, the solution is: 1 2 34, 3, and 2.x x x= = − =


1 1 5 31 0 2 1 .2 1 1 0

−⎡ ⎤⎢ ⎥−⎢ ⎥⎢ ⎥− −⎣ ⎦

Subtracting the first row from the second row yields a new second row.

1 1 5 30 1 3 22 1 1 0

−⎡ ⎤⎢ ⎥− −⎢ ⎥⎢ ⎥− −⎣ ⎦

Adding 2− times the first row to the third row yields a new third row.

1 1 5 30 1 3 20 3 9 6

−⎡ ⎤⎢ ⎥− −⎢ ⎥⎢ ⎥− −⎣ ⎦

Multiplying the second row by 1− yields a new second row.

1 1 5 30 1 3 20 3 9 6

−⎡ ⎤⎢ ⎥−⎢ ⎥⎢ ⎥− −⎣ ⎦

Adding 3 times the second row to the third row yields a new third row.

1 1 5 30 1 3 20 0 0 0

−⎡ ⎤⎢ ⎥−⎢ ⎥⎢ ⎥⎣ ⎦

Adding 1− times the second row to the first row yields a new first row.

1 0 2 10 1 3 20 0 0 0

−⎡ ⎤⎢ ⎥−⎢ ⎥⎢ ⎥⎣ ⎦

Converting back to a system of linear equations produces 1 3

2 3

2 13 2.

x xx x

− =− =

Finally, choosing 3x t= as the free variable you can describe the solution as 1 1 2 ,x t= + 2 2 3 ,x t= + and

3 ,x t= where t is any real number.




4 12 7 20 22

.3 9 5 28 30

− −⎡ ⎤⎢ ⎥− −⎣ ⎦

Dividing the first row by 4 yields a new first row.

7 11

241 3 5

3 9 5 28 30

−⎡ ⎤−⎢ ⎥

− −⎢ ⎥⎣ ⎦


7 11

241 274 2

1 3 5

0 0 13

−⎡ ⎤−⎢ ⎥

−⎢ ⎥⎣ ⎦

Multiplying the second row by 4 yields a new second row.

7 11

241 3 5

0 0 1 52 54

−⎡ ⎤−⎢ ⎥

−⎢ ⎥⎣ ⎦

Converting back to a system of linear equations produces

7 114 23 5

52 54.

x y z w

z w

+ − − =

− =

Choosing y s= and w t= as the free variables you can describe the solution as 100 3 96 ,x s t= − +

, 54 52 ,y s z t= = + and ,w t= where s and t are any real numbers.


3 3 12 61 1 4 2

.2 5 20 101 2 8 4

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢− ⎥⎣ ⎦


3 3 12 6 1 1 4 21 1 4 2 1 1 4 22 5 20 10 2 5 20 101 2 8 4 1 2 8 4

1 1 4 2 1 1 4 20 0 0 0 0 3 12 60 3 12 6 0 0 0 00 3 12 6 0 0 0 0

1 1 4 2 1 0 0 00 1 4 2 0 1 4 20 0 0 0 0 0 0 00 0 0 0 0 0 0 0

⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⇒⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢− ⎥ ⎢− ⎥⎣ ⎦ ⎣ ⎦

⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⇒ ⇒⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⇒ ⇒⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

Letting z t= be the free variable, the solution is: 0, 2 4 , ,x y t z t= = − = where t is any real number.

37. Using a computer software program or graphing utility, the augmented matrix reduces to

1 0 0 0.5278 23.53610 1 0 4.1111 18.5444 .0 0 1 2.1389 7.4306

−⎡ ⎤⎢ ⎥−⎢ ⎥⎢ ⎥−⎣ ⎦

Letting 4x t= be the free variable, you obtain

1

2

3

4

23.5361 0.527818.5444 4.11117.43062 2.1389, where is any real number.

x tx tx tx t t

= +

= +

= +

=


1 0 0 0 0 20 1 0 0 0 2

.0 0 1 0 0 30 0 0 1 0 50 0 0 0 1 1

⎡ ⎤⎢ ⎥−⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥−⎢ ⎥⎣ ⎦

So, the solution is: 1 2 3 42, 2, 3, 5,x x x x= = − = = − and 5 1.x =


1 0 0 0 0 0 10 1 0 0 0 0 20 0 1 0 0 0 6

.0 0 0 1 0 0 30 0 0 0 1 0 40 0 0 0 0 1 3

⎡ ⎤⎢ ⎥−⎢ ⎥⎢ ⎥⎢ ⎥⎢ − ⎥⎢ ⎥

−⎢ ⎥⎢ ⎥⎣ ⎦

So, the solution is: 1 2 31, 2, 6,x x x= = − =

4 53, 4,x x= − = − and 6 3.x =

43. The corresponding system of equations is 1

2 3

00

0 0

xx x

=

+ =

=

Letting 3x t= be the free variable, the solution is:

1 2 30, , ,x x t x t= = − = where t is any real number.

45. The corresponding system of equations is 1 4

3

00

0 0

x xx

+ =

=

=

Letting 4x t= and 3x s= be the free variables, the solution is: 1 ,x t= − 2 ,x s= 3 0,x = 4 .x t=



47. (a) Because there are two rows and three columns, there are two equations and two variables. (b) Gaussian elimination produces the following.

1 21 2 1 2

73 4 1 0 4 3 7 0 14 3

kk kk

k

⎡ ⎤⎡ ⎤ ⎡ ⎤ ⎢ ⎥⇒ ⇒⎢ ⎥ ⎢ ⎥ ⎢ ⎥− +⎣ ⎦ ⎣ ⎦ ⎢ ⎥+⎣ ⎦

The system is consistent if 4 3 0,k+ ≠ or if 43.k ≠ −

(c) Because there are two rows and three columns, there are two equations and three variables. (d) Gaussian elimination produces the following.

1 2 01 2 0 1 2 0

73 4 1 0 0 4 3 7 0 0 1 04 3

kk kk

k

⎡ ⎤⎡ ⎤ ⎡ ⎤ ⎢ ⎥⇒ ⇒⎢ ⎥ ⎢ ⎥ ⎢ ⎥− +⎣ ⎦ ⎣ ⎦ ⎢ ⎥+⎣ ⎦

Notice that 4 3 0,k+ ≠ or 43.k ≠ − But if 4

3− is substituted for k in the original matrix. Guassian elimination

produces the following:

4 43 31 2 0 1 2 0

3 4 1 0 0 0 7 0

⎡ ⎤ ⎡ ⎤− −⇒⎢ ⎥ ⎢ ⎥

−⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

The system is consistent. So, the original system is consistent for all real k.

49. Begin by forming the augmented matrix for the system

1 1 0 20 1 1 2

.1 0 1 2

0a b c

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

Then use Gauss-Jordan elimination as follows.

1 1 0 2 1 1 0 20 1 1 2 0 1 1 20 1 1 0 0 1 1 0

0 0 2

1 1 0 2 1 1 0 20 1 1 2 0 1 1 20 0 2 2 0 0 2 20 2 0 0 2

1 1 0 2 1 1 0 20 1 1 2 0 1 1 20 0 1 1 0 0 1 10 0 2 0 0 0

a b c b a c a

b a c a a b c b

a b c b a b c

⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⇒⎢ ⎥ ⎢ ⎥− −⎢ ⎥ ⎢ ⎥

− −⎣ ⎦ ⎣ ⎦

⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⇒ ⇒⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥

− − − + −⎣ ⎦ ⎣ ⎦

⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⇒ ⇒⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥

− + − + +⎣ ⎦ ⎣ ⎦

⇒

1 1 0 2 1 0 0 10 1 0 1 0 1 0 10 0 1 1 0 0 1 10 0 0 0 0 0a b c a b c

⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⇒⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥

+ + + +⎣ ⎦ ⎣ ⎦

Converting back to a system of linear equations 1

11

0 .

xyz

a b c

==== + +

The system (a) will have a unique solution if 0,a b c+ + = (b) will have no solution if 0,a b c+ + ≠ and (c) cannot have an infinite number of solutions.



51. Solve each pair of equations by Gaussian elimination as follows. (a) Equations 1 and 2:

5 86 35 86 3

1 04 2 5 160 11 1 0 0

⎡ ⎤−⎡ ⎤⇒ ⇒⎢ ⎥⎢ ⎥ − −⎢ ⎥⎣ ⎦ ⎣ ⎦

8 53 6

8 53 6

,

,

x t

y t

z t

= −

= − +

=

(b) Equations 1 and 3:

3611

14 1413 4014 14

1 04 2 5 160 11 3 2 6

⎡ ⎤−⎡ ⎤⇒ ⇒⎢ ⎥⎢ ⎥ − −− − ⎢ ⎥⎣ ⎦ ⎣ ⎦

18 117 14

20 137 14

,

,

x t

y t

z t

= −

= − +

=

(c) Equations 2 and 3:

1 1 0 0 1 0 1 31 3 2 6 0 1 1 3

⎡ ⎤ ⎡ ⎤⇒ ⇒⎢ ⎥ ⎢ ⎥− − − −⎣ ⎦ ⎣ ⎦

3 ,3 ,

x ty tz t

= −

= − +

=

(d) Each of these systems has an infinite number of solutions.

53. Use Gauss-Jordan elimination as follows.

1 2 1 2 1 2 1 01 2 0 4 0 1 0 1

⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤⇒ ⇒ ⇒⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦

55. Begin by finding all possible first rows.

[ ] [ ] [ ] [ ]0 0 , 0 1 , 1 0 , 1 .k

For each of these, examine the possible second rows.

0 0 0 1 1 0 1

, , ,0 0 0 0 0 1 0 0

k⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦

These represent all possible 2 2× reduced row-echelon matrices.

57. (a) True. In the notation ,m n m× is the number of rows of the matrix. So, a 6 3× matrix has six rows.

(b) True. On page 19, after Example 4, the sentence reads, “It can be shown that every matrix is row-equivalent to a matrix in row-echelon form.”

(c) False. Consider the row-echelon form

1 0 0 0 00 1 0 0 10 0 1 0 20 0 0 1 3

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

which gives the solution 1 2 30, 1, 2,x x x= = =

4and 3.x =

(d) True. Theorem 1.1 states that if a homogeneous system has fewer equations than variables, then it must have an infinite number of solutions.

59. First, a and c cannot both be zero. So, assume 0,a ≠ and use row reduction as follows.

00

a ba b a bcbc d ad bcda

⎡ ⎤⎡ ⎤ ⎡ ⎤⎢ ⎥⇒ ⇒−⎢ ⎥ ⎢ ⎥⎢ ⎥ −+⎣ ⎦ ⎣ ⎦⎢ ⎥⎣ ⎦

So, 0.ad bc− ≠ Similarly, if 0,c ≠ interchange rows and proceed as above. So the original matrix is row equivalent to the identity if and only if 0.ad bc− ≠

61. Form the augmented matrix for this system

2 1 0

1 2 0λ

λ−⎡ ⎤

⎢ ⎥−⎣ ⎦

and reduce the system using elementary row operations.

2

1 2 0 1 2 02 1 0 0 4 3 0

λ λλ λ λ

− −⎡ ⎤ ⎡ ⎤⇒⎢ ⎥ ⎢ ⎥− − +⎣ ⎦ ⎣ ⎦

To have a nontrivial solution you must have

( )( )

2 4 3 0

1 3 0.

λ λ

λ λ

− + =

− − =

So, if 1λ = or 3,λ = the system will have nontrivial solutions.

63. To show that it is possible you need give only one example, such as

1 2 3

1 2 3

01

x x xx x x

+ + =

+ + =

which has fewer equations than variables and obviously has no solution.

Section 1.3 Applications of Systems of Linear Equations 15


65. a b a c b d a c b d c d c dc d c d a b a b a b

− − − − − −⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤⇒ ⇒ ⇒ ⇒⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥

⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦

The rows have been interchanged. In general, the second and third elementary row operations can be used in this manner to interchange two rows of a matrix. So, the first elementary row operation is, in fact, redundant.

67. When a system of linear equations is inconsistent, the row-echelon form of the corresponding augmented matrix will have a row that is all zeros except for the last entry.

69. A matrix in reduced row-echelon form has zeros above each row’s leading 1, which may not be true for a matrix in row-echelon form.

Section 1.3 Applications of Systems of Linear Equations

1. (a) Because there are three points, choose a second- degree polynomial, ( ) 2

0 1 2 .p x a a x a x= + + Then

substitute 2, 3,x = and 4 into ( )p x and equate the

results to 5, 2,y = and 5, respectively.

( ) ( )( ) ( )( ) ( )

20 1 2 0 1 2

20 1 2 0 1 2

20 1 2 0 1 2

2 2 2 4 5

3 3 3 9 2

4 4 4 16 5

a a a a a a

a a a a a a

a a a a a a

+ + = + + =

+ + = + + =

+ + = + + =

Form the augmented matrix

1 2 4 51 3 9 21 4 16 5

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

and use Gauss-Jordan elimination to obtain the equivalent reduced row-echelon matrix

1 0 0 290 1 0 18 .0 0 1 3

⎡ ⎤⎢ ⎥−⎢ ⎥⎢ ⎥⎣ ⎦

So, ( ) 229 18 3 .p x x x= − +

(b)

3. (a) Because there are three points, choose a second- degree polynomial, ( ) 2

0 1 2 .p x a a x a x= + + Then

substitute 2, 3,x = and 5 into ( )p x and equate the

results to 4, 6,y = and 10, respectively.

( ) ( )( ) ( )( ) ( )

20 1 2 0 1 2

20 1 2 0 1 2

20 1 2 0 1 2

2 2 2 4 4

3 3 3 9 6

5 5 5 25 10

a a a a a a

a a a a a a

a a a a a a

+ + = + + =

+ + = + + =

+ + = + + =

Use Gauss-Jordan elimination on the augmented matrix for this system.

1 2 4 4 1 0 0 01 3 9 6 0 1 0 21 5 25 10 0 0 1 0

⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥⇒⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

So, ( ) 2 .p x x=

(b)

x

1

1

2

2

3

3

4

4

5

5

6

6

(2, 5) (4, 5)

(3, 2)

y

x

2

1 2 3 4 5

4

6

8

10

(2, 4)(3, 6)

(5, 10)

y



5. (a) Using the translation 2007,z x= − the points ( ),z y are ( ) ( )1, 5 , 0, 7 ,− and ( )1, 12 . Because there are three points,

choose a second-degree polynomial ( ) 20 1 2 .p z a a z a z= + + Then substitute 1, 0,z = − and 1 into ( )p z and equate the

results to 5, 7,y = and 12 respectively.

( ) ( )( ) ( )( ) ( )

20 1 2 0 1 2

20 1 2 0

20 1 2 0 1 2

1 1 5

0 0 7

1 9 1 12

a a a a a a

a a a a

a a a a a

+ − + − = − + =

+ + = =

+ + = + + =


7232

1 1 1 5 1 0 0 71 0 0 7 0 1 01 1 1 12 0 0 1

⎡ ⎤−⎡ ⎤⎢ ⎥⎢ ⎥ ⇒ ⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦ ⎣ ⎦

So, ( ) 27 32 27 .p z z z= + +

Letting 2007,z x= − you have ( ) ( ) ( )27 32 27 2007 2007 .p x x x= + − + −

(b)

7. Choose a fourth-degree polynomial and substitute 1, 2, 3,x = and 4 into

( ) 2 3 40 1 2 3 4 .p x a a x a x a x a x= + + + + However,

when you substitute 3x = into ( )p x and equate it to

2y = and 3y = you get the contradictory equations

0 1 2 3 4

0 1 2 3 4

3 9 27 81 23 9 27 81 3

a a a a aa a a a a

+ + + + =

+ + + + =

and must conclude that the system containing these two equations will have no solution. Also, y is not a function of x because the x-value of 3 is repeated. By similar reasoning, you cannot choose ( ) 2 3 4

0 1 2 3 4p y b b y b y b y b y= + + + + because

1y = corresponds to both 1x = and 2.x =

9. Letting ( ) 20 1 2 ,p x a a x a x= + + substitute 0, 2,x =

and 4 into ( )p x and equate the results to 1, 3,y = and

5, respectively.

( ) ( )( ) ( )( ) ( )

20 1 2 0

20 1 2 0 1 2

20 1 2 0 1 2

0 0 1

2 2 2 4 3

4 4 4 16 5

a a a a

a a a a a a

a a a a a a

+ + = =

+ + = + + =

+ + = + + =


1 0 0 1 1 0 0 11 2 4 3 0 1 0 11 4 16 5 0 0 1 0

⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥⇒⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

So, ( ) 1 .p x x= +

The graphs of ( ) ( )1 1 1y p x x= = + and that of the

function 27 115 151y x x= − + are shown below.

z

3

9

12

−1 1(2006) (2008)(2007)

(−1, 5)

(0, 7)

(1, 12)

y

−1 2 4

3

3

4

5

x(0, 1)

2, 4,1 13 5

y = − + 1 x2 7x15 15

) )) )

y

1

y = 11 + x



11. To begin, substitute 1x = − and 1x = into ( ) 2 30 1 2 3p x a a x a x a x= + + + and equate the results

to 2y = and 2,y = − respectively. 0 1 2 3

0 1 2 3

22

a a a aa a a a

− + − =+ + + = −

Then, differentiate p, yielding ( ) 21 2 32 3 .p x a a x a x′ = + + Substitute 1x = − and 1x = into ( )p x′ and equate the results to 0.

1 2 3

1 2 3

2 3 02 3 0

a a aa a a

− + =

+ + =

Combining these four equations into one system and forming the augmented matrix, you obtain

1 1 1 1 21 1 1 1 2

.0 1 2 3 00 1 2 3 0

− −⎡ ⎤⎢ ⎥−⎢ ⎥⎢ ⎥−⎢ ⎥⎣ ⎦

Use Gauss-Jordan elimination to find the equivalent reduced row-echelon matrix

1 0 0 0 00 1 0 0 3

.0 0 1 0 00 0 0 1 1

⎡ ⎤⎢ ⎥−⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

So, ( ) 33 .p x x x= − + The graph of ( )y p x= is shown below.

13. (a) Because you are given three points, choose a second-degree polynomial, ( ) 20 1 2 .p x a a x a x= + + Because the x-values

are large, use the translation 1990z x= − to obtain ( ) ( ) ( )10, 227 , 0, 249 , 10, 281 .− Substituting the given points into

( )p x produces the following system of linear equations.

( ) ( )( ) ( )( ) ( )

20 1 2 0 1 2

20 1 2 0

20 1 2 0 1 2

10 10 10 100 227

0 0 249

10 10 10 100 281

a a a a a a

a a a a

a a a a a a

+ − + − = − + =

+ + = =

+ + = + + =

Form the augmented matrix

1 10 100 2271 0 0 2491 10 100 281

−⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦


1 0 0 2490 1 0 2.7 .0 0 1 0.05

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

So, ( ) 2249 2.7 0.05p z z z= + + and ( ) ( ) ( )2249 2.7 1990 0.05 1990 .p x x x= + − + − To predict the population in 2010

and 2020, substitute these values into ( ).p x

( ) ( ) ( )22010 249 2.7 20 0.05 20 323 millionp = + + =

( ) ( ) ( )22020 249 2.7 30 0.05 30 375 millionp = + + =

−1 2

−2

−1

x

(−1, 2)

(1, −2)

y

1



15. (a) Letting 2000z x= − the four points are ( )1, 10,003 , ( )3, 10,526 , ( )5, 12,715 , and ( )7, 14,410 .

Let ( ) 2 30 1 2 3 .p z a a z a z a z= + + +

( ) ( ) ( )( ) ( ) ( )( ) ( ) ( )( ) ( ) ( )

2 30 1 2 3 0 1 2 3

2 30 1 2 3 0 1 2 3

2 30 1 2 3 0 1 2 3

2 30 1 2 3 0 1 2 3

1 1 1 10,003

3 3 3 3 9 27 10,526

5 5 5 5 25 125 12,715

7 7 7 7 49 343 14,410

a a a a a a a a

a a a a a a a a

a a a a a a a a

a a a a a a a a

+ + + = + + + =

+ + + = + + + =

+ + + = + + + =

+ + + = + + + =

(b) Use Gauss-Jordan elimination to solve the system.

1 1 1 1 10,003 1 0 0 0 11,041.251 3 9 27 10,526 0 1 0 0 1606.51 5 25 125 12,715 0 0 1 0 613.251 7 49 343 14,410 0 0 0 1 45

⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⇒⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ − ⎥⎣ ⎦ ⎣ ⎦

So, ( ) 2 311,041.25 1606.5 613.25 45 .p z z z z= + + −

Letting 2000,z x= − ( ) ( ) ( ) ( )2 311,041.25 1606.5 2000 613.25 2000 45 2000 .p x x x x= + − + − − −

Because the actual net profit increased each year from 2000 2007− except for 2006 and the predicted values decrease each year after 2008, the solution does not produce a reasonable model for predicting future net profits.

17. Choosing a second-degree polynomial approximation,

( ) 20 1 2 ,p x a a x a x= + + substitute 0, ,

2x π= and

π into ( )p x and equate the results to 0, 1, and 0,y =

respectively.

0

22

0 1

20 1 2

0

12 4

0

a

a a a

a a a

π π

π π

=

+ + =

+ + =

Then form the augmented matrix,

2

2

1 0 0 0

1 12 4

1 0

π π

π π

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦


2

1 0 0 040 1 0 .

40 0 1

π

π

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥

−⎢ ⎥⎢ ⎥⎣ ⎦

So, ( ) ( )2 22 2

4 4 4 .p x x x x xππ π π

= − = −

Furthermore,

sin3π

≈2

2

2

2

43 3 3

4 2 8 0.889.9 9

p π π πππ

ππ

⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎛ ⎞= −⎢ ⎥⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎢ ⎥⎣ ⎦

⎡ ⎤= = ≈⎢ ⎥

⎣ ⎦

Note that sin 3 0.866π = to three significant digits.

19. (i) ( ) ( ) ( )( ) ( ) ( )( ) ( ) ( )

20 1 2 0 1 2

20 1 2 0

20 1 2 0 1 2

1 1 1

0 0 0

1 1 1

p a a a a a a

p a a a a

p a a a a a a

− = + − + − = − +

= + + =

= + + = + +

(ii) 0 1 2

0

0 1 2

000

a a aaa a a

− + =

=

+ + =

(iii) From the augmented matrix

1 1 1 01 0 0 01 1 1 0

−⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦


1 0 0 00 1 0 0 .0 0 1 0

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

So, 0 10, 0,a a= = and 2 0.a =



21. (a) Each of the network’s six junctions gives rise to a linear equation as shown below. input = output 1 3

1 2 4

2 5

3 6

4 7 6

5 7

600

500600

500

x xx x x

x xx xx x x

x x

= +

= +

+ =

+ =

+ =

= +

Rearrange these equations, form the augmented matrix, and use Gauss-Jordan elimination.

1 0 1 0 0 0 0 600 1 0 0 0 0 1 0 01 1 0 1 0 0 0 0 0 1 0 0 0 0 1 00 1 0 0 1 0 0 500 0 0 1 0 0 1 0 6000 0 1 0 0 1 0 600 0 0 0 1 0 1 1 00 0 0 1 0 1 1 0 0 0 0 0 1 0 1 5000 0 0 0 1 0 1 500 0 0 0 0 0 0 0 0

−⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥− − −⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⇒⎢ ⎥ ⎢ − ⎥⎢ ⎥ ⎢ ⎥

−⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

Letting 7x t= and 6x s= be the free variables, you have

1

2

3

4

5

6

600

500

x sx tx sx s tx tx s

=

=

= −

= −

= −

=

7 ,x t= where s and t are any real numbers.

(b) If 6 7 0,x x= = then the solution is 1 0,x = 2 0,x = 3 600,x = 4 0,x = 5 500,x = 6 0,x = 7 0.x =

(c) If 5 1000x = and 6 0,x = then the solution is 1 0,x = 2 500,x = − 3 600,x = 4 500,x = 5 1000,x = 6 0,x = and

7 500.x = −

23. (a) Each of the network’s four junctions gives rise to a linear equation, as shown below. input = output 2 1

4 2

3 4

1 3

200100200

100

x xx xx x

x x

+ =

= +

= +

+ =

Rearranging these equations and forming the augmented matrix, you obtain

1 1 0 0 2000 1 0 1 100

.0 0 1 1 2001 0 1 0 100

−⎡ ⎤⎢ ⎥− −⎢ ⎥⎢ ⎥−⎢ ⎥⎢ − − ⎥⎣ ⎦

Gauss-Jordan elimination produces the matrix

1 0 0 1 1000 1 0 1 100

.0 0 1 1 2000 0 0 0 0

−⎡ ⎤⎢ ⎥− −⎢ ⎥⎢ ⎥−⎢ ⎥⎢ ⎥⎣ ⎦

Letting 4 ,x t= you have 1 100 ,x t= + 2 100 ,x t= − + 3 200 ,x t= + and 4 ,x t= where t is a real number.

(b) When 4 0,x t= = then 1 100,x = 2 100,x = − 3 200.x =

(c) When 4 100,x t= = then 1 2 3200, 0, 300.x x x= = =



25. Applying Kirchoff’s first law to either junction produces 1 3 2I I I+ =

and applying the second law to the two paths produces 1 1 2 2 1 2

2 2 3 3 2 3

4 3 33 4

R I R I I IR I R I I I

+ = + =

+ = + =


1 1 1 0 1 0 0 04 3 0 3 0 1 0 10 3 1 4 0 0 1 1

−⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥⇒⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

So, 1 20, 1,I I= = and 3 1.I =

27. (a) To find the general solution, let A have a volts and B have b volts. Applying Kirchoff’s first law to either

junction produces 1 3 2I I I+ =

and applying the second law to the two paths produces

1 1 2 2 1 2

2 2 3 3 2 3

22 3

R I R I I I aR I R I I b

+ = + =

+ = + =

Rearrange these three equations and form the augmented matrix.

1 1 1 01 2 00 2 4

ab

−⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

Gauss-Jordan elimination produces the matrix

( )( )( )

1 0 0 3 70 1 0 4 14 .0 0 1 3 2 14

a ba b

b a

⎡ − ⎤⎢ ⎥

+⎢ ⎥⎢ ⎥−⎣ ⎦

When 5a = and 8,b = then 1 2 31, 2, 1.I I I= = =

(b) When 2a = and 6,b = then

1 2 30, 1, 1.I I I= = =

29. ( ) ( ) ( )

2

2 24

1 11 1 1x A B C

x xx x x= + +

− ++ − +

( ) ( )( ) ( )

( ) ( )

22

2 2 2

2 2

4 1 1 1 1

4 2

4 2

x A x B x x C x

x Ax Ax A Bx B Cx C

x A B x A C x A B C

= + + + − + −

= + + + − + −

= + + + + − −

So, 42 0

.0

A BA CA B C

+ =

+ =

− − =

Use Gauss-Jordan elimination to solve the system.

1 1 0 4 1 0 0 12 0 1 0 0 1 0 31 1 1 0 0 0 1 2

⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥⇒⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥− − −⎣ ⎦ ⎣ ⎦

The solution is: 1, 3, and 2A B C= = = −

So, ( ) ( ) ( )

2

2 24 1 3 2 .

1 11 1 1x

x xx x x= + −

− ++ − +

31. ( )( ) ( )

2

2 220

2 22 2 2x A B C

x xx x x−

= + =+ −+ − −

( ) ( )( ) ( )

( ) ( )

22

2 2 2

2 2

20 2 2 2 2

20 4 4 4 2

20 4 4 4 2

x A x B x x C x

x Ax Ax A Bx B Cx C

x A B x A C x A B C

− = − + + − + +

− = − + + − + +

− = + + − + + − +

So, 14 04 4 2 20.

A BA CA B C

+ = −

− + =

− + =


1 1 0 1 1 0 0 14 0 1 0 0 1 0 24 4 2 20 0 0 1 4

−⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥− ⇒ −⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦

The solution is: 1, 2, and 4.A B C= = − =

So, ( )( ) ( )

2

2 220 1 2 4 .

2 22 2 2x

x xx x x−

= − ++ −+ − −

33. Use Gauss-Jordan elimination to solve the system

2 0 1 0 1 0 0 20 2 1 0 0 1 0 21 1 0 4 0 0 1 4

⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥⇒⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦

So, 2x = 2y = and 4.λ = −

Review Exercises for Chapter 1 21


35. Let 1x = number of touchdowns, 2x = number of extra-point kicks, and 3x = number of field goals.

1 2 3

1 2 3

2 3

136 3 46

0

x x xx x x

x x

+ + =

+ + =

− =


1 1 1 13 1 0 0 56 1 3 46 0 1 0 40 1 1 0 0 0 1 4

⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥⇒⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦

Because 1 5,x = 2 4,x = and 3 4,x = there were 5 touchdowns, 4 extra-point kicks, and 4 field goals.

Review Exercises for Chapter 1





9. Choosing y and z as the free variables and letting y s= and ,z t= you have

31 14 2 2

4 2 6 14 1 2 6

.

x s tx s t

x s t

− + − =

− = − +

= − + −

So, the solution set can be described as 31 14 2 2 ,x s t= − + −

, ,y s z t= = where s and t are real numbers.

11. Row reduce the augmented matrix for this system.

12

3 32 2

1 1 2 1 01 1 2 1 1 20 1 0 13 1 0 0 4 6

⎡ ⎤⎡ ⎤⎡ ⎤ ⎡ ⎤⇒ ⇒ ⇒ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥− − − ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦

Converting back to a linear system, the solution is: 12x = and 3

2.y =

13. Rearrange the equations as shown below. 4

2 3 0x y

x y− = −

− =

Row reduce the augmented matrix for this system.

1 1 4 1 1 4 1 1 4 1 0 122 3 0 0 1 8 0 1 8 0 1 8

− − − − − − −⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤⇒ ⇒ ⇒⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥− − − −⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦

Converting back to a linear system, the solution is: 12x = − and 8.y = −

15. Row reduce the augmented matrix for this system.

1 1 0 1 1 0 1 1 0 1 0 02 1 0 0 1 0 0 1 0 0 1 0⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤

⇒ ⇒ ⇒⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦

Converting back to a linear system, the solution is: 0x = and 0.y =




1 1 91 1 1

−⎡ ⎤⎢ ⎥−⎣ ⎦

which is equivalent to the reduced row-echelon matrix

1 1 0

.0 0 1

−⎡ ⎤⎢ ⎥⎣ ⎦

Because the second row corresponds to 0 1,= which is a false statement, you can conclude that the system has no solution.

19. Multiplying both equations by 100 and forming the augmented matrix produces

20 30 14

.40 50 20⎡ ⎤⎢ ⎥⎣ ⎦

Use Gauss-Jordan elimination as shown below.

3 7 13 7 3 7

22 102 10 2 104455

1 011 10 10 140 50 20 0 10 8

⎡ ⎤ ⎡ ⎤−⎡ ⎤ ⎡ ⎤⇒ ⇒ ⇒⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥

− −⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦⎣ ⎦

So, the solution is: 112x = − and 2

45.x =

21. Expanding the second equation, 3 2 0,x y+ = the augmented matrix for this system is

1 12 3 03 2 0

⎡ ⎤−⎢ ⎥⎢ ⎥⎣ ⎦


1 0 0

.0 1 0⎡ ⎤⎢ ⎥⎣ ⎦

So, the solution is: 0x = and 0.y =


25. This matrix has the characteristic stair step pattern of leading 1’s so it is in row-echelon form. However, the leading 1 in row three of column four has 1’s above it, so the matrix is not in reduced row-echelon form.

27. Because the first row begins with 1,− this matrix is not in row-echelon form.

29. This matrix corresponds to the system 1 2

3

2 00.

x xx

+ =

=

Choosing 2x t= as the free variable you can describe the solution as 1 2 ,x t= − 2 ,x t= and 3 0,x = where t is a real number.


1 1 2 12 3 1 25 4 2 4

−⎡ ⎤⎢ ⎥−⎢ ⎥⎢ ⎥⎣ ⎦


1 0 0 20 1 0 3 .0 0 1 3

⎡ ⎤⎢ ⎥−⎢ ⎥⎢ ⎥⎣ ⎦

So, the solution is: 2, 3,x y= = − and 3.z =

33. Use Gauss-Jordan elimination on the augmented matrix.

1213

2 3 3 3 1 0 06 6 12 13 0 1 0

12 9 1 2 0 0 1 1

⎡ ⎤⎡ ⎤⎢ ⎥⎢ ⎥ ⇒ −⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥−⎣ ⎦ ⎣ ⎦

So, 12,x = 1

3,y = − and 1.z =


1 2 1 62 3 0 71 3 3 11

− −⎡ ⎤⎢ ⎥− −⎢ ⎥⎢ ⎥− −⎣ ⎦


1 0 3 40 1 2 5 .0 0 0 0

−⎡ ⎤⎢ ⎥−⎢ ⎥⎢ ⎥⎣ ⎦

Choosing z t= as the free variable you find that the solution set can be described by 4 3 ,x t= +

5 2 ,y t= + and ,z t= where t is a real number.



37. Use the Gauss-Jordan elimination on the augmented matrix for this system.

322 1 2 4 1 0 2

2 2 0 5 0 1 2 12 1 6 2 0 0 0 0

⎡ ⎤⎡ ⎤⎢ ⎥⎢ ⎥ ⇒ −⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥−⎣ ⎦ ⎣ ⎦

So, the solution is: 32 2 ,x t= − 1 2 ,y t= + ,z t=

where t is any real number.


2 1 1 2 15 2 1 3 01 3 2 2 13 2 3 5 12

−⎡ ⎤⎢ ⎥− −⎢ ⎥⎢ ⎥−⎢ ⎥⎢ − ⎥⎣ ⎦


1 0 0 0 10 1 0 0 40 0 1 0 30 0 0 1 2

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥−⎢ ⎥⎢ − ⎥⎣ ⎦

So, the solution is: 1 21, 4,x x= = 3 3,x = − and

4 2.x = −

41. Using a graphing utility, the augmented matrix reduces to

1 0 0 00 1 4 20 0 0 00 0 0 0

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

Choosing z t= as the free variable, you find that the solution set can be described by 0,x = 4 ,y z t= − and ,z t= where t is a real number.


1 0 0 0 10 1 0 0 00 0 1 0 40 0 0 1 2

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ − ⎥⎣ ⎦

So, the solution is: 1, 0,x y= = 4,z = and 2.w = −


1 0 2 0 00 1 1 0 00 0 0 1 0

⎡ ⎤⎢ ⎥−⎢ ⎥⎢ ⎥⎣ ⎦

Choosing z t= as the free variable you find that the solution set can be described by 2 ,x t= − ,y t=

,z t= and 0,w = where t is a real number.

47. Use Guass-Jordan elimination on the augmented matrix.

1 2 8 0 1 0 0 03 2 0 0 0 1 0 01 1 7 0 0 0 1 0

− −⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥⇒⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦

So, the solution is 1 2 3 0.x x x= = =


2 8 4 03 10 7 00 10 5 0

−⎡ ⎤⎢ ⎥−⎢ ⎥⎢ ⎥⎣ ⎦


12

1 0 4 00 1 00 0 0 0

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

Choosing 3x t= as the free variable, you find that the

solution set can be described by 1 2124 , ,x t x t= − = −

and 3 ,x t= where t is a real number.

51. Forming the augmented matrix

1 0

1 1k

k⎡ ⎤⎢ ⎥⎣ ⎦

and using Gauss-Jordan elimination, you obtain

2

22

22

11 1 1 01 1 1 1 1 , 1 0.1 0 0 1 0 1 0 11 1

kk k k kkk k k kk k

−⎡ ⎤⎡ ⎤ ⎢ ⎥⎡ ⎤ ⎡ ⎤ −⎢ ⎥ ⎢ ⎥⇒ ⇒ ⇒ − ≠⎢ ⎥ ⎢ ⎥ ⎢ ⎥− − ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎢ ⎥−⎣ ⎦ ⎢ ⎥−⎣ ⎦

So, the system is inconsistent if 1.k = ±

53. Row reduce the augmented matrix.

( ) ( )

1 2 31 2 30 2 9 39 b a aa b⎡ ⎤⎡ ⎤

⇒ ⎢ ⎥⎢ ⎥ − − −−⎣ ⎦ ⎣ ⎦

(a) There will be no solution if 2 0b a− = and 9 3 0.a− − ≠ That is if 2b a= and 3.a = − (b) There will be exactly one solution if 2 .b a≠ (c) There will be an infinite number of solutions if 2b a= and 3.a = − That is, if 3a = − and 6.b = −



55. You can show that two matrices of the same size are row equivalent if they both row reduce to the same matrix. The two given matrices are row equivalent because each is row equivalent to the identity matrix.

57. Adding a multiple of row one to each row yields the following matrix.

( )( )

( ) ( ) ( )( )

1 2 30 2 10 2 4 2 1

0 1 2 1 1 1

nn n n nn n n n

n n n n n n n

⎡ ⎤⎢ ⎥− − − −⎢ ⎥⎢ ⎥− − − −⎢ ⎥⎢ ⎥⎢ ⎥− − − − − − −⎣ ⎦

Every row below row two is a multiple of row two. Therefore, reduce these rows to zeros.

( )

1 2 30 2 10 0 0 0

0 0 0 0

nn n n n

⎡ ⎤⎢ ⎥− − − −⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

Dividing row two by n− yields a new second row.

1 2 30 1 2 10 0 0 0

0 0 0 0

nn

⎡ ⎤⎢ ⎥−⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

Adding 2− times row two to row one yields a new first row.

1 0 1 20 1 2 10 0 0 0

0 0 0 0

nn

− −⎡ ⎤⎢ ⎥−⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

This matrix is in reduced row-echelon form.

59. (a) False. See page 3, following Example 2. (b) True. See page 5, Example 4(b).

61. (a) Let 1x = number of three-point baskets, 2x = number of two-point baskets, and 3x = number of one-point free throws.

1 2 3

1 2

2 3

3 2 593 0

1

x x xx x

x x

+ + =

− =

− =

(b) Use Gauss-Jordan elimination to solve the system.

3 2 1 59 1 0 0 53 1 0 0 0 1 0 150 1 1 1 0 0 1 14

⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥− ⇒⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦

Because 5,x = 15,y = and 14,z = there were 5 three-point baskets, 15 two-point baskets, and 14 one-point free throws.

63. ( )( ) ( )

2

2 23 3 2

2 22 2 2x x A B C

x xx x x− −

= + ++ −+ − −

( ) ( )( ) ( )

( ) ( )

22

2 2 2

2 2

3 3 2 2 2 2 2

3 3 2 4 4 4 23 3 2 4 4 4 2

x x A x B x x C x

x x Ax Ax A Bx B Cx Cx x A B x A C x A B C

− − = − + + − + +

− − = − + + − + +

− − = + + − + + − +

So, 34 3

.4 4 2 2

A BA CA B C

+ =− + = −

− + = −


1 1 0 3 1 0 0 14 0 1 3 0 1 0 24 4 2 2 0 0 1 1

⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥− − ⇒⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥− −⎣ ⎦ ⎣ ⎦

The solution is: 1, 2,A B= = and 1.C = So, ( )( ) ( )

2

2 23 3 2 1 2 1 .

2 22 2 2x x

x xx x x− −

= + ++ −+ − −



65. (a) Because there are three points, choose a second degree polynomial, ( ) 2

0 1 2 .p x a a x a x= + + By

substituting the values at each point into this equation you obtain the system

0 1 2

0 1 2

0 1 2

2 4 53 9 0

.4 16 20

a a aa a aa a a

+ + =

+ + =

+ + =

Forming the augmented matrix

1 2 4 51 3 9 01 4 16 20

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

and using Gauss-Jordan elimination you obtain

1352252

1 0 0 900 1 0 .0 0 1

⎡ ⎤⎢ ⎥

−⎢ ⎥⎢ ⎥⎣ ⎦

So, ( ) 2135 252 290 .p x x x= − +

(b)

67. Establish the first year as 0x = and substitute the values at each point into ( ) 2

0 1 2p x a a x a x= + + to

obtain the system 0

0 1 2

0 1 2

5060

2 4 75.

aa a aa a a

=

+ + =

+ + =

Forming the augmented matrix

1 0 0 501 1 1 601 2 4 75

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

and using Gauss-Jordan elimination, you obtain

15252

1 0 0 500 1 0 .0 0 1

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

So, ( ) 215 52 250 .p x x x= + + To predict the sales in the

fourth year, evaluate ( )p x when 3.x =

( ) ( ) ( )215 52 23 50 3 3 $95.p = + + =

69. (a) There are three points: (0, 80), (4, 68), and (80, 30). Because you are given three points, choose a second-

degree polynomial, ( ) 20 1 2 .p x a a x a x= + +

Substituting the given points into ( )p x produces the

following system of linear equations.

( ) ( )( ) ( )( ) ( )

20 1 2 0

20 1 2 0

20 1 2 0

0 0

4 4

80 80

a a a a

a a a a

a a a a

+ + =

+ + =

+ + =

1 2

1 2

80

4 16 68

80 6400 30

a a

a a

=

+ + =

+ + =

(b) Form the augmented matrix

1 0 0 801 4 16 681 80 6400 30

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦


258132

1 0 0 800 1 0 .0 0 1

⎡ ⎤⎢ ⎥

−⎢ ⎥⎢ ⎥⎣ ⎦

So, ( ) 225 18 3280 .p x x x= − +

(c) The graphing utility gives 0 80,a = 1258 ,a = − and

2132.a = In other words ( ) 225 1

8 3280 .p x x x= − +

(d) The results to (b) and (c) are the same. (e) There is precisely one polynomial function of degree

1n − (or less) that fits n distinct points.

71. Applying Kirchoff’s first law to either junction produces 1 3 2I I I+ = and applying the second law to the two

paths produces 1 1 2 2 1 2

2 2 3 3 2 3

3 4 34 2 2.

R I R I I IR I R I I I

+ = + =

+ = + =


5136

131

13

1 1 1 0 1 0 03 4 0 3 0 1 00 4 2 2 0 0 1

⎡ ⎤−⎡ ⎤⎢ ⎥⎢ ⎥ ⇒ ⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦ ⎣ ⎦

So, the solution is 15

13,I = 26

13,I = and 31

13.I =

1 2 3 4 5

5

10

15

20

25

x

(2, 5)(3, 0)

(4, 20)

y

CHAPTER 1 Systems of Linear Equations - Cengage...CHAPTER 1 Systems of Linear Equations Section 1.1...

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Transcript of CHAPTER 1 Systems of Linear Equations - Cengage...CHAPTER 1 Systems of Linear Equations Section 1.1...