2 Systems of Linear Equations: ✦ An Introduction ✦ Unique Solutions ✦ Underdetermined and...

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Systems of Linear Equations: Systems of Linear Equations: ✦ An IntroductionAn Introduction✦ Unique SolutionsUnique Solutions✦ Underdetermined and Underdetermined and

Overdetermined SystemsOverdetermined Systems Multiplication of MatricesMultiplication of Matrices The Inverse of a Square MatrixThe Inverse of a Square Matrix Leontief Input-Output ModelLeontief Input-Output Model

Systems of Linear Equations and Matrices Systems of Linear Equations and Matrices

2.12.1Systems of Linear Equations: Systems of Linear Equations: An IntroductionAn Introduction

11 22 33 44 55 66

66

55

44

33

22

11

––1 1

yy

xx

(2, 3)(2, 3)

2 1x y 2 1x y

3 2 12x y 3 2 12x y

Systems of EquationsSystems of Equations

Recall that a Recall that a system of two linear equations in two system of two linear equations in two variablesvariables may be written in the general form may be written in the general form

where where aa, , bb, , cc, , dd, , hh, and , and kk are are real numbersreal numbers and neither and neither aa and and bb nor nor cc and and dd are both zero. are both zero.

Recall that the graph of each equation in the system is a Recall that the graph of each equation in the system is a straight linestraight line in the plane, so that geometrically, the in the plane, so that geometrically, the solutionsolution to the system is the to the system is the point(s) of intersectionpoint(s) of intersection of the of the two straight lines two straight lines LL11 and and LL22, represented by the first and , represented by the first and

second equations of the system.second equations of the system.

ax by hcx dy k

ax by hcx dy k


Given the two straight lines Given the two straight lines LL11 and and LL22, , one and only oneone and only one of of

the following may occur:the following may occur:

1.1. LL11 and and LL22 intersect at intersect at exactly one pointexactly one point..

yy

xx

LL11

LL22

Unique Unique solutionsolution

((xx11, , yy11))((xx11, , yy11))

xx11

yy11




2.2. LL11 and and LL22 are are coincidentcoincident..

yy

xx

LL11, , LL22

Infinitely Infinitely many many

solutionssolutions




3.3. LL11 and and LL22 are are parallelparallel..

yy

xx

LL11

LL22

No No solutionsolution

Example: Example: A System of Equations With Exactly One SolutionA System of Equations With Exactly One Solution

Consider the systemConsider the system

SolvingSolving the the first equationfirst equation for for yy in terms of in terms of xx, we obtain, we obtain

SubstitutingSubstituting this expression for this expression for yy into the into the second equationsecond equation yieldsyields

2 13 2 12

x yx y

2 13 2 12

x yx y

2 1y x 2 1y x

3 2(2 1) 12

3 4 2 12

7 14

2

x x

x x

x

x

3 2(2 1) 12

3 4 2 12

7 14

2

x x

x x

x

x


Finally, Finally, substitutingsubstituting this value of this value of xx into the into the expression forexpression for yy obtained earlier gives obtained earlier gives

Therefore, the Therefore, the unique solutionunique solution of the system is given by of the system is given by xx = 2 = 2 and and yy = 3 = 3..

2 1

2(2) 1

3

y x

2 1

2(2) 1

3

y x

11 22 33 44 55 66

66

55

44

33

22

11

––1 1


GeometricallyGeometrically, the , the two linestwo lines represented by the two represented by the two equations that make up the system equations that make up the system intersectintersect at the at the pointpoint (2, 3)(2, 3)::

yy

xx

(2, 3)(2, 3)

2 1x y 2 1x y

3 2 12x y 3 2 12x y

Example: Example: A System of Equations With Infinitely Many SolutionsA System of Equations With Infinitely Many Solutions




which is a which is a true statementtrue statement.. This result follows from the fact that the This result follows from the fact that the second equationsecond equation

is is equivalentequivalent to the to the firstfirst..

2 16 3 3

x yx y

2 16 3 3

x yx y

2 1y x 2 1y x

6 3(2 1) 3

6 6 3 3

0 0

x x

x x

6 3(2 1) 3

6 6 3 3

0 0

x x

x x


Thus, any Thus, any order pair of numbersorder pair of numbers ((xx, , yy)) satisfying the satisfying the equation equation y y == 22x x – 1 – 1 constitutes a constitutes a solution to the systemsolution to the system..

By By assigning the valueassigning the value t t to to xx, where , where tt is any real number, is any real number, we find that we find that y y == 22t t – 1 – 1 and so the ordered pair and so the ordered pair ((tt, 2, 2t t – 1) – 1) is a is a solution to the systemsolution to the system..

The variable The variable tt is called a is called a parameterparameter..

For example:For example:

✦ Setting Setting t t = 0= 0, gives the point , gives the point (0, –1)(0, –1) as as aa solutionsolution of the of the system.system.

✦ Setting Setting t t = 1= 1, gives the point , gives the point (1, 1) (1, 1) as as another solutionanother solution of of the system.the system.

66

55

44

33

22

11

––1 1 11 22 33 44 55 66


Since Since tt represents represents any real numberany real number, there are , there are infinitely infinitely many solutionsmany solutions of the system. of the system.

Geometrically, the Geometrically, the two equationstwo equations in the system represent in the system represent the same linethe same line, and , and all solutionsall solutions of the system are of the system are points points lying on the linelying on the line::

yy

xx

2 16 3 3

x yx y

2 16 3 3

x yx y

Example: Example: A System of Equations That Has No SolutionA System of Equations That Has No Solution




which is which is clearlyclearly impossibleimpossible.. Thus, there is Thus, there is no solutionno solution to the system of equations. to the system of equations.

2 16 3 12

x yx y

2 16 3 12

x yx y

2 1y x 2 1y x

6 3(2 1) 12

6 6 3 12

0 9

x x

x x

6 3(2 1) 12

6 6 3 12

0 9

x x

x x

11 22 33 44 55 66

Example: Example: A System of Equations That Has No SolutionA System of Equations That Has No Solution

To interpret the situation To interpret the situation geometricallygeometrically, cast both , cast both equations in the equations in the slope-intercept formslope-intercept form, obtaining , obtaining

yy = 2 = 2xx – 1 – 1 and and yy = 2 = 2x x – 4– 4which shows that the lines are which shows that the lines are parallelparallel..

Graphically:Graphically: 66

55

44

33

22

11

––1 1

yy

xx

2 1x y 2 1x y

6 3 12x y 6 3 12x y

2.22.2Systems of Linear Equations:Systems of Linear Equations:

Unique SolutionsUnique Solutions

3 2 8 9

2 2 1 3

1 2 3 8

3 2 8 9

2 2 1 3

1 2 3 8

3 2 8 9

2 2 3

2 3 8

x y z

x y z

x y z

3 2 8 9

2 2 3

2 3 8

x y z

x y z

x y z

1 0 0 3

0 1 0 4

0 0 1 1

1 0 0 3

0 1 0 4

0 0 1 1

The Gauss-Jordan MethodThe Gauss-Jordan Method

The Gauss-Jordan elimination method is a The Gauss-Jordan elimination method is a techniquetechnique for for solving systems of linear equationssolving systems of linear equations of any size. of any size.

The operations of the Gauss-Jordan method areThe operations of the Gauss-Jordan method are1.1. InterchangeInterchange any two equations. any two equations.2.2. ReplaceReplace an equation by a an equation by a nonzero constant multiplenonzero constant multiple of of

itself.itself.3.3. ReplaceReplace an equation by the an equation by the sumsum of that equation and a of that equation and a

constant multiple of any other equationconstant multiple of any other equation..

ExampleExample Solve the following system of equations:Solve the following system of equations:

SolutionSolution First, we transform this system into an equivalent system First, we transform this system into an equivalent system

in which the coefficient of in which the coefficient of x x in the in the first equationfirst equation is is 11::

2 4 6 22

3 8 5 27

2 2

x y z

x y z

x y z

2 4 6 22

3 8 5 27

2 2

x y z

x y z

x y z

2 4 6 22

3 8 5 27

2 2

x y z

x y z

x y z

2 4 6 22

3 8 5 27

2 2

x y z

x y z

x y z

Multiply the Multiply the equation by equation by 1/21/2

Toggle slides Toggle slides back and forth to back and forth to compare before compare before

and changesand changes

Example 1, page 77


SolutionSolution First, we transform this system into an equivalent system First, we transform this system into an equivalent system

in which the coefficient of in which the coefficient of x x in the in the first equationfirst equation is is 11::

2 4 6 22

3 8 5 27

2 2

x y z

x y z

x y z

2 4 6 22

3 8 5 27

2 2

x y z

x y z

x y z

2 3 11

3 8 5 27

2 2

x y z

x y z

x y z

2 3 11

3 8 5 27

2 2

x y z

x y z

x y z

Multiply the first Multiply the first equation by equation by 1/21/2




SolutionSolution Next, we Next, we eliminateeliminate the variable the variable xx from all equations except from all equations except

the first:the first:

2 4 6 22

3 8 5 27

2 2

x y z

x y z

x y z

2 4 6 22

3 8 5 27

2 2

x y z

x y z

x y z

2 3 11

3 8 5 27

2 2

x y z

x y z

x y z

2 3 11

3 8 5 27

2 2

x y z

x y z

x y z

Replace by the sum of Replace by the sum of – 3 – 3 XX the first the first

equation equation ++ the the second equationsecond equation






2 4 6 22

3 8 5 27

2 2

x y z

x y z

x y z

2 4 6 22

3 8 5 27

2 2

x y z

x y z

x y z

2 3 11

2 4 6

2 2

x y z

y z

x y z

2 3 11

2 4 6

2 2

x y z

y z

x y z

Replace by the sum of Replace by the sum of – 3 – 3 ☓☓ the first equation the first equation

++ the second the second equationequation






2 4 6 22

3 8 5 27

2 2

x y z

x y z

x y z

2 4 6 22

3 8 5 27

2 2

x y z

x y z

x y z

2 3 11

2 4 6

2 2

x y z

y z

x y z

2 3 11

2 4 6

2 2

x y z

y z

x y z

Replace by the sum Replace by the sum of the first equation of the first equation + + the third equationthe third equation






2 4 6 22

3 8 5 27

2 2

x y z

x y z

x y z

2 4 6 22

3 8 5 27

2 2

x y z

x y z

x y z

2 3 11

2 4 6

3 5 13

x y z

y z

y z

2 3 11

2 4 6

3 5 13

x y z

y z

y z

Replace by the sum Replace by the sum

of the first equation of the first equation ++ the third equation the third equation




SolutionSolution Then we transform so that the coefficient ofThen we transform so that the coefficient of y y in the in the

second equationsecond equation is is 11::

2 4 6 22

3 8 5 27

2 2

x y z

x y z

x y z

2 4 6 22

3 8 5 27

2 2

x y z

x y z

x y z

2 3 11

2 4 6

3 5 13

x y z

y z

y z

2 3 11

2 4 6

3 5 13

x y z

y z

y z

Multiply the second Multiply the second equation by equation by 1/21/2




SolutionSolution Then we transform so that the coefficient ofThen we transform so that the coefficient of y y in the in the

second equationsecond equation is is 11::

2 4 6 22

3 8 5 27

2 2

x y z

x y z

x y z

2 4 6 22

3 8 5 27

2 2

x y z

x y z

x y z

2 3 11

2 3

3 5 13

x y z

y z

y z

2 3 11

2 3

3 5 13

x y z

y z

y z

Multiply the second Multiply the second equation by equation by 1/21/2




SolutionSolution We now We now eliminateeliminate yy from all equations except the second: from all equations except the second:

2 4 6 22

3 8 5 27

2 2

x y z

x y z

x y z

2 4 6 22

3 8 5 27

2 2

x y z

x y z

x y z

2 3 11

2 3

3 5 13

x y z

y z

y z

2 3 11

2 3

3 5 13

x y z

y z

y z

Replace by the sum of Replace by the sum of the first equation the first equation ++ (–2) (–2) ☓☓ the second the second equationequation





2 4 6 22

3 8 5 27

2 2

x y z

x y z

x y z

2 4 6 22

3 8 5 27

2 2

x y z

x y z

x y z

7 17

2 3

3 5 13

x z

y z

y z

7 17

2 3

3 5 13

x z

y z

y z

Replace by the sum of Replace by the sum of the first equationthe first equation + + (–2) (–2) ☓☓ the second the second equationequation





2 4 6 22

3 8 5 27

2 2

x y z

x y z

x y z

2 4 6 22

3 8 5 27

2 2

x y z

x y z

x y z

7 17

2 3

3 5 13

x z

y z

y z

7 17

2 3

3 5 13

x z

y z

y z

Replace by the sum of Replace by the sum of

the third equation the third equation ++ (–3) (–3) ☓☓ the second the second equationequation





2 4 6 22

3 8 5 27

2 2

x y z

x y z

x y z

2 4 6 22

3 8 5 27

2 2

x y z

x y z

x y z

7 17

2 3

11 22

x z

y z

z

7 17

2 3

11 22

x z

y z

z

Replace by the sum of Replace by the sum of the third equationthe third equation + + (–3) (–3) ☓☓ the second the second equationequation




SolutionSolution Now we transform so that the coefficient ofNow we transform so that the coefficient of z z in the in the third third

equationequation is is 11::

2 4 6 22

3 8 5 27

2 2

x y z

x y z

x y z

2 4 6 22

3 8 5 27

2 2

x y z

x y z

x y z

7 17

2 3

11 22

x z

y z

z

7 17

2 3

11 22

x z

y z

z

Multiply the third Multiply the third equation by equation by 1/111/11




SolutionSolution Now we transform so that the coefficient ofNow we transform so that the coefficient of z z in the in the third third

equationequation is is 11::

2 4 6 22

3 8 5 27

2 2

x y z

x y z

x y z

2 4 6 22

3 8 5 27

2 2

x y z

x y z

x y z

7 17

2 3

2

x z

y z

z

7 17

2 3

2

x z

y z

z

Multiply the third Multiply the third equation by equation by 1/111/11




SolutionSolution We now We now eliminateeliminate zz from all equations except the third: from all equations except the third:

2 4 6 22

3 8 5 27

2 2

x y z

x y z

x y z

2 4 6 22

3 8 5 27

2 2

x y z

x y z

x y z

Replace by the sum of Replace by the sum of the first equation the first equation + + (–7) (–7) ☓☓ the third equation the third equation

7 17

2 3

2

x z

y z

z

7 17

2 3

2

x z

y z

z





2 4 6 22

3 8 5 27

2 2

x y z

x y z

x y z

2 4 6 22

3 8 5 27

2 2

x y z

x y z

x y z

3

2 3

2

x

y z

z

3

2 3

2

x

y z

z

Replace by the sum of Replace by the sum of the first equation the first equation + + (–7) (–7) ☓☓ the third equation the third equation





2 4 6 22

3 8 5 27

2 2

x y z

x y z

x y z

2 4 6 22

3 8 5 27

2 2

x y z

x y z

x y z

3

2 3

2

x

y z

z

3

2 3

2

x

y z

z

Replace by the sum of Replace by the sum of the second equation the second equation + + 22 ☓☓ the third the third equationequation





2 4 6 22

3 8 5 27

2 2

x y z

x y z

x y z

2 4 6 22

3 8 5 27

2 2

x y z

x y z

x y z

3

1

2

x

y

z

3

1

2

x

y

z

Replace by the sum of Replace by the sum of the second equationthe second equation + + 22 ☓☓ the third the third equationequation




SolutionSolution Thus, the Thus, the solutionsolution to the system is to the system is xx = 3 = 3, , yy = 1 = 1, and , and zz = 2 = 2..

2 4 6 22

3 8 5 27

2 2

x y z

x y z

x y z

2 4 6 22

3 8 5 27

2 2

x y z

x y z

x y z

3

1

2

x

y

z

3

1

2

x

y

z

Augmented MatricesAugmented Matrices

Matrices are Matrices are rectangular arrays of numbersrectangular arrays of numbers that can aid that can aid us by us by eliminating the need to write the variableseliminating the need to write the variables at each at each step of the reduction.step of the reduction.

For example, the For example, the systemsystem

may be represented by the may be represented by the augmentedaugmented matrixmatrixCoefficient Coefficient

MatrixMatrix

2 4 6 22

3 8 5 27

2 2

x y z

x y z

x y z

2 4 6 22

3 8 5 27

2 2

x y z

x y z

x y z

2 4 6 22

3 8 5 27

1 1 2 2

2 4 6 22

3 8 5 27

1 1 2 2

Matrices and Gauss-JordanMatrices and Gauss-Jordan

Every stepEvery step in the in the Gauss-Jordan elimination methodGauss-Jordan elimination method can be can be expressed with expressed with matricesmatrices, rather than systems of equations, , rather than systems of equations, thus simplifying the whole process:thus simplifying the whole process:

Steps expressed as Steps expressed as systems of equationssystems of equations::

Steps expressed as Steps expressed as augmented matricesaugmented matrices::

2 4 6 22

3 8 5 27

2 2

x y z

x y z

x y z

2 4 6 22

3 8 5 27

2 2

x y z

x y z

x y z

2 4 6 22

3 8 5 27

1 1 2 2

2 4 6 22

3 8 5 27

1 1 2 2



Example 2, page 78





1 2 3 11

3 8 5 27

1 1 2 2

1 2 3 11

3 8 5 27

1 1 2 2

2 3 11

3 8 5 27

2 2

x y z

x y z

x y z

2 3 11

3 8 5 27

2 2

x y z

x y z

x y z



Example 2, page 78





1 2 3 11

0 2 4 6

1 1 2 2

1 2 3 11

0 2 4 6

1 1 2 2

2 3 11

2 4 6

2 2

x y z

y z

x y z

2 3 11

2 4 6

2 2

x y z

y z

x y z



Example 2, page 78





1 2 3 11

0 2 4 6

0 3 5 13

1 2 3 11

0 2 4 6

0 3 5 13

2 3 11

2 4 6

3 5 13

x y z

y z

y z

2 3 11

2 4 6

3 5 13

x y z

y z

y z



Example 2, page 78





1 2 3 11

0 1 2 3

0 3 5 13

1 2 3 11

0 1 2 3

0 3 5 13

2 3 11

2 3

3 5 13

x y z

y z

y z

2 3 11

2 3

3 5 13

x y z

y z

y z



Example 2, page 78





1 0 7 17

0 1 2 3

0 3 5 13

1 0 7 17

0 1 2 3

0 3 5 13

7 17

2 3

3 5 13

x z

y z

y z

7 17

2 3

3 5 13

x z

y z

y z



Example 2, page 78





1 0 7 17

0 1 2 3

0 0 11 22

1 0 7 17

0 1 2 3

0 0 11 22

7 17

2 3

11 22

x z

y z

z

7 17

2 3

11 22

x z

y z

z



Example 2, page 78





1 0 7 17

0 1 2 3

0 0 1 2

1 0 7 17

0 1 2 3

0 0 1 2

7 17

2 3

2

x z

y z

z

7 17

2 3

2

x z

y z

z



Example 2, page 78





1 0 0 3

0 1 2 3

0 0 1 2

1 0 0 3

0 1 2 3

0 0 1 2

3

2 3

2

x

y z

z

3

2 3

2

x

y z

z



Example 2, page 78





1 0 0 3

0 1 0 1

0 0 1 2

1 0 0 3

0 1 0 1

0 0 1 2

3

1

2

x

y

z

3

1

2

x

y

z

Row Reduced Form Row Reduced Form of the Matrixof the Matrix



Example 2, page 78

Row-Reduced Form of a MatrixRow-Reduced Form of a Matrix

Each row consisting entirely of Each row consisting entirely of zeroszeros lies lies belowbelow all all rows having rows having nonzero entriesnonzero entries..

The The firstfirst nonzero entrynonzero entry in each nonzero row is in each nonzero row is 1 1 (called a (called a leadingleading 11).).

In any two successive (nonzero) rows, the In any two successive (nonzero) rows, the leadingleading 11 in the lower row lies in the lower row lies to the rightto the right of the of the leadingleading 11 in in the the upper rowupper row..

If a column contains a If a column contains a leadingleading 11, then the other , then the other entries in that column are entries in that column are zeroszeros..

Row OperationsRow Operations

1.1. Interchange any two rows.Interchange any two rows.

2.2. Replace any row by a nonzero constant Replace any row by a nonzero constant multiple of itself.multiple of itself.

3.3. Replace any row by the sum of that row Replace any row by the sum of that row and a constant multiple of any other row.and a constant multiple of any other row.

Terminology for theTerminology for theGauss-Jordan Elimination MethodGauss-Jordan Elimination Method

Unit ColumnUnit Column A column in a coefficient matrix is in unit form A column in a coefficient matrix is in unit form

if if oneone of the entries in the column is a of the entries in the column is a 11 and the and the otherother entries are entries are zeroszeros..

PivotingPivoting The The sequencesequence of of row operationsrow operations that that transformstransforms

a a given columngiven column in an augmented matrix into a in an augmented matrix into a unit columnunit column..

Notation for Row OperationsNotation for Row Operations

Letting Letting RRii denote the denote the iithth row of a matrix, we writerow of a matrix, we write

Operation 1:Operation 1: RRii ↔ R↔ Rjj to mean: to mean: InterchangeInterchange row row ii with row with row jj..

Operation 2:Operation 2: cRcRii to mean: to mean: replacereplace row row ii with with cc timestimes row row ii..

Operation 3:Operation 3: RRii + aR + aRjj to mean: to mean:

ReplaceReplace row row ii with the with the sumsum of row of row i i and and aa timestimes row row jj..

ExampleExample

Pivot the matrix about the circled elementPivot the matrix about the circled element

SolutionSolution

3 5 9

2 3 5

3 5 9

2 3 5

3 5 9

2 3 5

3 5 9

2 3 5

113 R113 R 5

3 31

52 3

53 31

52 3

2 12R R2 12R R 53

13

1 3

0 1

53

13

1 3

0 1

Example 4, page 81

The Gauss-Jordan Elimination MethodThe Gauss-Jordan Elimination Method

1.1. Write the Write the augmented matrixaugmented matrix corresponding to corresponding to the linear system.the linear system.

2.2. Interchange rowsInterchange rows, if necessary, to obtain an , if necessary, to obtain an augmented matrix in which the augmented matrix in which the first entryfirst entry in in the the first rowfirst row is is nonzerononzero. Then . Then pivotpivot the matrix the matrix about this entry.about this entry.

3.3. Interchange Interchange thethe second rowsecond row with any row below with any row below it, if necessary, to obtain an augmented matrix it, if necessary, to obtain an augmented matrix in which the in which the second entrysecond entry in the in the second rowsecond row is is nonzerononzero. . PivotPivot the matrix about this entry. the matrix about this entry.

4.4. ContinueContinue until the final matrix is in until the final matrix is in row-row-reduced formreduced form..

ExampleExample

Use the Gauss-Jordan elimination method to solve the Use the Gauss-Jordan elimination method to solve the system of equationssystem of equations

SolutionSolution

3 2 8 9

2 2 3

2 3 8

x y z

x y z

x y z

3 2 8 9

2 2 3

2 3 8

x y z

x y z

x y z

3 2 8 9

2 2 1 3

1 2 3 8

3 2 8 9

2 2 1 3

1 2 3 8

1 2R R1 2R R

Toggle slides back and Toggle slides back and forth to compare before forth to compare before and after matrix changesand after matrix changes

Example 5, page 82-83

ExampleExample


SolutionSolution

3 2 8 9

2 2 3

2 3 8

x y z

x y z

x y z

3 2 8 9

2 2 3

2 3 8

x y z

x y z

x y z

1 0 9 12

2 2 1 3

1 2 3 8

1 0 9 12

2 2 1 3

1 2 3 8

2 12R R2 12R R

3 1R R3 1R R1 2R R1 2R R



ExampleExample


SolutionSolution

3 2 8 9

2 2 3

2 3 8

x y z

x y z

x y z

3 2 8 9

2 2 3

2 3 8

x y z

x y z

x y z

1 0 9 12

0 2 19 27

0 2 12 4

1 0 9 12

0 2 19 27

0 2 12 4

2 12R R2 12R R

3 1R R3 1R R2 3R R2 3R R



ExampleExample


SolutionSolution

3 2 8 9

2 2 3

2 3 8

x y z

x y z

x y z

3 2 8 9

2 2 3

2 3 8

x y z

x y z

x y z

2 3R R2 3R R122 R122 R

1 0 9 12

0 2 12 4

0 2 19 27

1 0 9 12

0 2 12 4

0 2 19 27



ExampleExample


SolutionSolution

3 2 8 9

2 2 3

2 3 8

x y z

x y z

x y z

3 2 8 9

2 2 3

2 3 8

x y z

x y z

x y z

122 R122 R

1 0 9 12

0 1 6 2

0 2 19 27

1 0 9 12

0 1 6 2

0 2 19 27

3 2R R3 2R R



ExampleExample


SolutionSolution

3 2 8 9

2 2 3

2 3 8

x y z

x y z

x y z

3 2 8 9

2 2 3

2 3 8

x y z

x y z

x y z

1 0 9 12

0 1 6 2

0 0 31 31

1 0 9 12

0 1 6 2

0 0 31 31

3 2R R3 2R R1331 R1331 R



ExampleExample


SolutionSolution

3 2 8 9

2 2 3

2 3 8

x y z

x y z

x y z

3 2 8 9

2 2 3

2 3 8

x y z

x y z

x y z

1 0 9 12

0 1 6 2

0 0 1 1

1 0 9 12

0 1 6 2

0 0 1 1

1331 R1331 R

1 39R R1 39R R

2 36R R2 36R R



ExampleExample


SolutionSolution

3 2 8 9

2 2 3

2 3 8

x y z

x y z

x y z

3 2 8 9

2 2 3

2 3 8

x y z

x y z

x y z

1 0 0 3

0 1 0 4

0 0 1 1

1 0 0 3

0 1 0 4

0 0 1 1

2 36R R2 36R R

1 39R R1 39R R



ExampleExample


SolutionSolution

The The solutionsolution to the system is thus to the system is thus xx = 3 = 3, , yy = 4 = 4, and , and zz = 1 = 1..

3 2 8 9

2 2 3

2 3 8

x y z

x y z

x y z

3 2 8 9

2 2 3

2 3 8

x y z

x y z

x y z

1 0 0 3

0 1 0 4

0 0 1 1

1 0 0 3

0 1 0 4

0 0 1 1


2.32.3Systems of Linear Equations:Systems of Linear Equations:Underdetermined and Overdetermined systemsUnderdetermined and Overdetermined systems

2 3 2

3 2 1

2 3 5 3

x y z

x y z

x y z

2 3 2

3 2 1

2 3 5 3

x y z

x y z

x y z

1 2 3 2

3 1 2 1

2 3 5 3

1 2 3 2

3 1 2 1

2 3 5 3

1

x z

y z

1

x z

y z

0

1

x z

y z

0

1

x z

y z

A System of Equations A System of Equations with an Infinite Number of Solutionswith an Infinite Number of Solutions

Solve the system of equations given bySolve the system of equations given by

SolutionSolution

2 3 2

3 2 1

2 3 5 3

x y z

x y z

x y z

2 3 2

3 2 1

2 3 5 3

x y z

x y z

x y z

1 2 3 2

3 1 2 1

2 3 5 3

1 2 3 2

3 1 2 1

2 3 5 3

2 13R R2 13R R

3 12R R3 12R R


Example 1, page 91



SolutionSolution

2 3 2

3 2 1

2 3 5 3

x y z

x y z

x y z

2 3 2

3 2 1

2 3 5 3

x y z

x y z

x y z

1 2 3 2

0 7 7 7

0 1 1 1

1 2 3 2

0 7 7 7

0 1 1 1

2 13R R2 13R R

3 12R R3 12R R1

27 R 127 R


Example 1, page 91



SolutionSolution

2 3 2

3 2 1

2 3 5 3

x y z

x y z

x y z

2 3 2

3 2 1

2 3 5 3

x y z

x y z

x y z

1 2 3 2

0 1 1 1

0 1 1 1

1 2 3 2

0 1 1 1

0 1 1 1

127 R 127 R

1 22R R1 22R R

3 2R R3 2R R


Example 1, page 91



SolutionSolution

2 3 2

3 2 1

2 3 5 3

x y z

x y z

x y z

2 3 2

3 2 1

2 3 5 3

x y z

x y z

x y z

1 0 1 0

0 1 1 1

0 0 0 0

1 0 1 0

0 1 1 1

0 0 0 0

1 22R R1 22R R

3 2R R3 2R R


Example 1, page 91



SolutionSolution

Observe that Observe that row threerow three reads reads 0 = 00 = 0, which is , which is truetrue but but of no useof no use to us. to us.

2 3 2

3 2 1

2 3 5 3

x y z

x y z

x y z

2 3 2

3 2 1

2 3 5 3

x y z

x y z

x y z

1 0 1 0

0 1 1 1

0 0 0 0

1 0 1 0

0 1 1 1

0 0 0 0

Example 1, page 91



SolutionSolution This last augmented matrix is in This last augmented matrix is in row-reduced formrow-reduced form.. Interpreting it as a Interpreting it as a system of equations system of equations gives a system of gives a system of

two equationstwo equations in in three variables three variables xx, , yy, and , and zz::

2 3 2

3 2 1

2 3 5 3

x y z

x y z

x y z

2 3 2

3 2 1

2 3 5 3

x y z

x y z

x y z

1 0 1 0

0 1 1 1

0 0 0 0

1 0 1 0

0 1 1 1

0 0 0 0

0

1

x z

y z

0

1

x z

y z

Example 1, page 91



SolutionSolution Lets Lets single outsingle out a single variable – a single variable – say, say, zz – and – and solvesolve for for xx

and and yy in terms of it. in terms of it. If we assign a If we assign a particular valueparticular value of of zz – – say, say, zz = 0 = 0 – we obtain – we obtain

xx = 0 = 0 and and yy = –1 = –1, giving the , giving the solutionsolution (0, –1, 0)(0, –1, 0)..

2 3 2

3 2 1

2 3 5 3

x y z

x y z

x y z

2 3 2

3 2 1

2 3 5 3

x y z

x y z

x y z

1

x z

y z

1

x z

y z

0

1

x z

y z

0

1

x z

y z

(0) 0

(0) 1 1

(0) 0

(0) 1 1

Example 1, page 91




and and yy in terms of it. in terms of it. If we instead assign If we instead assign zz = 1 = 1, we obtain the , we obtain the solutionsolution (1, 0, 1)(1, 0, 1)..

2 3 2

3 2 1

2 3 5 3

x y z

x y z

x y z

2 3 2

3 2 1

2 3 5 3

x y z

x y z

x y z

1

x z

y z

1

x z

y z

0

1

x z

y z

0

1

x z

y z

(1) 1

(1) 1 0

(1) 1

(1) 1 0

Example 1, page 91




and and yy in terms of it. in terms of it. In generalIn general, we set , we set zz = = tt, where , where t t represents represents any real numberany real number

(called the (called the parameterparameter) to obtain the ) to obtain the solutionsolution ((tt, , tt – 1, – 1, tt))..

2 3 2

3 2 1

2 3 5 3

x y z

x y z

x y z

2 3 2

3 2 1

2 3 5 3

x y z

x y z

x y z

1

x z

y z

1

x z

y z

0

1

x z

y z

0

1

x z

y z

( )

( ) 1 1

t t

t t

( )

( ) 1 1

t t

t t

Example 1, page 91

A System of Equations That Has No SolutionA System of Equations That Has No Solution


SolutionSolution

1

3 4

5 5 1

x y z

x y z

x y z

1

3 4

5 5 1

x y z

x y z

x y z

1 1 1 1

3 1 1 4

1 5 5 1

1 1 1 1

3 1 1 4

1 5 5 1

2 13R R2 13R R

3 1R R3 1R R


Example 2, page 92



SolutionSolution

1 1 1 1

0 4 4 1

0 4 4 2

1 1 1 1

0 4 4 1

0 4 4 2

2 13R R2 13R R

3 1R R3 1R R3 2R R3 2R R

1

3 4

5 5 1

x y z

x y z

x y z

1

3 4

5 5 1

x y z

x y z

x y z


Example 2, page 92



SolutionSolution

1 1 1 1

0 4 4 1

0 0 0 1

1 1 1 1

0 4 4 1

0 0 0 1

3 2R R3 2R R

1

3 4

5 5 1

x y z

x y z

x y z

1

3 4

5 5 1

x y z

x y z

x y z


Example 2, page 92



SolutionSolution

Observe that row three reads Observe that row three reads 00xx + 0 + 0yy + 0 + 0zz = –1 = –1 or or 0 = –10 = –1!! We therefore conclude We therefore conclude the system is the system is inconsistentinconsistent and has and has

no solutionno solution..

1 1 1 1

0 4 4 1

0 0 0 1

1 1 1 1

0 4 4 1

0 0 0 1

1

3 4

5 5 1

x y z

x y z

x y z

1

3 4

5 5 1

x y z

x y z

x y z

Systems with no SolutionSystems with no Solution

If there is a If there is a rowrow in the augmented matrix in the augmented matrix containing containing all zerosall zeros to the to the leftleft of the of the vertical linevertical line and a and a nonzerononzero entry to the entry to the rightright of the of the lineline, then , then the system of equations has the system of equations has no solutionno solution..

Theorem 1Theorem 1

a.a. If the If the number of equationsnumber of equations is is greatergreater than or than or equal to the equal to the number of variablesnumber of variables in a linear in a linear system, then one of the following is true:system, then one of the following is true:

i.i. The system has The system has no solutionno solution..

ii.ii. The system has The system has exactly one solutionexactly one solution..

iii.iii. The system has The system has infinitely many solutionsinfinitely many solutions..

b.b. If there are If there are fewerfewer equations than variablesequations than variables in in a linear system, then the system either has noa linear system, then the system either has no solution solution or it has or it has infinitely many solutionsinfinitely many solutions..

2.42.4MatricesMatrices

2 3

2 3

X B A

X A B

2 3

2 3

X B A

X A B

3 4 3 23

1 2 1 2

3 4 3 23

1 2 1 2

9 12 3 2

3 6 1 2

9 12 3 2

3 6 1 2

6 10

2 4

6 10

2 4

6 101

2 42X

6 101

2 42X

3 5

1 2

3 5

1 2

MatrixMatrix

A A matrixmatrix is an is an ordered rectangular array of numbersordered rectangular array of numbers.. A matrix with A matrix with mm rowsrows and and nn columnscolumns has size has size m m ☓☓ n n.. The The entryentry in the in the iithth rowrow and and jjthth columncolumn is denoted by is denoted by aaijij..

Applied Example:Applied Example: Organizing Production Data Organizing Production Data The Acrosonic Company manufactures The Acrosonic Company manufactures four differentfour different

loudspeaker systemsloudspeaker systems at at threethree separate separate locationslocations.. The company’s The company’s MayMay output is as follows: output is as follows:

If we agree to preserve the If we agree to preserve the relative locationrelative location of each of each entryentry in the table, we can summarize the set of data as follows:in the table, we can summarize the set of data as follows:

Model AModel A Model BModel B Model CModel C Model DModel D

Location ILocation I 320320 280280 460460 280280

Location IILocation II 480480 360360 580580 00

Location IIILocation III 540540 420420 200200 880880

320 280 460 280

480 360 580 0

540 420 200 880

320 280 460 280

480 360 580 0

540 420 200 880

Applied Example 1, page 102

Applied Example:Applied Example: Organizing Production Data Organizing Production Data

We have Acrosonic’s We have Acrosonic’s MayMay output expressed as a matrix: output expressed as a matrix:

a.a. What is the What is the sizesize of the matrix of the matrix PP??

SolutionSolution

Matrix Matrix PP has has three rowsthree rows and and four columnsfour columns and hence and hence has has sizesize 3 3 ☓ ☓ 44..

320 280 460 280

480 360 580 0

540 420 200 880

P

320 280 460 280

480 360 580 0

540 420 200 880

P




b.b. Find Find aa2424 (the entry in (the entry in row 2row 2 and and column 4column 4 of the of the matrix matrix PP) and give an ) and give an interpretationinterpretation of this number. of this number.

SolutionSolution

The required entry lies in The required entry lies in row 2row 2 and and column 4column 4, and is , and is the the number 0number 0. This means that . This means that no model Dno model D loudspeaker system was manufactured at location II loudspeaker system was manufactured at location II in May.in May.

320 280 460 280

480 360 580 0

540 420 200 880

P

320 280 460 280

480 360 580 0

540 420 200 880

P




c.c. Find the Find the sumsum of the of the entriesentries that make up that make up rowrow 11 of of P P and and interpretinterpret the result. the result.

SolutionSolution

The required The required sumsum is given by is given by

320 + 280 + 460 + 280 = 1340320 + 280 + 460 + 280 = 1340

which gives the which gives the total numbertotal number of loudspeaker systems of loudspeaker systems manufactured at manufactured at location Ilocation I in May as in May as 13401340 units. units.

320 280 460 280

480 360 580 0

540 420 200 880

P

320 280 460 280

480 360 580 0

540 420 200 880

P




d.d. Find the Find the sumsum of the of the entriesentries that make up that make up columncolumn 44 of of P P and and interpretinterpret the result. the result.

SolutionSolution

The required The required sumsum is given by is given by

280 + 0 + 880 = 1160280 + 0 + 880 = 1160

giving the giving the outputoutput of of Model D Model D loudspeaker systems at loudspeaker systems at allall locations locations in May as in May as 11601160 units. units.

320 280 460 280

480 360 580 0

540 420 200 880

P

320 280 460 280

480 360 580 0

540 420 200 880

P


Equality of MatricesEquality of Matrices

Two matrices are equal if they have the same size Two matrices are equal if they have the same size and their corresponding entries are equal.and their corresponding entries are equal.

ExampleExample

Solve the following matrix equation for Solve the following matrix equation for xx, , yy, and , and zz::

SolutionSolution Since the Since the corresponding elementscorresponding elements of the two matrices of the two matrices must must

be equalbe equal, we find that , we find that xx = 4 = 4, , zz = 3 = 3, and , and yy – 1 = 1 – 1 = 1, or , or yy = 2 = 2..

1 3 1 4

2 1 2 2 1 2

x z

y

1 3 1 4

2 1 2 2 1 2

x z

y

Example 2, page 103

Addition and Subtraction of MatricesAddition and Subtraction of Matrices

If If AA and and BB are two matrices of the are two matrices of the same sizesame size, then:, then:

1.1. The The sumsum A A ++ B B is the matrix obtained by is the matrix obtained by adding adding the corresponding entriesthe corresponding entries in the two matrices. in the two matrices.

2.2. The The differencedifference A A – – BB is the matrix obtained by is the matrix obtained by subtracting the corresponding entriessubtracting the corresponding entries in in BB from from those in those in AA..


The The total outputtotal output of Acrosonic for of Acrosonic for MayMay is is

The The total outputtotal output of Acrosonic for of Acrosonic for JuneJune is is

Find the Find the total outputtotal output of the company for of the company for MayMay and and JuneJune..











SolutionSolution Expressing the Expressing the outputoutput for for MayMay and and JuneJune as matrices: as matrices:

✦ The The total outputtotal output of Acrosonic for of Acrosonic for MayMay is is

✦ The The total outputtotal output of Acrosonic for of Acrosonic for JuneJune is is

320 280 460 280

480 360 580 0

540 420 200 880

A

320 280 460 280

480 360 580 0

540 420 200 880

A

210 180 330 180

400 300 450 40

420 280 180 740

B

210 180 330 180

400 300 450 40

420 280 180 740

B



SolutionSolution The The total outputtotal output of the company for of the company for MayMay and and JuneJune is is

given by the matrixgiven by the matrix

320 280 460 280 210 180 330 180

480 360 580 0 400 300 450 40

540 420 200 880 420 280 180 740

530 460 790 460

880 660 1030 40

960 700 380 1620

A B

320 280 460 280 210 180 330 180

480 360 580 0 400 300 450 40

540 420 200 880 420 280 180 740

530 460 790 460

880 660 1030 40

960 700 380 1620

A B


Laws for Matrix AdditionLaws for Matrix Addition

If If AA, , BB, and , and CC are are matricesmatrices of the of the same sizesame size, then, then

1.1. A A + + B B == B B ++ A A Commutative Commutative lawlaw

2.2. ((A A + + BB) +) + C C == A A + (+ (B B ++ C C)) Associative lawAssociative law

Transpose of a MatrixTranspose of a Matrix

If If AA is an is an m m ☓☓ n n matrix with elements matrix with elements aaijij, ,

then the then the transposetranspose of of AA is the is the n n ☓☓ m m matrix matrix AATT with elements with elements aajiji..

ExampleExample

Find the Find the transposetranspose of the matrix of the matrix

SolutionSolution The The transposetranspose of the matrix of the matrix A A isis

1 2 3

4 5 6

7 8 9

A

1 2 3

4 5 6

7 8 9

A

1 4 7

2 5 8

3 6 9

TA

1 4 7

2 5 8

3 6 9

TA

Scalar ProductScalar Product

If If AA is a matrix and is a matrix and c c is a is a real numberreal number, then , then the the scalar productscalar product cA cA is the matrix obtained is the matrix obtained by by multiplyingmultiplying each entryeach entry of of AA by by cc..

ExampleExample

GivenGiven

find the matrix find the matrix XX that satisfies that satisfies 22X X ++ B B = 3= 3AASolutionSolution

2 3

2 3

X B A

X A B

2 3

2 3

X B A

X A B

3 4 3 23

1 2 1 2

3 4 3 23

1 2 1 2

9 12 3 2

3 6 1 2

9 12 3 2

3 6 1 2

6 10

2 4

6 10

2 4

6 101

2 42X

6 101

2 42X

3 5

1 2

3 5

1 2

3 4 3 2

1 2 1 2A B

and 3 4 3 2

1 2 1 2A B

and

Example 4, page 106

Applied Example:Applied Example: Production Planning Production Planning The management of Acrosonic has decided to increase its The management of Acrosonic has decided to increase its

JulyJuly production of production of loudspeaker systemsloudspeaker systems by by 10%10% (over (over JuneJune output). output).

Find a Find a matrixmatrix giving the targeted production for giving the targeted production for JulyJuly..SolutionSolution We have seen that Acrosonic’s total output for We have seen that Acrosonic’s total output for JuneJune may may

be represented by the matrixbe represented by the matrix

210 180 330 180

400 300 450 40

420 280 180 740

B

210 180 330 180

400 300 450 40

420 280 180 740

B


Applied Example:Applied Example: Production Planning Production Planning The management of Acrosonic has decided to increase its The management of Acrosonic has decided to increase its

JulyJuly production of production of loudspeaker systemsloudspeaker systems by by 10%10% (over (over JuneJune output). output).

Find a Find a matrixmatrix giving the targeted production for giving the targeted production for JulyJuly..SolutionSolution The required matrix is given byThe required matrix is given by

210 180 330 180

(1.1) 1.1 400 300 450 40

420 280 180 740

B

210 180 330 180

(1.1) 1.1 400 300 450 40

420 280 180 740

B

231 198 363 198

440 330 495 44

462 308 198 814

231 198 363 198

440 330 495 44

462 308 198 814

2.52.5Multiplication of MatricesMultiplication of Matrices

11 12 13 1411 12 13

21 22 23 2421 22 23

31 32 33 34

b b b ba a a

A B b b b ba a a

b b b b

11 12 13 1411 12 13

21 22 23 2421 22 23

31 32 33 34

b b b ba a a

A B b b b ba a a

b b b b

Size of A (2 ☓ 3) (3 ☓ 4) Size of B(2 ☓ 4)

Size of AB

Same

Multiplying a Row Matrix by a Column MatrixMultiplying a Row Matrix by a Column Matrix

If we have a If we have a row matrixrow matrix of size of size 11☓☓ nn,,

And a And a column matrixcolumn matrix of size of size nn ☓☓ 1 1,,

Then we may define the Then we may define the matrix productmatrix product of of AA and and B B, written , written ABAB, by, by

1 2 3[ ]nA a a a a 1 2 3[ ]nA a a a a

1

2

3

n

b

b

B b

b

1

2

3

n

b

b

B b

b

1

2

1 2 3 3 1 1 2 2 3 3[ ]n n n

n

b

b

AB a a a a b a b a b a b a b

b

1

2

1 2 3 3 1 1 2 2 3 3[ ]n n n

n

b

b

AB a a a a b a b a b a b a b

b

ExampleExample

LetLet

Find the matrix product Find the matrix product ABAB..

SolutionSolution

2

3[1 2 3 5]

0

1

a d nA B

2

3[1 2 3 5]

0

1

a d nA B

2

3[1 2 3 5] (1)(2) ( 2)(3) (3)(0) (5)( 1) 9

0

1

AB

2

3[1 2 3 5] (1)(2) ( 2)(3) (3)(0) (5)( 1) 9

0

1

AB

Example 1, page 114

Dimensions Requirement Dimensions Requirement for Matrices Being Multipliedfor Matrices Being Multiplied

Note from the last example that for the multiplication to Note from the last example that for the multiplication to be be feasiblefeasible, the , the number of columnsnumber of columns of the of the row matrixrow matrix AA must be equalmust be equal to the to the number of rowsnumber of rows of the of the column column matrixmatrix BB..

Dimensions of the Product MatrixDimensions of the Product Matrix

From last example, note that the From last example, note that the product matrixproduct matrix ABAB has has size size 1 1 ☓☓ 11..

This has to do with the fact that we are multiplying a This has to do with the fact that we are multiplying a row row matrixmatrix with a with a column matrixcolumn matrix. .

We can establish the We can establish the dimensionsdimensions of a of a product matrixproduct matrix schematically: schematically:

Size ofSize of A A (1 (1 ☓☓ nn)) ((nn ☓ ☓ 11) ) Size ofSize of BB

Size ofSize of ABAB

((11 ☓ ☓ 11))

SameSame

Dimensions of the Product MatrixDimensions of the Product Matrix

More generally, if More generally, if AA is a matrix of size is a matrix of size mm ☓☓ nn and and BB is a is a matrix of size matrix of size nn ☓☓ pp, then the , then the matrix productmatrix product of of AA and and BB, , ABAB, is defined and is a matrix of , is defined and is a matrix of size mm ☓☓ pp..

Schematically: Schematically:

The The number of columnsnumber of columns of of AA must be the must be the samesame as the as the number of rowsnumber of rows of of BB for the multiplication to be for the multiplication to be feasiblefeasible..

Size ofSize of A A ((mm ☓☓ nn)) ((nn ☓ ☓ pp) ) Size ofSize of B B

Size ofSize of ABAB((mm ☓ ☓ pp))

SameSame

Mechanics of Matrix MultiplicationMechanics of Matrix Multiplication

To see how to compute the To see how to compute the productproduct of a of a 2 2 ☓ ☓ 33 matrix matrix A A and and a a 3 3 ☓☓ 4 4 matrix matrix BB, suppose, suppose

From the schematicFrom the schematic

we see that the matrix product we see that the matrix product C = ABC = AB is is feasible feasible (since the (since the number of number of columnscolumns of of AA equalsequals the number of the number of rowsrows of of BB) and ) and has sizehas size 2 2 ☓ ☓ 44..

11 12 13 1411 12 13

21 22 23 2421 22 23

31 32 33 34

b b b ba a a

A B b b b ba a a

b b b b

11 12 13 1411 12 13

21 22 23 2421 22 23

31 32 33 34

b b b ba a a

A B b b b ba a a

b b b b

Size ofSize of A A (2 3☓(2 3☓ )) (3 4☓(3 4☓ )) Size of Size of BB

Size ofSize of ABAB((22 ☓ ☓ 44))

SameSame



Thus, Thus,

To see how to To see how to calculate the entriescalculate the entries of of C C consider consider entryentry cc1111::

11 12 13 14

21 22 23 24

c c c cC

c c c c

11 12 13 14

21 22 23 24

c c c cC

c c c c

11

11 11 12 13 21 11 11 12 21 13 31

31

[ ]

b

c a a a b a b a b a b

b

11

11 11 12 13 21 11 11 12 21 13 31

31

[ ]

b


b

11 12 13 1411 12 13

21 22 23 2421 22 23

31 32 33 34

b b b ba a a

A B b b b ba a a

b b b b

11 12 13 1411 12 13

21 22 23 2421 22 23

31 32 33 34

b b b ba a a

A B b b b ba a a

b b b b



Thus, Thus,

Now consider calculating the Now consider calculating the entryentry cc1212::

11 12 13 14

21 22 23 24

c c c cC

c c c c

11 12 13 14

21 22 23 24

c c c cC

c c c c

12

12 11 12 13 22 11 12 12 22 13 32

32

[ ]

b


b

12

12 11 12 13 22 11 12 12 22 13 32

32

[ ]

b


b

11 12 13 1411 12 13

21 22 23 2421 22 23

31 32 33 34

b b b ba a a

A B b b b ba a a

b b b b

11 12 13 1411 12 13

21 22 23 2421 22 23

31 32 33 34

b b b ba a a

A B b b b ba a a

b b b b



Thus, Thus,

Now consider calculating the Now consider calculating the entryentry cc2121::

11 12 13 14

21 22 23 24

c c c cC

c c c c

11 12 13 14

21 22 23 24

c c c cC

c c c c

11

21 21 22 23 21 21 11 22 21 23 31

31

[ ]

b


b

11

21 21 22 23 21 21 11 22 21 23 31

31

[ ]

b


b

11 12 13 1411 12 13

21 22 23 2421 22 23

31 32 33 34

b b b ba a a

A B b b b ba a a

b b b b

11 12 13 1411 12 13

21 22 23 2421 22 23

31 32 33 34

b b b ba a a

A B b b b ba a a

b b b b



Thus, Thus,

Other entries are computed in a Other entries are computed in a similar mannersimilar manner..

11 12 13 14

21 22 23 24

c c c cC

c c c c

11 12 13 14

21 22 23 24

c c c cC

c c c c

11 12 13 1411 12 13

21 22 23 2421 22 23

31 32 33 34

b b b ba a a

A B b b b ba a a

b b b b

11 12 13 1411 12 13

21 22 23 2421 22 23

31 32 33 34

b b b ba a a

A B b b b ba a a

b b b b

ExampleExample

LetLet

Compute Compute ABAB..

SolutionSolution Since the Since the number of columnsnumber of columns of of AA is is equalequal to the to the number number

of rowsof rows of of BB, the matrix product , the matrix product C = ABC = AB is definedis defined.. The The sizesize of of CC is is 2 2 ☓ ☓ 33..

1 3 33 1 4

4 1 21 2 3

2 4 1

A B

1 3 33 1 4

4 1 21 2 3

2 4 1

A B

Example 3, page 116

ExampleExample

LetLet


SolutionSolution Thus,Thus,

Calculate all entries for Calculate all entries for CC::

1 3 33 1 4

4 1 21 2 3

2 4 1

A B

1 3 33 1 4

4 1 21 2 3

2 4 1

A B

11 12 13

21 22 23

1 3 33 1 4

4 1 21 2 3

2 4 1

c c cC AB

c c c

11 12 13

21 22 23

1 3 33 1 4

4 1 21 2 3

2 4 1

c c cC AB

c c c

11

1

[3 1 4] 4 (3)(1) (1)(4) (4)(2) 15

2

c

11

1

[3 1 4] 4 (3)(1) (1)(4) (4)(2) 15

2

c

Example 3, page 116

ExampleExample

LetLet




1 3 33 1 4

4 1 21 2 3

2 4 1

A B

1 3 33 1 4

4 1 21 2 3

2 4 1

A B

12 13

21 22 23

1 3 3153 1 4

4 1 21 2 3

2 4 1

c cC AB

c c c

12 13

21 22 23

1 3 3153 1 4

4 1 21 2 3

2 4 1

c cC AB

c c c

12

3

[3 1 4] 1 (3)(3) (1)( 1) (4)(4) 24

4

c

12

3

[3 1 4] 1 (3)(3) (1)( 1) (4)(4) 24

4

c

Example 3, page 116

ExampleExample

LetLet




1 3 33 1 4

4 1 21 2 3

2 4 1

A B

1 3 33 1 4

4 1 21 2 3

2 4 1

A B

13

21 22 23

1 3 315 243 1 4

4 1 21 2 3

2 4 1

cC AB

c c c

13

21 22 23

1 3 315 243 1 4

4 1 21 2 3

2 4 1

cC AB

c c c

13

3

[3 1 4] 2 (3)( 3) (1)(2) (4)(1) 3

1

c

13

3

[3 1 4] 2 (3)( 3) (1)(2) (4)(1) 3

1

c

Example 3, page 116

ExampleExample

LetLet




1 3 33 1 4

4 1 21 2 3

2 4 1

A B

1 3 33 1 4

4 1 21 2 3

2 4 1

A B

21 22 23

1 3 315 24 33 1 4

4 1 21 2 3

2 4 1

C ABc c c

21 22 23

1 3 315 24 33 1 4

4 1 21 2 3

2 4 1

C ABc c c

21

1

[ 1 2 3] 4 ( 1)(1) (2)(4) (3)(2) 13

2

c

21

1

[ 1 2 3] 4 ( 1)(1) (2)(4) (3)(2) 13

2

c

Example 3, page 116

ExampleExample

LetLet




1 3 33 1 4

4 1 21 2 3

2 4 1

A B

1 3 33 1 4

4 1 21 2 3

2 4 1

A B

22 23

1 3 315 24 33 1 4

4 1 2131 2 3

2 4 1

C ABc c

22 23

1 3 315 24 33 1 4

4 1 2131 2 3

2 4 1

C ABc c

22

3

[ 1 2 3] 1 ( 1)(3) (2)( 1) (3)(4) 7

4

c

22

3

[ 1 2 3] 1 ( 1)(3) (2)( 1) (3)(4) 7

4

c

Example 3, page 116

ExampleExample

LetLet




1 3 33 1 4

4 1 21 2 3

2 4 1

A B

1 3 33 1 4

4 1 21 2 3

2 4 1

A B

23

1 3 315 24 33 1 4

4 1 213 71 2 3

2 4 1

C ABc

23

1 3 315 24 33 1 4

4 1 213 71 2 3

2 4 1

C ABc

23

3

[ 1 2 3] 2 ( 1)( 3) (2)(2) (3)(1) 10

1

c

23

3

[ 1 2 3] 2 ( 1)( 3) (2)(2) (3)(1) 10

1

c

Example 3, page 116

ExampleExample

LetLet



1 3 33 1 4

4 1 21 2 3

2 4 1

A B

1 3 33 1 4

4 1 21 2 3

2 4 1

A B

1 3 33 1 4 15 24 3

4 1 21 2 3 13 7 10

2 4 1

C AB

1 3 33 1 4 15 24 3

4 1 21 2 3 13 7 10

2 4 1

C AB

Example 3, page 116

Laws for Matrix MultiplicationLaws for Matrix Multiplication

If the If the productsproducts and and sumssums are defined for the are defined for the matrices matrices AA, , BB, and , and CC, then, then

1.1. ((ABAB))CC = = AA((BCBC)) Associative lawAssociative law

2.2. AA((BB + + CC) = ) = ABAB + + ACAC Distributive lawDistributive law

Identity MatrixIdentity Matrix

The The identity matrixidentity matrix of of sizesize nn is given by is given by

nn rowsrows

nn columnscolumns

1 0 0

0 1 0

0 0 1

nI

1 0 0

0 1 0

0 0 1

nI

Properties of the Identity MatrixProperties of the Identity Matrix

The The identity matrixidentity matrix has the has the propertiesproperties that that

✦ IIn n A A == A A for any for any nn ☓☓ rr matrix matrix AA..

✦ BIBInn == B B for any for any s s ☓☓ nn matrix matrix BB..

✦ In particular, if In particular, if AA is a is a square matrixsquare matrix of of size size nn, then , then

n nI A AI A n nI A AI A

ExampleExample LetLet

ThenThen

So, So, II33A A = = AIAI33 = = AA..

1 3 1

4 3 2

1 0 1

A

1 3 1

4 3 2

1 0 1

A

3

1 0 0 1 3 1 1 3 1

0 1 0 4 3 2 4 3 2

0 0 1 1 0 1 1 0 1

I A A

3

1 0 0 1 3 1 1 3 1

0 1 0 4 3 2 4 3 2

0 0 1 1 0 1 1 0 1

I A A

3

1 3 1 1 0 0 1 3 1

4 3 2 0 1 0 4 3 2

1 0 1 0 0 1 1 0 1

AI A

3

1 3 1 1 0 0 1 3 1

4 3 2 0 1 0 4 3 2

1 0 1 0 0 1 1 0 1

AI A

Example 5, page 118

Matrix RepresentationMatrix Representation

A system of linear equationsA system of linear equations can be expressed in the form can be expressed in the form of of an equation of matricesan equation of matrices. Consider the system . Consider the system

The The coefficientscoefficients on the left-hand side of the equation can on the left-hand side of the equation can be expressed as be expressed as matrixmatrix AA below, the below, the variablesvariables as as matrixmatrix XX, , and the and the constantsconstants on right-hand side of the equation as on right-hand side of the equation as matrixmatrix BB::

2 4 6

3 6 5 1

3 7 0

x y z

x y z

x y z

2 4 6

3 6 5 1

3 7 0

x y z

x y z

x y z

2 4 1 6

3 6 5 1

1 3 7 0

x

A X y B

z

2 4 1 6

3 6 5 1

1 3 7 0

x

A X y B

z



The The matrix representationmatrix representation of the system of linear of the system of linear equations is given by equations is given by AX AX == B B, or, or

2 4 6

3 6 5 1

3 7 0

x y z

x y z

x y z

2 4 6

3 6 5 1

3 7 0

x y z

x y z

x y z

2 4 1 6

3 6 5 1

1 3 7 0

x

y

z

2 4 1 6

3 6 5 1

1 3 7 0

x

y

z



To confirm this, we can To confirm this, we can multiply the two matricesmultiply the two matrices on the on the left-hand sideleft-hand side of the equation, obtaining of the equation, obtaining

which, by matrix equality, is easily seen to be which, by matrix equality, is easily seen to be equivalentequivalent to to the given the given system of linear equationssystem of linear equations..

2 4 6

3 6 5 1

3 7 0

x y z

x y z

x y z

2 4 6

3 6 5 1

3 7 0

x y z

x y z

x y z

2 4 6

3 6 5 1

3 7 0

x y z

x y z

x y z

2 4 6

3 6 5 1

3 7 0

x y z

x y z

x y z

2.62.6The Inverse of a Square MatrixThe Inverse of a Square Matrix

(3)(1) ( 1)(2) ( 1)( 1) 2

( 4)(1) (2)(2) (1)( 1) 1

( 1)(1) (0)(2) (1)( 1) 2

(3)(1) ( 1)(2) ( 1)( 1) 2

( 4)(1) (2)(2) (1)( 1) 1

( 1)(1) (0)(2) (1)( 1) 2

x

y

z

x

y

z

Inverse of a MatrixInverse of a Matrix

Let Let A A be a be a square matrixsquare matrix of size of size nn.. A A square matrixsquare matrix AA–1–1 of size of size nn such that such that

is called the is called the inverse ofinverse of AA.. Not every matrix has an inverse.Not every matrix has an inverse.

✦ A square matrix that A square matrix that hashas an inverse is an inverse is said to be said to be nonsingularnonsingular..

✦ A square matrix that A square matrix that does not havedoes not have an an inverse is said to be inverse is said to be singularsingular..

1 1nA A AA I 1 1nA A AA I

Example: Example: A Nonsingular MatrixA Nonsingular Matrix

The matrixThe matrix hashas a matrix a matrix

as its as its inverseinverse.. This can be demonstrated by This can be demonstrated by multiplying themmultiplying them::

1 2

3 4A

1 2

3 4A

1

3 12 2

2 1A

1

3 12 2

2 1A

1

3 12 2

2 11 2 1 0

3 4 0 1AA I

1

3 12 2

2 11 2 1 0

3 4 0 1AA I

1

3 12 2

2 1 1 2 1 0

3 4 0 1A A I

1

3 12 2

2 1 1 2 1 0

3 4 0 1A A I

Example:Example: A Singular Matrix A Singular Matrix

The matrixThe matrix does does notnot have have an an inverseinverse..

If If BB had an inverse given by had an inverse given by wherewhere

aa, , bb, , cc, and , and dd are some appropriate numbers, then are some appropriate numbers, then by by definitiondefinition of an of an inverseinverse we would have we would have BBBB–1–1 = = I I. .

That isThat is

implyingimplying that that 0 = 1,0 = 1, which is which is impossibleimpossible!!

0 1

0 0B

0 1

0 0B

0 1 1 0

0 0 0 1

1 0

0 0 0 1

a b

c d

c d

0 1 1 0

0 0 0 1

1 0

0 0 0 1

a b

c d

c d

1 a bB

c d

1 a bB

c d

Finding the Inverse of a Square MatrixFinding the Inverse of a Square Matrix

Given the Given the n n ☓☓ nn matrix matrix AA::

1.1. Adjoin the Adjoin the n n ☓☓ n n identity matrix identity matrix I I to obtain to obtain the the augmented matrixaugmented matrix [[AA || II ]]..

2.2. Use a sequence of Use a sequence of row operationsrow operations to to reducereduce [[AA || II ]] to the form to the form [[II || BB]] if possible. if possible.

Then the matrix Then the matrix B B is the is the inverseinverse of of AA..

ExampleExample

Find the inverse of the matrixFind the inverse of the matrix

SolutionSolution We form the We form the augmented matrixaugmented matrix

2 1 1

3 2 1

2 1 2

A

2 1 1

3 2 1

2 1 2

A

2 1 1 1 0 0

3 2 1 0 1 0

2 1 2 0 0 1

2 1 1 1 0 0

3 2 1 0 1 0

2 1 2 0 0 1

Example 1, page 130

ExampleExample


SolutionSolution And use the And use the Gauss-Jordan elimination methodGauss-Jordan elimination method to to reduce itreduce it

to the form to the form [[II || BB]]::

2 1 1

3 2 1

2 1 2

A

2 1 1

3 2 1

2 1 2

A

2 1 1 1 0 0

3 2 1 0 1 0

2 1 2 0 0 1

2 1 1 1 0 0

3 2 1 0 1 0

2 1 2 0 0 1

1 2R R1 2R R



Example 1, page 130

ExampleExample




2 1 1

3 2 1

2 1 2

A

2 1 1

3 2 1

2 1 2

A

1 1 0 1 1 0

3 2 1 0 1 0

2 1 2 0 0 1

1 1 0 1 1 0

3 2 1 0 1 0

2 1 2 0 0 1

1 2R R1 2R R1

2 3

3 1

3

2

R

R R

R R

1

2 3

3 1

3

2

R

R R

R R



Example 1, page 130

ExampleExample




2 1 1

3 2 1

2 1 2

A

2 1 1

3 2 1

2 1 2

A

1 1 0 1 1 0

0 1 1 3 2 0

0 1 2 2 2 1

1 1 0 1 1 0

0 1 1 3 2 0

0 1 2 2 2 1

1

2 3

3 1

3

2

R

R R

R R

1

2 3

3 1

3

2

R

R R

R R

1 2

2

3 2

R R

R

R R

1 2

2

3 2

R R

R

R R



Example 1, page 130

1 2

2

3 2

R R

R

R R

1 2

2

3 2

R R

R

R R

ExampleExample




2 1 1

3 2 1

2 1 2

A

2 1 1

3 2 1

2 1 2

A

1 0 1 2 1 0

0 1 1 3 2 0

0 0 1 1 0 1

1 0 1 2 1 0

0 1 1 3 2 0

0 0 1 1 0 1

1 3

2 3

R R

R R

1 3

2 3

R R

R R



Example 1, page 130

ExampleExample




2 1 1

3 2 1

2 1 2

A

2 1 1

3 2 1

2 1 2

A

1 0 0 3 1 1

0 1 0 4 2 1

0 0 1 1 0 1

1 0 0 3 1 1

0 1 0 4 2 1

0 0 1 1 0 1

1 3

2 3

R R

R R

1 3

2 3

R R

R R

IInn BB



Example 1, page 130

ExampleExample


SolutionSolution Thus, the Thus, the inverseinverse of of AA is the matrix is the matrix

2 1 1

3 2 1

2 1 2

A

2 1 1

3 2 1

2 1 2

A

3 1 1

4 2 1

1 0 1

3 1 1

4 2 1

1 0 1

1 A 1 A

Example 1, page 130

A Formula for the Inverse of a A Formula for the Inverse of a 2 2 ☓ ☓ 22 Matrix Matrix

LetLet

Suppose Suppose D D == ad – bc ad – bc is is notnot equal to equal to zerozero.. ThenThen AA–1–1 exists and is given byexists and is given by

a bA

c d

a bA

c d

1 1 d bA

c aD

1 1 d b

Ac aD

ExampleExample

Find the inverse of Find the inverse of

SolutionSolution We first We first identifyidentify aa, , bb, , cc, and , and dd as being as being 11, , 22, , 33, and , and 44

respectively.respectively. We then We then computecompute

DD = = ad – bc ad – bc = (1)(4) – (2)(3) = 4 – 6 = –= (1)(4) – (2)(3) = 4 – 6 = – 22

1 2

3 4A

1 2

3 4A

Example 3, page 132

ExampleExample


SolutionSolution Next, we Next, we substitutesubstitute the the valuesvalues 11, , 22, , 33, and , and 44 instead of instead of

aa, , bb, , cc, and , and dd, respectively, in the formula , respectively, in the formula matrixmatrix

to obtainto obtain the the matrixmatrix

1 2

3 4A

1 2

3 4A

4 2

3 1

4 2

3 1

d b

c a

d b

c a

Example 3, page 132

ExampleExample


SolutionSolution Finally, Finally, multiplyingmultiplying this matrix by this matrix by 1/1/DD, we obtain , we obtain

1 2

3 4A

1 2

3 4A

1

3 12 2

2 14 21 1

3 12

d bA

c aD

1

3 12 2

2 14 21 1

3 12

d bA

c aD

Example 3, page 132

Using Inverses to Solve Systems of EquationsUsing Inverses to Solve Systems of Equations

If If AXAX = = BB is a is a linear systemlinear system of of nn equationsequations in in nn unknownsunknowns and if and if AA–1–1 exists, then exists, then

XX = = AA–1–1BB

is the is the unique solutionunique solution of the system. of the system.

ExampleExample

Solve the system of linear equationsSolve the system of linear equations

SolutionSolution Write the system of equations in the form Write the system of equations in the form AXAX = = B B wherewhere

2 1

3 2 2

2 2 1

x y z

x y z

x y z

2 1

3 2 2

2 2 1

x y z

x y z

x y z

1

2

1

x

X y B

z

1

2

1

x

X y B

z

2 1 1

3 2 1

2 1 2

A

2 1 1

3 2 1

2 1 2

A

Example 4, page 133

ExampleExample


SolutionSolution Find the inverse matrix of Find the inverse matrix of AA::

2 1

3 2 2

2 2 1

x y z

x y z

x y z

2 1

3 2 2

2 2 1

x y z

x y z

x y z

1

2

1

x

X y B

z

1

2

1

x

X y B

z

2 1 1

3 2 1

2 1 2

A

2 1 1

3 2 1

2 1 2

A

1

3 1 1

4 2 1

1 0 1

A

1

3 1 1

4 2 1

1 0 1

A

Example 4, page 133

ExampleExample


SolutionSolution Finally, we write the matrix equation Finally, we write the matrix equation XX = = AA–1–1BB and multiply: and multiply:

2 1

3 2 2

2 2 1

x y z

x y z

x y z

2 1

3 2 2

2 2 1

x y z

x y z

x y z

x

y

z

x

y

z

3 1 1 1

4 2 1 2

1 0 1 1

3 1 1 1

4 2 1 2

1 0 1 1

Example 4, page 133

ExampleExample


SolutionSolution Finally, we write the matrix equation Finally, we write the matrix equation XX = = AA–1–1B B and multiply:and multiply:

Thus, the solution is Thus, the solution is xx = 2 = 2, , yy = –1 = –1, and , and zz = –2 = –2..

2 1

3 2 2

2 2 1

x y z

x y z

x y z

2 1

3 2 2

2 2 1

x y z

x y z

x y z

(3)(1) ( 1)(2) ( 1)( 1) 2

( 4)(1) (2)(2) (1)( 1) 1

( 1)(1) (0)(2) (1)( 1) 2

(3)(1) ( 1)(2) ( 1)( 1) 2

( 4)(1) (2)(2) (1)( 1) 1

( 1)(1) (0)(2) (1)( 1) 2

x

y

z

x

y

z

Example 4, page 133

2.72.7Leontief Input-Output ModelLeontief Input-Output Model

0.2 0.2 0.1

0.2 0.4 0.1

0.1 0.2 0.3

A M S

A

M

S

0.2 0.2 0.1

0.2 0.4 0.1

0.1 0.2 0.3

A M S

A

M

S

Output (amount produced)

Input (amount used in production)

Input-Output AnalysisInput-Output Analysis

One of the many important One of the many important applicationsapplications of of matrix theorymatrix theory to to the field of the field of economicseconomics is the study of the relationship between is the study of the relationship between industrial production industrial production andand consumer demand consumer demand..

At the heart of this analysis is the At the heart of this analysis is the Leontief input-output Leontief input-output model model pioneered by Wassily Leontief.pioneered by Wassily Leontief.

Consider an Consider an economyeconomy consisting of only consisting of only three sectorsthree sectors: : agriculture (agriculture (AA)),, manufacturing ( manufacturing (MM)),, andand service ( service (SS)). .

In general, part of the output of one sector is absorbed by In general, part of the output of one sector is absorbed by another sector through another sector through interindustry purchasesinterindustry purchases, with the , with the excess excess available to fulfill available to fulfill consumer demandsconsumer demands..


The The relationshiprelationship governing both governing both intraindustrialintraindustrial and and interindustrialinterindustrial sales and purchases is conveniently sales and purchases is conveniently represented by means of an represented by means of an input-output matrixinput-output matrix::

For example:For example: The amount of The amount of agricultural products agricultural products consumedconsumed in the in the

productionproduction of of $100$100 million worth of million worth of manufactured goodsmanufactured goods is is

(100)(0.2) 20 $20 , or million.(100)(0.2) 20 $20 , or million.

0.2 0.2 0.1

0.2 0.4 0.1

0.1 0.2 0.3

A M S

A

M

S

0.2 0.2 0.1

0.2 0.4 0.1

0.1 0.2 0.3

A M S

A

M

S

Output (amount produced)Output (amount produced)

Input Input (amount used (amount used in production)in production)



The The relationshiprelationship governing both governing both intraindustrialintraindustrial and and interindustrialinterindustrial sales and purchases is conveniently sales and purchases is conveniently represented by means of an represented by means of an input-output matrixinput-output matrix::

For example:For example: The amount of The amount of manufactured goodsmanufactured goods requiredrequired to produce to produce

1 1 unitunit of of all goods and servicesall goods and services in the economy is in the economy is

0.2 0.2 0.1

0.2 0.4 0.1

0.1 0.2 0.3

A M S

A

M

S

0.2 0.2 0.1

0.2 0.4 0.1

0.1 0.2 0.3

A M S

A

M

S


0.2 0.4 0.1 0.7 , or unit.0.2 0.4 0.1 0.7 , or unit.




Now suppose the Now suppose the total outputtotal output of of agricultureagriculture, , manufacturingmanufacturing, , and and servicesservices sectors in the economy are sectors in the economy are representedrepresented by the by the variables variables xx, , yy, and , and z z respectively.respectively.

The value of The value of agricultural agricultural products products consumedconsumed in the in the internal process of internal process of producingproducing this total output of this total output of various various goods and servicesgoods and services is is

0.2 0.2 0.1

0.2 0.4 0.1

0.1 0.2 0.3

A M S

A

M

S

0.2 0.2 0.1

0.2 0.4 0.1

0.1 0.2 0.3

A M S

A

M

S


0.2 0.2 0.1 units.x y z 0.2 0.2 0.1 units.x y z




The value of The value of manufacturing manufacturing products products consumedconsumed in the in the internal process of internal process of producingproducing this total output of this total output of various various goods and servicesgoods and services is is

0.2 0.2 0.1

0.2 0.4 0.1

0.1 0.2 0.3

A M S

A

M

S

0.2 0.2 0.1

0.2 0.4 0.1

0.1 0.2 0.3

A M S

A

M

S






The value of The value of services services consumedconsumed in the internal process of in the internal process of producingproducing this total output of this total output of various goods and servicesvarious goods and services is is

0.2 0.2 0.1

0.2 0.4 0.1

0.1 0.2 0.3

A M S

A

M

S

0.2 0.2 0.1

0.2 0.4 0.1

0.1 0.2 0.3

A M S

A

M

S





These results could also be obtained using These results could also be obtained using matrix matrix multiplicationmultiplication..

To see this, write the To see this, write the total outputtotal output of goods and services of goods and services xx, , yy, , and and zz as a as a 3 3 ☓ ☓ 11 matrix called the matrix called the gross production matrixgross production matrix::

Letting Letting AA denote the denote the input-output matrixinput-output matrix, we have, we have

x

X y

z

x

X y

z

0.2 0.2 0.1

0.2 0.4 0.1

0.1 0.2 0.3

A

0.2 0.2 0.1

0.2 0.4 0.1

0.1 0.2 0.3

A


Then, the productThen, the product

is a is a 3 3 ☓ ☓ 11 matrix whose entries represent the respective values matrix whose entries represent the respective values of the of the agricultural agricultural products, products, manufacturing manufacturing products, and products, and servicesservices consumedconsumed in the internal process of in the internal process of productionproduction..

The matrix The matrix AXAX is called the is called the internal consumption matrixinternal consumption matrix..

0.2 0.2 0.1

0.2 0.4 0.1

0.1 0.2 0.3

0.2 0.2 0.1

0.2 0.4 0.1

0.1 0.2 0.3

x

AX y

z

x y z

x y z

x y z

0.2 0.2 0.1

0.2 0.4 0.1

0.1 0.2 0.3

0.2 0.2 0.1

0.2 0.4 0.1

0.1 0.2 0.3

x

AX y

z

x y z

x y z

x y z


Since Since XX gives the gives the total productiontotal production of of goods and servicesgoods and services in the in the economy and economy and AXAX gives the amount of gives the amount of goods and servicesgoods and services consumedconsumed in the production of these goods and services, it in the production of these goods and services, it follows that follows that X X –– AX AX gives the gives the net outputnet output of of goods and servicesgoods and services that is exactly enough to satisfy that is exactly enough to satisfy consumer demandsconsumer demands..

Letting Letting DD represent represent consumer demandsconsumer demands, we obtain, we obtain

where where I I is the is the 3 3 ☓ ☓ 33 identity matrixidentity matrix..

( )

X AX D

I A X D

( )

X AX D

I A X D

Leontief Input-Output ModelLeontief Input-Output Model

In this model, the In this model, the matrix equationmatrix equation giving the giving the net outputnet output of of goods and services needed to satisfy consumer demand isgoods and services needed to satisfy consumer demand is

TotalTotal InternalInternal ConsumerConsumeroutputoutput consumptionconsumption demanddemand

XX –– AXAX == DDwherewhere X X is the is the gross output matrixgross output matrix, , AA is the is the input-output input-output matrixmatrix, and , and DD is the is the matrixmatrix representing representing consumer consumer demanddemand..

The The solutionsolution to this equation is to this equation is

XX = ( = (II – – AA))–1–1DD Assuming thatAssuming that ((II – – AA))–1–1

existsexists

which gives the amount of goods and services that must be which gives the amount of goods and services that must be produced to satisfy produced to satisfy consumer demandconsumer demand..

Applied Example:Applied Example: A Three-Sector Economy A Three-Sector Economy For the three-sector economy with input-output matrix For the three-sector economy with input-output matrix

Find the Find the gross outputgross output of goods and services of goods and services needed to needed to satisfysatisfy a a consumer demandconsumer demand of of $100$100 million worth of million worth of agricultural agricultural products, products, $80$80 million worth of million worth of manufacturedmanufactured products, and products, and $50$50 million worth of million worth of servicesservices..

0.2 0.2 0.1

0.2 0.4 0.1

0.1 0.2 0.3

0.2 0.2 0.1

0.2 0.4 0.1

0.1 0.2 0.3



SolutionSolution

0.2 0.2 0.1

0.2 0.4 0.1

0.1 0.2 0.3

0.2 0.2 0.1

0.2 0.4 0.1

0.1 0.2 0.3

x

X y

z

x

X y

z

100

80

50

D

100

80

50

D

We are required to determine We are required to determine the the gross production matrixgross production matrix::

The matrix representing the The matrix representing the consumer demandconsumer demand is: is:



SolutionSolution Next, we computeNext, we compute

0.2 0.2 0.1

0.2 0.4 0.1

0.1 0.2 0.3

0.2 0.2 0.1

0.2 0.4 0.1

0.1 0.2 0.3

1 0 0 0.2 0.2 0.1 0.8 0.2 0.1

0 1 0 0.2 0.4 0.1 0.2 0.6 0.1

0 0 1 0.1 0.2 0.3 0.1 0.2 0.7

I A

1 0 0 0.2 0.2 0.1 0.8 0.2 0.1

0 1 0 0.2 0.4 0.1 0.2 0.6 0.1

0 0 1 0.1 0.2 0.3 0.1 0.2 0.7

I A



SolutionSolution We now obtain the We now obtain the inverseinverse of matrix of matrix I I –– A A

0.2 0.2 0.1

0.2 0.4 0.1

0.1 0.2 0.3

0.2 0.2 0.1

0.2 0.4 0.1

0.1 0.2 0.3

1

1.43 0.57 0.29

( ) 0.54 1.96 0.36

0.36 0.64 1.57

I A

1

1.43 0.57 0.29

( ) 0.54 1.96 0.36

0.36 0.64 1.57

I A



SolutionSolution Finally, we use Finally, we use Leontief’s formulaLeontief’s formula to find: to find:

Thus, Thus, to fulfill consumer demandto fulfill consumer demand, , $203$203 million worth of million worth of agricultural productsagricultural products, , $229$229 million worth of million worth of manufactured manufactured productsproducts, and , and $166$166 million worth of million worth of servicesservices should be should be produced.produced.

0.2 0.2 0.1

0.2 0.4 0.1

0.1 0.2 0.3

0.2 0.2 0.1

0.2 0.4 0.1

0.1 0.2 0.3

1

1.43 0.57 0.29 100 203.1

( ) 0.54 1.96 0.36 80 228.8

0.36 0.64 1.57 50 165.7

X I A D

1

1.43 0.57 0.29 100 203.1

( ) 0.54 1.96 0.36 80 228.8

0.36 0.64 1.57 50 165.7

X I A D


End of End of Chapter Chapter

2 Systems of Linear Equations: ✦ An Introduction ✦ Unique Solutions ✦ Underdetermined and...

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