Chapter 8

Matrices and Determinants

Matrix Solutions to Linear Systems

A matrix gives us a shortened way of writing a system of equations. The first step in solving a system of linear equations using matrices is to write the augmented matrix. An augmented matrix has a vertical bar separating the columns of the matrix into two groups. The coefficients of each variable are placed to the left of the vertical line, and the constants are placed to the right. If any variable is missing, its coefficient is 0. Here are two examples.

252311921131213

25=2z+3y+x19=2z+y+x31=2z+y+3x

41009310

-19-521

25=z19=3z+y31=5z–2y+x

Solving Linear Systems by Using Matrices

Example

2134

6503

2132

Write the augmented matrix for the following system:

2x-3y+z = -23x-5z = 64x-3y+z = 2

Solution:

Write the solution set for a system of equations represented by the matrix

41009310

-19-521

Solution The system represented by the matrix is

This system can be simplified as follows.

41009310

-19-521 1x + 2y – 5z = -190x + 1y + 3z = 90x + 0y + 1z = 4

x + 2y – 5z = -19 y + 3z = 9 z = 4

Equation 1Equation 2

Equation 3

Text Example

Solution

y + 3z = 9 Equation 2

y + 3(4) = 9 Substitute 4 for z.

y + 12 = 9 Multiply.

y = -3 Subtract 12 from both sides.

The value of z is known. We can find y by back-substitution.

With values for y and z, we can now use back-substitution to find x.

We see that x = 7, y = -3, and z = 4. The solution set for the system is {(7, -3, 4)}.

x + 2y – 5z = -19 Equation 1

x + 2(-3) – 5(4) = -19 Substitute -3 for y and 4 for z.

x – 6 – 20 = -19 Multiply.

x – 26 = -19 Add.

x = 7 Add 26 to both sides.

Text Example cont.

Matrix Row Operations• These row operations produce matrices that lead to systems with the

same solution set as the original system.1. Two rows of a matrix may be interchanged. This is the same as

interchanging two equations in the linear system.2. The elements in any row may be multiplied by a nonzero number.

This is the same as multiplying both sides of an equation by a nonzero number.

3. The elements in any row may be multiplied by a nonzero number, and these products may be added to the corresponding elements in any other row. This is the same as multiplying both sides of an equation by a nonzero number and then adding equations to eliminate a variable.

• Two matrices are row equivalent if one can be obtained from the other by a sequence of row operations.

Use the matrix

And perform each indicated row operation:a. R1 R2 b. 3R1 c. 2R2 + R3

-64-3-25-321

21-12183

Solution

a. The notation R1 R2 means to interchange the elements in row 1 and row 2. This results in the row-equivalent matrix.

-64-3-221-121835-321 This was row 2; now it’s row 1.

This was row 1; now it’s row 2.

Text Example

Solution

b. The notation 3R1 means to multiply each element in row 1 by 3. This results in the row-equivalent matrix.

-64-3-221-121833•53•(-3)3•23•1

-64-3-221-121835-321

= .

c. The notation 2R2 + R3 means to add 2 times the elements in row 2 to the corresponding elements in row 3. Replace the elements in row 3 by these sums. First, we find 2 times the elements in row 2:

2(1) or 2, 2(2) or 4, 2(-3) or –6, 2(5) or 10.Now we add these products to the corresponding elements in row 3. Although we use row 2 to find the products, row 2 does not change. It is the elements in row 3 that change, resulting in the row-equivalent matrix

4-2105-321

21-12183

-6+104-6 -3+4-2+221-121835-321

=

Text Example cont.

Solving Linear Systems Using Gaussian Elimination

• Write the augmented matrix for the system

• Use matrix row operations to simplify the matrix to one with 1s down the diagonal from upper left to lower right, and 0s below the 1s

• Write the system of linear equations corresponding to the matrix in step 2, and use back-substitution to find the system’s solutions

Use matrices to solve the system

25=2z+3y+x19=2z+y+x31=2z+y+3x

SolutionStep 1 Write the augmented matrix for the system.

Linear System Augmented Matrix

25=2z+3y+x19=2z+y+x31=2z+y+3x

252311921131213

Text Example

SolutionStep 2 Use matrix row operations to simplify the matrix to one with 1s down the diagonal from upper left to lower right, and 0s below the 1s. Our goal is to obtain a matrix of the form

f100ed10cba1

.

Our first step in achieving this goal is to get 1 in the top position of the first column.

252311921131213

To get 1 in this position, we interchange rows 1 and 2. (We could also interchange rows 1 and 3 to attain our goal.)

252313121319211 This was row 2; now it’s row 1.

This was row 1; now it’s row 2.

We want 1 in this position.

Text Example cont.

SolutionNow we want to get 0s below the 1 in the first column.

252313121319211

Now add these products to the corresponding numbers in row 2. Notice that although we use row 1 to find the products, row 1 does not change.

We want 0 in these positions.

Let’s first get a 0 where there is now a 3. If we multiply the top row of numbers by –3 and add these products to the second row of numbers, we will get 0 in this position. The top row of numbers multiplied by –3 gives

-3(1) or –3, -3(1) or –3, -3(2) or –6, -3(19) or –57.

2523131 + (-3)2 + (-3)1 + (-3)3 + (-3)

19211

25231-26-4-2019211

=

We want 0 in this position.

Text Example cont.

Solution

Now add these products to the corresponding numbers in row 3.

We are not yet done with the first column. If we multiply the top row of numbers by –1 and add these products to the third row of numbers, we will get 0 in this position. The top row of numbers multiplied by –1 gives

-1(1) or –1, -1(1) or –1, -1(2) or –2, -1(19) or –19.

25+(-19)2 + (-2)3 + (-1)1 + (-1)-26-4-2019211

6020-26-4-2019211

=

We want 1 in this position.

6020-26-4-2019211

We move on to the second column. We want 1 in the second row, second column.

Text Example cont.

SolutionTo get 1 in the desired position, we multiply –2 by its reciprocal, -1/2. Therefore, we multiply all the numbers in the second row by –1/2 to get

6020› (-26)› (-4)› (-2)› (0)

19211

60201321019211

=

We want 0 in this position.

Now add these products to the corresponding numbers in row 3.

We are not yet done with the second column. If we multiply the top row of numbers by –2 and add these products to the third row of numbers, we will get 0 in this position. The second row of numbers multiplied by –2 gives

-2(0) or 0, -2(1) or –2, -2(2) or –4, -2(13) or –26.

6+(-26)0 + (-4)2 + (-2)0 + 01321019211

-20-4001321019211

=

Text Example cont.

Solution

We want 1 in this position.

-20-4001321019211

We move on to the third column. We want 1 in the third row, third column.

To get 1 in the desired position, we multiply –4 by its reciprocal, -1/4. Therefore, we multiply all the numbers in the third row by –1/4 to get

-1/4(-20)-1/4(-4)-1/4(0)-1/4(0)1321019211

51001321019211

=

We now have the desired matrix with 1s down the diagonal and 0s below the 1s.

Step 3 Write the system of linear equations corresponding to the matrix in step 2, and use back-substitution to find the system’s solution. The system represented by the matrix in step 2 is

Text Example cont.

Solution

We immediately see that the value for z is 5. To find y, we back-substitute 5 for z in the second equation.

5=z13=2z+y19=2z+y+x

252311921131213

y + 2z = 13 Equation 2

y + 2(5) = 13 Substitute 5 for x.

y = 3 Solve for y.

Finally, back-substitute 3 for y and 5 for z in the first equation:

x + y + 2z = 19 Equation 1

x + 3 + 2(5) = 19 Substitute 3 for y and 5 for x.

x + 13 = 19 Multiply and add.

x = 6 Subtract 13 from both sides.

The solution set for the original system is {(6, 3, 5)}.

Text Example cont.

Solving Linear Systems Using Gauss-Jordan Elimination

1. Write the augmented matrix for the system.

2. Use matrix row operations to simplify the matrix to one with 1s down the diagonal from upper left to lower right, and 0s above and below the 1s.

3. Use the reduced row-echelon form of the matrix in step 2 to write the system’s solutions set. (Back-substitution is not necessary.)

Inconsistent and Dependent Systems

and Their Applications

Use Gaussian elimination to solve the system

12=7z+4y–3x5=6z+3y–2x2=2z–y–x

124-4356-322-2-11

12=4z+4y–3x5=6z+3y–2x2=2z–y–x

Solution

Step 1 Write the augmented matrix for the system.

Linear System Augmented Matrix

Text Example

SolutionStep 2 Attempt to simplify the matrix to one with 1s down the diagonal and 0s below the 1s. Notice that the augmented matrix already has a 1in the top position of the first column. Now we want 0s below the 1. To get the first 0, multiply row 1 by 2 and add these products to row 2. To get the second 0, multiply row 1 by 3 and add these products to row 3. Performing these operations, we obtain the following matrix.

610-10110-102-2-11We

want 1 in this positio

n.

Use the Previous matrix and:

Replace row 2 by -2R1 + R2.

Replace row 3 by -3R1 + R3.

Text Example cont.

SolutionStep 2 Moving on to the second column, we obtain 1 in the desired position by multiplying row 2 by 1.

Text Example cont.

1 1 2

1(0) 1( 1) 1(10)

0 1 10

2

1(1)

6

1 1 2

0 1 10

0 1 10

2

1

6

It is impossible to convert this last matrix to the desired form of 1s down the main diagonal. If we translate the last row back into equation form, we get

 0x 0y 0z 5,

which is false. Regardless of which values we select for x, y, and z, the last equation can never be a true statement. Consequently, the system has no solution. The solution set is , the empty set.

SolutionNow we want a 0 below the 1 in column 2. To get the 0, multiply row 2 by 1 and add these products to row 3. (Equivalently, add row 2 to row 3.) We obtain the following matrix.

1 1 2

0 1 10

0 0 0

2

1

5

Text Example cont.

Up to this point, we have encountered only square systems in which the number of equations is equal to the number of variables. In a nonsquare system, the number of variables differs from the number of equations.

Nonsquare Systems

ExampleUse Gaussian elimination to solve the following

system: 3x-y+z = 2

2x+3y-z = 7 z = t

Solution:

First, write the system in augmented matrix form:

t100

7132

2113

Example cont.

t100

7132

2113

• (1/3)R1

t100

7132

3/23/13/11

Example cont.

t100

7132

3/23/13/11

• -2R1 + R2

t100

3/173/53/110

3/23/13/11

Example cont.

t100

3/173/53/110

3/23/13/11

• 3/11 R2

t100

11/1711/510

3/23/13/11

Example cont.

x - (1/3)y + (1/3)z = 2/3y - (5/11)z = 17/11z = t

y - (5/11)t = 17/11

y =

11

517 t

x - (1/3)(17+5t)/11 + (1/3)t = 2/333x - 17 - 5t + 11t = 2233x + 6t = 3933x = 39 - 6t

11

213

33

639 ttx

11

213 tx

11

517 ty

tz

Use Gaussian elimination to solve the system

8=z+2y+x26=6z+7y+3x

Text Example

Solution We begin with the augmented matrix.

Because we now have 1s down the diagonal that begins with the upperleft

entry and a 0 below this 1, we translate the matrix back into equation form.

x 2y z 8 Equation 1

y 3z 2 Equation 2

3 7 6

1 2 1

26

8

1 2 1

3 7 6

8

26

1 2 1

0 1 3

8

2

We can let z equal any real number and use backsubstitution to express x and y in terms of z.

Equation 2 Equation 1

y 3z 2 x 2y z 8y 3z 2 x 2(3z 2) z 8

x 6z 4 z 8 x 5z 4 8

x 5z 4

Solution

Text Example cont.

With z = t, the ordered solution (x, y, z) enables us to express the system's solution set as

{(5t 4, 3t 2, t)}

where t is any real number.

Solution

Text Example cont.

Matrix Operations and Their

Applications

We have seen that an array of numbers, arranged in rows and columns and placed in brackets, is called a matrix. We can represent the matrix in two different ways.

• A capital letter, such as A, B, or C, can denote a matrix.• A lowercase letter enclosed in brackets, such as that shown below, can

denote a matrix.

A [aij]

A general element in matrix A is denoted by aij. This refers to the element in the ith row and jth column. For example, a32 is the

element of A located in the third row, second column.

A matrix of order m X n has m rows and n columns. If m n, a matrix has the same number of rows as columns and is called a square matrix.

Matrix A with

elements ai j

Notations for Matrices

Let

a. What is the order of A?b. If A [aij], identify a23 and a12.

Solution

a. The matrix has 2 rows and 3 columns, so it is of order 2 X 3.

b. The element a23 is in the second row and third column. Thus, a23

1/5. The element a12 is in the first row and second column, and

consequently a12 2.

A 3 2 0

4 5 1/ 5

Text Example

Definition of Equality of Matrices

• Two matrices A and B are equal if and only if they have the same order m X n and aij = bij for i = 1, 2, …, m and j = 1, 2, …, n

Properties of Matrix Addition• If A, B, and C are m X n matrices and 0 is

an m X n zero matrix, then the following properties are true.

1. A+B = B+A Commutative Property

2. (A+B)+C = A+(B+C) Associative Property

3. A + 0 = 0 + A Additive Identity

4. A+(-A) = (-A) + A = 0 Additive Inverse

Example

131

012

504

312• Add the following matrices

Solution:

635

300

)1)5()3(014

031)1()2(2

131

012

504

312

Definition of Scalar Multiplication

If A [aij] is a matrix of order m X n and c is

a scalar, then the matrix cA is the m X n matrix given by

cA [caij].

This matrix is obtained by multiplying each element of A by the real number c. We call cA a scalar multiple of A.

AfindAIf 320

31

Example

Example cont.

60

93

2*30*3

3*31*3

20

313

320

31AfindAIf

Solution:

Properties of Scalar Multiplication

• If A and B are m X n matrices, and c and d scalars, then the following properties are true.

1. (cd)A = c(dA) Associative Property

2. 1A = A Scalar Identity Property

3. c(A+B) = cA + cB Distributive Property

4. (c+d)A = cA + dA Distributive Property

Definition of Matrix Multiplication: 2X2 Matrices

dhcfdgce

bhafbgae

hg

fe

dc

baAB

510

402

123

713

201

Example

• Find the product:

5*74*11*31*70*12*30*72*13*3

5*24*01*11*20*02*10*22*03*1

510

402

123

713

201

Example cont.Solution:

3 0 9

11 13 34

Definition of Matrix Multiplication

The product of an m X n matrix, A, and an n X p matrix, B, is an m X p matrix, AB, whose elements are found as follows. The element in the ith row and jth column of AB is found by multiplying the each element in the ith row of A by the corresponding element in the jth column of B and adding the products.

Properties of Matrix Multiplication

• (AB)C = A(BC) Associative Property

• A(B+C) = AB+AC

• (A+B)C = AC + BC Distributive Properties

• c(AB) = (cA)B Associative Property of Scalar Multiplication

Multiplicative Inverses of Matrices

and Matrix Equations

Definition of the Multiplicative Inverse of a Square Matrix

Let A be an n X n matrix. If there exists an n X n matrix A-1 such that

AA-1 = In and -1 A = In

then A -1 is the multiplicative inverse of A.

Show that B is the multiplicative inverse of A, where

A = and B =-523-1

1235

Text Example

Solution To show that B is the multiplicative inverse of A, we must find the products AB and BA. If B is the multiplicative inverse of A, then AB will be the multiplicative identity matrix and BA will be the multiplicative identity matrix. Because A and B are 2 X 2 matrices, n 2. Thus, we denote the multiplicative identity matrix as I2; it is also a 2 X 2

matrix. We must show that

AB I2 1 0

0 1

BA I2 1 0

0 1

Show that B is the multiplicative inverse of A, where

A = and B =-523-1

1235

Text Example cont.

SolutionLet’s first show AB=I2

Both products give the multiplicative identity matrix. Thus, B is the multiplicative inverse of A and we can designate B as A-1

AB 1 3

2 5

5 3

2 1

1(5) 3(2) 1(3) 3(1)

2(5) ( 5)(2) 1(3) ( 5)(1)

1 0

0 1

Let’s now show BA=I2

BA 5 3

2 1

1 3

2 5

5( 1) 3(2) 5(3) 3( 5)

2( 1) 1(2) 2(3) 1( 5)

1 0

0 1

5 3

2 1

Example

41

23A

• Find the multiplicative inverse of A.

Solution

wz

yxALet 1

10

01

41

231

wz

yxAA

10

01

4141

2323

wyzx

wyzx

Example cont.

41

23A

• Find the multiplicative inverse of A.

Solution

14

04

023

123

10

01

4141

2323

wy

zx

wy

zx

wyzx

wyzx

Example cont.

41

23A

• Find the multiplicative inverse of A.

Solution

1404

023123

wyzx

wyzx

)14(3)04(3

023123

wyzx

wyzx

31230123

023123

wyzx

wyzx

Example cont.

41

23A

• Find the multiplicative inverse of A.

Solution

10

3

10

1

310110

31230123

023123

wz

wz

wyzx

wyzx

Example cont.

41

23A

• Find the multiplicative inverse of A.

Solution

10

2

10

4

140410

3

10

1

yx

wyzx

wz

Multiplicative Inverse of a 2x2 Matrix

ac

bd

bcadA

thendc

baAIf

1

,

1

The matrix A is invertible if and only if ad-bc0. Ifad-bc=0, then A does not have a multiplicative inverse.

Procedure for Finding the Multiplicative Inverse of an Invertible Matrix

• To find A-1 for any n X n matrix A for which A-1 exists:

1. Form the augmented matrix [A | I ], where I is the multiplicative identity matrix of the same order as the given matrix A.

2. Perform row transformations on [A | I ] to obtain a matrix of the form [I | B ]. This is equivalent to using Gauss-Jordan elimination to change A into the identity matrix.

3. Matrix B is A-1.

4. Verify the result by showing that AA-1 I and A-1A I.

Solving a System Using A-1

• If AXB has a unique solution, X A-1B. To solve a linear system of equations, multiply A-1 and B to find X.

Solve the system by using A-1, the inverse of the coefficient matrix

Solution The linear system can be written as

1/2=3y–-2x2=z+2y2=z+ y–x

0-3-21-201-11

zyx

1/222

=

The solution is given by X A-1B. Consequently, we must find A-1. Using the inverse of matrix A that we found previously,

Thus, x 1/2, y -1/2, and z 1. The solution set is {(1/2, -1/2, 1)}.

X A 1B 3 3 1

0 2 1

2 3 0

2

2

1/ 2

1/ 2

1/ 2

1

Text Example

Encoding a Word or Message1. Express the word or message numerically.2. List the numbers in step 1 by columns and form a square

matrix. If you do not have enough numbers to form a square matrix, put zeros in any remaining spaces in the last column.

3. Select any square invertible matrix, called the coding matrix, the same size as the matrix in step 2. Multiply the coding matrix by the square matrix that expresses the message numerically. The resulting matrix is the coded matrix.

4. Use the numbers, by columns, from the coded matrix in step 3 to write the encoded message.

Decoding a Word or Message That Was Encoded

1. Find the multiplicative inverse of the coding matrix.

2. Multiply the multiplicative inverse of the coding matrix and the coded matrix.

3. Express the numbers, by columns, from the matrix in step 2 as letters.

Determinants and Cramer’s Rule

Definition of the Determinant of a 2 x 2 Matrix

122122

11 bababa

ba

We also say that the value of the second-order determinant is a1b2 – a2b1.

The determinant of the matrix is denoted by And is defined by

a1 b1

a2 b2

a1 b1

a2 b2

a1 b1

a2 b2

Example

50

23

• Evaluate the determinant of:

15015

2*05*350

23

Solution:

Solving a Linear System in Two Variables Using Determinants

22

11

22

11

22

11

22

11

222

111

ba

ba

ca

ca

yand

ba

ba

bc

bc

x

then

cybxa

cybxa

• Cramer’s Rule

Example

534

823

yx

yx• Use Cramer’s rule to solve the system:

Solution:

34

23

54

83

34

23

35

28

yandx

Example cont.

534

823

yx

yx• Use Cramer’s rule to solve the system:

Solution:

117

17

89

3215

)2(*43*3

)8(*4)5(*3

217

34

89

1024

)2(*43*3

)5(*)2(3*8

y

x

Definition of a Third-Order Determinant

123123123321321321

333

222

111

bacacbcbabacacbcba

cba

cba

cba

Definition of a 3 X 3 Matrix

• A third-order determinant is defined by

a1 b1 c1

a2 b2 c2

a3 b3 c3

a1

b2 c2

b3 c3

a2

b1 c1

b3 c3

a3

b1 c1

b2 c2

Evaluating the Determinant of a 3 X 3 Matrix

1. Each of the three terms in the definition contains two factors - a numerical factor and a second-order determinant.

2. The numerical factor in each term is an element from the first column of the third-order determinant.

3. The minus sign precedes the second term.4. The second-order determinant that appears in

each term is obtained by crossing out the row and the column containing the numerical factor.

Evaluate:

2410-3-2059

SolutionNote that the last column has two 0s. We will expand the determinant about the elements in that column.

9 5 0

2 3 0

1 4 2

0 2 3

1 4 0

9 5

1 4 2

9 5

2 3

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

2( 27 10) 2( 17) 34

Text Example

Determinants: Inconsistent and Dependent-Systems

1. If D 0 and at least one of the determinants in the numerator is not 0, then the system is inconsistent. The solution set is Ø.

2. If D 0 and all the determinants in the numerators are 0, then the equations in the system are dependent.