Objectives

� Find the multiplicative inverseof a square matrix.

� Use inverses to solve matrixequations.

� Encode and decodemessages.

Multiplicative Inverses of Matrices and Matrix Equations

In 1939, Britain’s secret servicehired top chess players, mathe-

maticians, and other masters oflogic to break the code used bythe Nazis in communicationsbetween headquarters and troops.The project, which employed over

10,000 people, broke the code less than a year later, providing the Allies with infor-mation about Nazi troop movements throughout World War II.

S e c t i o n 8.4

8 Mid-Chapter Check PointWhat You Know: We learned to use matrices to solvesystems of linear equations. Gaussian elimination requiredsimplifying the augmented matrix to one with 1s down themain diagonal and 0s below the 1s. Gauss-Jordan elimina-tion simplified the augmented matrix to one with 1s downthe main diagonal and 0s above and below each 1. Such amatrix, in reduced row-echelon form, did not require back-substitution to solve the system.We applied Gaussian elimi-nation to systems with no solution, as well as to representthe solution set for systems with infinitely many solutions,including nonsquare systems. We learned how to performoperations with matrices, including matrix addition, matrixsubtraction, scalar multiplication, and matrix multiplication.

In Exercises 1–5, use matrices to find the complete solution to eachsystem of equations, or show that none exists.

1. 2. c 2x + 4y + 5z = 2x + y + 2z = 1

3x + 5y + 7z = 4c x + 2y - 3z = -7

3x - y + 2z = 82x - y + z = 5

C h a p t e r

842 Chapter 8 Matrices and Determinants

Preview Exercises

Exercises 85–87 will help you prepare for the material covered inthe next section.

85. Multiply:

After performing the multiplication, describe what happensto the elements in the first matrix.

Ba11 a12

a21 a22R B1 0

0 1R .

86. Use Gauss-Jordan elimination to solve the system:

87. Multiply and write the linear system represented by thefollowing matrix multiplication:

Ca1 b1 c1

a2 b2 c2

a3 b3 c3

S Cx

y

z

S = Cd1

d2

d3

S .

c -x - y - z = 14x + 5y = 0

y - 3z = 0.

This 1941 RCA radiogram shows anencoded message from the Japanesegovernment.

3. 4. dw + x + y + z = 6w - x + 3y + z = -14w + 2x - 3z = 12

2w + 3x + 6y + z = 1

b x - 2y + 2z = -22x + 3y - z = 1

5.

In Exercises 6–10, perform the indicated matrix operations orsolve the matrix equation for given that and are definedas follows. If an operation is not defined, state the reason.

6. 7. 8.

9. 10. 2X - 3C = BA + C

A1BC2A1B + C22C -12 B

A = C 0 2-1 3

1 0S B = B 4 1

-6 -2R C = B -1 0

0 1R

CA, B,X

c 2x - 2y + 2z = 5x - y + z = 2

2x + y - z = 1

P-BLTZMC08_805-872-hr 21-11-2008 13:26 Page 842

Section 8.4 Multiplicative Inverses of Matrices and Matrix Equations 843

� Find the multiplicative inverse of a square matrix.

Messages must often be sent in such a way that the real meaning is hiddenfrom everyone but the sender and the recipient. In this section, we will look at therole that matrices and their inverses play in this process.

The Multiplicative Identity MatrixFor the real numbers, we know that 1 is the multiplicative identity because

Is there a similar property for matrix multiplication? That is, is there amatrix such that and The answer is yes. A square matrix with 1sdown the main diagonal from upper left to lower right and 0s elsewhere does not changethe elements in a matrix in products with that matrix. In the case of matrices,

The square matrix with 1s down the main diagonal from upper left tolower right and 0s elsewhere is called the multiplicative identity matrix of order This matrix is designated by For example,

and so on.

The Multiplicative Inverse of a MatrixThe multiplicative identity matrix, will help us to define a new concept: themultiplicative inverse of a matrix.To do so, let’s consider a similar concept, the multi-plicative inverse of a nonzero number, Recall that the multiplicative inverse of

is The multiplicative inverse has the following property:

We can define the multiplicative inverse of a square matrix in a similar manner.

a # 1a = 1 and 1

a# a = 1.

1a .a

a.

In ,

I2 = B1 00 1

R , I3 = C1 0 00 1 00 0 1

S ,

In .n.

n * n

The elements in the matrixdo not change.

The elements in the matrixdo not change.

B Ra11

a21

a12

a22B Ra11

a21

a12

a22B R1

0

0

1=

B Ra11

a21

a12

a22B R1

0

0

1and = .B Ra11

a21

a12

a22

2 * 2

IA = A?AI = AIa # 1 = 1 # a = a.

Definition of the Multiplicative Inverse of a Square MatrixLet be an matrix. If there exists an matrix (read:“ inverse”)such that

then is the multiplicative inverse of A.A-1AA-1

= In and A-1 A = In ,

AA-1n * nn * nA

We have seen that matrix multiplication is not commutative.Thus, to show thata matrix is the multiplicative inverse of the matrix find both and If is the multiplicative inverse of both products ( and ) will be the multiplicativeidentity matrix,

The Multiplicative Inverse of a Matrix

Show that is the multiplicative inverse of where

Solution To show that is the multiplicative inverse of we must find theproducts and If is the multiplicative inverse of then will be theABA,BBA.AB

A,B

A = B -1 32 -5

R and B = B5 32 1

R .

A,B

EXAMPLE 1

In .BAABA,

BBA.ABA,B

P-BLTZMC08_805-872-hr 21-11-2008 13:26 Page 843

844 Chapter 8 Matrices and Determinants

multiplicative identity matrix and will be the multiplicative identity matrix.Because and are matrices, Thus, we denote the multiplicativeidentity matrix as it is also a matrix. We must show that

• and

•

Let’s first show

Let’s now show

Both products give the multiplicative identity matrix. Thus, is the multiplicative

inverse of and we can designate as

Check Point 1 Show that is the multiplicative inverse of where

One method for finding the multiplicative inverse of a matrix is to begin bydenoting the elements in with variables. Using the equation we canfind a value for each element in the multiplicative inverse that is represented by avariable. Example 2 shows how this is done.

Finding the Multiplicative Inverse of a Matrix

Find the multiplicative inverse of

Solution Let us denote the multiplicative inverse by

Because is a matrix, we use the equation to find values forand

B 2w + y 2x + z

5w + 3y 5x + 3zR = B1 0

0 1R

B R =2

51

3B R1

00

1

A

B Rw

yx

z

A−1 I2

z.w, x, y,AA-1

= I22 * 2A

A-1= Bw x

y zR .

A = B2 15 3

R .

EXAMPLE 2

AA-1= In ,A-1

A

A = B2 11 1

R and B = B 1 -1-1 2

R .

A,B

A-1= B5 3

2 1R .BA

B

= B51-12 + 3122 5132 + 31-5221-12 + 1122 2132 + 11-52

R = B1 00 1

R BA = B5 3

2 1R B -1 3

2 -5R

BA = I2 .

= B -1152 + 3122 -1132 + 31122152 + 1-52122 2132 + 1-52112

R = B1 00 1

R AB = B -1 3

2 -5R B5 3

2 1R

AB = I2 .

BA = I2 = B1 00 1

R .

AB = I2 = B1 00 1

R2 * 2I2 ;

n = 2.2 * 2BABA

Use row-by-column matrix multiplication on the

left side of B2 15 3

R Bw xy z

R = B 1 00 1

R .

P-BLTZMC08_805-872-hr 21-11-2008 13:26 Page 844

Section 8.4 Multiplicative Inverses of Matrices and Matrix Equations 845

We now equate corresponding elements to obtain the following two systems oflinear equations:

Each of these systems can be solved using the addition method.

Add:

Use back-substitution.

Add:

Use back-substitution.

Using these values, we have

Check Point 2 Find the multiplicative inverse of

Only square matrices of order have multiplicative inverses, but not everysquare matrix possesses a multiplicative inverse. For example, suppose that you

apply the procedure of Example 2 to

Multiplying matrices on the left and equating corresponding elements results ininconsistent systems with no solutions. There are no values for and Thisshows that matrix does not have a multiplicative inverse.

A nonsquare matrix, one with a different number of rows than columns, cannothave a multiplicative inverse. If is an matrix and is an matrix

then the products and are of different orders.This means that theycould not be equal to each other, so that and could not both equal themultiplicative identity matrix.

If a square matrix has a multiplicative inverse, that inverse is unique. Thismeans that the square matrix has no more than one inverse. If a square matrix has amultiplicative inverse, it is said to be invertible or nonsingular. If a square matrix hasno multiplicative inverse, it is called singular.

BAABBAAB1n Z m2,

n * mBm * nA

Az.w, x, y,

B R =–6

–34

2B R1

00

1

This isA.

This representsA−1.

This is themultiplicative

identity matrix.

B Rw

yx

z.

A = B -6 4-3 2

R :

n * n

A = B5 72 3

R .

A-1= Bw x

y zR = B 3 -1

-5 2R .

z = 2

x = -1

- x = 1

5x + 3z = 1 No change " 5x + 3z = 1

-6x - 3z = 0 Multiply by -3. " 2x + z = 0

y = -5 w = 3

-w = -3

5w + 3y = 0 No change " 5w + 3y = 0

-6w - 3y = -3 Multiply by -3. " 2w + y = 1

e2w + y = 15w + 3y = 0

and e2x + z = 05x + 3z = 1.

TechnologyYou can use a graphing utility to findthe inverse of the matrix in Example 2.Enter the matrix and name it Thescreens show and Verify thatthis is correct by showing that

A-1.AA.

AA-1= I2 and A-1

A = I2 .

e

e

P-BLTZMC08_805-872-hr 21-11-2008 13:26 Page 845

846 Chapter 8 Matrices and Determinants

Study TipTo find the matrix that appears as thesecond factor for the inverse of

• Reverse and the numbersin the diagonal from upper leftto lower right.

• Negate and the numbers inthe other diagonal.

c,b

d,a

A = Ba b

c dR :

Using the Quick Method to Find a Multiplicative Inverse

Find the multiplicative inverse of

Solution

BA= R–1

3–2

4

a b

c d

A = B -1 -23 4

R .

EXAMPLE 3

Multiplicative Inverse of a Matrix

If then

The matrix is invertible if and only if If then does not have a multiplicative inverse.

Aad - bc = 0,ad - bc Z 0.A

A-1=

1ad - bc

B d -b

-c ad .A = Ba b

c dR ,

2 : 2

Perform the scalar multiplication by multiplying each element in the matrix by 1

2 .

Study TipWhen using the formula to find themultiplicative inverse, start by com-puting If the computedvalue is 0, there is no need to continue.The given matrix is singular—that is, itdoes not have a multiplicative inverse.

ad - bc.

This is the given matrix.We’ve designated theelementsand d.

a, b, c,

A-1=

1ad - bc

B d -b

-c aR This is the formula for

the inverse of Ba bc d

R .

=

11-12142 - 1-22132

B 4 -1-22-3 -1

R Apply the formula with

and d = 4.a = - 1, b = -2, c = 3,

Simplify.

= B 2 1-

32 -

12R

=

12

B 4 2-3 -1

R

The inverse of is

We can verify this result by showing that and

Check Point 3 Find the multiplicative inverse of

A = B 3 -2-1 1

R .

A-1 A = I2 .AA-1

= I2

A-1= B 2 1

- 32 -

12R .A = B -1 -2

3 4R

A Quick Method for Finding the Multiplicative Inverse of a MatrixThe same method used in Example 2 can be used to develop the general form of themultiplicative inverse of a matrix. The following rule enables us to calculatethe multiplicative inverse, if there is one:

2 * 2

2 : 2

P-BLTZMC08_805-872-hr 21-11-2008 13:26 Page 846

Section 8.4 Multiplicative Inverses of Matrices and Matrix Equations 847

Finding Multiplicative Inverses of Matrices with Greater Than 2To find the multiplicative inverse of a invertible matrix, we begin by denotingthe elements in the multiplicative inverse with variables. Here is an example:

We multiply the matrices on the left, using the row-by-column definition of matrixmultiplication.

We now equate corresponding entries to obtain the following three systems oflinear equations:

Notice that the variables on the left of the equal signs have the same coefficients ineach system.We can use Gauss-Jordan elimination to solve all three systems at once.Form an augmented matrix that contains the coefficients of the three systems to theleft of the vertical line and the constants for the systems to the right.

To solve all three systems using Gauss-Jordan elimination,we must obtain

to the left of the vertical line. Use matrix row operations, working one column at atime. Obtain 1 in the required position. Then obtain 0s in the other two positions.Using these operations, we obtain the matrix

This augmented matrix provides the solutions to the three systems of equations.They are given by

and and

C1 0 00 1 00 0 1

3 15-12-4S x1 = 15

y1 = -12z1 = -4

C1 0 00 1 00 0 1

3 15 4 -5-12 -3 4-4 -1 1

S .

C1 0 00 1 00 0 1

SCoefficients of the

three systemsConstants on the right ineach of the three systems

C 3 S–1

4

0

–1

5

1

–1

0

–3

0

0

1

0

1

0

1

0

0

c -x1 - y1 - z1 = 14x1 + 5y1 + 0z1 = 00x1 + y1 - 3z1 = 0

c -x2 - y2 - z2 = 04x2 + 5y2 + 0z2 = 10x2 + y2 - 3z2 = 0

c -x3 - y3 - z3 = 04x3 + 5y3 + 0z3 = 00x3 + y3 - 3z3 = 1.

C -x1 - y1 - z1 -x2 - y2 - z2 -x3 - y3 - z3

4x1 + 5y1 + 0z1 4x2 + 5y2 + 0z2 4x3 + 5y3 + 0z3

0x1 + 1y1 - 3z1 0x2 + 1y2 - 3z2 0x3 + 1y3 - 3z3

S = C1 0 00 1 00 0 1

S

This is the multiplicativeidentity matrix, I3.

This is matrix A whoseinverse we wish to find.

This represents A−1.

C S–1

4

0

–1

5

1

–1

0

–3

C= S1

0

0

0

1

0

0

0

1

C Sx1

y1

z1

x2

y2

z2

x3

y3

z3

.

3 * 3

nn : n

C1 0 00 1 00 0 1

3 -541S x3 = -5

y3 = 4z3 = 1.

C1 0 00 1 00 0 1

3 4-3-1S x2 = 4

y2 = -3z2 = -1

P-BLTZMC08_805-872-hr 21-11-2008 13:26 Page 847

848 Chapter 8 Matrices and Determinants

Using the nine values from the previous page, the inverse matrix is

Take a second look at the matrix obtained at the point where Gauss-Jordanelimination was completed.This matrix is shown, again, below. Notice that the matrix to the right of the vertical bar is the multiplicative inverse of Also noticethat the multiplicative identity matrix, is the matrix that appears to the left of thevertical bar.

The observations in the voice balloons and the procedures followed above give us ageneral method for finding the multiplicative inverse of an invertible matrix.

This is themultiplicativeidentity, I3.

This is themultiplicativeinverse of A.

C 3 S15

–12

–4

4

–3

–1

–5

4

1

0

0

1

0

1

0

1

0

0

.

I3

A.3 * 3

Cx1 x2 x3

y1 y2 y3

z1 z2 z3

S = C 15 4 -5-12 -3 4-4 -1 1

S .

Procedure for Finding the Multiplicative Inverse of an Invertible MatrixTo find for any matrix for which exists,

1. Form the augmented matrix where is the multiplicative identitymatrix of the same order as the given matrix

2. Perform row operations on to obtain a matrix of the form This is equivalent to using Gauss-Jordan elimination to change into theidentity matrix.

3. Matrix is

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

= In

A-1.B

A3In ƒ B4.3A ƒ In4

A.In3A ƒ In4,

A-1An * nA-1

Finding the Multiplicative Inverse of a Matrix

Find the multiplicative inverse of

SolutionStep 1 Form the augmented matrix

Step 2 Perform row operations on to obtain a matrix of the form Tothe left of the vertical dividing line, we want 1s down the diagonal from upper left tolower right and 0s elsewhere.

[I3 �B].[A �I3]

This ismatrix A.

This is I3, the multiplicative identitymatrix, with 1s down the main diagonal

and 0s elsewhere.

C 3 S1

0

–2

–1

–2

–3

1

1

0

0

0

1

0

1

0

1

0

0

[A �I3].

A = C 1 -1 10 -2 1

-2 -3 0S .

3 : 3EXAMPLE 4

Study TipBecause we have a quick method forfinding the multiplicative inverse of a

matrix, the procedure on theright is recommended for matrices oforder or greater when agraphing utility is not being used.

3 * 3

2 * 2

P-BLTZMC08_805-872-hr 21-11-2008 13:26 Page 848

Section 8.4 Multiplicative Inverses of Matrices and Matrix Equations 849

This is themultiplicativeidentity, I3.

This is themultiplicativeinverse of A.

C 3 S3

–2

–4

–3

2

5

1

–1

–2

0

0

1

0

1

0

1

0

0

Step 3 Matrix is The last matrix shown is in the form The multi-plicative identity matrix is on the left of the vertical bar. Matrix the multiplicativeinverse of is on the right. Thus, the multiplicative inverse of is

Step 4 Verify the result by showing that and Try confirmingthe result by multiplying and to obtain Do you obtain if you reverse theorder of the multiplication?

We have seen that not all square matrices have multiplicative inverses. Ifthe row operations in step 2 result in all zeros in a row or column to the left of thevertical line, the given matrix does not have a multiplicative inverse.

Check Point 4 Find the multiplicative inverse of

A = C 1 0 2-1 2 3

1 -1 0S .

I3I3 .A-1AA�1

A � I3.AA�1 � I3

A-1= C 3 -3 1

-2 2 -1-4 5 -2

S .

AA,B,3I3 ƒ B4.A�1.B

TechnologyWe can use a graphing utility to verifythe inverse matrix we found inExample 4. Enter the elements inmatrix and press to display A-1.�x-1�A

Summary: Finding Multiplicative Inverses for Invertible MatricesUse a graphing utility with matrix capabilities, or

a. If the matrix is The inverse of is

b. If the matrix is where Use the procedure on page 848.Form and use row transformations to obtain Then A-1

= B.3In ƒ B4.3A ƒ In4n 7 2:n * nA

A-1=

1ad - bc

B d -b

-c ad .

A = Ba b

c dR2 * 2:

Replace row 1 by -

12 R3 + R1 .

Replace row 2 by 12 R3 + R2 .

" .C1 0 1

2

0 1 - 12

0 0 1 3 1 -

12 0

0 - 12 0

-4 5 -2S

-2R3 "C1 0

12

0 1 - 12

0 0 - 12

3 1 - 12 0

0 - 12 0

2 - 52 1

S Replace row 1 by 1R2 + R1 .Replace row 3 by 5R2 + R3 .

"C1 -1 1

0 1 - 12

0 -5 2 3 1 0 0

0 - 12 0

2 0 1S

- 12 R2 "C1 -1 1

0 -2 10 -5 2

3 1 0 00 1 02 0 1

S Replace row 3by 2R1 + R3 .

"C 1 -1 1

0 -2 1-2 -3 0

3 1 0 00 1 00 0 1

S

TechnologyThe matrix

has no multiplicative inverse because

When we try to find the inverse witha graphing utility, an ERROR messageoccurs, indicating the matrix issingular.

= 12 - 12 = 0.ad - bc = 4 # 3 - 6 # 2

A = B4 62 3

R = Ba b

c dR

P-BLTZMC08_805-872-hr 21-11-2008 13:26 Page 849

850 Chapter 8 Matrices and Determinants

You can work with the matrix form of the system and obtain the form of the linearsystem on the left. To do so, perform the matrix multiplication on the left side of thematrix equation. Then equate the corresponding elements.

The matrix equation

is abbreviated as where is the coefficient matrix of the system, and and are matrices containing one column, called column matrices. The matrix iscalled the constant matrix.

Here is a specific example of a linear system and its matrix form:

BBXAAX = B,

T T ∂ T

A X = B

Ca1 b1 c1

a2 b2 c2

a3 b3 c3

S Cx

y

z

S = Cd1

d2

d3

S

Linear System Matrix Form of the System

c a1x + b1y + c1z = d1

a2x + b2y + c2z = d2

a3x + b3y + c3z = d3

The matrixcontains the

system’scoefficients.

The matrixcontains the

system’svariables.

The matrixcontains the

system’sconstants.

C Sa1

a2

a3

b1

b2

b3

c1

c2

c3

C Sd1

d2

d3

C =Sx

y

z

Linear System Matrix Form

c x - y + z = 2- 2y + z = 2

-2x - 3y =12

A, thecoefficent

matrix

Coefficients Constants

X B, theconstantmatrix

C S1

0

–2

–1

–2

–3

1

1

0

C S22C =

=

Sx

y

z12

The matrix equation can be solved using if it exists.

This is the matrix equation.

Multiply both sides by Because matrix multiplication is not commutative, put in the same left position on both sides.

The multiplicative inverse property tells us that

Because is the multiplicative identity,

We see that if then X = A-1 B.AX = B,

In X = X.In X = A-1 B

A-1 A = In .

InX = A-1 B

A-1A-1. A-1AX = A-1B

AX = B

A-1AX = B

Solving a System Using If has a unique solution, then To solve a linear system ofequations, multiply and to find X.BA-1

X = A-1 B.AX = B

A�1� Use inverses to solve matrixequations.

Solving Systems of Equations Using MultiplicativeInverses of Matrices

Matrix multiplication can be used to represent a system of linear equations.

P-BLTZMC08_805-872-hr 21-11-2008 13:26 Page 850

Section 8.4 Multiplicative Inverses of Matrices and Matrix Equations 851

Using the Inverse of a Matrix to Solve a System

Solve the system by using the inverse of the coefficient matrix:

Solution The linear system can be written as

The solution is given by Consequently, we must find We found theinverse of matrix in Example 4. Using this result,

Thus, and The solution set is E A12 , - 12 , 1 B F .z = 1.x =

12 , y = -

12 ,

X = A-1 B = C 3 -3 1

-2 2 -1-4 5 -2

S C2212

S = C 3 # 2 + 1-32 # 2 + 1 # 12

-2 # 2 + 2 # 2 + 1-12 # 12

-4 # 2 + 5 # 2 + 1-22 # 12

S = C 12

- 12

1S .

AA-1.X = A-1

B.

X BA

C S1

0

–2

–1

–2

–3

1

1

0

C S22 .C =Sx

y

z12

c x - y + z = 2- 2y + z = 2

-2x - 3y =12 .

A-1,

EXAMPLE 5

TechnologyWe can use a graphing utility to solvea linear system with a unique solutionby entering the elements in thecoefficient matrix, and the columnmatrix. Then find the product of and The screen below verifies oursolution in Example 5.

This verifies thatx = .5, or y = −.5, or −and z = 1.

12 1

2

,,

B.A-1

B,A,

Check Point 5 Solve the system by using the inverse of the coefficient matrixthat you found in Check Point 4:

Applications of Matrix Inverses to CodingA cryptogram is a message written so that no one other than the intended recipientcan understand it.To encode a message, we begin by assigning a number to each letterin the alphabet: and a For example,the numerical equivalent of the word MATH is 13, 1, 20, 8. The numerical equivalentof the message is then converted into a matrix. Finally, an invertible matrix can beused to convert the message into code.The multiplicative inverse of this matrix can beused to decode the message.

space = 0.A = 1, B = 2, C = 3, Á , Z = 26,

c x + 2z = 6-x + 2y + 3z = -5

x - y = 6.

A-1,

Encoding a Word or Message

1. Express the word or message numerically.

2. List the numbers in step 1 by columns and form a square matrix. If you do nothave enough numbers to form a square matrix, put zeros in any remainingspaces in the last column.

3. Select any square invertible matrix, called the coding matrix, the same size asthe matrix in step 2. Multiply the coding matrix by the square matrix thatexpresses 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 theencoded message.

� Encode and decode messages.

P-BLTZMC08_805-872-hr 21-11-2008 13:26 Page 851

852 Chapter 8 Matrices and Determinants

Encoding a Word

Use matrices to encode the word MATH.

SolutionStep 1 Express the word numerically. As shown previously, the numericalequivalent of MATH is 13, 1, 20, 8.

Step 2 List the numbers in step 1 by columns and form a square matrix. The matrix for the numerical equivalent of MATH, 13, 1, 20, 8, is

Step 3 Multiply the matrix in step 2 by a square invertible matrix. We will use

as the coding matrix.

Step 4 Use the numbers, by columns, from the coded matrix in step 3 to write theencoded message. The encoded message is 43, 92.

Check Point 6 Use the coding matrix in Example 6, to encode the

word BASE.

The inverse of a coding matrix can be used to decode a word or message thatwas encoded.

B -2 -33 4

R ,

-64,-29,

Codingmatrix

Codedmatrix

Numericalrepresentation of

MATH

B R–2

3–3

4B R13

120

8B R–2(13)-3(1)

3(13)+4(1)

–2(20)-3(8)

3(20)+4(8)=

B R–29

43

–64

92=

B -2 -33 4

RB13 20

1 8R .

2 * 2

EXAMPLE 6

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.

Decoding a Word

Decode 43, 92 from Example 6.

SolutionStep 1 Find the inverse of the coding matrix. The coding matrix in Example 6

was We use the formula for the multiplicative inverse of a matrix

to find the multiplicative inverse of this matrix. It is B 4 3-3 -2

R .

2 * 2B -2 -33 4

R .

-64,-29,

EXAMPLE 7

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Section 8.4 Multiplicative Inverses of Matrices and Matrix Equations 853

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

Step 3 Express the numbers, by columns, from the matrix in step 2 as letters. Thenumbers are 13, 1, 20, and 8. Using letters, the decoded message is MATH.

Check Point 7 Decode the word that you encoded in Check Point 6.

Decoding is simple for an authorized receiver who knows the coding matrix.Because any invertible matrix can be used for the coding matrix, decoding a cryptogramfor an unauthorized receiver who does not know this matrix is extremely difficult.

Multiplicative inverseof the coding matrix

Codedmatrix

B R4

–33

–2B R–29

43–64

92B R4(–29)+3(43)

–3(–29)-2(43)

4(–64)+3(92)

–3(–64)-2(92)=

B R13

1

20

8=

Exercise Set 8.4

Practice ExercisesIn Exercises 1–12, find the products and to determinewhether is the multiplicative inverse of

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11. A = D0 0 -2 1

-1 0 1 10 1 -1 01 0 0 -1

T , B = D1 2 0 30 1 1 10 1 0 11 2 0 2

T

A = C 0 2 03 3 22 5 1

S , B = C -3.5 -1 20.5 0 04.5 2 -3

S

A = C 1 2 31 3 41 4 3

S , B = C 72 -3 1

2

- 12 0 1

2

- 12 1 -

12

S

A = C -2 1 -1-5 2 -1

3 -1 1S , B = C 1 0 1

2 1 3-1 1 1

S

A = C 0 1 00 0 11 0 0

S , B = C0 0 11 0 00 1 0

SA = B4 5

2 3R , B = B 3

2 - 52

-1 2R

A = B -2 132 -

12R , B = B1 2

3 4R

A = B -2 41 -2

R , B = B 1 2-1 -2

RA = B -4 0

1 3R , B = B -2 4

0 1R

A = B -2 -1-1 1

R , B = B1 11 2

RA = B 4 -3

-5 4R , B = B4 3

5 4R

A.BBAAB 12.

In Exercises 13–18, use the fact that if then

to find the inverse of each matrix, if

possible. Check that and

13. 14.

15. 16.

17. 18.

In Exercises 19–28, find by forming and then using rowoperations to obtain where Check that and

19. 20.

21. 22.

23. 24.

25. 26.

27. 28. A = D2 0 0 10 1 0 00 0 -1 00 0 0 2

TA = D1 0 0 00 -1 0 00 0 3 01 0 0 1

T

A = C3 2 61 1 22 2 5

SA = C 5 0 22 2 1

-3 1 -1S

A = C2 4 -41 3 -42 4 -3

SA = C 2 2 -10 3 -1

-1 -2 1S

A = C1 -1 10 2 -12 3 0

SA = C 1 2 -1-2 0 1

1 -1 0S

A = C3 0 00 6 00 0 9

SA = C2 0 00 4 00 0 6

SA-1

A = I.AA-1

= IA-1= 3B4.3I ƒ B4,3A ƒ I4A-1

A = B 6 -3-2 1

RA = B 10 -2-5 1

RA = B2 -6

1 -2RA = B 3 -1

-4 2R

A = B0 34 -2

RA = B 2 3-1 2

RA-1

A = I2 .AA-1= I2

A-1=

1ad - bc

cd -b

-c ad

A = Ba b

c dR ,

A = D1 -2 1 00 1 -2 10 0 1 -20 0 0 1

T , B = D1 2 3 40 1 2 30 0 1 20 0 0 1

T

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854 Chapter 8 Matrices and Determinants

In Exercises 29–32, write each linear system as a matrix equation inthe form where is the coefficient matrix and is theconstant matrix.

29. 30.

31. 32.

In Exercises 33–36, write each matrix equation as a system of linearequations without matrices.

33.

34.

35.

36.

In Exercises 37–42,

a. Write each linear system as a matrix equation in theform

b. Solve the system using the inverse that is given for thecoefficient matrix.

37. c 2x + 6y + 6z = 82x + 7y + 6z = 102x + 7y + 7z = 9

AX = B.

C -1 0 10 -1 00 1 1

S Cx

y

z

S = C -424S

C2 0 -10 3 01 1 0

S Cx

y

z

S = C695S

B 3 0-3 1

R Bx

yR = B 6

-7R

B4 -72 -3

R Bx

yR = B -3

1R

c x + 4y - z = 3x + 3y - 2z = 5

2x + 7y - 5z = 12c x + 3y + 4z = -3

x + 2y + 3z = -2x + 4y + 3z = -6

b7x + 5y = 233x + 2y = 10

b6x + 5y = 135x + 4y = 10

BAAX = B,42.

The inverse of

Practice PlusIn Exercises 43–44, find and check.

43. 44.

In Exercises 45–46, if I is the multiplicative identity matrix oforder 2, find for the given matrix

45. 46.

In Exercises 47–48, find and What doyou observe?

47.

48.

49. Prove the following statement:

If

then

50. Prove the following statement:

If

then

(Hint: Use the method of Example 2 on page 844.)

Application ExercisesIn Exercises 51–52, use the coding matrix

to encode and then decode the given message.

51. HELP 52. LOVE

In Exercises 53–54, use the coding matrix

A-1= C 0 1 2

-1 1 20 -1 -3

S to write a cryptogram for each

A = C 1 -1 03 0 2

-1 0 -1S and its inverse

A = B 4 -1-3 1

R and its inverse A-1= B1 1

3 4R

A-1=

1ad - bc

B d -b

-c ad .

A = Ba b

c dR and ad - bc Z 0,

A-1= C 1

a 0 00 1

b 00 0 1

c

S .

A = Ca 0 00 b 00 0 c

S , a Z 0, b Z 0, c Z 0,

A = B2 -91 -4

R B = B9 57 4

RA = B2 1

3 1R B = B4 7

1 2R

B-1 A-1.1AB2-1, A-1

B-1,

B 7 -5-4 3

RB 8 -5-3 2

RA.1I - A2-1

A = Be2x-ex

e3x e2xRA = B ex e3x

-e3x e5xRA-1

D2 0 1 13 0 0 1

- 1 1 -2 14 - 1 1 0

T is D- 1 2 - 1 - 1-4 9 -5 -6

0 1 - 1 - 13 -5 3 3

T .

d2w + y + z = 63w + z = 9-w + x - 2y + z = 44w - x + y = 6

The inverse of

C2 6 62 7 62 7 7

S is C 72 0 -3

- 1 1 00 - 1 1

S .

38. c x + 2y + 5z = 22x + 3y + 8z = 3-x + y + 2z = 3

The inverse of

C 1 2 52 3 8

- 1 1 2S is C 2 - 1 - 1

12 -7 -2-5 3 1

S .

39. c x - y + z = 82y - z = -7

2x + 3y = 1

The inverse of

C 1 - 1 10 2 - 12 3 0

S is C 3 3 - 1-2 -2 1-4 -5 2

S .

40. c x - 6y + 3z = 112x - 7y + 3z = 144x - 12y + 5z = 25

The inverse of

C 1 -6 32 -7 34 - 12 5

S is C 1 -6 32 -7 34 - 12 5

S .

41.

The inverse of

D1 - 1 2 0

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

T is D0 0 - 1 - 11 4 1 31 2 1 2

0 - 1 0 - 1

T .

dw - x + 2y = -3

x - y + z = 4-w + x - y + 2z = 2

-x + y - 2z = -4

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Section 8.4 Multiplicative Inverses of Matrices and Matrix Equations 855

message. Check your result by decoding the cryptogram.

53. S E N D _ C A S H

19 5 14 4 0 3 1 19 8

54. S T A Y _ W E L L

19 20 1 25 0 23 5 12 12

Writing in Mathematics55. What is the multiplicative identity matrix?

56. If you are given two matrices, and explain how todetermine if is the multiplicative inverse of

57. Explain why a matrix that does not have the same number ofrows and columns cannot have a multiplicative inverse.

58. Explain how to find the multiplicative inverse for a invertible matrix.

59. Explain how to find the multiplicative inverse for a invertible matrix.

60. Explain how to write a linear system of three equations inthree variables as a matrix equation.

61. Explain how to solve the matrix equation

62. What is a cryptogram?

63. It’s January 1, and you’ve written down your major goal forthe year. You do not want those closest to you to see whatyou’ve written in case you do not accomplish your objective.Consequently, you decide to use a coding matrix to encodeyour goal. Explain how this can be accomplished.

64. A year has passed since Exercise 63. (Time flies whenyou’re solving exercises in algebra books.) It’s been aterrific year and so many wonderful things have happenedthat you can’t remember your goal from a year ago. Youconsult your personal journal and you find the encodedmessage and the coding matrix. How can you use these tofind your original goal?

Technology Exercises

In Exercises 65–70, use a graphing utility to find the multiplicativeinverse of each matrix. Check that the displayed inverse is correct.

65. 66.

67. 68.

69. 70. D1 2 0 00 0 1 01 3 0 14 0 0 2

TD7 -3 0 2

-2 1 0 -14 0 1 -2

-1 1 0 -1

T

C 1 1 -1-3 2 -1

3 -3 2SC -2 1 -1

-5 2 -13 -1 1

SB -4 1

6 -2RB 3 -1

-2 1R

AX = B.

3 * 3

2 * 2

A.BB,A

Use C 19 25 520 0 121 23 12

S .

Use C 19 4 15 0 19

14 3 8S .

In Exercises 71–76, write each system in the form Thensolve the system by entering and into your graphing utilityand computing

71. 72.

73. 74.

75.

76.

In Exercises 77–78, use a coding matrix of your choice. Use agraphing utility to find the multiplicative inverse of your codingmatrix.Write a cryptogram for each message. Check your result bydecoding the cryptogram. Use your graphing utility to perform allnecessary matrix multiplications.

77. A R R I V E D _ S A F E L Y

1 18 18 9 22 5 4 0 19 1 6 5 12 25

78. A R T _ E N R I C H E S

1 18 20 0 5 14 18 9 3 8 5 19

Critical Thinking ExercisesMake Sense? In Exercises 79–82, determine whether eachstatement makes sense or does not make sense, and explainyour reasoning.

79. I found the multiplicative inverse of a matrix.

80. I used Gauss-Jordan elimination to find the multiplicativeinverse of a matrix.

81. I used matrix multiplication to represent a system of linearequations.

82. I made an encoding error by selecting the wrong squareinvertible matrix.

In Exercises 83–88, determine whether each statement is true orfalse. If the statement is false, make the necessary change(s) toproduce a true statement.

83. All square matrices have inverses because there is aformula for finding these inverses.

84. Two invertible matrices can have a matrix sum that isnot invertible.

85. To solve the matrix equation for multiply andthe inverse of

86. assuming and are invertible.

87. assuming and areinvertible.

88.

89. Give an example of a matrix that is its own inverse.2 * 2

B 1 -3-1 3

R is an invertible matrix.

A + BA, B,1A + B2-1= A-1

+ B-1,

ABA, B,1AB2-1= A-1

B-1,

B.AX,AX = B

2 * 2

2 * 2

3 * 3

2 * 3

A

dw + x + y + z = 4w + 3x - 2y + 2z = 7

2w + 2x + y + z = 3w - x + 2y + 3z = 5

ey -3x + z = -3

w + y = -1x + z = 7

y + w - x + 4y = -8y + w + x + y + z = 8

c x - y = 16x + y + 20z = 14

y + 3z = 1c 3x - 2y + z = -2

4x - 5y + 3z = -92x - y + 5z = -5

c y + 2z = 0-x + y = 12x - y + z = -1

c x - y + z = -64x + 2y + z = 94x - 2y + z = -3

A-1 B.

BAAX = B.

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8.5

856 Chapter 8 Matrices and Determinants

Determinants and Cramer’s Rule

A s cyberspace absorbs more andmore of our work, play, shopping,

and socializing, where will it all end?Which activities will still be offline in2025?

Our technologically transformedlives can be traced back to the Englishinventor Charles Babbage (1792–1871).Babbage knew of a method for

solving linear systems called Cramer’s rule, in honor of the Swiss geometer GabrielCramer (1704–1752). Cramer’s rule was simple, but involved numerousmultiplications for large systems. Babbage designed a machine, called the“difference engine,” that consisted of toothed wheels on shafts for performingthese multiplications. Despite the fact that only one-seventh of the functions everworked, Babbage’s invention demonstrated how complex calculations could behandled mechanically. In 1944, scientists at IBM used the lessons of the differenceengine to create the world’s first computer.

Those who invented computers hoped to relegate the drudgery of repeatedcomputation to a machine. In this section, we look at a method for solving linearsystems that played a critical role in this process. The method uses real numbers,called determinants, that are associated with arrays of numbers. As with matrixmethods, solutions are obtained by writing down the coefficients and constants of alinear system and performing operations with them.

The Determinant of a Matrix

Associated with every square matrix is a real number, called its determinant. Thedeterminant for a square matrix is defined as follows:2 * 2

2 : 2

Objectives

� Evaluate a second-orderdeterminant.

� Solve a system of linearequations in two variablesusing Cramer’s rule.

� Evaluate a third-orderdeterminant.

� Solve a system of linearequations in three variablesusing Cramer’s rule.

� Use determinants to identifyinconsistent systems andsystems with dependentequations.

� Evaluate higher-orderdeterminants.

S e c t i o n

90. If find

91. Find values of for which the following matrix is not invertible:

Group Exercise92. Each person in the group should work with one partner.

Send a coded word or message to each other by giving yourpartner the coded matrix and the coding matrix that youselected. Once messages are sent, each person should decodethe message received.

B 1 a + 1a - 2 4

R .

a

1A-12-1.A = B3 5

2 4R ,

Preview ExercisesExercises 93–95 will help you prepare for the material covered inthe next section. Simplify the expression in each exercise.

93.

94.

95. 21-30 - 1-322 - 316 - 92 + 1-1211 - 152

21-52 - 11-42

51-52 - 61-42

21-52 - 1-32142

A portion of Charles Babbage’s unrealizedDifference Engine

� Evaluate a second-orderdeterminant.

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