Download Multiplicative Inverses of Matrices and Matrix Equations

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842 Chapter 8 Matrices and Determinants
86. Use Gauss-Jordan elimination to solve the system:
Preview Exercises
-x - y - z = 1
c 4x + 5y
= 0
y - 3z = 0.
Exercises 85–87 will help you prepare for the material covered in
the next section.
85. Multiply:
B
a11
a21
a12 1
RB
a22 0
0
R.
1
87. Multiply and write the linear system represented by the
following matrix multiplication:
After performing the multiplication, describe what happens
to the elements in the first matrix.
Chapter
What You Know: We learned to use matrices to solve
systems of linear equations. Gaussian elimination required
simplifying the augmented matrix to one with 1s down the
main diagonal and 0s below the 1s. Gauss-Jordan elimination simplified the augmented matrix to one with 1s down
the main diagonal and 0s above and below each 1. Such a
matrix, in reduced row-echelon form, did not require backsubstitution to solve the system. We applied Gaussian elimination to systems with no solution, as well as to represent
the solution set for systems with infinitely many solutions,
including nonsquare systems. We learned how to perform
operations with matrices, including matrix addition, matrix
subtraction, scalar multiplication, and matrix multiplication.
In Exercises 1–5, use matrices to find the complete solution to each
system of equations, or show that none exists.
x + 2y - 3z = - 7
1. c 3x - y + 2z = 8
2x - y + z = 5
2x + 4y + 5z = 2
2. c x + y + 2z = 1
3x + 5y + 7z = 4
8.4
Objectives
�
�
�
c1
x
d1
c2 S C y S = C d2 S.
c3
z
d3
Mid-Chapter Check Point
8
Section
b1
b2
b3
a1
C a2
a3
Find the multiplicative inverse
of a square matrix.
Use inverses to solve matrix
equations.
Encode and decode
messages.
3. b
x - 2y + 2z = - 2
2x + 3y - z = 1
4.
w
w
d
w
2w
+
+
+
x + y + z
x + 3y + z
2x
- 3z
3x + 6y + z
=
6
= - 14
= 12
=
1
2x - 2y + 2z = 5
5. c x - y + z = 2
2x + y - z = 1
In Exercises 6–10, perform the indicated matrix operations or
solve the matrix equation for X given that A, B, and C are defined
as follows. If an operation is not defined, state the reason.
0
A = C -1
1
6. 2C - 12 B
9. A + C
2
3S
0
B = B
4
-6
1
R
-2
7. A1B + C2
C = B
-1
0
0
R
1
8. A1BC2
10. 2X - 3C = B
Multiplicative Inverses of Matrices and Matrix Equations
This 1941 RCA radiogram shows an
encoded message from the Japanese
government.
I
n 1939, Britain’s secret service
hired top chess players, mathematicians, and other masters of
logic to break the code used by
the Nazis in communications
between headquarters and troops.
The project, which employed over
10,000 people, broke the code less than a year later, providing the Allies with information about Nazi troop movements throughout World War II.
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Page 843
Section 8.4 Multiplicative Inverses of Matrices and Matrix Equations
843
Messages must often be sent in such a way that the real meaning is hidden
from everyone but the sender and the recipient. In this section, we will look at the
role that matrices and their inverses play in this process.
The Multiplicative Identity Matrix
For the real numbers, we know that 1 is the multiplicative identity because
a # 1 = 1 # a = a. Is there a similar property for matrix multiplication? That is, is there a
matrix I such that AI = A and IA = A? The answer is yes. A square matrix with 1s
down the main diagonal from upper left to lower right and 0s elsewhere does not change
the elements in a matrix in products with that matrix. In the case of 2 * 2 matrices,
B
a11
a21
a12
1
R B
a22
0
0
a11
R=B
a21
1
a12
R
a22
and B
The elements in the matrix
do not change.
1
0
0
a
R B 11
1
a21
a12
a
R = B 11
a22
a21
a12
R.
a22
The elements in the matrix
do not change.
The n * n square matrix with 1s down the main diagonal from upper left to
lower right and 0s elsewhere is called the multiplicative identity matrix of order n.
This matrix is designated by In . For example,
1
I2 = B
0
0
R,
1
1
I3 = C 0
0
0
1
0
0
0 S,
1
and so on.
�
Find the multiplicative inverse
of a square matrix.
The Multiplicative Inverse of a Matrix
The multiplicative identity matrix, In , will help us to define a new concept: the
multiplicative inverse of a matrix. To do so, let’s consider a similar concept, the multiplicative inverse of a nonzero number, a. Recall that the multiplicative inverse of
a is a1 . The multiplicative inverse has the following property:
a # a1 = 1 and
1
a
#a
= 1.
We can define the multiplicative inverse of a square matrix in a similar manner.
Definition of the Multiplicative Inverse of a Square Matrix
Let A be an n * n matrix. If there exists an n * n matrix A-1 (read: “A inverse”)
such that
AA-1 = In and A-1 A = In ,
then A-1 is the multiplicative inverse of A.
We have seen that matrix multiplication is not commutative. Thus, to show that
a matrix B is the multiplicative inverse of the matrix A, find both AB and BA. If B
is the multiplicative inverse of A, both products (AB and BA) will be the multiplicative
identity matrix, In .
EXAMPLE 1
The Multiplicative Inverse of a Matrix
Show that B is the multiplicative inverse of A, where
A = B
-1
2
3
R
-5
and B = B
5
2
3
R.
1
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
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844 Chapter 8 Matrices and Determinants
multiplicative identity matrix and BA will be the multiplicative identity matrix.
Because A and B are 2 * 2 matrices, n = 2. Thus, we denote the multiplicative
identity matrix as I2 ; it is also a 2 * 2 matrix. We must show that
• AB = I2 = B
1
0
0
R and
1
• BA = I2 = B
1
0
0
R.
1
Let’s first show AB = I2 .
AB = B
= B
3 5
RB
-5 2
-1
2
3
R
1
-1152 + 3122
2152 + 1- 52122
- 1132 + 3112
1
R = B
2132 + 1-52112
0
0
R
1
Let’s now show BA = I2 .
BA = B
= B
5
2
3
-1
RB
1
2
3
R
-5
51-12 + 3122
21- 12 + 1122
5132 + 31-52
1
R = B
2132 + 11-52
0
0
R
1
Both products give the multiplicative identity matrix. Thus, B is the multiplicative
5 3
inverse of A and we can designate B as A-1 = B
R.
2 1
Check Point
1
Show that B is the multiplicative inverse of A, where
A = B
2
1
1
R
1
B = B
and
1
-1
-1
R.
2
One method for finding the multiplicative inverse of a matrix A is to begin by
denoting the elements in A-1 with variables. Using the equation AA-1 = In , we can
find a value for each element in the multiplicative inverse that is represented by a
variable. Example 2 shows how this is done.
EXAMPLE 2
Finding the Multiplicative Inverse of a Matrix
Find the multiplicative inverse of
A = B
2
5
1
R.
3
Solution Let us denote the multiplicative inverse by
w x
R.
y z
Because A is a 2 * 2 matrix, we use the equation AA-1 = I2 to find values for
w, x, y, and z.
A-1 = B
A−1
A
B
2w + y
B
5w + 3y
2
5
2x + z
1
R = B
5x + 3z
0
w
1
R B
y
3
0
R
1
I2
1
x
R=B
0
z
0
R
1
Use row-by-column matrix multiplication on the
2 1 w x
1 0
left side of B
RB
R = B
R.
5 3
y z
0
1
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Section 8.4 Multiplicative Inverses of Matrices and Matrix Equations
845
We now equate corresponding elements to obtain the following two systems of
linear equations:
e
Technology
You can use a graphing utility to find
the inverse of the matrix in Example 2.
Enter the matrix and name it A. The
screens show A and A-1. Verify that
this is correct by showing that
2w + y = 1
5w + 3y = 0
e
and
2x + z = 0
5x + 3z = 1.
Each of these systems can be solved using the addition method.
e
2w + y = 1
Multiply by - 3.
"
- 6w - 3y = - 3
5w + 3y = 0
No change
"
5w + 3y = 0
-w
= -3
w = 3
y = -5
Add:
AA-1 = I2 and A-1 A = I2 .
Use back-substitution.
e
2x + z = 0
Multiply by - 3.
"
-6x - 3z =
5x + 3z = 1
No change
"
5x + 3z = 1
- x
= 1
x = -1
z = 2
Add:
Use back-substitution.
0
Using these values, we have
A-1 = B
Check Point
2
w
y
x
3
R = B
z
-5
-1
R.
2
Find the multiplicative inverse of A = B
5
2
7
R.
3
Only square matrices of order n * n have multiplicative inverses, but not every
square matrix possesses a multiplicative inverse. For example, suppose that you
apply the procedure of Example 2 to A = B
This is
A.
B
–6
–3
This represents
A−1.
w
4
R B
y
2
-6
-3
4
R:
2
This is the
multiplicative
identity matrix.
1
x
R=B
0
z
0
R.
1
Multiplying matrices on the left and equating corresponding elements results in
inconsistent systems with no solutions. There are no values for w, x, y, and z. This
shows that matrix A does not have a multiplicative inverse.
A nonsquare matrix, one with a different number of rows than columns, cannot
have a multiplicative inverse. If A is an m * n matrix and B is an n * m matrix
1n Z m2, then the products AB and BA are of different orders. This means that they
could not be equal to each other, so that AB and BA could not both equal the
multiplicative identity matrix.
If a square matrix has a multiplicative inverse, that inverse is unique. This
means that the square matrix has no more than one inverse. If a square matrix has a
multiplicative inverse, it is said to be invertible or nonsingular. If a square matrix has
no multiplicative inverse, it is called singular.
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846 Chapter 8 Matrices and Determinants
A Quick Method for Finding the Multiplicative Inverse
of a 2 : 2 Matrix
The same method used in Example 2 can be used to develop the general form of the
multiplicative inverse of a 2 * 2 matrix. The following rule enables us to calculate
the multiplicative inverse, if there is one:
Study Tip
Multiplicative Inverse of a 2 : 2 Matrix
To find the matrix that appears as the
second factor for the inverse of
A = B
a
c
b
R:
d
• Reverse a and d, the numbers
in the diagonal from upper left
to lower right.
• Negate b and c, the numbers in
the other diagonal.
If A = B
a
c
-b
d.
a
b
1
d
R , then A-1 =
B
d
ad - bc -c
The matrix A is invertible if and only if ad - bc Z 0. If ad - bc = 0, then A
does not have a multiplicative inverse.
Using the Quick Method to Find a Multiplicative Inverse
EXAMPLE 3
Find the multiplicative inverse of
A = B
-1
3
-2
R.
4
Solution
Study Tip
a
When using the formula to find the
multiplicative inverse, start by computing ad - bc. If the computed
value is 0, there is no need to continue.
The given matrix is singular—that is, it
does not have a multiplicative inverse.
b
This is the given matrix.
We’ve designated the
elements a, b, c,
and d.
–1 –2
A=B
R
3
4
c
A-1 =
d
1
d
B
ad - bc -c
This is the formula for
a b
the inverse of B
R.
c d
-b
R
a
1
4
=
B
1-12142 - 1- 22132 -3
=
1
4
B
2 -3
= B
The inverse of A = B
2
- 32
-1
3
-1- 22
R
-1
2
R
-1
Apply the formula with
a = - 1, b = - 2, c = 3,
and d = 4.
Simplify.
Perform the scalar
multiplication by
multiplying each
element in the
matrix by 21 .
1
R
- 12
-2
2
R is A-1 = B 3
4
-2
1
R.
- 12
We can verify this result by showing that AA-1 = I2 and A-1 A = I2 .
Check Point
3
Find the multiplicative inverse of
A = B
3
-1
-2
R.
1
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847
Section 8.4 Multiplicative Inverses of Matrices and Matrix Equations
Finding Multiplicative Inverses of n : n Matrices
with n Greater Than 2
To find the multiplicative inverse of a 3 * 3 invertible matrix, we begin by denoting
the elements in the multiplicative inverse with variables. Here is an example:
–1
C 4
0
–1 –1
x1
5
0 S C y1
1 –3
z1
This is matrix A whose
inverse we wish to find.
x2
y2
z2
x3
1
y3 S = C 0
z3
0
This
represents A−1.
0
1
0
0
0S .
1
This is the multiplicative
identity matrix, I3.
We multiply the matrices on the left, using the row-by-column definition of matrix
multiplication.
- x1 - y1 - z1
C 4x1 + 5y1 + 0z1
0x1 + 1y1 - 3z1
-x2 - y2 - z2
4x2 + 5y2 + 0z2
0x2 + 1y2 - 3z2
-x3 - y3 - z3
1
4x3 + 5y3 + 0z3 S = C 0
0x3 + 1y3 - 3z3
0
0
1
0
0
0S
1
We now equate corresponding entries to obtain the following three systems of
linear equations:
-x1 - y1 - z1 = 1
c 4x1 + 5y1 + 0z1 = 0
0x1 + y1 - 3z1 = 0
-x2 - y2 - z2 = 0
c 4x2 + 5y2 + 0z2 = 1
0x2 + y2 - 3z2 = 0
-x3 - y3 - z3 = 0
c 4x3 + 5y3 + 0z3 = 0
0x3 + y3 - 3z3 = 1.
Notice that the variables on the left of the equal signs have the same coefficients in
each 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 the
left of the vertical line and the constants for the systems to the right.
–1
C 4
0
–1
5
1
Coefficients of the
three systems
–1 1
0 3 0
–3 0
0
1
0
0
0S
1
Constants on the right in
each of the three systems
1 0 0
To solve all three systems using Gauss-Jordan elimination, we must obtain C 0 1 0 S
0 0 1
to the left of the vertical line. Use matrix row operations, working one column at a
time. Obtain 1 in the required position. Then obtain 0s in the other two positions.
Using these operations, we obtain the matrix
1
C0
0
0
1
0
0
15
3
0 -12
1
-4
4
-3
-1
-5
4 S.
1
This augmented matrix provides the solutions to the three systems of equations.
They are given by
1
C0
0
1
and C 0
0
0
1
0
0
4
3
0 -3 S
1 -1
0
1
0
0
15
3
0 -12 S
1
-4
x2 = 4
y2 = - 3
z2 = - 1
x1 = 15
y1 = - 12
z1 = - 4
1
and C 0
0
0
1
0
0 -5
0 3 4S
1
1
x3 = - 5
y3 = 4
z3 = 1.
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Page 848
848 Chapter 8 Matrices and Determinants
Using the nine values from the previous page, the inverse matrix is
x1
C y1
z1
x2
y2
z2
15
x3
y3 S = C -12
z3
-4
4
-3
-1
-5
4 S.
1
Take a second look at the matrix obtained at the point where Gauss-Jordan
elimination was completed. This matrix is shown, again, below. Notice that the 3 * 3
matrix to the right of the vertical bar is the multiplicative inverse of A. Also notice
that the multiplicative identity matrix, I3 is the matrix that appears to the left of the
vertical bar.
1
C0
0
0
1
0
0
15
0 3 –12
1
–4
This is the
multiplicative
identity, I3.
4
–3
–1
–5
4S .
1
This is the
multiplicative
inverse of A.
The observations in the voice balloons and the procedures followed above give us a
general method for finding the multiplicative inverse of an invertible matrix.
Study Tip
Because we have a quick method for
finding the multiplicative inverse of a
2 * 2 matrix, the procedure on the
right is recommended for matrices of
order 3 * 3 or greater when a
graphing utility is not being used.
Procedure for Finding the Multiplicative Inverse of an Invertible Matrix
To find A-1 for any n * n matrix A for which A-1 exists,
1. Form the augmented matrix 3A ƒ In4, where In is the multiplicative identity
matrix of the same order as the given matrix A.
2. Perform row operations on 3A ƒ In4 to obtain a matrix of the form 3In ƒ B4.
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 = In and A-1 A = In .
EXAMPLE 4
Finding the Multiplicative Inverse of a 3 : 3 Matrix
Find the multiplicative inverse of
1
A = C 0
-2
-1
-2
-3
1
1 S.
0
Solution
Step 1 Form the augmented matrix [A 円I3].
1
C 0
–2
–1
–2
–3
This is
matrix A.
1 1
1 3 0
0 0
0
1
0
0
0S
1
This is I3, the multiplicative identity
matrix, with 1s down the main diagonal
and 0s elsewhere.
Step 2 Perform row operations on [A 円I3] to obtain a matrix of the form [I3 円B]. To
the left of the vertical dividing line, we want 1s down the diagonal from upper left to
lower right and 0s elsewhere.
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Section 8.4 Multiplicative Inverses of Matrices and Matrix Equations
1
C 0
-2
1
1
0
-1
-2
-3
1
3 0
0
1
C0
0
-1
1
-5
1 1
- 12 3 0
2 2
1
C0
0
0
1
0
1
2
1
2
1
3
0
1 -4
0
1
0
0
0S
1
0
-
1
2
0
- 12
- 12
5
Replace row 3
by 2R1 + R3 . "
1
C0
0
-1
-2
-5
1
1
2
1
3 0
2
0
1
0
0
0S
1
Replace row 1 by 1R2 + R1 .
Replace row 3 by 5R2 + R3 .
1
C0
0
0
1
0
1
2
1
2
1
2
1
3 0
2
- 12
- 12
- 52
0
0S
1
0
0S
-2
Replace row 1 by - 21 R3 + R1 .
Replace row 2 by 21 R3 + R2 .
1
C0
0
0
1
0
0
3
0 3 –2
1 –4
–3
2
5
1
–1 S
–2
"
" .
This is the
multiplicative
identity, I3.
Technology
We can use a graphing utility to verify
the inverse matrix we found in
Example 4. Enter the elements in
matrix A and press 冷x -1冷 to display A-1.
A
"
- 2R3 "
This is the
multiplicative
inverse of A.
3
= C -2
-4
-3
2
5
1
- 1 S.
-2
Step 4 Verify the result by showing that AAⴚ1 ⴝ I3 and Aⴚ1 A ⴝ I3. Try confirming
the result by multiplying A and A-1 to obtain I3 . Do you obtain I3 if you reverse the
order of the multiplication?
Technology
The matrix
4
2
- 21 R2
Step 3 Matrix B is Aⴚ1. The last matrix shown is in the form 3I3 ƒ B4. The multiplicative identity matrix is on the left of the vertical bar. Matrix B, the multiplicative
inverse of A, is on the right. Thus, the multiplicative inverse of A is
-1
A = B
-
0
0S
1
849
6
a
R = B
3
c
b
R
d
has no multiplicative inverse because
ad - bc = 4 # 3 - 6 # 2
= 12 - 12 = 0.
When we try to find the inverse with
a graphing utility, an ERROR message
occurs, indicating the matrix is
singular.
We have seen that not all square matrices have multiplicative inverses. If
the row operations in step 2 result in all zeros in a row or column to the left of the
vertical line, the given matrix does not have a multiplicative inverse.
Check Point
4
Find the multiplicative inverse of
1
A = C -1
1
0
2
-1
2
3 S.
0
Summary: Finding Multiplicative Inverses for Invertible Matrices
Use a graphing utility with matrix capabilities, or
a. If the matrix is 2 * 2: The inverse of A = B
A-1 =
d
1
B
ad - bc -c
a
c
b
R is
d
-b
d.
a
b. If the matrix A is n * n where n 7 2: Use the procedure on page 848.
Form 3A ƒ In4 and use row transformations to obtain 3In ƒ B4. Then A-1 = B.
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Page 850
850 Chapter 8 Matrices and Determinants
Solving Systems of Equations Using Multiplicative
Inverses of Matrices
Matrix multiplication can be used to represent a system of linear equations.
Linear System
Matrix Form of the System
a1x + b1y + c1z = d1
c a2x + b2y + c2z = d2
a3x + b3y + c3z = d3
a1
C a2
a3
b1
b2
b3
c1
x
d1
c2 S C y S= C d2 S
c3
z
d3
The matrix
contains the
system’s
coefficients.
The matrix
contains the
system’s
variables.
The matrix
contains the
system’s
constants.
You can work with the matrix form of the system and obtain the form of the linear
system on the left. To do so, perform the matrix multiplication on the left side of the
matrix equation. Then equate the corresponding elements.
The matrix equation
a1
C a2
a3
b1
b2
b3
c1
x
d1
c2 S C y S = C d2 S
z
c3
d3
T
A
T ∂
X =
T
B
is abbreviated as AX = B, where A is the coefficient matrix of the system, and X
and B are matrices containing one column, called column matrices. The matrix B is
called the constant matrix.
Here is a specific example of a linear system and its matrix form:
Linear System
c
Matrix Form
x - y + z = 2
- 2y + z = 2
- 2x - 3y
= 12
1
C 0
–2
Coefficients
–1
–2
–3
A, the
coefficent
matrix
1
x
2
1 S C y S= C 2 S
1
0
z
2
X
=
Constants
B, the
constant
matrix
The matrix equation AX = B can be solved using A-1 if it exists.
AX = B
A-1AX = A-1B
InX = A-1 B
X = A-1 B
This is the matrix equation.
Multiply both sides by A-1. Because matrix
multiplication is not commutative, put A-1
in the same left position on both sides.
The multiplicative inverse property tells
us that A-1 A = In .
Because In is the multiplicative identity, In X = X.
We see that if AX = B, then X = A B.
-1
�
Use inverses to solve matrix
equations.
Solving a System Using Aⴚ1
If AX = B has a unique solution, then X = A-1 B. To solve a linear system of
equations, multiply A-1 and B to find X.
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Section 8.4 Multiplicative Inverses of Matrices and Matrix Equations
EXAMPLE 5
851
Using the Inverse of a Matrix to Solve a System
Solve the system by using A-1, the inverse of the coefficient matrix:
c
Technology
Solution The linear system can be written as
We can use a graphing utility to solve
a linear system with a unique solution
by entering the elements in A, the
coefficient matrix, and B, the column
matrix. Then find the product of A-1
and B. The screen below verifies our
solution in Example 5.
This verifies that
x = .5, or 21 ,
y = −.5, or − 21 ,
and z = 1.
x - y + z = 2
- 2y + z = 2
-2x - 3y
= 12 .
1
C 0
–2
–1
–2
–3
A
1
x
2
1 S C y S = C 2S .
1
0
z
2
X
B
The solution is given by X = A-1 B. Consequently, we must find A-1. We found the
inverse of matrix A in Example 4. Using this result,
3
-1
X = A B = C -2
-4
-3
2
5
1
1
2
3 # 2 + 1-32 # 2 + 1 # 12
2
-1 S C 2 S = C -2 # 2 + 2 # 2 + 1 -12 # 12 S = C - 12 S.
1
-2
-4 # 2 + 5 # 2 + 1 -22 # 12
1
2
Thus, x = 12 , y = - 12 , and z = 1. The solution set is E A 12 , - 12 , 1 B F .
Check Point
5 Solve the system by using A-1, the inverse of the coefficient matrix
that you found in Check Point 4:
x
+ 2z = 6
c - x + 2y + 3z = - 5
x - y
= 6.
�
Encode and decode messages.
Applications of Matrix Inverses to Coding
A cryptogram is a message written so that no one other than the intended recipient
can understand it. To encode a message, we begin by assigning a number to each letter
in the alphabet: A = 1, B = 2, C = 3, Á , Z = 26, and a space = 0. For example,
the numerical equivalent of the word MATH is 13, 1, 20, 8. The numerical equivalent
of the message is then converted into a matrix. Finally, an invertible matrix can be
used to convert the message into code. The multiplicative inverse of this matrix can be
used to decode the message.
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 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.
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852 Chapter 8 Matrices and Determinants
Encoding a Word
EXAMPLE 6
Use matrices to encode the word MATH.
Solution
Step 1 Express the word numerically. As shown previously, the numerical
equivalent of MATH is 13, 1, 20, 8.
Step 2 List the numbers in step 1 by columns and form a square matrix. The 2 * 2
matrix for the numerical equivalent of MATH, 13, 1, 20, 8, is
B
13
1
20
R.
8
Step 3 Multiply the matrix in step 2 by a square invertible matrix. We will use
-2 - 3
B
R as the coding matrix.
3
4
B
–2
3
13
–3
R B
1
4
Coding
matrix
–2(13)-3(1)
20
R= B
3(13)+4(1)
8
Numerical
representation of
MATH
–29
=B
43
–2(20)-3(8)
R
3(20)+4(8)
–64
R
92
Coded
matrix
Step 4 Use the numbers, by columns, from the coded matrix in step 3 to write the
encoded message. The encoded message is -29, 43, -64, 92.
Check Point
6
Use the coding matrix in Example 6, B
-2
3
-3
R , to encode the
4
word BASE.
The inverse of a coding matrix can be used to decode a word or message that
was encoded.
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.
EXAMPLE 7
Decoding a Word
Decode -29, 43, -64, 92 from Example 6.
Solution
Step 1 Find the inverse of the coding matrix. The coding matrix in Example 6
-2 - 3
was B
R . We use the formula for the multiplicative inverse of a 2 * 2 matrix
3
4
4
3
to find the multiplicative inverse of this matrix. It is B
R.
-3 -2
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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.
B
4
–3
–29
3
R B
43
–2
Multiplicative inverse
of the coding matrix
4(–29)+3(43)
4(–64)+3(92)
–64
R= B
R
–3(–29)-2(43) –3(–64)-2(92)
92
Coded
matrix
=B
13 20
R
1 8
Step 3 Express the numbers, by columns, from the matrix in step 2 as letters. The
numbers 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 cryptogram
for an unauthorized receiver who does not know this matrix is extremely difficult.
Exercise Set 8.4
Practice Exercises
In Exercises 1–12, find the products AB and BA to determine
whether B is the multiplicative inverse of A.
4
-5
-3
R,
4
B = B
2. A = B
-2
-1
-1
R,
1
B = B
3. A = B
-4
1
0
R,
3
-2
1
4
R,
-2
B = B
5. A = B
-2
1
R,
- 12
B = B
6. A = B
4
2
1. A = B
4. A = B
3
2
0
7. A = C 0
1
5
R,
3
1
0
0
B = B
3
R
4
1
1
1
R
2
-2
0
4
R
1
1
-1
1
3
3
2
-1
7
3
2
4 S , B = C - 12
3
- 12
0
10. A = C 3
2
2
3
5
0
- 3.5
2 S , B = C 0.5
1
4.5
0
-1
11. A = D
0
1
0
0
1
0
-2
1
-1
0
4
3
T
2
1
b
R , then
d
-3
0
1
-1
0
2
1
1
0
1
T, B = D
0
0
1
-1
13. A = B
2
-1
3
R
2
14. A = B
0
4
3
R
-2
15. A = B
3
-4
-1
R
2
16. A = B
2
1
-6
R
-2
17. A = B
10
-5
-2
R
1
18. A = B
6
-2
-3
R
1
In Exercises 19–28, find A-1 by forming 3A ƒ I4 and then using row
operations to obtain 3I ƒ B4, where A-1 = 3B4. Check that AA-1 = I
and A-1 A = I.
1
0S
0
-1
1
-1 S, B = C 2
1
-1
2
3
4
a
c
3
2
1
0
1
d -b
c
d to find the inverse of each matrix, if
ad - bc -c
a
possible. Check that AA-1 = I2 and A-1 A = I2 .
- 52
R
2
1
9. A = C 1
1
2
1
0
0
A-1 =
2
R
-2
0
0
1
1
0
0
1
T, B = D
0
-2
0
1
1
-2
1
0
-2
1
0
0
In Exercises 13–18, use the fact that if A = B
2
R
4
0
0
1 S, B = C 1
0
0
1
2
-1
-2
8. A = C -5
3
B = B
4
5
1
0
12. A = D
0
0
0
1
1
2
19. A = C 0
0
1
3S
1
-
1
2
1
2S
1
2
2
0S
-3
2
1
1
2
0
1
0
0
3
1
T
1
2
0
4
0
0
0S
6
3
20. A = C 0
0
0
6
0
0
0S
9
1
21. A = C -2
1
2
0
-1
-1
1S
0
1
22. A = C 0
2
-1
2
3
2
23. A = C 0
-1
2
3
-2
-1
-1 S
1
2
24. A = C 1
2
4
3
4
-4
-4 S
-3
5
25. A = C 2
-3
0
2
1
2
1S
-1
3
26. A = C 1
2
2
1
2
6
2S
5
0
0
3
0
2
0
28. A = D
0
0
0
1
0
0
0
0
-1
0
1
0
27. A = D
0
1
0
-1
0
0
0
0
T
0
1
1
-1 S
0
1
0
T
0
2
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854 Chapter 8 Matrices and Determinants
In Exercises 29–32, write each linear system as a matrix equation in
the form AX = B, where A is the coefficient matrix and B is the
constant matrix.
29. b
6x + 5y = 13
5x + 4y = 10
30. b
x + 3y + 4z = - 3
31. c x + 2y + 3z = - 2
x + 4y + 3z = - 6
7x + 5y = 23
3x + 2y = 10
x + 4y - z = 3
32. c x + 3y - 2z = 5
2x + 7y - 5z = 12
In Exercises 33–36, write each matrix equation as a system of linear
equations without matrices.
4
33. B
2
34. B
-7 x
-3
RB R = B R
-3 y
1
2
35. C 0
1
-1
36. C 0
0
0
3
1
45. B
a. Write each linear system as a matrix equation in the
form AX = B.
b. Solve the system using the inverse that is given for the
coefficient matrix.
The inverse of
7
6
2
6 S is C -1
7
0
2x + 6y + 6z = 8
37. c 2x + 7y + 6z = 10
2x + 7y + 7z = 9
2
C2
2
x + 2y + 5z = 2
38. c 2x + 3y + 8z = 3
- x + y + 2z = 3
The inverse of
1 2 5
2
C 2 3 8 S is C 12
-1 1 2
-5
6
7
7
The inverse of
1 -1
1
3
C0
2 -1 S is C -2
2
3
0
-4
The inverse of
1
-6 3
1
x - 6y + 3z = 11
- 7 3 S is C 2
40. c 2x - 7y + 3z = 14 C 2
4
4x - 12y + 5z = 25 4 - 12 5
w - x
x
41. d
-w + x
-x
The inverse of
1 -1
2
0
1 -1
D
-1
1 -1
0 -1
1
+
+
-1
-6
T.
-1
3
ex
-e3x
e3x
R
e5x
44. A = B
e2x
e3x
- ex
R
e2x
0
4
2
-1
8
-3
-5
R
2
46. B
7
-4
-5
R
3
0
1
-1
-3
0 S.
1
-1
-7
3
-1
- 2 S.
1
47. A = B
2
3
1
R
1
48. A = B
2
1
-9
R
-4
-6
-7
- 12
-1
1 S.
2
3
3 S.
5
4
1
B = B
7
R
2
9
7
5
R
4
49. Prove the following statement:
a
If A = C 0
0
then A
-1
0
b
0
0
0 S, a Z 0, b Z 0, c Z 0,
c
1
a
0
= C0
0
0
1
b
0
0 S.
1
c
50. Prove the following statement:
If A = B
a
c
b
R and ad - bc Z 0,
d
1
d
B
ad - bc -c
then A-1 =
3
-2
-5
B = B
-b
d.
a
(Hint: Use the method of Example 2 on page 844.)
Application Exercises
In Exercises 51–52, use the coding matrix
A = B
4
-3
-1
1
R and its inverse A-1 = B
1
3
1
R
4
to encode and then decode the given message.
2y
= -3
y + z = 4
y + 2z = 2
y - 2z = - 4
0
0
1
1
T is D
2
1
-2
0
-1
-5
-1
3
In Exercises 47–48, find 1AB2-1, A-1 B-1, and B-1 A-1. What do
you observe?
1
x
-4
0S CyS = C 2S
1
z
4
x - y + z = 8
2y - z = - 7
2x + 3y
= 1
2
9
1
-5
In Exercises 45–46, if I is the multiplicative identity matrix of
order 2, find 1I - A2-1 for the given matrix A.
In Exercises 37–42,
39. c
1
-1
1
-4
T is D
1
0
0
3
Practice Plus
43. A = B
-1
x
6
0S CyS = C9S
0
z
5
0
-1
1
The inverse of
2
0
1
3
0
0
D
-1
1 -2
4 -1
1
In Exercises 43–44, find A-1 and check.
0 x
6
RB R = B R
1 y
-7
3
-3
2w
+ y + z = 6
3w
+ z = 9
42. d
-w + x - 2y + z = 4
4w - x + y
= 6
51. HELP
52. LOVE
In Exercises 53–54, use the coding matrix
-1
1
1
0
-1
3
T.
2
-1
1
A = C 3
-1
0
A-1 = C - 1
0
-1
0
0
1
1
-1
0
2 S and its inverse
-1
2
2 S to write a cryptogram for each
-3
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Page 855
Section 8.4 Multiplicative Inverses of Matrices and Matrix Equations
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
19
Use C 5
14
4
0
3
1
19 S .
8
54. S
T
A
Y
_
W
E
L
L
19 20
1
25
0
23
5
12
12
19
Use C 20
1
25
0
23
5
12 S .
12
855
In Exercises 71–76, write each system in the form AX = B. Then
solve the system by entering A and B into your graphing utility
and computing A-1 B.
x - y + z = -6
71. c 4x + 2y + z = 9
4x - 2y + z = -3
y + 2z = 0
= 1
72. c - x + y
2x - y + z = - 1
3x - 2y + z = - 2
73. c 4x - 5y + 3z = - 9
2x - y + 5z = - 5
x - y
= 1
74. c 6x + y + 20z = 14
y + 3z = 1
+ z = -3
= -1
x
+ z = 7
75. e
y + w - x + 4y
= -8
y + w + x + y + z = 8
- 3x
y
w
Writing in Mathematics
55. What is the multiplicative identity matrix?
56. If you are given two matrices, A and B, explain how to
determine if B is the multiplicative inverse of A.
57. Explain why a matrix that does not have the same number of
rows and columns cannot have a multiplicative inverse.
58. Explain how to find the multiplicative inverse for a 2 * 2
invertible matrix.
59. Explain how to find the multiplicative inverse for a 3 * 3
invertible matrix.
60. Explain how to write a linear system of three equations in
three variables as a matrix equation.
61. Explain how to solve the matrix equation AX = B.
w
w
76. d
2w
w
+
+
+
-
x
3x
2x
x
+ y
+
+
+
y
2y
y
2y
+
+
+
+
z
2z
z
3z
=
=
=
=
4
7
3
5
In Exercises 77–78, use a coding matrix A of your choice. Use a
graphing utility to find the multiplicative inverse of your coding
matrix. Write a cryptogram for each message. Check your result by
decoding the cryptogram. Use your graphing utility to perform all
necessary matrix multiplications.
77. A R R I
V
E
D _
1 18 18 9 22
5
4
R I
78. A R
S A F E L
0 19 1
T _
E
N
1 18 20 0
5
14 18 9
6
C H E
3
8
Y
5 12 25
S
5 19
62. What is a cryptogram?
63. It’s January 1, and you’ve written down your major goal for
the year. You do not want those closest to you to see what
you’ve written in case you do not accomplish your objective.
Consequently, you decide to use a coding matrix to encode
your goal. Explain how this can be accomplished.
64. A year has passed since Exercise 63. (Time flies when
you’re solving exercises in algebra books.) It’s been a
terrific year and so many wonderful things have happened
that you can’t remember your goal from a year ago. You
consult your personal journal and you find the encoded
message and the coding matrix. How can you use these to
find your original goal?
Technology Exercises
In Exercises 65–70, use a graphing utility to find the multiplicative
inverse of each matrix. Check that the displayed inverse is correct.
65. B
3
-2
-1
R
1
66. B
-2
67. C - 5
3
1
2
-1
-1
-1 S
1
1
68. C -3
3
7
-2
69. D
4
-1
-3
1
0
1
0
0
1
0
1
0
70. D
1
4
2
-1
T
-2
-1
1
R
-2
-4
6
1
2
-3
2
0
3
0
0
1
0
0
Critical Thinking Exercises
Make Sense? In Exercises 79–82, determine whether each
statement makes sense or does not make sense, and explain
your reasoning.
79. I found the multiplicative inverse of a 2 * 3 matrix.
80. I used Gauss-Jordan elimination to find the multiplicative
inverse of a 3 * 3 matrix.
81. I used matrix multiplication to represent a system of linear
equations.
82. I made an encoding error by selecting the wrong square
invertible matrix.
In Exercises 83–88, determine whether each statement is true or
false. If the statement is false, make the necessary change(s) to
produce a true statement.
83. All square 2 * 2 matrices have inverses because there is a
formula for finding these inverses.
84. Two 2 * 2 invertible matrices can have a matrix sum that is
not invertible.
-1
-1 S
2
0
0
T
1
2
85. To solve the matrix equation AX = B for X, multiply A and
the inverse of B.
86. 1AB2-1 = A-1 B-1, assuming A, B, and AB are invertible.
87. 1A + B2-1 = A-1 + B-1, assuming A, B, and A + B are
invertible.
88. B
1
-1
-3
R is an invertible matrix.
3
89. Give an example of a 2 * 2 matrix that is its own inverse.
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856 Chapter 8 Matrices and Determinants
90. If A = B
3
2
Preview Exercises
5
-1
R , find 1A-12 .
4
91. Find values of a for which the following matrix is not invertible:
B
1
a - 2
a + 1
R.
4
Exercises 93–95 will help you prepare for the material covered in
the next section. Simplify the expression in each exercise.
93. 21-52 - 1-32142
94.
Group Exercise
92. Each person in the group should work with one partner.
Send a coded word or message to each other by giving your
partner the coded matrix and the coding matrix that you
selected. Once messages are sent, each person should decode
the message received.
Section
8.5
21- 52 - 11- 42
51- 52 - 61- 42
95. 21- 30 - 1- 322 - 316 - 92 + 1 - 1211 - 152
Determinants and Cramer’s Rule
Objectives
A portion of Charles Babbage’s unrealized
Difference Engine
� Evaluate a second-order
�
�
�
�
�
�
determinant.
Solve a system of linear
equations in two variables
using Cramer’s rule.
Evaluate a third-order
determinant.
Solve a system of linear
equations in three variables
using Cramer’s rule.
Use determinants to identify
inconsistent systems and
systems with dependent
equations.
Evaluate higher-order
determinants.
Evaluate a second-order
determinant.
s cyberspace absorbs more and
more of our work, play, shopping,
and socializing, where will it all end?
Which activities will still be offline in
2025?
Our technologically transformed
lives can be traced back to the English
inventor Charles Babbage (1792–1871).
Babbage knew of a method for
solving linear systems called Cramer’s rule, in honor of the Swiss geometer Gabriel
Cramer (1704–1752). Cramer’s rule was simple, but involved numerous
multiplications for large systems. Babbage designed a machine, called the
“difference engine,” that consisted of toothed wheels on shafts for performing
these multiplications. Despite the fact that only one-seventh of the functions ever
worked, Babbage’s invention demonstrated how complex calculations could be
handled mechanically. In 1944, scientists at IBM used the lessons of the difference
engine to create the world’s first computer.
Those who invented computers hoped to relegate the drudgery of repeated
computation to a machine. In this section, we look at a method for solving linear
systems that played a critical role in this process. The method uses real numbers,
called determinants, that are associated with arrays of numbers. As with matrix
methods, solutions are obtained by writing down the coefficients and constants of a
linear system and performing operations with them.
A
The Determinant of a 2 : 2 Matrix
Associated with every square matrix is a real number, called its determinant. The
determinant for a 2 * 2 square matrix is defined as follows: