Download Chapter 9 Section 3 - Canton Local Schools

Chapter 9
Matrices and
Determinants
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All rights reserved
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1
SECTION 9.3 The Matrix Inverse
OBJECTIVES
1
2
3
4
5
Verify the multiplicative inverse of a matrix.
Find the inverse of a matrix.
Find the inverse of a 2 × 2 matrix.
Use matrix inverses to solve systems of
linear equations.
Use matrix inverses in applied problems.
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2
INVERSE OF A MATRIX
If A be an n × n matrix and let I be the n × n
identity matrix that has 1’s on the main diagonal
and 0s elsewhere. If there is an n × n matrix B
such that
AB  I
and
BA  I ,
then B is called the inverse of A and we write
B = A–1 (read “A inverse”).
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3
EXAMPLE 1
Verifying the Inverse of a Matrix
Show that B is the inverse of A:
Solution
You need to verify that AB = I and BA = I.
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EXAMPLE 1
Verifying the Inverse of a Matrix
Solution continued
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5
EXAMPLE 1
Verifying the Inverse of a Matrix
Solution continued
Since AB = I and BA = I it follows that B = A−1.
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EXAMPLE 2
Proving That a Particular Nonzero Matrix
Has No Inverse
Show that the matrix A does not have an inverse.
Solution
Suppose A has an inverse B, where
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EXAMPLE 2
Proving That a Particular Nonzero Matrix
Has No Inverse
Solution continued
Then you have
AB = I
Multiply the two
matrices on the left
side.
Since these two matrices are equal, you must
have 0 = 1. Because this is a false statement,
the matrix A does not have an inverse.
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8
PROCEDURE FOR FINDING THE
INVERSE OF A MATRIX
Let A be an n × n matrix.
1. Form the n × 2n augmented matrix [A|I],
where I is the n × n identity matrix.
2. If there is a sequence of row operations that
transforms A into I, then this same sequence
of row operations will transform [A|I] into
[I|B], where B = A–1.
3. Check your work by showing that AA–1 = I.
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PROCEDURE FOR FINDING THE
INVERSE OF A MATRIX
If it is not possible to transform A into I by
row operations, then A does not have an
inverse. (This occurs if, at any step in the
process, you obtain a matrix [C|D] in which
C has a row of zeros.)
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EXAMPLE 4
Finding the Inverse of a Matrix
Find the inverse (if it exists) of the matrix
1 1 0 
A  0 3 1  .


 2 3 3 
Solution
Step 1 Start with the matrix
1 1 0 1 0 0 


 A I    0 3 1 0 1 0 
 2 3 3 0 0 1 
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EXAMPLE 4
Finding the Inverse of a Matrix
Solution continued Step 2
1 1
1

R2  R2
0 1
3

2 R1  R3  R3 
0 1
Use row operations.
0 1
1
0
3
3 2
0 0

1
0
3

0 1 


1 1 0 1
0 0


1
1 

0
0
 1 R2  R3  R3 0 1
3
3 


8
1
0 0
2 
1
3
3


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12
EXAMPLE 4
Finding the Inverse of a Matrix
Solution continued

1
0
1 1 0

1
0 1 1
0

3
3
3 
6
1
R3 0 0 1 

8 
8
8

0

0

3


8 1 1 0
1
0
0


1
2
3
1

 R3  R2  R2 0 1 0


3
8
8
8

6
1
3
0 0 1 


8
8
8

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EXAMPLE 4
Finding the Inverse of a Matrix
Solution continued

 1 R2  R1  R1  1 0

0 1


0 0
3
1
 6


 8
8
8
 6 3


2
3
1 1 
1

A 
 
2 3
 8
8
8 8 
 6 1
 6

1
3



8
8
 8
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0
8
2
0
8
6
1 
8
1
1

3
3
1

8
8

3
1

8
8
1
3


8
8
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A RULE FOR FINDING THE
INVERSE OF 2 × 2 MATRIX
a b 
The matrix A  
is invertible

c d 
if and only if ad – bc ≠ 0. Moreover, if
ad − bc ≠ 0, then the inverse is given by
1  d b 
A 
.


ad  bc  c a 
1
If ad – bc = 0, the matrix does not have
an inverse.
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EXAMPLE 5
Finding the Inverse of a 2 × 2 Matrix
Find the inverse (if it exists) of each matrix.
Solution
a. For the matrix A, a = 5, b = 2, c = 4, and
d = 3. Here
ad – bc = (5)(3) – (2)(4) = 15 – 8 = 7 ≠ 0
so A is invertible.
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EXAMPLE 5
Finding the Inverse of a 2 × 2 Matrix
Solution continued
A−1 =
You should verify
that AA−1 = I.
b. For the matrix B, a = 4, b = 6, c = 2, and
d = 3. Here ad – bc = (4)(3) – (6)(2) = 0 so B
does not have an inverse.
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17
SOLVING SYSTEMS OF LINEAR EQUATIONS
BY USING MATRIX INVERSES
Matrix multiplication can be used to write a
system of linear equations in matrix form.
 3x  2y  4
 3 2   x   4 
 





 4 3   y   5 
 4x  3y  5
Solving a system of linear equations amounts
to solving the matrix equation of the form
AX  B.
The solution to this equation is
1
X  A B.
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EXAMPLE 6
Solving a Linear System by Using an
Inverse Matrix
Use a matrix to solve the linear system.
4
 x y

3y  z  7

2 x  3 y  3z  21

Solution
Write the linear system in matrix form.
1 1 0   x   4 
Use zeros for
0 3 1   y    7 
coefficients
of

   
missing variables.
 2 3 3   z   21
A
X = B
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EXAMPLE 6
Solving a Linear System by Using an
Inverse Matrix
Solution continued
Since the matrix is invertible (see Example 4)
the system has a unique solution X = A–1B.
Computed in
Example 4
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20
EXAMPLE 6
Solving a Linear System by Using an
Inverse Matrix
Solution continued
 6 3 1  4 
1
1
X A B
2 3 1  7 
 
8
 6 1 3  21
 6  4   3  7   1 21 
 24   3 
1
 1   
  2  4   3  7   1 21  
8  1
8
8   
6  4   1 7   3  21 
 32   4 
The solution set is {(3, 1, 4)}, which you can
check in the original system.
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21
THE LEONTIEF INPUT-OUTPUT MODEL
Suppose a simplified economy depends on two
products: energy (E) and food (F). To produce
1
1
one unit of E requires unit of E and unit of F.
4
2
To produce one unit of F
1
requires unit of E and
3
1
unit of F. Then the
4
interindustry consumption
is given by the matrix to
the right.
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THE LEONTIEF INPUT-OUTPUT MODEL
The matrix A is the interindustry technology
input-output matrix, or simply the technology
matrix, of the system. If the system is producing
x1 units of energy and x2 units of food, then the
 x1 
column matrix X    is called the gross
 x2 
production matrix.
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THE LEONTIEF INPUT-OUTPUT MODEL
Consequently,
1 
1 1
1
x

x
1
2
 4 3   x1   4
3
AX  
   

1
1
x
1
1


2


 x  x 
 2 4 
 2 1 4 2 
units consumed by E
units consumed by F
represents the interindustry consumption.
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THE LEONTIEF INPUT-OUTPUT MODEL
 d1 
If the column matrix D   
 d2 
represents consumer demand, then
D  X  AX  IX  AX   I  A  X
 I  A
1
D   I  A
1
 I  A X  IX  X
where I is the identity matrix.
The gross production matrix is: X = (I – A)–1D
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EXAMPLE 7
Using the Leontief Input-Output Model
In the preceding discussion, suppose the consumer
demand for energy is 1000 units and that for food
is 3000 units. Find the level of production (X) that
will meet interindustry and consumer demand.
Solution
For the matrix A, we have
1
1 0   4
I  A


0 1   1
 2
1  3
3  4

1  1

4   2
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1
 
3

3
4 
26
EXAMPLE 7
Using the Leontief Input-Output Model
Solution continued
Use the formula for the inverse of a 2 × 2 matrix
3 1
4 3
1
1
I  A   3 3


 1  1  1 3 
    
4 4  3   2   2 4 
3
48  4


19  1
 2
1
3  1  36 16 
 
3  19  24 36 
4 
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EXAMPLE 7
Using the Leontief Input-Output Model
Solution continued
Hence, the gross production matrix X is given by
X   I  A D
1
1 36 16  1000 
 



19  24 36  3000 
 84, 000 
1  84, 000   19 
 



19 132, 000   132, 000 
 19 
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EXAMPLE 7
Using the Leontief Input-Output Model
Solution continued
Thus, to meet the consumer demand for 1000
units of energy and 3000 units of food, the
84, 000
energy produced must be
units and
19
132, 000
units.
food production must be
19
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29