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CHAPTER 9:
Systems of Equations
and Matrices
9.1
9.2
9.3
9.4
9.5
9.6
9.7
9.8
Systems of Equations in Two Variables
Systems of Equations in Three Variables
Matrices and Systems of Equations
Matrix Operations
Inverses of Matrices
Determinants and Cramer’s Rule
Systems of Inequalities and Linear Programming
Partial Fractions
Copyright © 2009 Pearson Education, Inc.
9.5
Inverses of Matrices


Find the inverse of a square matrix, if it exists.
Use inverses of matrices to solve systems of
equations.
Copyright © 2009 Pearson Education, Inc.
The Identity Matrix
Copyright © 2009 Pearson Education, Inc.
Slide 9.5-4
Example
1 3
For A  
and I =

4 6
1 0 
0 1  find each of the following.


a) AI
b) IA
1 3 1 0 

 0 1 
4
6



 1(1)  3(0) 1(0)  3(1) 


4(
1
)

6(
0
)
4(
0
)

6(
1)


1 3

A

4 6
Copyright © 2009 Pearson Education, Inc.
1 0   1 3 

 4 6
0
1



(1)(1)  (0)(4) (1)(3)  (0)(6) 


(
0
)(1)

(
1
)(
4)
(
0
)(3)

(
1
)(6)


1 3

A

4 6
Slide 9.5-5
Inverse of a Matrix
For an n  n matrix A, if there is a matrix A1 for which
A1 • A = I = A • A1, then A1 is the inverse of A.
3 4 
 7 4
Verify that B  
is the inverse of A  
.

5 7 
 5 3 
We show that BA = I = AB.
3 4   7 4  1
BA = 




5 7   5 3  0
 7 4  3 4  1
AB  




 5 3  5 7  0
Copyright © 2009 Pearson Education, Inc.
0
1 
0
1 
Slide 9.5-6
Finding an Inverse Matrix
To find an inverse, we first form an augmented
matrix consisting of A on the left side and the
identity matrix on the right side.
3 4 1 0 
5 7 0 1 


The 2  2
The 2  2
matrix A
identity matrix
Then we attempt to transform the augmented matrix
to one of the form
1 0 a b 
0 1 c d  .


Copyright © 2009 Pearson Education, Inc.
Slide 9.5-7
Example
Find
A1,
3 4 
where A = 
.

5 7 
3 4 1 0 
5 7 0 1 


1 43 13 0 new row 1 = 13 (row 1)


5 7 0 1 
1

0
4
3
1
3
1
3
5
3
0

1  new row 2 =  5(row 1) + row 2
Copyright © 2009 Pearson Education, Inc.
Slide 9.5-8
Example continued
1 43 13 0
0 1 5 3 new row 2 = 3(row 2)


1 0 7 4 new row 1 = - 43 (row 2) + row 1
0 1 5 3 


Thus,
A1
 7 4
= 
.

 5 3 
Copyright © 2009 Pearson Education, Inc.
Slide 9.5-9
Notes
If a matrix has an inverse, we say that it is
invertible, or nonsingular.
When we cannot obtain the identity matrix on the
left using the Gauss-Jordan method, then no inverse
exists.
Copyright © 2009 Pearson Education, Inc.
Slide 9.5-10
Solving Systems of Equations
Matrix Solutions of Systems of Equations
For a system of n linear equations in n variables,
AX = B, if A is an invertible matrix, then the unique
solution of the system is given by X = A1B.
Since matrix multiplication is not commutative in
general, care must be taken to multiply on the left by
A1.
Copyright © 2009 Pearson Education, Inc.
Slide 9.5-11
Example
Use an inverse matrix to solve the following system
of equations:
3x + 4y = 5
5x + 7y = 9
We write an equivalent matrix, AX = B:
3 4   x  5
5 7    y   9 

    
A  X = B
In the previous example we found
Copyright © 2009 Pearson Education, Inc.
A1
 7 4
=
.

 5 3 
Slide 9.5-12
Example continued
We now have X = A1B.
 x   7 4 5 7(5)  (4)(9)   1
 y    5 3  9   5(5)  3(9)    2 
  
  
  
The solution of the system of equations is (1, 2).
Copyright © 2009 Pearson Education, Inc.
Slide 9.5-13