Ch. 7 – Matrices and Systems of
Equations
7.6 – Inverses of Square Matrices
1 0 0 
0 1 0 


0 0 1 
Inverses


Identity Matrix (I)= a square matrix with 1s in the main diagonal
and 0s everywhere else
Two matrices are inverses of each other if they multiply to the
identity matrix



The inverse of A is denoted A-1
A is said to be invertible if it has an inverse
Ex: Prove that B is the inverse of A.

Multiply A and B and see if you get the identity matrix!
1 2  1 2
AB  
  1 1 
1

1



1

0
0
1

1 2 
A

1

1


 1 2 
B


1
1


 2 6 
A

 2 7 
Inverses

Ex 1: Find the inverse of A.


We want to solve A A-1 = I for A-1, but how?
Matrix multiplication tells us that we want to solve the following two
augmented matrices simultaneously:
 2 6 1 
 2 7 0 




 2 6 0 
 2 7 1 


…so we do that by reducing the adjoined matrix [ A : I ] !
When finding inverses of square matrices, RREF the
augmented matrix  A I  !

In this problem, we need to RREF the following matrix:
 2 6 1 0 
 2 7 0 1 


 2 6 
A

 2 7 
Inverses

Ex 1 (cont’d): Find the inverse of A.
 2 6 1 0 
 2 7 0 1 


2
6 1 0
2
7
0
0 1
1 1 1
Add E2 and E1 and replace for E2…
 2 6 1 0
0 1 1 1


Add 6(E2) and E1 and replace for E1…
2 6 1 0
0
6
6 6
2
0
7
6
 2 6 
A

 2 7 
Inverses

Ex 1 (cont’d): Find the inverse of A.
2 0 7 6
0 1 1 1 


1 0 3.5 3
0 1 1 1


Divide E1 by 2…

So the inverse of A is the right side of the RREF’d matrix.
3.5 3
 1 1



1 1 0 
A  1 0 1
6 2 3
Ex 2 : Find the inverse of A.

RREF the adjoined matrix!
1 1 0 1 0 0 
1 0 1 0 1 0 


6 2 3 0 0 1 
1
1 0 1
1
0
Add -6(E1) and E3 and replace for E3…
0
1 0 1 0
0 1 1 1 1 0
Add -(E2) and E1 and replace for E2…
1 1 0 1 0 0 
0 1 1 1 1 0 


6 2 3 0 0 1 
0
6
6
6
2 3
0
6 0 0
0
0 1
0 4 3 6 0 1

Ex 2 (cont’d):

RREF the adjoined matrix!
1 1 0 1 0 0
0 1 1 1 1 0 


0 4 3 6 0 1 
Add -(E1) and E2 and replace for E1…
 1 0 1 0 1 0 
 0 1 1 1 1 0


 0 4 3 6 0 1 
Add 4(E2) and E3 and replace for E3…
1
1
0
1 1
1
0
0 1
1 0 1
4
0
3 6
0
0
0
1 0
1
0 4
4
0
4
0
4 0
0
1
1 2 4
1

Ex 2 (cont’d):

RREF the adjoined matrix!
 1 0 1 0 1 0 
 0 1 1 1 1 0 


 0 0 1 2 4 1 
Add -(E1) and E3 and replace for E1…
1 0 0 2 3 1 
0 1 1 1 1 0 


0 0 1 2 4 1 
Add -(E2) and E3 and replace for E2…
1 0 1
0
0 0
2 4 1
1
1
1
0 0 2 3
0 1 1 1
0 0
1
1
0
1
0
2 4 1
0 1 0 3 3
1

Ex 2 (cont’d):
1 0 0 2 3 1
0 1 0 3 3 1


0 0 1 2 4 1
Separate the inverse matrix from the identity to get…
 2 3 1
A1   3 3 1
 2 4 1

Check your solution with your calculator!
x  2 y  3z  9


An alternative method to solving systems
of equations involves using inverse functions.
y  3z  5
4x  2z  0
Ex 3: Use an inverse matrix to solve the system.
 If we write the system in AX = B form, where A is the
coefficient matrix, X is the variable matrix, and B is the
constant term matrix, it will look like this: 1 2 3   x  9 
0

 4

1
0
3   y   5
2   z  0
By multiplying A-1 in front of both sides of the equation, we get
A to cancel and an equation in X = A-1B form:
x
9 
 y   A 1  5 
 
 
 z 
 0 
 x   .05 .11 .23  9 
 y    .31 .37 .08  5
  
 
 z   .11 .21 .03 0
x  1 
 y    1
   
 z   2 