6.6 Inverses of Matrices
 we know that
1  a = a  1 = a (for all real numbers a)
 multiplying any number by 1 yields the identical number
 1 is called the multiplicative identity element for the
real number system
Now note:
a

 c
b  1

d  0
0 1

1 0
0  a

1  c
b  a

d   c
b

d 
so the system of 2  2 matrices also has an identity
element, called the identity matrix, and denoted by I
 also, for any real number a  0
 there is an element a-1, such that a  a-1 = a-1  a = 1
 a-1 is called the multiplicative inverse of a
Q: does a square matrix A have a multiplicative inverse?
That is, a matrix A-1 such that AA-1 = A-1A = I ?
A: Sometimes
6.6-1
Our job:
a
 given matrix A = 
 c
b
 (or any square matrix)
d 
c
 find, if possible, a matrix A = 
e
-1
a

 c
b  c

d  e
d  c

f  e
d  a

f   c
Here’s how!
1

-1
Example: find A for A = 0

2
The procedure:
(1) start with:
1

0

2
1
1
1
0
2
1
0
1
3
0
0
0
d
 such that
f 
b  1

d  0
1
2
3
0

1
1

 1

0 
0

0

1
(2) do row operations on the whole matrix to put the left
matrix in RREF form
(3) if the left matrix is now the identity matrix, the right
matrix will be A-1
6.6-2
Matrix Equations
Consider the system of equations:
x1 x2 +
x3 =
2x2 x3 =
2x1 +
3x2
=
1
1
1
We can write this system in matrix form, as follows:
 x1 
1
 

X =  x2 
B = 1
 

 x3 
1
Now the system of equations can be expressed by:
x 
1
1  1 1   1 

0 2  1  x  = 1  matrix equation AX = B

  2

2 3 0   x 
1
 3
1  1 1 
let A = 0 2  1


2 3 0 
 if A-1 exists
 we can do some algebra with our matrix equation
 much like solving ordinary scalar equations:
AX = B
A-1AX = A-1B
(mult both sides on the left by A-1)
IX = A-1B
X = A-1B
 you only need to remember this!
Voila! We have solved our equation (but we have to find
A-1 to carry out this solution). Let’s do one.
6.6-3