2.2 The Inverse of a Matrix
Recall: The multiplicative inverse of a
real number a is denoted by a-1.
Example: 7 and 7-1.
7(7-1) = (7-1)7 = 1, the multiplicative
identity.
Definition: An n x n matrix A is said to be
invertible if  an n x n matrix C such that
CA  AC  I n
Note that we have commutativity here.
We call C the inverse of A, and write
C  A1 .
1
1
A
A

AA
I
So,
Fact: If A is invertible, then the inverse is
unique.
Proof: Assume B and C are both
inverses of A. We will show that B and C
are equal.
We can write
B  BI
 BCA
 BAC
 IC
C
So, the inverse is unique, since any two
inverses coincide.
Note: not all square matrices are
invertible. A matrix that is not invertible
is called singular. An invertible matrix is
sometimes called non-singular.
Definition: If A is a 2 x 2 matrix,
a c 
b d  , then the determinant of A is


ad  bc .
Theorem 4: Let
a c 
A

b d  .
If ad  bc  0 , then A is invertible, and
1  d  c
1
A 


ad  bc  b a  .
If
ad  bc  0 , then A is not invertible.
Example: Let
 7 3 
A

5

2


1
A
Find
.
 2  3
1
A 
 2 7   53   5  7 
  2  3
 1


5

7


1
 2 3


5 7 
If we want to solve the equation Ax = b,
we can use A-1.
Ax  b
A1 Ax  A1b
Ix  A1b
x  A1b
Theorem 2.5.: If A is an invertible n x n
matrix, then for each b in Rn, the
equation Ax = b has the unique solution
x  A1b .
Proof: Suppose w is another solution:
Aw  b
A1 Aw  A1b
w  A1b
w  A1b  x
So, w is the same as x, and there is
only one solution.
Note: this gives both existence and
uniqueness. The solution exists
1
because we can display it: x  A b .
Example: Use the inverse of
 7 3 
A
 to solve
5

2


 7 x1  3x2  2
5 x1  2 x2  1
We write Ax = b as
 7 3   x1  2
 5  2  x   1

 2   
Earlier, we found
 2 3
1
A 

5
7


 2 3  2  7 
x A b
 



Now
5 7 1 17
1
Theorem 2.6: Suppose A and B are
invertible, then the following hold.
a. A is invertible and A   A
-1
-1 -1


AB

B
A .
b. AB is invertible, and
-1 1
-1
c. A is invertible, and A
Proof of b:
 AB B -1 A-1   ABB -1 A-1
 AIA-1
 AA-1
I
also
B
-1
  A  .
T 1
T

A-1  AB   B -1 A-1 AB
 BIB -1
 BB
I
-1
1 T
Part b can be generalized to the product
of any number of invertible matrices.
 ABC 
-1
 C -1B -1 A-1
Finding the inverse of a general n x n
matrix:
First we look at Elementary matrices.
Definition: an elementary matrix is one
that is obtained by performing a single
elementary row operation on an identity
matrix.
Let
1 0 0
1 0
E1  0 2 0 E2  0 0

0 0 1 ,
0 1
1 0 0
a

E3  0 1 0
A  d
3 0 1 and
 g
0
1
0 ,
b
e
c

f
i 
h
The Ei are elementary matrices.
What row operations were used to get
them from the 3x3 identity matrix?
Look at what they do to the matrix A.
1 0 0  a b c 


 0 2 0 d e f 


E1 A 
,



0
0
1
g
h
i

 

E2 A 
E3 A 
Note: Elementary matrices are invertible
because row operations are reversible.
Example:
 1 0 0
1
E3   0 1 0
- 3 0 1
Note: If an elementary row operation is
performed on an m x n matrix A, the
resulting matrix can be written as EA,
where the m x m matrix E is created by
performing the same row ops on Im.
Example: Let
 1 0 0
 3
1
A  
0

2
2
 0 1 0 .


Find A-1 by using row operations to turn
this into the 3x3 identity matrix I3.
1 0 0  1 0 0   1 0 0
 3
1 



E1 A  0 2 0 
0
  3 0 1

2
2
0 0 1  0 1 0   0 1 0
1 0 0  1 0 0




E2 E1 A  0 0 1  3 0 1
0 1 0  0 1 0
 1 0 0
  0 1 0
 3 0 1
R3 + 3R1
1 0 0   1 0 0 
E3 E2 E1 A  0 1 0  0 1 0
3 0 1  3 0 1
1 0 0 
 0 1 0
0 0 1
From this, we see:
E3E2 E1 A  I3
Multiply by A-1 on the right to get
E3 E2 E1 AA1  I3 A1
E3 E2 E1   A1
So, we have created A-1 out of
elementary matrices.
The elementary row operations that row
reduce A to In are the same elementary
row operations that transform In into A-1.
Theorem 2.7: An n x n matrix A is
invertible if and only if A is row
equivalent to In; and in this case, any
sequence of elementary row operations
that reduces A to In will also transform In
into A-1.
This gives us an algorithm for finding A-1
1. Place A and I side by side to form an
augmented matrix [A I].
2. Then perform row operations on this
matrix. So, by Theorem 7, [A I] reduces
to [I A-1].
Example: Find the inverse of
 2 0 0
A   3 0 1
 0 1 0 if it exists.
Step 1:
 2 0 0 1 0 0
  3 0 1  0 1 0


 0 1 0 0 0 1
½ R1
1


 1 0 0 2 0 0
  3 0 1  0 1 0


 0 1 0 0 0 1


R2 ↔ R3
1


 1 0 0 2 0 0
 0 1 0  0 0 1


 3 0 1 0 1 0


R3 + 3R1
1


1 0 0 2 0 0
0 1 0  0 0 1 


3
1 0
0 0 1
2




 A1  



1
2
0
3
2

0 0
0 1

1 0

►The order of the multiplication is
important.
Example: Suppose A, B, C , and D are
invertivle n x n matrices and
A  BD  I n C
Solve for D.
Undo multiplication on the right by B and
on the left by C:
B AC  B BD  I n CC
1
1
1
1
1
B AC  D  I n
Add In to both sides:
B 1 AC 1  I n  D
1