Section 4 Inverse Matrix
4.1 Definition:
Definition of inverse matrix:
An n  n matrix A is called nonsingular or invertible if there exists an n  n
matrix B such that
AB  BA  I n ,
In
where
is a n  n identity matrix. The matrix B is called an inverse of A.
If there exists no such matrix B, then A is called singular or noninvertible.
is called a odd permutation.
Theorem:
If A is an invertible matrix, then its inverse is unique.
[proof:]
Suppose B and C are inverses of A. Then,
BA  CA  I n  B  BI n  B( AC)  ( BA)C  I n C  C .
Note:
Since the inverse of a nonsingular matrix A is unique, we denoted the
inverse of A as
A 1 .
Note:
If A is not a square matrix, then
1
 there might be more than one matrix L such that
LA  I (or AL  I ) .
 there might be some matrix U such that
UA  I but AU  I
Example:
Let
1
A   1
 3
1
0  .
 1
Then,

there are infinite number of matrices L such that LA  I , for example
1
L
2

As
1
L
2
3 1
5 1
4
L
7
or
15
25
4
6 .
3 1
5 1 ,
LA  I
but
8
2
3
AL   1  3  1  0 .
 1
4
2 
4.2 Calculation of Inverse Matrix:
1. Using Gauss-Jordan reduction:
The procedure for computing the inverse of a n  n matrix A:
2
1.
Form the n  2n augmented matrix
 a11 a12
a
a
A  I n    21 22
 


a n1 a n 2
 a1n  1
 a2n  0
   
 a nn  0
0  0
1  0
  

0  1
and transform the augmented matrix to the matrix
C
D

in reduced row echelon form via elementary row operations.
2.
If
(a) C  I n , then A1  D .
(b) C  I n , then A is singular and A 1 does not exist.
Example:
1
 1
To find the inverse of A   2

 1

 2
, we can employ the procedure
 5


5 
3
3
introduced above.
1.
1
2

 1
(3)(3)(1)
( 2)( 2)2*(1)

( 2 )1*( 2 )

1
2  1
0
3
3
5 
5 
1
0
0
0
1
0

0
1
2 
1
2
1
3
1
0

0
1  2 
1


2
1
1
2
1
3


3
1
0
2
1
1
0
0
0 .
1
0
0
1
0
 1 0
0 1
0
(1)(1)( 2)
(3)(3)2*( 2)

1
0

0
(1)(1)(3)
( 2)( 2)(3)

2.
1
0

0
0
1 
1
0


1
1
 1 0
 1 0
2 1
3
2
3
0
0 
0
1
1
0
0 
1 
5
3
3
2
1
 1
1 
The inverse of A is
 0
 5


 3
1
3
2
1
1
.
1

Example:
1
Find the inverse of A  0


5
1
2
5
1
if it exists.
3

1

[solution:]
1. Form the augmented matrix
1
1

A | I 3    2  3
 1 3
2  1
0
5  0
5  0
1
0
0
0 .
1
And the transformed matrix in reduced row echelon form is
1
0

0
0
0 
13 / 8
1/ 2
1
0
0 
1 
 15 / 8
5/ 4
1/ 2
0
2. The inverse of A is
4
 1 / 8
3 / 8 
 1 / 4
1/ 2
 13 / 8
 15 / 8


 5/ 4
 1 / 8
3/8 
.
 1 / 4

1/ 2
0
Example:
1
Find the inverse of A  1


5
2
2
2
 3
if it exists.
1 

 3

[solution:]
1. Form the augmented matrix
1
A | I 3   1
5
2
3  1
0
2
2
1  0
3  0
1
0
0
0 .
1
And the transformed matrix in reduced row echelon form is
1
0

0
0
1  1/ 2
1/ 2
1
0
1  1/ 4
0  2
 1/ 4
3
0
0
1
2. A is singular!!
2. Using the adjoint
As
adj ( A)
of a matrix:
det( A)  0 , then
A 1 
adj ( A)
.
det( A)
Note:
adj ( A) A  det( A) I n
is always true.
5
Note:
As
det( A)  0
 A is nonsingular.
4.3 Properties of The Inverse Matrix:
The inverse matrix of an n  n nonsingular matrix A has the following
important properties:
1.
1.
A 
A 
1
1
1
t
 A.

 A1

t
2.
If A is symmetric, So is its inverse.
3.
 AB 1  B 1 A1
4.
5.
If C is an invertible matrix, then
 AC  BC  A  B.

CA  CB  A  B .
As
 A  I 1 exists, then



I  A  A 2    A n1  A n  I  A  I    A  I  A n  I
1
1
[proof of 2]
A  A
1 t
t

 AA

1 t
 It  I
similarly,
t
   A A  I
1 t
A A
1
[proof of 3:]
6
t
t
I
.
.
By property 2,
A   A 
t 1
1 t
 A1 .
[proof of 4:]
B 1 A1  AB  B 1 A1 AB  B 1IB  I .
Similarly,
 ABB 1 A1  ABB 1 A1  AIA1  I
.
[proof of 5:]
Multiplied by the inverse of C, then
ACC 1  AI  A  BCC 1  BI  B .
Similarly,
C 1CA  IA  A  C 1CB  IB  B .
[proof of 6:]
I  A  A
2


   An1  A  I   A  A 2    An  I  A  A 2    An1
 A I .
n
Multiplied by
 A  I  1
on both sides, we have


  A  I  A  I 
1  A  A 2    A n 1  A n  I  A  I 
I  A  A2    An1
can be obtained by using similar procedure.
Example:
Prove that
I  AB1  I  AI  BA 1 B .
7
1
n
1
.

[proof:]
I  AI  BA BI  AB  I  AB  AI  BA B  AI  BA  BAB
 I  AB  AI  BA   I  BA  BA B
 I  AB  AI  BA  I  BA B
1
1
1
1
1
1
 I  AB  AIB  I  AB  AB  I
Similar procedure can be used to obtain
I


 AB  I  AI  BA  B  I
1
Left and Right Inverses:
Definition of left inverse:
For a matrix A,
LA  I but AL  I ,
with more than one such L. Then, the matrices L are called left inverse of
A.
Definition of right inverse:
For a matrix A,
AR  I but RA  I ,
with more than one such R. Then, the matrices R are called left inverse of
A.
Theorem:
A r  c matrix
Arc
has left inverses only if
r  c.
[proof:]
We prove that a contradictory result can be obtained as
8
r  c and
Arc
having a left inverse. For
r  c , let

Ar c  X r r Yr ( c  r )

Then, suppose
Lcr
is the left inverse of
 M r r 


N
 ( c  r ) r 
Arc . Then,
 M r r 
Lcr Ar c  
X r r Yr (c  r )

 N ( c  r ) r 
0 
 I r r
MX MY 
.

 I cc  


 NX NY 
 0 I ( c  r ) ( c  r ) 


Thus,
MX  I , MY  0, NX  0, NY  I .
Since
MX  I and both M and X are square matrices, then
M  X 1 .
Therefore,
multiplied by X
MY  X 1Y  0 
 XX 1Y  Y  X 0  0 .
However,
NY  N 0  0  I .
It is contradictory. Therefore, as r  c , Arc has no left inverse.
9
Theorem:
A r  c matrix
Arc
has left inverses only if
10
r c.