Let A be a given square matrix. If there exists another matrix B in such a way that, AB = BA = I, where I is the identity matrix whose order is same as A. Then, B is said to be the inverse matrix of A and written as A-1. Thus, AA-1 = In= A-1A.

In general, inverse is the reciprocal matrix of given square matrix. A square matrix A has its inverse, if and only if the determinant of the matrix is non-zero, i.e. IAI $\neq$ 0.

In order to find the inverse of a matrix, we have some operations as follows:

  • Adjoint Method
  • Elementary Row Transformation

Adjoint Method:


In this method, we can use adjoint of the given matrix in order to find inverse of matrix, but for this determinant of the matrix should be non-zero.
So, A-1 = $\frac{adj A}{I A I}$

Elementary Row Transformation:


In elementary transformation, first we write, A = IA, where I be the identity matrix. After that, we can apply some operations on A(left hand side) and on I(right hand side) not on A(right hand side). We continue this process till the A gets converted into the identity matrix and at the same point, I is also converted into some different form of matrix which is said to be the inverse matrix of A.

Solved Examples

Question 1: Let A = $\begin{bmatrix}
3 &-4 \\
2 &1
\end{bmatrix}$ be a square matrix, find its inverse.
Solution:
We know that A-1 = $\frac{adj A}{I A I}$, where IAI $\neq$ 0.

Then IAI = $\begin{vmatrix}
3 &-4 \\
2 &1
\end{vmatrix}$ = (3 + 8) = 11 $\neq$ 0. Hence A-1 exists.

adj A = [ cofactor matrix of A ]T
= $\begin{bmatrix}
1&-2 \\
4 &3
\end{bmatrix}^T$ = $\begin{bmatrix}
1 &4 \\
-2 &3
\end{bmatrix}$

So A-1 = $\frac{1}{I A I}$ x adj A

or A-1 = $\frac{1}{11}$ x $\begin{bmatrix}
1 &4 \\
-2 &3
\end{bmatrix}$

or A-1 = $\begin{bmatrix}
1/11&4/11 \\
-2/11 &3/11
\end{bmatrix}$ is the inverse of A.

Clarification: If we multiply A and A-1, we get identity matrix, then

A . A-1 = $\begin{bmatrix}
3 &-4 \\
2 &1
\end{bmatrix}$ x $\begin{bmatrix}
1/11&4/11 \\
-2/11 &3/11
\end{bmatrix}$

or A .A-1 = $\begin{bmatrix}
3/11+8/11&12/11 - 12/11 \\
-2/11 + 2/11 &8/11 + 3/11
\end{bmatrix}$ = $\begin{bmatrix}
1&0\\
0 &1
\end{bmatrix}$

Hence A-1 = $\begin{bmatrix}
1/11&4/11 \\
-2/11 &3/11
\end{bmatrix}$.

Question 2: If A = $\begin{bmatrix}
1 &1 &3 \\
1 &3 &-3 \\
-2 &-4 &-4
\end{bmatrix}$ be a non-singular matrix. Find the inverse of the matrix by the use of elementary row transformation.
Solution:
First we write,
A = In A
or $\begin{bmatrix}
1 &1 &3 \\
1 &3 &-3 \\
-2 &-4 &-4
\end{bmatrix}$ = $\begin{bmatrix}
1 &0 &0 \\
0 &1&0 \\
0 &0 &1
\end{bmatrix}$ A

Using operations, R2 $\rightarrow $ R2 - R1 and R3 $\rightarrow $ R3 + 2R1, we get,

$\begin{bmatrix}
1 &1 &3 \\
0 &2 &-6 \\
0 &-2 &2
\end{bmatrix}$ = $\begin{bmatrix}
1 &0 &0 \\
-1 &1&0 \\
2 &0 &1
\end{bmatrix}$ A

Apply, R3$\rightarrow $ R3 + R2

$\begin{bmatrix}
1 &1 &3 \\
0 &2 &-6 \\
0 &0 &-4
\end{bmatrix}$ = $\begin{bmatrix}
1 &0 &0 \\
-1 &1&0 \\
1 &1&1
\end{bmatrix}$ A

Using R3$\rightarrow $ R3/-4 and R1 $\rightarrow $ 2R1- R2

$\begin{bmatrix}
2 &0 &12\\
0 &2 &-6 \\
0 &0 &1
\end{bmatrix}$ = $\begin{bmatrix}
3 &-1 &0 \\
-1 &1&0 \\
-1/4 &-1/4&-1/4
\end{bmatrix}$ A

Using, R2$\rightarrow $ R2/2 and R1$\rightarrow $ R1- 12R3

$\begin{bmatrix}
2 &0 &0\\
0 &1 &-3 \\
0 &0 &1
\end{bmatrix}$ = $\begin{bmatrix}
6 &2&3 \\
-1/2 &1/2&0 \\
-1/4 &-1/4&-1/4
\end{bmatrix}$ A

lastly, R1$\rightarrow $ R1/2 and R2 $\rightarrow $ R2 + 3R3

$\begin{bmatrix}
1 &0 &0\\
0 &1 &0 \\
0 &0 &1
\end{bmatrix}$ = $\begin{bmatrix}
3 &1&3/2 \\
-5/4 &-1/4&-3/4 \\
-1/4 &-1/4&-1/4
\end{bmatrix}$ A

Hence A-1 = $\begin{bmatrix}
3 &1&3/2 \\
-5/4 &-1/4&-3/4 \\
-1/4 &-1/4&-1/4
\end{bmatrix}$

A square matrix say B of order nxn is said to be invertible, if there exists a matrix C of order nxn such that BC = Inxn = CB, where I be the identity matrix of order nxn and C is called the inverse of B.
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To find the inverse of 2x2 matrix, lets take A = $\begin{bmatrix}
7 &9 \\
5 &6
\end{bmatrix}$
We know that, A-1 = $\frac{adj A}{\left | A \right |}$

First, we check IAI = 7 x 6 - 9 x 5 = 42 - 45 = -3 $\neq$ 0
IAI is non zero. Hence, inverse of matrix exists.

Now, calculate cofactor matrix of A say C.
Then, C = $\begin{bmatrix}
6 &-5 \\
-9&7
\end{bmatrix}$

adj A = CT = $\begin{bmatrix}
6 &-9 \\
-6&7
\end{bmatrix}$

Then, A-1 = $\frac{1}{\left | A \right |}$ x adj A

or A-1 = $\frac{1}{-3}$ x $\begin{bmatrix}
6 &-9 \\
-6&7
\end{bmatrix}$

A -1 = $\begin{bmatrix}
-2 &3 \\
2&-7/3
\end{bmatrix}$
Let A3x3 = $\begin{bmatrix}
1 &2 &1 \\
3 &0 &4 \\
-1 &6 &-2
\end{bmatrix}$

First, calculate IAI = 1(0 -24) - 2(-6 + 4 ) + 1(18 - 0)
= -24 + 4 + 18
= -2 $\neq$ 0
Let B be the cofactor matrix of A, then
B = $\begin{bmatrix}
-24 &2 &18 \\
8 &-1 &-8 \\
8 &-1 &6
\end{bmatrix}$

Hence, adj A = BT = $\begin{bmatrix}
-24 &8 &8 \\
2 &-1 &-1 \\
18 &-8 &6
\end{bmatrix}$

Therefore, A-1 = $\frac{1}{-2}$ x $\begin{bmatrix}
-24 &8 &8 \\
2 &-1 &-1 \\
18 &-8 &6
\end{bmatrix}$

= $\begin{bmatrix}
12 &-4 &-4 \\
-1&1/2 &1/2 \\
-9 &4 &-3
\end{bmatrix}$

Hence, A-1 = $\begin{bmatrix}
12 &-4 &-4 \\
-1&1/2 &1/2 \\
-9 &4 &-3
\end{bmatrix}$
Listed below are some of the properties of Inverse Matrix.

Property 1:             

If a matrix A is invertible, then (AT)-1 = (A-1)T.

Proof:
 

Let A is any square matrix of order 2x2
        A = $\begin{bmatrix}
4 &-3 \\
-2 &1
\end{bmatrix}$.

Now, AT = $\begin{bmatrix}
4 &-2 \\
-3&1
\end{bmatrix}$

Calculate, (AT)-1 = $\frac{adjA^{T}}{\left | A^T \right |}$
Find Iadj ATI= $\begin{vmatrix}
4 &-2 \\
-3&1
\end{vmatrix}$ = (4 - 6) = -2

adj AT = transpose matrix of cofactor matrix of AT 
          = $\begin{bmatrix}
1 &3\\
2&4
\end{bmatrix}^T$ = $\begin{bmatrix}
1 &2 \\
3&4
\end{bmatrix}$

(AT)-1 = $\frac{1}{-2}$ .$\begin{bmatrix}
1 &2 \\
3&4
\end{bmatrix}$

        = $\begin{bmatrix}
-1/2 &-1 \\
-3/2&-2
\end{bmatrix}$                          ................................................(1)

Now, find A-1  =$\frac{adjA^{-1}}{\left | A \right |}$
IA-1I = $\begin{vmatrix}
4 &-3 \\
-2 &1
\end{vmatrix}$ = (4 - 6) = -2

adj A-1 = $\begin{bmatrix}
1 &2\\
3 &4
\end{bmatrix}^T$ = $\begin{bmatrix}
1&3 \\
2&4
\end{bmatrix}$

Then, A-1 = $\frac{1}{-2}$ . $\begin{bmatrix}
1 &2 \\
3&4
\end{bmatrix}$
          
             =  $\begin{bmatrix}
-1/2 &-1 \\
-3/2&-2
\end{bmatrix}$                    ................................................(2)

Hence, (AT)-1 = (A-1)T

Property 2:           

If A and B are two matrices, then (AB)-1 = B-1 A-1.

Proof:


Let A = $\begin{bmatrix}
1&1 \\
1 &2
\end{bmatrix}$ and B = $\begin{bmatrix}
2 &-3 \\
1 &1
\end{bmatrix}$

(AB) = $\begin{bmatrix}
1&1 \\
1 &2
\end{bmatrix}$ . $\begin{bmatrix}
2 &-3 \\
1 &1
\end{bmatrix}$
 
     = $\begin{bmatrix}
3 &-2 \\
4 &-1
\end{bmatrix}$

Now (AB)-1= $\frac{adj AB} {I ABI}$

                 = $\frac{1}{5}$ . $\begin{bmatrix}
-1 &-4 \\
2&3
\end{bmatrix}^T$

               =  $\frac{1}{5}$ . $\begin{bmatrix}
-1 &2 \\
-4&3
\end{bmatrix}$         ........................................................(a)

Now, calculate A-1 and B-1 

A-1 = $\frac{adj A}{I AI}$
     = $\frac{1}{1}$ . $\begin{bmatrix}
2 &-1 \\
-1 &1
\end{bmatrix}$
 
    = $\begin{bmatrix}
2 &-1 \\
-1 &1
\end{bmatrix}$        ..........................................................(b)

B-1= $\frac{adj B}{I BI}$
    = $\frac{1}{5}$ . $\begin{bmatrix}
1 &3 \\
-1&2
\end{bmatrix}$      ..........................................................(c)

So, B-1 A-1 = $\frac{1}{5}$ . $\begin{bmatrix}
1 &3 \\
-1&2
\end{bmatrix}$ . $\begin{bmatrix}
2 &-1 \\
-1 &1
\end{bmatrix}$

              = $\frac{1}{5}$ . $\begin{bmatrix}
-1 &2 \\
-4&3
\end{bmatrix}$      ..........................................................(d)

Therefore, it is clear that (AB)-1 = B-1 A-1.

Property 3:

If A is a square matrix, then (A-1)-1 = A, i.e. inverse of an inverse matrix is the matrix itself.

Proof:


Let A = $\begin{bmatrix}
3&5\\
-1&2
\end{bmatrix}$

Then, A-1 = $\frac{adj A} {I AI}$

             = $\frac{1}{11}$ . $\begin{bmatrix}
2&-5\\
1&3
\end{bmatrix}$

 Now, (A-1)-1  = $\frac{adjA^{-1}}{\left | A \right |}$

                       = $\frac{1}{1}$ x adj A-1 

                       = $\begin{bmatrix}
3&5\\
-1&2
\end{bmatrix}$ = A
 
Hence, (A-1)-1 = A