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.

## Finding the Inverse of a Matrix

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

• Elementary Row Transformation

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