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