# Inverse of a Matrix

The multiplicative inverse of a square matrix is called its inverse matrix. If a matrix $A$ has an inverse, then $A$ is said to be nonsingular or invertible. A singular matrix does not have an inverse. To find the inverse of a square matrix $A$ , you need to find a matrix ${A}^{-1}$ such that the product of $A$ and ${A}^{-1}$ is the identity matrix.

In other words, for every square matrix $A$ which is nonsingular there exist an inverse matrix, with the property that, $A{A}^{-1}={A}^{-1}A=I$ , where $I$ is the identity matrix of the appropriate size.

You can use either of the following method to find the inverse of a square matrix.

Method 1:

Let $A$ be an $n×n$ matrix.

1. Write the doubly augmented matrix $\left[A\text{\hspace{0.17em}}|\text{\hspace{0.17em}}{I}_{n}\right]$ .

2. Apply elementary row operations to write the matrix in reduced row-echelon form.

3. Decide whether the matrix $A$ is invertible (nonsingular).

4. If $A$ can be reduced to the identity matrix ${I}_{n}$ , then ${A}^{-1}$ is the matrix on the right of the transformed augmented matrix.

5. If $A$ cannot be reduced to the identity matrix, then $A$ is singular.

Method 2:

You may use the following formula when finding the inverse of $n×n$ matrix.

If $A$ is non-singular matrix, there exists an inverse which is given by , where $|\text{\hspace{0.17em}}A\text{\hspace{0.17em}}|$ is the determinant of the matrix.

Example :

Find ${A}^{-1}$ , if it exists. If ${A}^{-1}$ does not exist, write singular.

$A=\left[\begin{array}{rr}\hfill 1& \hfill 2\\ \hfill 1& \hfill 1\end{array}\right]$

Step 1:

Write the doubly augmented matrix $\left[A\text{\hspace{0.17em}}|\text{\hspace{0.17em}}{I}_{n}\right]$ .

$\left[A\text{\hspace{0.17em}}|\text{\hspace{0.17em}}I\right]=\left[\begin{array}{rrrr}\hfill 1& \hfill 2& \hfill 1& \hfill 0\\ \hfill 1& \hfill 1& \hfill 0& \hfill 1\end{array}\right]$

Step 2:

Apply elementary row operations to write the matrix in reduced row-echelon form.

$\begin{array}{rr}\hfill \left[\begin{array}{rrrr}\hfill 1& \hfill 2& \hfill 1& \hfill 0\\ \hfill 0& \hfill 1& \hfill 1& \hfill -1\end{array}\right]& \hfill {R}_{2}={R}_{1}-{R}_{2}\\ \hfill \left[\begin{array}{rrrr}\hfill 1& \hfill 0& \hfill -1& \hfill 2\\ \hfill 0& \hfill 1& \hfill 1& \hfill -1\end{array}\right]& \hfill {R}_{1}=-2{R}_{2}+{R}_{1}\\ \hfill \left[\begin{array}{rrrr}\hfill 1& \hfill 0& \hfill -1& \hfill 2\\ \hfill 0& \hfill 1& \hfill 1& \hfill -1\end{array}\right]& \hfill =\left[I\text{\hspace{0.17em}}|\text{\hspace{0.17em}}{A}^{-1}\right]\end{array}$

The system has a solution.

Therefore, $A$ is invertible and ${A}^{-1}=\left[\begin{array}{rr}\hfill -1& \hfill 2\\ \hfill 1& \hfill -1\end{array}\right]$