# Find Matrix Inverse Using Row Operations

   

## Introduction

We present examples on how to find the inverse of a matrix using the three row operations listed below:

1. Interchange two rows
2. Add a multiple of one row to another
3. Multiply a row by a non zero constant
Examples with detailed solutions are also included.
An
Inverse of a Matrix Using Row Reduction - Calculator - Calculator

## Inverse of a Matrix

Let $$A$$ be an $$n \times n$$ matrix. If matrix $$A^{-1}$$ is the inverse of matrix $$A$$ , then we have $A A^{-1} = I_n = A^{-1} A$ where $$I_n$$ is the $$n \times n$$ identity matrix
Consider the matrix equation $$A A^{-1} = I_n$$ where $$A^{-1}$$ is the unknown. To find the inverse $$A^{-1}$$, we start with the augmented matrix $$[ A | I_n ]$$ and then row reduce it. If matrix $$A$$ is invertible, the row reduction will end with an augmented matrix in the form
$[ I_n | A^{-1} ]$ where the inverse $$A^{-1}$$ is the $$n \times n$$ on the right side of $$[ I_n | A^{-1} ]$$

## Examples with Solutions

Example 1
Find the inverse of matrix $$A = \begin{bmatrix} 1 & -1 \\ 2 & -1 \end{bmatrix}$$.

Solution to Example 1
Write the augmented matrix $$[ A | I_2 ]$$
$$\begin{bmatrix} 1 & -1 & | & 1 & 0 \\ 2 & -1 & | & 0 & 1 \end{bmatrix}$$
Write the above augmented matrix in reduced row echelon form .
$$\begin{matrix} \\ \color{red}{R_2 - 2 R_1} \end{matrix} \begin{bmatrix} 1 & -1 & | & 1 & 0 \\ 0 & 1 & | & -2 & 1 \end{bmatrix}$$

$$\begin{matrix} \color{red}{R_1 + R_2} \\ \\ \end{matrix} \begin{bmatrix} 1 & 0 & | & -1 & 1 \\ 0 & 1 & | & -2 & 1 \end{bmatrix}$$
The above augmented matrix has the form $$[ I_2 | A^{-1} ]$$ and therefore $$A^{-1}$$ is given by
$$A^{-1} = \begin{bmatrix} -1 & 1 \\ -2 & 1 \end{bmatrix}$$

Example 2
Find the inverse of matrix $$A = \begin{bmatrix} 2 & 1 & -1\\ 3 & 1 & 0 \\ -1 & 0 & 2 \end{bmatrix}$$.

Solution to Example 2
Write the augmented matrix $$[ A | I_3 ]$$
$$\begin{bmatrix} 2 & 1 & -1 & | & 1 & 0 & 0\\ 3 & 1 & 0 & | & 0 & 1 & 0\\ -1 & 0 & 2 & | & 0 & 0 & 1 \end{bmatrix}$$
Write the above augmented matrix in reduced row echelon form .
$$\begin{matrix} \color{red}{-R_3} \\ \color{red}{R_1} \\ \color{red}{R_2}\\ \end{matrix} \begin{bmatrix} 1 & 0 & -2 & | & 0 & 0 & -1\\ 2 & 1 & -1 & | & 1 & 0 & 0 \\ 3 & 1 & 0 & | & 0 & 1 & 0 \\ \end{bmatrix}$$

$$\begin{matrix} \\ \color{red}{R_2 - 2 R_1} \\ \color{red}{R_3 - 3 R_1}\\ \end{matrix} \begin{bmatrix} 1 & 0 & -2 & | & 0 & 0 & -1\\ 0 & 1 & 3 & | & 1 & 0 & 2 \\ 0 & 1 & 6 & | & 0 & 1 & 3 \\ \end{bmatrix}$$

$$\begin{matrix} \\ \\ \color{red}{R_3 - R_2}\\ \end{matrix} \begin{bmatrix} 1 & 0 & -2 & | & 0 & 0 & -1\\ 0 & 1 & 3 & | & 1 & 0 & 2 \\ 0 & 0 & 3 & | & -1 & 1 & 1 \\ \end{bmatrix}$$

$$\begin{matrix} \color{red}{R_1 + \frac{2}{3} R_3} \\ \color{red}{R_2 - R_3}\\ \color{red}{\frac{1}{3} R_3}\\ \end{matrix} \begin{bmatrix} 1 & 0 & 0 & | & -\frac{2}{3} & \frac{2}{3} & -\frac{1}{3}\\ 0 & 1 & 0 & | & 2 & -1 & 1 \\ 0 & 0 & 1 & | & -\frac{1}{3} & \frac{1}{3} & \frac{1}{3} \\ \end{bmatrix}$$
The above augmented matrix has the form $$[ I_3 | A^{-1} ]$$ and therefore $$A^{-1}$$ is given by
$$A^{-1} = \begin{bmatrix} -\frac{2}{3} & \frac{2}{3} & -\frac{1}{3} \\ 2 & -1 & 1 \\ -\frac{1}{3} & \frac{1}{3} & \frac{1}{3} \end{bmatrix}$$

Example 3
Find the inverse of matrix $$A = \begin{bmatrix} 0 & 1 & -1 & 1\\ 2 & 2 & 0 & -2\\ 1 & 1 & -2 & 0 \\ 0 & 1 & 2 & 0 \end{bmatrix}$$.

Solution to Example 3
Write the augmented matrix $$[ A | I_4 ]$$
$$\begin{bmatrix} 0 & 1 & -1 & 1 & | & 1 & 0 & 0 & 0\\ 2 & 2 & 0 & -2 & | & 0 & 1 & 0 & 0 \\ 1 & 1 & -2 & 0 & | & 0 & 0 & 1 & 0\\ 0 & 1 & 2 & 0 & | & 0 & 0 & 0 & 1 \\ \end{bmatrix}$$
Write the above augmented matrix in
reduced row echelon form .
Interchange $$R_1$$ and $$R_3$$
$$\begin{matrix} \color{red}{R_3} \\ \\ \color{red}{R_1}\\ \\ \end{matrix} \begin{bmatrix} 1 & 1 & -2 & 0 & | & 0 & 0 & 1 & 0\\ 2 & 2 & 0 & -2 & | & 0 & 1 & 0 & 0 \\ 0 & 1 & -1 & 1 & | & 1 & 0 & 0 & 0\\ 0 & 1 & 2 & 0 & | & 0 & 0 & 0 & 1 \\ \end{bmatrix}$$

$$\begin{matrix} \\ \color{red}{R_2 - 2 R_1} \\ \\ \\ \end{matrix} \begin{bmatrix} 1 & 1 & -2 & 0 & | & 0 & 0 & 1 & 0\\ 0 & 0 & 4 & -2 & | & 0 & 1 & -2 & 0 \\ 0 & 1 & -1 & 1 & | & 1 & 0 & 0 & 0\\ 0 & 1 & 2 & 0 & | & 0 & 0 & 0 & 1 \\ \end{bmatrix}$$

Interchange $$R_2$$ and $$R_4$$
$$\begin{matrix} \\ \color{red}{R_4} \\ \\ \color{red}{R_2} \\ \end{matrix} \begin{bmatrix} 1 & 1 & -2 & 0 & | & 0 & 0 & 1 & 0\\ 0 & 1 & 2 & 0 & | & 0 & 0 & 0 & 1 \\ 0 & 1 & -1 & 1 & | & 1 & 0 & 0 & 0\\ 0 & 0 & 4 & -2 & | & 0 & 1 & -2 & 0 \\ \end{bmatrix}$$

$$\begin{matrix} \\ \\ \color{red}{R_3- R_2} \\ \\ \end{matrix} \begin{bmatrix} 1 & 1 & -2 & 0 & | & 0 & 0 & 1 & 0\\ 0 & 1 & 2 & 0 & | & 0 & 0 & 0 & 1 \\ 0 & 0 & -3 & 1 & | & 1 & 0 & 0 & -1\\ 0 & 0 & 4 & -2 & | & 0 & 1 & -2 & 0 \\ \end{bmatrix}$$

$$\begin{matrix} \\ \\ \color{red}{R_3 + R_4} \\ \\ \end{matrix} \begin{bmatrix} 1 & 1 & -2 & 0 & | & 0 & 0 & 1 & 0\\ 0 & 1 & 2 & 0 & | & 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & -1 & | & 1 & 1 & -2 & -1\\ 0 & 0 & 4 & -2 & | & 0 & 1 & -2 & 0 \\ \end{bmatrix}$$

$$\begin{matrix} \\ \\ \\ \color{red}{R_4 - 4 R_3} \\ \end{matrix} \begin{bmatrix} 1 & 1 & -2 & 0 & | & 0 & 0 & 1 & 0\\ 0 & 1 & 2 & 0 & | & 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & -1 & | & 1 & 1 & -2 & -1\\ 0 & 0 & 0 & 2 & | & -4 & -3 & 6 & 4 \\ \end{bmatrix}$$

$$\begin{matrix} \\ \\ \\ \color{red}{\frac{1}{2} R_4} \\ \end{matrix} \begin{bmatrix} 1 & 1 & -2 & 0 & | & 0 & 0 & 1 & 0\\ 0 & 1 & 2 & 0 & | & 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & -1 & | & 1 & 1 & -2 & -1\\ 0 & 0 & 0 & 1 & | & -2 & -3/2 & 3 & 2 \\ \end{bmatrix}$$

$$\begin{matrix} \\ \\ \color{red}{R_3 + R_4} \\ \\ \end{matrix} \begin{bmatrix} 1 & 1 & -2 & 0 & | & 0 & 0 & 1 & 0\\ 0 & 1 & 2 & 0 & | & 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 & | & -1 & -1/2 & 1 & 1\\ 0 & 0 & 0 & 1 & | & -2 & -3/2 & 3 & 2 \\ \end{bmatrix}$$

$$\begin{matrix} \color{red}{R_1 + 2 R_3} \\ \color{red}{R_2 -2 R_3} \\ \\ \\ \end{matrix} \begin{bmatrix} 1 & 1 & 0 & 0 & | & -2 & -1 & 3 & 2\\ 0 & 1 & 0 & 0 & | & 2 & 1 & -2 & -1 \\ 0 & 0 & 1 & 0 & | & -1 & -1/2 & 1 & 1\\ 0 & 0 & 0 & 1 & | & -2 & -3/2 & 3 & 2 \\ \end{bmatrix}$$

$$\begin{matrix} \color{red}{R_1 - R_2} \\ \\ \\ \\ \end{matrix} \begin{bmatrix} 1 & 0 & 0 & 0 & | & -4 & -2 & 5 & 3\\ 0 & 1 & 0 & 0 & | & 2 & 1 & -2 & -1 \\ 0 & 0 & 1 & 0 & | & -1 & -1/2 & 1 & 1\\ 0 & 0 & 0 & 1 & | & -2 & -3/2 & 3 & 2 \\ \end{bmatrix}$$
The above augmented matrix has the form $$[ I_4 | A^{-1} ]$$ and therefore $$A^{-1}$$ is given by
$$A^{-1} = \begin{bmatrix} -4 & -2 & 5 & 3 \\ 2 & 1 & -2 & -1 \\ -1 & -1/2 & 1 & 1 \\ -2 & -3/2 & 3 & 2 \end{bmatrix}$$

Example 4
Find the inverse of matrix $$A = \begin{bmatrix} 1 & 2 & 0\\ 0 & 1 & 3 \\ 1 & 4 & 6 \end{bmatrix}$$.

Solution to Example 4
Write the augmented matrix $$[ A | I_3 ]$$
$$\begin{bmatrix} 1 & 2 & 0 & | & 1 & 0 & 0\\ 0 & 1 & 3 & | & 0 & 1 & 0\\ 1 & 4 & 6 & | & 0 & 0 & 1 \end{bmatrix}$$
Write the above augmented matrix in reduced row echelon form .
$$\begin{matrix} \\ \\ \color{red}{R_3 - R_1}\\ \end{matrix} \begin{bmatrix} 1 & 2 & 0 & | & 1 & 0 & 0\\ 0 & 1 & 3 & | & 0 & 1 & 0\\ 0 & 2 & 6 & | & -1 & 0 & 1 \\ \end{bmatrix}$$

$$\begin{matrix} \\ \\ \color{red}{R_3 - 2 R_2}\\ \end{matrix} \begin{bmatrix} 1 & 2 & 0 & | & 1 & 0 & 0\\ 0 & 1 & 3 & | & 0 & 1 & 0\\ 0 & 0 & 0 & | & -1 & -2 & 1 \\ \end{bmatrix}$$

The last row of the original matrix (on the left side) is all zeros and therefore the rows in the orgiginal matrix $$A$$ are not linearly independent and hence the given matrix is NOT invertible.