# 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 � n matrix. If matrix A-1 is the inverse of matrix A , then we have

A A-1 = In = A-1 A

where In is the n � n identity matrix
Consider the matrix equation
A A-1 = In where A-1 is the unknown. To find the inverse A-1 , we start with the augmented matrix [ A | In ] and then row reduce it. If matrix A is invertible, the row reduction will end with an augmented matrix in the form

[ In | A-1 ]

where the inverse A-1 is the n � n on the right side of the augmented matrix [ In | A-1 ].

## Examples with Solutions

Example 1
Find the inverse of matrix Solution to Example 1
Write the augmented matrix [ A | I2 ] Let R1 and R2 be the first and the second rows of the above augmented matrix.
Write the above augmented matrix in reduced row echelon form . The above augmented matrix has the form [ I2 | A-1 ] and therefore A-1 is given by Example 2
Find the inverse of matrix Solution to Example 2
Write the augmented matrix [ A | I3 ] Let R1, R2 and R3 be the first, the second and the third rows respectively of the above augmented matrix.
Write the above augmented matrix in reduced row echelon form . 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 $$R_3$$ Interchange $$R_2$$ and $$R_4$$ 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.