# 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:

- Interchange two rows
- Add a multiple of one row to another
- Multiply a row by a non zero constant

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} = I_{n} = A^{-1} A

_{n}is the n × 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} ]

^{-1}is the n × n on the right side of the augmented matrix [ I

_{n}| A

^{-1}].

## Examples with Solutions

__Example 1__

Find the inverse of matrix

__Solution to Example 1__

Write the augmented matrix [ A | I_{2} ]

Let R_{1} and R_{2} 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 [ I_{2} | 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 | I_{3} ]

Let R_{1}, R_{2} and R_{3} 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.