# Inverse of 3 x 3 Matrix – Explanation & Examples

The inverse of a matrix is significant in linear algebra. It helps us solve a system of linear equations. We can find the inverse of square matrices only. Some matrices do not have inverses. So, what is the inverse of a matrix?

The inverse of a matrix $A$ is $A^{ – 1 }$, such that multiplying the matrix with its inverse results in the identity matrix, $I$.

In this lesson, we will take a brief look at what an inverse matrix is, how to find the inverse of a $3 \times 3$ matrix, and the formula for the inverse of a $3 \times 3$ matrix. We will look at a couple of examples and some practice problems for you to try out!

## What is the Inverse of a Matrix?

In matrix algebra, matrix inverse plays the same role as a reciprocal in number systems. Inverse matrix is the matrix with which we can multiply another matrix to get the identity matrix (the matrix equivalent of the number $1$)To know more about the identity matrix, please check here.

Consider the $3 \times 3$ matrix shown below:

$B = \begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end {bmatrix}$

We denote the inverse of this matrix as $B^{ – 1 }$.

The multiplicative inverse (reciprocal) in the number system and the inverse matrix in matrices play the same role. Also, the identity matrix ($I$ ) (in matrices domain) plays the same role as the number one ( $1$ ).

## How to Find the Inverse of a 3 x 3 Matrix

So how do we find the inverse of a $3 \times 3$ matrix?

To find the inverse of a matrix, we can use a formula that requires a few points to be satisfied before its usage.

For a matrix to have an inverse, it has to satisfy $2$ conditions:

1. The matrix needs to be a square matrix (the number of rows must be equal to the number of columns).
2. The determinant of the matrix (this is a scalar value of a matrix from a few operations done on its elements) must not be $0$.

Remember, not all matrices that are square matrices have an inverse. A matrix whose determinant is $0$ is not invertible (doesn’t have an inverse) and is known as a singular matrix.

The formula for the inverse of a $3 \times 3$ matrix is quite messy! Nonetheless, let’s tackle it!!

## 3 x 3 Inverse Matrix Formula

Consider the $3 \times 3$ matrix shown below:

$A = \begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end {bmatrix}$

The formula for the inverse of a $3 \times 3$ matrix (Matrix $A$) is given as:

$A^{ – 1 } = \frac{ 1 }{ det ( A ) } \begin{bmatrix} { (ei – fh) } & { – (bi – ch) } & {(bf – ce)} \\ { – (di- fg) } & { (ai – cg)} & { – (af – cd)} \\ { (dh – eg)} & { – (ah – bg)} & {(ae – bd)} \end {bmatrix}$

Where $det( A )$ is the determinant of the $3\times 3$ matrix given as:

$det(A) = a(ei – fh) – b(di – fg) + c(dh – eg)$

Tough!
Tough!
But don’t worry, after working out several questions, it will come to you naturally!

Let’s calculate the inverse of  a $3 \times 3$ matrix ( Matrix $C$ ) shown below:

$C = \begin{bmatrix} 1 & 2 & 1 \\ 3 & 4 & 1 \\ { – 1 } & 2 & { – 1 } \end {bmatrix}$

Before we calculate the inverse, we have to check the $2$ conditions outlined above.

• Is it a square matrix?

Yes, it is a $3 \times 3$ square matrix!

• Is the determinant equal to $0$?

Let’s calculate the determinant of Matrix $C$ by using the determinant formula for a $3 \times 3$ matrix.

$| C | = a(ei – fh) – b(di – fg) + c(dh – eg)$

$= 1( – 4 – 2 ) – 2(- 3 – ( – 1 ) ) + 1(6 – ( – 4 ) )$

$= 1( – 6 ) – 2( – 2 ) + 1 ( 10 )$

$= 8$

The determinant isn’t $0$. So, we can go ahead and calculate the inverse using the formula we just learned. Shown below:

$C^{ – 1} = \frac{ 1 }{ det( C ) } \begin{bmatrix} { ( ei – fh ) } & { – ( bi – ch ) } & { ( bf – ce ) } \\ { – ( di – fg ) } & { ( ai – cg ) } & { – ( af – cd ) } \\ { ( dh – eg ) } & { – ( ah – bg ) } & { ( ae – bd ) } \end {bmatrix}$

$C^{ – 1} = \frac{ 1 }{ 8 } \begin{bmatrix} { – 6 } & { 4 } & { – 2 } \\ { 2 } & { 0 } & { 2 } \\ { 10 } & { – 4 } & { – 2 } \end {bmatrix}$

$C^{ – 1 } = \begin{bmatrix} { – \frac{ 6 }{ 8 } } & { \frac{ 4 }{ 8 } } & { – \frac{ 2 }{ 8 } }\\ { \frac{ 2 }{ 8 } } & { 0 } & { \frac{ 2 }{ 8 } } \\ { \frac{ 10 }{8} } & { – \frac{ 4 }{ 8 } } & { – \frac{ 2 }{ 8 } } \end{bmatrix}$

Note: We multiplied the scalar constant, $\frac{ 1 }{ 8 }$, with each element of the matrix. This is the scalar multiplication of a matrix.

Let’s reduce the fractions and write the final answer:

$C^{ – 1 } = \begin{bmatrix} { – \frac{ 3 }{ 4 } } & { \frac{ 1 }{ 2 } } & {- \frac{ 1 }{ 4 } } \\ { \frac{ 1 }{ 4 } } & 0 & { \frac{ 1 }{ 4 } } \\ { \frac{ 5 }{ 4 } } & {- \frac{ 1 }{ 2 } } & {- \frac{ 1 }{ 4 }} \end {bmatrix}$

Let us look at some examples to enhance our understanding further!

#### Example 1

Given $A = \begin{bmatrix} 0 & 1 & 4 \\ { – 1 } & { – 1 } & 1 \\ 4 & { – 2 } & 0 \end{bmatrix}$, find $A^{ – 1 }$.

Solution

We will use the formula for the inverse of a $3 \times 3$ matrix to find the inverse of Matrix $A$. Shown below:

$A^{- 1} = \frac{ 1 }{a(ei – fh) – b(di – fg) + c(dh – eg)} \begin{bmatrix} { ( ei – fh ) } & { – ( bi – ch ) } & { ( bf – ce ) } \\ { – ( di – fg ) } & { ( ai – cg ) } & { – ( af – cd ) } \\ { ( dh – eg ) } & { – ( ah – bg ) } & { ( ae – bd ) } \end {bmatrix}$

$A^{ – 1 } = \frac{ 1 }{0( 2 ) – 1( -4 ) + 4( 6 ) } \begin{bmatrix} 2 & -8 & 5 \\ 4 & -16 & -4 \\ 6 & 4 & 1 \end {bmatrix}$

$A^{ – 1 } = \frac{ 1 }{ 28 } \begin{bmatrix} 2 & -8 & 5 \\ 4 & -16 & -4 \\ 6 & 4 & 1 \end {bmatrix}$

$A^{ – 1 } = \begin{bmatrix} \frac{ 1 }{ 14 } & – \frac{ 2 }{ 7 } & \frac{ 5 }{ 28 } \\ \frac{ 1 }{ 7 } & -\frac{ 4 }{ 7 } & -\frac{ 1 }{ 7 } \\ \frac{ 3 }{ 14 } & \frac{ 1 }{ 7 } & \frac{ 1 }{ 28 } \end {bmatrix}$

#### Example 2

Given $A= \begin{bmatrix} 2 & 2 & 1 \\ 0 & 1 & 0 \\ 1 & 2 & 1 \end {bmatrix}$ and $B= \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & { – 2 } & 2 \end {bmatrix}$, confirm if  Matrix $B$ is the inverse of Matrix $A$.

Solution

For Matrix $B$ to be the inverse of Matrix $, A$, the matrix multiplication between these two matrices should result in an identity matrix ($3 \times 3$ identity matrix). If so, $B$ is the inverse of $A$.

Let’s check:

$A\times B= \begin{bmatrix} 2 & 2 & 1 \\ 0 & 1 & 0 \\ 1 & 2 & 1 \end {bmatrix} \times \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & -2 & 2 \end {bmatrix}$

$=\begin{bmatrix} { (2)(1) + (2)(0) + (1)(1) } & { (2)(0) + (2)(1) + (1)(-2) } & { (2)(1) + (2)(0) + (1)(2) } \\ { (0)(1) + (1)(0) + (0)(1) } & { (0)(0) + (1)(1) + (0)(-2) } & { (0)(1) + (1)(0) + (0)(2) } \\ { (1)(1) + (2)(0) + (1)(1)} & { (1)(0) + (2)(1) + (1)(-2) } & {(1)(1) + (2)(0) + (1)(2) } \end {bmatrix}$

$= \begin{bmatrix} 3 & 0 & 4 \\ 0 & 1 & 0 \\ 2 & 0 & 3 \end{bmatrix}$

This is not the  $3 \times 3$ identity matrix

Thus, Matrix $B$ is not the inverse of Matrix $A$.

If you want to review matrix multiplication, please check this lesson out!

### Practice Questions

1. Given $K = \begin{bmatrix} 0 & 2 & -1 \\ 3 & -2 & 1 \\ 3 & 2 & -1 \end {bmatrix}$, find $K^{ – 1 }$.

2. Calculate $A^{ – 1 }$ for Matrix $A$ shown below:
$A = \begin{bmatrix} 1 & – 9 & 1 \\ – 3 & – 1 & 9 \end{bmatrix}$
3. Calculate the inverse of the $3 \times 3$ matrix shown below:
$D = \begin{bmatrix} 2 & 4 & 8 \\ 0 & 1 & 0 \\ 1 & -4 & 1 \end{bmatrix}$

1. This matrix does not have an inverse because this matrix’s determinant is equal to $0$!

Recall that the determinant cannot be $0$ for a matrix to have an inverse. Let’s check the value of the determinant:

$| K | = 0( 2 – 2 ) – 2( – 3 – 3 ) +( – 1 )( 6 + 6 )$
$| K | = 0( 0 ) – 2 ( – 6 ) – 1( 12 )$
$| K | = 12 – 12$
$| K | = 0$

Since the determinant is $0$, this matrix will not have an inverse!

2. If you look at this matrix carefully, you will see that it is not a square matrix!. It is a $2 \times 3$ matrix ( $2$ rows and $3$  columns). Recall that we cannot find the inverse of a non-square matrix.
Thus, Matrix $A$ doesn’t have an inverse!
3. We will use the formula for the inverse of a $3 \times 3$ matrix to find the inverse of Matrix $D$. Shown below:

$D^{ – 1 } = \frac{ 1 }{a(ei – fh) – b(di – fg) + c(dh – eg)} \begin{bmatrix} { ( ei – fh ) } & { – ( bi – ch ) } & { ( bf – ce ) } \\ { – ( di – fg ) } & { ( ai – cg ) } & { – ( af – cd ) } \\ { ( dh – eg ) } & { – ( ah – bg ) } & { ( ae – bd ) } \end {bmatrix}$

$D^{ – 1 } = \frac{ 1 }{2( 1 ) – 4( 0 ) +8( – 1 ) } \begin{bmatrix} 1 & – 36 & – 8 \\ 0 & – 6 & 0 \\ – 1 & 12 & 2 \end {bmatrix}$

$D^{ – 1 } = \frac{ 1 }{ – 6 } \begin{bmatrix} 1 & – 36 & – 8 \\ 0 & – 6 & 0 \\ – 1 & 12 & 2 \end {bmatrix}$

$D^{ – 1 } = \begin{bmatrix} – \frac{ 1 }{ 6 } & 6 & \frac{ 4 }{ 3 } \\ 0 & 1 & 0 \\ \frac{ 1 }{ 6 } & – 2 & – \frac{ 1 }{ 3 } \end {bmatrix}$