# Inverse of 2 x 2 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, find the inverse of a $2 \times 2$ matrix, and the formula for the inverse of a $2 \times 2$ matrix. There will be a lot of examples for you to look at. Practice problems will follow. Happy learning!

## 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 $2 \times 2$ matrix shown below:

$A = \begin{bmatrix} { a } & { b } \\ { c } & { d } \end {bmatrix}$

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

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 2 x 2 Matrix

So how do we find the inverse of a $2 \times 2$ 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:

• The matrix needs to be a square matrix (the number of rows must be equal to the number of columns).
• 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.

We will look at a nifty formula for finding the inverse of a $2 \times 2$ matrix below.

## 2 x 2 Inverse Matrix Formula

Consider the $2 \times 2$ matrix shown below:

$A = \begin{bmatrix} { a } & { b } \\ { c } & { d } \end {bmatrix}$

The formula for the inverse of a $2 \times 2$ matrix (Matrix $A$) is given as:

$A^{ – 1 } = \frac{ 1 }{ad – bc} \begin{bmatrix} d & { – b } \\ { – c } & a \end {bmatrix}$

The quantity $ad – bc$ is known as the determinant of the matrix. Read more about the determinant of $2 \times 2$ matrices here.

In other words, to calculate the inverse, we interchange $a$ and $d$, negate $b$ and $c$, and divide the result by the determinant of the matrix!

Let’s calculate the inverse of  a $2 \times 2$ matrix ( Matrix $B$ ) shown below:

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

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

• Is it a square matrix?

Yes, it is a $2 \times 2$ square matrix!

• Is the determinant equal to $0$?

Let’s calculate the determinant of Matrix $B$ by using the determinant formula for a $2 \times 2$ matrix.

$det( B ) = | B | = \begin{vmatrix} { 4 } & { – 2 } \\ { 3 } & { – 4 } \end {vmatrix}$

$= ( 4 ) ( – 4 ) – ( – 2 ) ( 3 )$

$= – 16 + 6$

$= – 10$

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

$B^{ – 1 } = \frac{ 1 }{ad – bc} \begin{bmatrix} d & { – b } \\ { – c } & a \end {bmatrix}$

$B^{ – 1 } = – \frac{ 1 }{ 10 } \begin{bmatrix} { – 4 } & { 2 } \\ { – 3 } & { 4 } \end {bmatrix}$

$B^{ – 1 } = \begin{bmatrix} { \frac{ 4 }{ 10 } } & { – \frac{ 2 }{ 10 } } \\ { \frac{ 3 }{ 10 } } & { – \frac{ 4 }{ 10 } } \end {bmatrix}$

Note: In the last step, we multiplied the scalar constant, $– \frac{1}{10}$, with each element of the matrix. This is the scalar multiplication of a matrix.

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

$B^{ – 1 } = \begin{bmatrix} { \frac{ 2 }{ 5 } } & { – \frac{ 1 }{ 5 } } \\ { \frac{ 3 }{ 10 } } & { – \frac{ 2 }{ 5 } } \end {bmatrix}$

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

#### Example 1

Given $C = \begin{bmatrix} { – 10 } & { – 5 } \\ { 6 } & { – \frac{ 2 }{ 5 } } \end {bmatrix}$, find $C^{ – 1 }$.

Solution

We will use the formula for the inverse of a $2 \times 2$ matrix to find the inverse of Matrix $C$. Shown below:

$C^{ – 1 } = \frac{ 1 }{ad – bc} \begin{bmatrix} d & { – b } \\ { – c } & a \end {bmatrix}$

$C^{ – 1 } = \frac{ 1 }{( -10 )( – \frac{ 2 }{ 5 } ) – ( – 5 )(6)} \begin{bmatrix} -1 & 2 \\ 3 & 1 \end {bmatrix}$

$C^{ – 1 } = \frac{1}{4 + 30}\begin{bmatrix} { – \frac{ 2 }{ 5 } } & { 5 } \\ { – 6 } & { – 10 } \end {bmatrix}$

$C^{ – 1 } = \frac{1}{34}\begin{bmatrix} { – \frac{ 2 }{ 5 } } & { 5 } \\ { – 6 } & { – 10 } \end {bmatrix}$

$C^{ – 1 } = \begin{bmatrix} { – \frac{ 1 }{ 85 } } & { \frac{ 5 }{ 34 } } \\ { – \frac{ 3 }{ 17 } } & { – \frac{ 5 }{ 17 } } \end {bmatrix}$

#### Example 2

Given $A= \begin{bmatrix} 0 & { – 4 } \\ { – 1 } & 1 \end {bmatrix}$ and $B= \begin{bmatrix} -\frac{ 1 }{ 4 } & -1 \\ -\frac{ 1 }{ 4 } & 0 \end {bmatrix}$, 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 ($2 \times 2$ identity matrix). If so, $B$ is the inverse of $A$.

Let’s check:

$A\times B= \begin{bmatrix} 0 & { – 4 } \\ { – 1 } & 1 \end {bmatrix} \times \begin{bmatrix} -\frac{ 1 }{ 4 } & -1 \\ -\frac{ 1 }{ 4 } & 0 \end {bmatrix}$

$= \begin{bmatrix} (0)(-\frac{1}{4}) + (-4)(-\frac{1}{4}) & (0)(-1) + (-4)(0) \\ (-1)(-\frac{1}{4}) + (1)(-\frac{1}{4}) & (-1)(-1) + (1)(0) \end {bmatrix}$

$= \begin{bmatrix} { 1 } & { 0 } \\ { 0 } & { 1 } \end {bmatrix}$

This is the $2 \times 2$ identity matrix!

Thus, Matrix $B$ is the inverse of Matrix $A$.

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

### Practice Questions

1. Given $A = \begin{bmatrix} { \frac{ 1 }{ 2 } } & { – \frac{ 1 }{ 2 } } \\ { \frac{ 3 }{ 2 } } & { \frac{ 1 }{ 12 } } \end {bmatrix}$, find $A^{ – 1 }$.

2. Given $B = \begin{bmatrix} { – 4 } & { 12 } \\ { – 2 } & { 6 } \end {bmatrix}$, find $B^{ – 1 }$.
3. Find the inverse of matrix $C$ shown below:
$C = \begin{bmatrix} { 2 } & { 1 } \\ { – 2 } & { 2 } \\ { 1 } & 7 \end {bmatrix}$
4. Given $J = \begin{bmatrix} 1 & { 3 } \\ { – 2} & – 10 \end {bmatrix}$ and $K = \begin{bmatrix} \frac{ 5 }{ 2 } & \frac{ 4 }{ 3 } \\ – \frac{ 1 }{ 2 } & – \frac{ 1 }{ 4 } \end {bmatrix}$, confirm if  Matrix $K$ is the inverse of Matrix $J$.

1. We will use the formula for the inverse of a $2 \times 2$ matrix to find the inverse of Matrix $A$. Shown below:

$A^{ – 1 } = \frac{ 1 }{ad – bc} \begin{bmatrix} d & { – b } \\ { – c } & a \end {bmatrix}$

$A^{ – 1 } = \frac{ 1 }{( \frac{ 1 }{ 2 } )( \frac{ 1 }{ 12 } ) – ( – \frac{1}{2} )(\frac{ 3 }{ 2 })} \begin{bmatrix} \frac{ 1 }{ 12 } & \frac{ 1 }{ 2 } \\ – \frac{ 3 }{ 2 } & \frac{ 1 }{ 2 } \end {bmatrix}$

$A^{ – 1 } = \frac{ 1 }{ \frac{ 1 }{ 24 } + \frac{ 3 }{ 4 } } \begin{bmatrix} \frac{ 1 }{ 12 } & \frac{ 1 }{ 2 } \\ – \frac{ 3 }{ 2 } & \frac{ 1 }{ 2 } \end {bmatrix}$

$A^{ – 1 } = \frac{1}{\frac{ 19 }{ 24 } } \begin{bmatrix} \frac{ 1 }{ 12 } & \frac{ 1 }{ 2 } \\ – \frac{ 3 }{ 2 } & \frac{ 1 }{ 2 } \end {bmatrix}$

$A^{ – 1 } = \frac{ 24 }{ 19 } \begin{bmatrix} \frac{ 1 }{ 12 } & \frac{ 1 }{ 2 } \\ – \frac{ 3 }{ 2 } & \frac{ 1 }{ 2 } \end {bmatrix}$

$A^{ – 1 } = \begin{bmatrix} \frac{ 2 }{ 19 } & \frac{ 12 }{ 19 } \\ – \frac{ 36 }{ 19 } & \frac{ 12 }{ 19 } \end {bmatrix}$

2. This matrix does not have an inverse.
Why?
Because its 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:

$| B | = ad – bc = ( – 4 )( 6 ) – ( 12 )( -2 ) = – 24 +24 = 0$

Thus, this matrix will not have an inverse!

3. This matrix does not have an inverse as well. Recall that only square matrices have inverses! This is not a square matrix. It is a $3 \times 2$ matrix with $3$ rows and $2$ columns. Thus, we can’t calculate the inverse of Matrix $C$.
4. For Matrix $K$ to be the inverse of Matrix $J$, the matrix multiplication between these two matrices should result in an identity matrix ($2 \times 2$ identity matrix). If so, $K$ is the inverse of $J$.

Let’s check:

$J\times K = \begin{bmatrix} 1 & { 3 } \\ { – 2} & – 10 \end {bmatrix} \times \begin{bmatrix} \frac{ 5 }{ 2 } & \frac{ 4 }{ 3 } \\ – \frac{ 1 }{ 2 } & – \frac{ 1 }{ 4 } \end {bmatrix}$

$= \begin{bmatrix} (1)( \frac{ 5 }{ 2 } ) + ( 3 )( – \frac{ 1 }{ 2 }) & ( 1 )(\frac{ 4 }{ 3 } ) + ( 3 )(- \frac{ 1 }{ 4 } ) \\ ( – 2 )( \frac{ 5 }{ 2 } ) + (- 10 )(-\frac{ 1 }{ 2 } ) & (- 2 )(\frac{ 4 }{ 3 } ) + (- 10 )(- \frac{ 1 }{ 4 } ) \end {bmatrix}$

$= \begin{bmatrix} { \frac{ 5 }{ 2 } – \frac{ 3 }{ 2 } } & { \frac{ 4 }{ 3 } – \frac{ 3 }{ 4 } } \\ { – 5 + 5 } & { – \frac{ 8 }{ 3 } + \frac{ 5 }{ 2 } } \end {bmatrix}$

$= \begin{bmatrix} { 1 } & { \frac{ 7 }{ 12 } } \\ { 0 } & { – \frac{ 1 }{ 6 } } \end {bmatrix}$

This is not the $2 \times 2$ identity matrix!

Thus, Matrix $K$ is NOT the inverse of Matrix $J$.