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.

Read more about singular matrices here!

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 $.

Answers

  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 $.

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