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Inverse Matrix – Explanation & Examples

The inverse of a matrix is widely used in linear algebra. It is of immense importance in higher-order linear algebra. For now, we will work on the basics of an inverse of a matrix. Let’s look at the formal definition of an inverse 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 look at what an inverse matrix is, how to find the inverse of a matrix, the formula for the inverse of a $ 2 \times 2 $ matrix and $ 3 \times 3 $ matrix, and examples to clarify our understanding of matrix inverses.

What is the Inverse of a Matrix?

Remember multiplicative inverses of numbers?

For example, take the number $ 8 $. What number can we multiply $ 8 $ by, to get $ 1 $?

Simple!

It is the reciprocal of $ 8 $, which is $ \frac { 1 }{ 8 } $!

$ 8 \times \frac{1}{8} = 1 $

Similarly, 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.

Check the picture below:

We denote the inverse of Matrix $ A $ 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 Matrix

So how do we find the inverse of matrices?

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

Does every matrix have an inverse?

Well, NO!

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 determinants and singular matrices!

We will look at a nifty formula for finding the inverse of a $ 2 \times 2 $ matrix below. Also, the formula for a $ 3 \times 3 $ matrix will be a real messy one! Be prepared!  🙂

Remember,

We can only calculate the inverse for square matrices! Also, the determinant of the matrix cannot be equal to $ 0 $ since the formula requires the division by the determinant. We know division by $ 0 $ is undefined!

Inverse Matrix Formula

We will find the inverse of $ 2 \times 2 $ and $ 3 \times 3 $ matrices in this section. 

Inverse of a 2 x 2 Matrix

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.

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

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

This is in fact a square matrix and let’s check first if the determinant is $ 0 $ or not. Let’s calculate the determinant:

$ det( B ) = | B | = \begin{vmatrix}  1   &  3  \\ { – 3 } &  2   \end {vmatrix} $

$ = ( 1 ) ( 2 ) – ( 3 ) ( – 3 ) $

$ = 2 + 9 $

$ = 11 $

The determinant isn’t $ 0 $. Now, let’s go ahead and calculate the inverse of matrix $ B $.

$ B^{ – 1 } = \frac{ 1 }{ 11 } \begin{bmatrix}  2   & { – 3 } \\  3  &  1   \end {bmatrix}  $

$ B^{ – 1 } =  \begin{bmatrix} {\frac{2}{11}}  & { – \frac{3}{11} } \\ { \frac{ 3 }{11}} & { \frac{1}{11} }  \end {bmatrix} $

Note: In the last step, we multiplied the scalar constant, $ \frac{1}{11} $, with each element of the matrix. This is the scalar multiplication of a matrix.

Inverse of a 3 x 3 Matrix

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

$ B = \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 $ B $ ) is given as:

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

Where $ det( B ) $ is the determinant of the $ 3\times3 $ matrix given as:

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

Woah! Really messy one!

But nonetheless, let’s calculate the inverse of  a $ 3 \times 3 $ matrix ( Matrix $ B $ ) shown below:

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

This is in fact a square matrix and let’s check first if the determinant is $ 0 $ or not. Let’s calculate the determinant:

$ det( B ) = | B | = 1 [ ( 3 )( 1 ) – ( – 4 )( 2 ) ] – 2 [ ( 0 )( 1 ) – ( – 4 )( – 1 ) ] + (-1) [ ( 0 )( 2 ) – ( 3 )( – 1 ) ] $
$ = 1 [ 3 + 8 ] – 2 [ 0 – 4 ] + (-1) [ 0 + 3 ] $
$ = 1 [ 11 ] – 2[ – 4 ] – 1[ 3 ] $
$ = 11 + 8 – 3 $
$ = 16 $

The determinant isn’t $ 0 $. Now, let’s go ahead and calculate the inverse of matrix $ B $. 

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

$ B^{ – 1 } = \frac{ 1 }{ 16 } \begin{bmatrix} { 11 }  & { – 4 } & { – 5 } \\ { 4 } & { 0 } & { 4 } \\ { 3 } & { – 4 } & { 3 }  \end {bmatrix}  $

$ B^{ – 1 } =  \begin{bmatrix} { \frac{ 11 }{ 16 } }  & { – \frac{1}{4} } & { – \frac{5}{16} } \\ { \frac{ 1 }{ 4 } } & { 0 } & { \frac{1}{4} } \\ { \frac{3 }{16} } & { – \frac{1}{4} } & { \frac{ 3 }{ 16 } }  \end {bmatrix}  $

Note: In the last step, we multiplied the scalar constant, $ \frac{ 1 }{ 16 } $, with each element of the matrix. This is the scalar multiplication of a matrix.

Let’s take a look at a couple of examples!

Example 1

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


Solution

We will use the formula 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 }{-7} \begin{bmatrix}  -1   & 2 \\ 3 &  1   \end {bmatrix}  $

$ C^{ – 1 } = \begin{bmatrix}  \frac{1}{7}   & – \frac{2}{7} \\ -\frac{3}{7} &  – \frac{1}{7}   \end {bmatrix}  $


Example 2

Given $ A= \begin{bmatrix}  0  & -2 \\ -1  & 1   \end {bmatrix} $ and $ B= \begin{bmatrix}  -\frac{1}{2}  & -1 \\ -\frac{1}{2}  & 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. If so, $B$ is the inverse of $A$. Let’s check:

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

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

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

This is the identity matrix.
Thus, Matrix $B$ is the inverse of Matrix $A$.

Practice Questions

  1. Given $A = \begin{bmatrix}  9   & 3  \\  -6  &  9  \end {bmatrix}$, find $A^{-1}$.

  2. Given $B = \begin{bmatrix} 1  & 6 & 4 \\ 2 & 4 & {-1} \\ {-1} & 2 & 5 \end {bmatrix}$, find $B^{-1}$.

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 }{(9)(9) – (3)(-6)} \begin{bmatrix}  9   & -3 \\ 6 &  9   \end {bmatrix}  $

    $ A^{ – 1 } = \frac{ 1 }{99} \begin{bmatrix}  9   & -3 \\ 6 &  9   \end {bmatrix}  $

    $ A^{ – 1 } = \begin{bmatrix}  \frac{1}{11}   & -\frac{1}{33} \\ \frac{2}{33} &   \frac{1}{11}   \end {bmatrix}  $

  2. This is a $ 3 \times 3 $ square matrix. Let’s calculate the determinant of this matrix.

    If we have a $ 3 \times 3 $ matrix $ A = \begin{bmatrix}  a   &  b  &  c  \\  d  &  e  &  f  \\  g  &  h  &  i  \end {bmatrix} $,

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

    Now, calculating the determinant of our matrix $ B $:

    $ det(B) = a(ei – fh) – b(di – fg) + c(dh – eg) $
    $ det(B) = 1(20+2) – 6(10-1) + 4(4+4) $
    $ det(B) = 1(22) – 6(9) + 4(8) $
    $ det(B) = 1(20+2) – 6(10-1) + 4(4+4) $
    $det(B) = 0$

    Since the determinant is $ 0 $, we can’t calculate the inverse of Matrix $B$. Thus, this matrix does not have an inverse!

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