# Inverse Matrices

Do you know that just as real numbers other than zero can have an inverse, matrices can have inverses too? Hereafter, you would understand how to calculate the inverse of matrices.

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## Definition of Inverse matrices

A matrix is said to be the inverse of another matrix if the product of both matrices results in an identity matrix. However, before going into inverse matrices we need to refresh our knowledge of identity matrix.

### What is an Identity matrix?

An identity matrix is a square matrix in which when multiplied by another square matrix equals to the same matrix. In this matrix, the elements from the topmost left diagonal to the downmost right diagonal is 1 while every other element in the matrix is 0. Below are examples of a 2 by 2 and 3 by 3 identity matrix respectively:

A 2 by 2 identity matrix:

$\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\phantom{\rule{0ex}{0ex}}$

A 3 by 3 identity matrix:

$\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$

Thus, the inverse of a matrix can be derived as:

Where I is the identity matrix and A is a square matrix, then:

$A×I=I×A=A$

To have a little insight on this, consider:

$A×I=A\phantom{\rule{0ex}{0ex}}I=A×{A}^{-1}$

A-1 is the inverse of matrix A. The equation:

$I=A×{A}^{-1}$

means that the product of matrix A and inverse matrix A would give I, the identity matrix.

Therefore, we can verify if two matrices being multiplied are inverse of each other.

Verify if the following are inverse matrices or not.

a.

$A=\left[\begin{array}{cc}2& 2\\ -1& 4\end{array}\right]andB=\left[\begin{array}{cc}\frac{1}{2}& \frac{1}{2}\\ -1& \frac{1}{4}\end{array}\right]$

b.

$M=\left[\begin{array}{cc}3& 4\\ 1& 2\end{array}\right]andN=\left[\begin{array}{cc}1& -2\\ -\frac{1}{2}& \frac{3}{2}\end{array}\right]$

Solution:

a. find the product between matrix A and B;

$A×B=\left[\begin{array}{cc}2& 2\\ -1& 4\end{array}\right]×\left[\begin{array}{cc}\frac{1}{2}& \frac{1}{2}\\ -1& \frac{1}{4}\end{array}\right]\phantom{\rule{0ex}{0ex}}A×B=\left[\begin{array}{cc}\left(2×\frac{1}{2}\right)+\left(2×\left(-1\right)\right)& \left(2×\frac{1}{2}\right)+\left(2×\frac{1}{4}\right)\\ \left(-1×\frac{1}{2}\right)+\left(4×\left(-1\right)\right)& \left(-1×\frac{1}{2}\right)+\left(4×\frac{1}{4}\right)\end{array}\right]\phantom{\rule{0ex}{0ex}}A×B=\left[\begin{array}{cc}1-2& 1+\frac{1}{2}\\ -\frac{1}{2}-4& -\frac{1}{2}+1\end{array}\right]\phantom{\rule{0ex}{0ex}}A×B=\left[\begin{array}{cc}-1& 1\frac{1}{2}\\ -4\frac{1}{2}& \frac{1}{2}\end{array}\right]$

Since the product of matrix A and B fails to give an identity matrix, hence, A is not an inverse of B and vice versa.

b.

$M×N=\left[\begin{array}{cc}3& 4\\ 1& 2\end{array}\right]×\left[\begin{array}{cc}1& -2\\ -\frac{1}{2}& \frac{3}{2}\end{array}\right]\phantom{\rule{0ex}{0ex}}M×N=\left[\begin{array}{cc}\left(3×1\right)+\left(4×\left(-\frac{1}{2}\right)\right)& \left(3×\left(-2\right)\right)+\left(4×\frac{3}{2}\right)\\ \left(1×1\right)+\left(2×\left(-\frac{1}{2}\right)& \left(1×\left(-2\right)\right)+\left(2×\frac{3}{2}\right)\end{array}\right]\phantom{\rule{0ex}{0ex}}M×N=\left[\begin{array}{cc}3-2& -6+6\\ 1-1& -2+3\end{array}\right]\phantom{\rule{0ex}{0ex}}M×N=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\phantom{\rule{0ex}{0ex}}$

Since the product of matrices M and N yields an identity matrix, it means matrix M is the inverse of matrix N.

## What methods are used in finding the inverse of matrices?

There are three ways of finding the inverse of matrices, namely:

1. Determinant method for 2 by 2 matrices.

2. Gaussian method or augmented matrix.

3. The adjoint method through the use of matrix cofactors.

However, at this level, we shall only learn the determinant method.

## Determinant method

In order to find the inverse of a 2 by 2 matrix, you should apply this formula:

$M=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]\phantom{\rule{0ex}{0ex}}{M}^{-1}=\frac{1}{ad-bc}\left[\begin{array}{cc}d& -b\\ -c& a\end{array}\right]$

Provided that:

$ad-bc\ne 0$

Where the determinant of a matrix is 0, there is no inverse.

Therefore, the inverse of a 2 by 2 matrix is the product of the inverse of the determinant and the matrix being altered. The altered matrix is gotten by swapping the diagonal elements with the cofactor sign on each.

Find the inverse of matrix B.

$B=\left[\begin{array}{cc}1& 0\\ 2& 3\end{array}\right]$

Solution:

$B=\left[\begin{array}{cc}1& 0\\ 2& 3\end{array}\right]$

Using;

${\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]}^{-1}=\frac{1}{ad-bc}\left[\begin{array}{cc}d& -b\\ -c& a\end{array}\right]$

Then;

${B}^{-1}=\frac{1}{\left(1×3\right)-\left(0×2\right)}\left[\begin{array}{cc}3& 0\\ -2& 1\end{array}\right]\phantom{\rule{0ex}{0ex}}{B}^{-1}=\frac{1}{3-0}\left[\begin{array}{cc}3& 0\\ -2& 1\end{array}\right]\phantom{\rule{0ex}{0ex}}{B}^{-1}=\frac{1}{3}\left[\begin{array}{cc}3& 0\\ -2& 1\end{array}\right]$

or,

${B}^{-1}=\frac{1}{3}\left[\begin{array}{cc}3& 0\\ -2& 1\end{array}\right]=\left[\begin{array}{cc}\frac{3}{3}& 0\\ -\frac{2}{3}& \frac{1}{3}\end{array}\right]\phantom{\rule{0ex}{0ex}}{B}^{-1}=\left[\begin{array}{cc}1& 0\\ -\frac{2}{3}& \frac{1}{3}\end{array}\right]$

Most importantly, once your determinant is calculated and your answer is equal to 0, it just means that the matrix has no inverse.

The inverse of 3 by 3 matrices can also be derived using:

${M}^{-1}=\frac{1}{\left|\begin{array}{c}M\end{array}\right|}adj\left(M\right)$

Where,

$\left|\begin{array}{c}M\end{array}\right|$is the determinant of a matrix M

adj(M) is the adjoint of matrix M

To achieve this, four basic steps are followed:

Step 1 - Find the determinant of the given matrix. If the determinant is equal to 0, it means no inverse.

Step 2 - Find the cofactor of the matrix.

Step 3 - Transpose of the cofactor matrix to give the adjoint of the matrix.

Step 4 - Divide the adjoint matrix by the determinant of the matrix.

## Examples of inverse matrices

Let's have some more examples to understand inverse matrices better.

Find the inverse of the matrix X.

$X=\left[\begin{array}{ccc}2& 1& -3\\ 5& 3& 0\\ -4& 2& 1\end{array}\right]$

Solution:

This is a 3 by 3 matrix.

Step1: Find the determinant of the given matrix.

$\left|\begin{array}{c}X\end{array}\right|=2\left|\begin{array}{cc}3& 0\\ 2& 1\end{array}\right|-1\left|\begin{array}{cc}5& 0\\ -4& 1\end{array}\right|-3\left|\begin{array}{cc}5& 3\\ -4& 2\end{array}\right|\phantom{\rule{0ex}{0ex}}\left|\begin{array}{c}X\end{array}\right|=2\left(3-0\right)-1\left(5-0\right)-3\left(10+12\right)\phantom{\rule{0ex}{0ex}}\left|\begin{array}{c}X\end{array}\right|=6-5-66\phantom{\rule{0ex}{0ex}}\left|\begin{array}{c}X\end{array}\right|=-65$

Since the determinant is not equal to 0, it means that the matrix X has an inverse.

Step2: Find the cofactor of the matrix.

The cofactor is calculated with

${C}_{ij}={\left(-1\right)}^{i+j}×{M}_{ij}$

The cofactor of 2 which is C11 is

${C}_{11}={\left(-1\right)}^{1+1}×\left|\begin{array}{cc}3& 0\\ 2& 1\end{array}\right|\phantom{\rule{0ex}{0ex}}{C}_{11}=1\left(3-0\right)\phantom{\rule{0ex}{0ex}}{C}_{11}=3$

The cofactor of 1 which is C12 is

${C}_{12}={\left(-1\right)}^{1+2}×\left|\begin{array}{cc}5& 0\\ -4& 1\end{array}\right|\phantom{\rule{0ex}{0ex}}{C}_{12}=-1\left(5-0\right)\phantom{\rule{0ex}{0ex}}{C}_{12}=-5$

The cofactor of -3 which is C13 is

${C}_{13}={\left(-1\right)}^{1+3}×\left|\begin{array}{cc}5& 3\\ -4& 2\end{array}\right|\phantom{\rule{0ex}{0ex}}{C}_{13}=1\left(10+12\right)\phantom{\rule{0ex}{0ex}}{C}_{13}=22$

The cofactor of 5 which is C21 is

${C}_{21}={\left(-1\right)}^{2+1}×\left|\begin{array}{cc}1& -3\\ 2& 1\end{array}\right|\phantom{\rule{0ex}{0ex}}{C}_{21}=-1\left(1+6\right)\phantom{\rule{0ex}{0ex}}{C}_{21}=-7$

The cofactor of 3 which is C22 is

${C}_{22}={\left(-1\right)}^{2+2}×\left|\begin{array}{cc}2& -3\\ -4& 1\end{array}\right|\phantom{\rule{0ex}{0ex}}{C}_{22}=1\left(2+12\right)\phantom{\rule{0ex}{0ex}}{C}_{22}=14$

The cofactor of 0 which is C23 is

${C}_{23}={\left(-1\right)}^{2+3}×\left|\begin{array}{cc}2& 1\\ -4& 2\end{array}\right|\phantom{\rule{0ex}{0ex}}{C}_{23}=-1\left(4+4\right)\phantom{\rule{0ex}{0ex}}{C}_{23}=-8$

The cofactor of -4 which is C31 is

${C}_{31}={\left(-1\right)}^{3+1}×\left|\begin{array}{cc}1& -3\\ 3& 0\end{array}\right|\phantom{\rule{0ex}{0ex}}{C}_{31}=1\left(0+9\right)\phantom{\rule{0ex}{0ex}}{C}_{31}=9$

The cofactor of 2 which is C32 is

${C}_{32}={\left(-1\right)}^{3+2}×\left|\begin{array}{cc}2& -3\\ 5& 0\end{array}\right|\phantom{\rule{0ex}{0ex}}{C}_{32}=-1\left(0+15\right)\phantom{\rule{0ex}{0ex}}{C}_{32}=-15$

The cofactor of 1 which is C33 is

${C}_{33}={\left(-1\right)}^{3+3}×\left|\begin{array}{cc}2& 1\\ 5& 3\end{array}\right|\phantom{\rule{0ex}{0ex}}{C}_{33}=1\left(6-5\right)\phantom{\rule{0ex}{0ex}}{C}_{33}=1$

So the cofactor of the matrix X is

${X}^{c}=\left[\begin{array}{ccc}3& -5& 22\\ -7& 14& -8\\ 9& -15& 1\end{array}\right]$

Step 3: Transpose of the cofactor matrix to give the adjoint of the matrix.

the transpose of Xc is

${\left({X}^{c}\right)}^{T}=Adj\left(X\right)=\left[\begin{array}{ccc}3& -7& 9\\ -5& 14& -15\\ 22& -8& 1\end{array}\right]$

Step 4: Divide the adjoint matrix by the determinant of the matrix.

Remember the determinant of matrix X is 65. This final stage gives us the inverse of matrix X which is X-1. Hence, we have

${X}^{-1}=\frac{1}{-65}\left[\begin{array}{ccc}3& -7& 9\\ -5& 14& -15\\ 22& -8& 1\end{array}\right]\phantom{\rule{0ex}{0ex}}{X}^{-1}=\left[\begin{array}{ccc}-\frac{3}{65}& \frac{7}{65}& -\frac{9}{65}\\ \frac{5}{65}& -\frac{14}{65}& \frac{15}{65}\\ -\frac{22}{65}& \frac{8}{65}& -\frac{1}{65}\end{array}\right]\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}{X}^{-1}=\mathbf{\left[}\begin{array}{ccc}\mathbf{-}\frac{\mathbf{3}}{\mathbf{65}}& \frac{\mathbf{7}}{\mathbf{65}}& \mathbf{-}\frac{\mathbf{9}}{\mathbf{65}}\\ \frac{\mathbf{1}}{\mathbf{13}}& \mathbf{-}\frac{\mathbf{14}}{\mathbf{65}}& \frac{\mathbf{3}}{\mathbf{13}}\\ \mathbf{-}\frac{\mathbf{22}}{\mathbf{65}}& \frac{\mathbf{8}}{\mathbf{65}}& \mathbf{-}\frac{\mathbf{1}}{\mathbf{65}}\end{array}\mathbf{\right]}$

Using matrix operations solve for x and y in the following:

$2x+3y=6\phantom{\rule{0ex}{0ex}}x-2y=-2$

Solution:

This equation can be represented in matrix form as

$\left[\begin{array}{cc}2& 3\\ 1& -2\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}6\\ -2\end{array}\right]$

Let the matrices be represented by P, Q and R respectively such that

$\left[P\right]×\left[Q\right]=\left[R\right]\phantom{\rule{0ex}{0ex}}$

We intend on finding matrix Q since it represents our unknowns x and y. So we make matrix Q the subject of the formula

${\left[P\right]}^{-1}×\left[P\right]×\left[Q\right]={\left[P\right]}^{-1}×\left[R\right]\phantom{\rule{0ex}{0ex}}{\left[P\right]}^{-1}×\left[P\right]=I\phantom{\rule{0ex}{0ex}}$

I is an Identity matrix and its determinant is 1.

$I\left[Q\right]=\left[R\right]×{\left[P\right]}^{-1}\phantom{\rule{0ex}{0ex}}\left[Q\right]=\left[R\right]×{\left[P\right]}^{-1}\phantom{\rule{0ex}{0ex}}$

${\left[P\right]}^{-1}={\left[\begin{array}{cc}2& 3\\ 1& -2\end{array}\right]}^{-1}\phantom{\rule{0ex}{0ex}}{\left[P\right]}^{-1}=\frac{1}{\left(-4-3\right)}\left[\begin{array}{cc}-2& -3\\ -1& 2\end{array}\right]\phantom{\rule{0ex}{0ex}}{\left[P\right]}^{-1}=\left[\begin{array}{cc}\frac{2}{7}& \frac{3}{7}\\ \frac{1}{7}& -\frac{2}{7}\end{array}\right]\phantom{\rule{0ex}{0ex}}$

Then,

$\left[Q\right]=\left[\begin{array}{cc}\frac{2}{7}& \frac{3}{7}\\ \frac{1}{7}& -\frac{2}{7}\end{array}\right]×\left[\begin{array}{c}6\\ -2\end{array}\right]\phantom{\rule{0ex}{0ex}}\left[Q\right]=\left[\begin{array}{c}\left(\frac{2}{7}×6\right)+\left(\frac{3}{7}×-2\right)\\ \left(\frac{1}{7}×6\right)+\left(\left(-\frac{2}{7}\right)×-2\right)\end{array}\right]\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\left[Q\right]=\left[\begin{array}{c}\frac{12}{7}-\frac{6}{7}\\ \frac{6}{7}+\frac{4}{7}\end{array}\right]\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\left[Q\right]=\left[\begin{array}{c}\frac{6}{7}\\ \frac{10}{7}\end{array}\right]\phantom{\rule{0ex}{0ex}}\left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}\frac{6}{7}\\ \frac{10}{7}\end{array}\right]\phantom{\rule{0ex}{0ex}}x=\frac{\mathbf{6}}{\mathbf{7}}\phantom{\rule{0ex}{0ex}}y=\frac{\mathbf{10}}{\mathbf{7}}\phantom{\rule{0ex}{0ex}}$

## Inverse Matrices - Key takeaways

• A matrix is said to be the inverse of another matrix if the product of both matrices results in an identity matrix.
• Inverse of a matrix is possible for a square matrix where the determinant is not equal to 0.
• The inverse of a two-by-two matrix is obtained using: ${\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]}^{-1}=\frac{1}{ad-bc}\left[\begin{array}{cc}d& -b\\ -c& a\end{array}\right]$

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##### Frequently Asked Questions about Inverse Matrices

How do you inverse the sum of two matrices?

You can calculate the inverse of the sum of two matrices by adding the two matrices, then applying the formula for inverse matrices on it.

What are the examples of matrices that can have an inverse?

Any matrix that has its determinant not equal to 0 is an example of a matrix that has an inverse.

How do you do the inverse of a  3x3 matrix?

To get the inverse of a 3 by 3 matrix, you need to find the determinant first. Then, divide the adjoint of the matrix by the determinant of the matrix.

How do you get the inverse of matrices in multiplication?

To get the inverse of matrices in multiplication, find the product of the matrices. Then, use the formula on the new matrix to find its inverse.

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