# Algebra 2 - Identity & Inverse Matrices

## Introduction

IDENTITY MATRIX

• A square matrix in which the numerical value of all the diagonal elements is 1 and the remaining elements are 0 (zero), is called an Identity Matrix.
• A $n×n$ identity matrix is represented as ${I}_{n}$.
• An identity matrix of order 2 is written as ${I}_{2}=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\begin{array}{}\end{array}$.
• An identity matrix of the order 3 is written as ${I}_{3}=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$.
• Properties of Identity Matrix:
• It is always a square matrix.
• If we multiply any matrix by the identity matrix, the result will be the matrix itself.
• The product of two inverse matrices yields an identity matrix.

INVERSE OF A MATRIX:

• The inverse of a matrix $A$ is denoted by ${A}^{-1}$.
• It is defined only for square matrices.
• If A is a square matrix of order $n$, and if there exists another square matrix B of the same order $n$, such that $AB=BA={I}_{n}$, then B is called the inverse of A or reciprocal of A. Here, ${I}_{n}$ is an identity matrix of order $n$.
• If B is the inverse of A, then A is also the inverse of B.
• The inverse of a square matrix exists if and only if it is non-singular i.e. $\left|A\right|\ne 0$.

## Finding Inverse of a Matrix

• We have seen that the inverse of a square matrix exists only if the given matrix is non-singular. So, the first step towards finding the inverse of a matrix must be to check whether the given matrix is non-singular or not.
• There are two methods for finding the inverse of a given non-singular matrix:
1. by elementary (row or column) operations

1. Finding the inverse of a matrix by elementary operations

• If A is a non-singular matrix, then in order to find ${A}^{-1}$ using elementary row operations, we first write $A=IA$ and then apply a sequence of row operations on $A=IA$ till we get, $I=BA$. The matrix B will be the inverse of A.
• Similarly, if we wish to find ${A}^{-1}$ using column operations, then, we write $A=IA$ and apply a sequence of column operations on $A=IA$ till we get, $I=BA$.
• Refer to example 2 for a better understanding.

2. Finding the inverse of a matrix by the adjoint method

• The formula to find the inverse of a non-singular matrix A is given by: , where  means adjoint of A.
• Refer to example 3 for a better understanding.

## Solved Examples

Example 1: Find the inverse of a matrix $A={\left[\begin{array}{cc}2& 5\\ 6& 15\end{array}\right]}_{2×2}$ using a suitable method.

Solution: First of all, we must check whether it is non-singular or not. To do so, we will find out $\left|A\right|$. We know that for a non-singular matrix $\left|A\right|\ne 0$.

$\left|A\right|=\left(2×15\right)-\left(5×6\right)=30-30=0$.

Since $\left|A\right|=0$, the inverse of the given matrix doesn't exist.

Example 2: Find ${A}^{-1}$ if it exists for a matrix $A=\left[\begin{array}{ccc}-1& 4& 2\\ 2& 1& 0\\ 1& 0& 2\end{array}\right]$.

Solution: Let us check whether the given matrix is non-singular or not. For that, we evaluate the value of $\left|A\right|$.

$\left|A\right|=\left(-1\right)\left[\left(1×2\right)-0\right]-\left(4\right)\left[\left(2×2\right)-\left(1×0\right)\right]+\left(2\right)\left[\left(2×0\right)-\left(1×1\right)\right]$$=-2-16-2=-20$

Since $\left|A\right|\ne 0$, inverse of the given matrix exists. Now, it is up to us to select a convenient method to proceed ahead.

Let us find the inverse by elementary row operations.

First, we write $A=IA$.

or $\left[\begin{array}{ccc}-1& 4& 2\\ 2& 1& 0\\ 1& 0& 2\end{array}\right]=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]·A$

Applying ${R}_{1}\to {R}_{1}-{R}_{3}$, we get $\left[\begin{array}{ccc}-2& 4& 0\\ 2& 1& 0\\ 1& 0& 2\end{array}\right]=\left[\begin{array}{ccc}1& 0& -1\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]·A$

Applying ${R}_{1}\to {R}_{1}-4{R}_{2}$, we get $\left[\begin{array}{ccc}-10& 0& 0\\ 2& 1& 0\\ 1& 0& 2\end{array}\right]=\left[\begin{array}{ccc}1& -4& -1\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]·A$

Applying ${R}_{1}\to -\frac{1}{10}{R}_{1}$, we get $\left[\begin{array}{ccc}1& 0& 0\\ 2& 1& 0\\ 1& 0& 2\end{array}\right]=\left[\begin{array}{ccc}\frac{-1}{10}& \frac{2}{5}& \frac{1}{10}\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]·A$

Applying ${R}_{2}\to {R}_{2}-2{R}_{1}$, we get $\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 1& 0& 2\end{array}\right]=\left[\begin{array}{ccc}\frac{-1}{10}& \frac{2}{5}& \frac{1}{10}\\ \frac{1}{5}& \frac{1}{5}& \frac{-1}{5}\\ 0& 0& 1\end{array}\right]·A$

Applying ${R}_{3}\to {R}_{3}-{R}_{1}$, we get $\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 2\end{array}\right]=\left[\begin{array}{ccc}\frac{-1}{10}& \frac{2}{5}& \frac{1}{10}\\ \frac{1}{5}& \frac{1}{5}& \frac{-1}{5}\\ \frac{1}{10}& \frac{-2}{5}& \frac{9}{10}\end{array}\right]·A$

Applying ${R}_{3}\to \frac{1}{2}{R}_{3}$, we get $\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]=\left[\begin{array}{ccc}\frac{-1}{10}& \frac{2}{5}& \frac{1}{10}\\ \frac{1}{5}& \frac{1}{5}& \frac{-1}{5}\\ \frac{1}{20}& \frac{-1}{5}& \frac{9}{20}\end{array}\right]·A$

Thus, ${A}^{-1}=\left[\begin{array}{ccc}\frac{-1}{10}& \frac{2}{5}& \frac{1}{10}\\ \frac{1}{5}& \frac{1}{5}& \frac{-1}{5}\\ \frac{1}{20}& \frac{-1}{5}& \frac{9}{20}\end{array}\right]$

Example 3: Find ${A}^{-1}$, if it exists, given $A=\left[\begin{array}{ccc}4& 5& 1\\ -1& 0& 2\\ 5& 3& 6\end{array}\right]$.

Solution: First, let us find $\left|A\right|$.

$\left|A\right|=\left(4\right)\left[0-\left(2×3\right)\right]-\left(5\right)\left[\left(-1×6\right)-\left(2×5\right)\right]+\left(1\right)\left[\left(-1×3\right)-0\right]$$=-24+80-3=53$

Since $\left|A\right|\ne 0$, the inverse of the given matrix exists. Let us find the inverse by the adjoint method.

We know that . We have already found that $\left|A\right|=53$. We need to compute .

Let us find the co-factors.

${A}_{11}={\left(-1\right)}^{2}\left(-6\right)=-6$

${A}_{12}={\left(-1\right)}^{3}\left(-6-10\right)=16$

${A}_{13}={\left(-1\right)}^{4}\left(-3\right)=-3$;

${A}_{21}={\left(-1\right)}^{3}\left(30-3\right)=-27$

${A}_{22}={\left(-1\right)}^{4}\left(24-5\right)=19$

${A}_{23}={\left(-1\right)}^{5}\left(12-25\right)=13$;

${A}_{31}={\left(-1\right)}^{4}\left(10\right)=10$;

${A}_{32}={\left(-1\right)}^{5}\left(8+1\right)=-9$

${A}_{33}={\left(-1\right)}^{6}\left(5\right)=5$

Now,

## Cheat Sheet

• The inverse of a matrix is unique.
• The inverse of a [2 x 2] matrix: ${\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]}^{-1}=\frac{1}{\left(ad-bc\right)}\left[\begin{array}{cc}d& -b\\ -c& a\end{array}\right]$
• ${\left(AB\right)}^{-1}={B}^{-1}{A}^{-1}$
• ${\left({A}^{-1}\right)}^{-1}=A$
• ${\left({A}^{T}\right)}^{-1}={\left({A}^{-1}\right)}^{T}$

## Blunder Areas

• An identity matrix can never be a rectangular matrix. It is always a square matrix.
• The inverse of a matrix exists only for non-singular matrices. So, one must verify the same before actually jumping toward finding the inverse of a given matrix.