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 $n\times n$ 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
  2. by adjoint method

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.

 

Solved Examples

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 $\left|A\right|\ne 0$.

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

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\times 2\right)-0\right]-\left(4\right)\left[\left(2\times 2\right)-\left(1\times 0\right)\right]+\left(2\right)\left[\left(2\times 0\right)-\left(1\times 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 $R_1\to R_1-4R_2$, 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 $R_2\to R_2-2R_1$, 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\times 3\right)\right]-\left(5\right)\left[\left(-1\times 6\right)-\left(2\times 5\right)\right]+\left(1\right)\left[\left(-1\times 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 $A^{-1}=\frac{adj A}{\left|A\right|}$. We have already found that $\left|A\right|=53$. We need to compute $adj A$.

Let us find the co-factors.

$A_{11}={\left(-1\right)}^2\left(-6\right)=-6$; 

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

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

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

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

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

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

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

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

$adj A=\left[\begin{array}{ccc}{A}_{11}& A_{12}& A_{13}\\ A_{21}& A_{22}& A_{23}\\ A_{31}& A_{32}& A_{33}\end{array}\right]=\left[\begin{array}{ccc}-6& -27& 10\\ 16& 19& -9\\ -3& 13& 5\end{array}\right]$

Now, $A^{-1}=\frac{adj A}{\left|A\right|}=\frac{1}{53}·\left[\begin{array}{ccc}-6& -27& 10\\ 16& 19& -9\\ -3& 13& 5\end{array}\right]=\left[\begin{array}{ccc}\frac{-6}{53}& \frac{-27}{53}& \frac{10}{53}\\ \frac{16}{53}& \frac{19}{53}& \frac{-9}{53}\\ \frac{-3}{53}& \frac{13}{53}& \frac{5}{53}\end{array}\right]$

Cheat Sheet

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

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.