Question

How to find the inverse of a matrix?

Answer 1

Step1. Inverse of a matrix

Let a square matrix of order $n\times n$, denoted by $A$ then it's inverse is denoted by $A^{-1}$but it not written as $A^{-1}\ne \frac{1}{A}$

Then $A^{-1}=\frac{1}{\left|A\right|}.adj\left(A\right)$, where C is determinant of given matrix.

Step2. Find the matrix of minor.

Consider, matrix $A=\left[\begin{array}{ccc}{a}_{11}& a_{12}& a_{13}\\ a_{21}& a_{22}& a_{23}\\ a_{31}& a_{32}& a_{33}\end{array}\right]$

Then, Minors of $A$ is given by

$M_{11}=\left|\begin{array}{cc}{a}_{22}& a_{23}\\ a_{32}& a_{33}\end{array}\right|$

For calculating minors of the matrix just the hide the row and column for which minors to be calculated.

Step3. Change of matrix into cofactors.
Cofactor of matrix is

$C_{ij}={(-1)}^{i+j}{M}_{i+j}$

$\begin{array}{rcl}{C}_{11}& =& {(-1)}^2{M}_{11}\\ & =& 1\times (a_{22}\times a_{33}-a_{23}\times a_{32})\\ & =& (a_{22}\times a_{33}-a_{23}\times a_{32})\end{array}$

Step4. Multiply by the reciprocal of determinant

Determinant of matrix is $\left|A\right|$and calculated as

$\begin{array}{rcl}\left|A\right|& =& \left|\begin{array}{ccc}{a}_{11}& a_{12}& a_{13}\\ a_{21}& a_{22}& a_{23}\\ a_{31}& a_{32}& a_{33}\end{array}\right|\\ & =& a_{11}(a_{22}\times a_{33}-a_{23}\times a_{32})-a_{12}(a_{21}\times a_{33}-a_{23}{a}_{31})+a_{13}(a_{21}\times a_{32}-a_{22}\times a_{31})\end{array}$

Step5. Ad joint of a matrix

Ad joint of matrix is transpose of cofactor of matrix.

Transpose of cofactor $=\left[\begin{array}{ccc}{C}_{11}& C_{12}& C_{13}\\ C_{21}& C_{22}& C_{23}\\ C_{31}& C_{32}& C_{33}\end{array}\right]$

Then,

$C^T=\left[\begin{array}{ccc}{C}_{11}& C_{21}& C_{31}\\ C_{12}& C_{22}& C_{32}\\ C_{13}& C_{23}& C_{33}\end{array}\right]$

Therefore,

$A^{-1}=\frac{AdjA}{\left|A\right|}=\frac{1}{\left|A\right|}\left[\begin{array}{ccc}{C}_{11}& C_{21}& C_{31}\\ C_{12}& C_{22}& C_{32}\\ C_{13}& C_{23}& C_{33}\end{array}\right]$

Hence, in this way we can find the inverse of a matrix.