Provide Solution for RD Sharma Class 12 Chapter 6 Adjoint and Inverse Matrix Exercise 6.1 Question 7 Subquestion (ii)

$A^{-1}=\left[\begin{array}{ll} 0 & 1 \\ 1 & 0 \end{array}\right]$

Hint:

Here, we use basic concept of inverse of matrix

Such that  $A^{-1}=\frac{1}{|A|} \times \operatorname{Adj}(A)$

Given:

$A=\left[\begin{array}{ll} 0 & 1 \\ 1 & 0 \end{array}\right]$

Solution:

$|A|=\left|\begin{array}{ll} 0 & 1 \\ 1 & 0 \end{array}\right|=0-1=-1$

Let’s find  $Adj\left ( A \right )$

\begin{aligned} &C_{11}=0, C_{12}=-1 \\ &C_{21}=-1, C_{22}=0 \end{aligned}

So,

$C_{i j}=\left[\begin{array}{cc} 0 & -1 \\ -1 & 0 \end{array}\right]$

$Adj \left ( A \right )$  Is transpose of  $C_{ij}$

So,

$\operatorname{Adj}(A)=\left[\begin{array}{cc} 0 & -1 \\ -1 & 0 \end{array}\right]$

So,

\begin{aligned} A^{-1} &=\frac{1}{|A|} \times \operatorname{Adj}(A) \\ A^{-1} &=-\frac{1}{1} \times\left[\begin{array}{cc} 0 & -1 \\ -1 & 0 \end{array}\right] \\ A^{-1} &=-1 \times\left[\begin{array}{cc} 0 & -1 \\ -1 & 0 \end{array}\right] \\ A^{-1} &=\left[\begin{array}{ll} 0 & 1 \\ 1 & 0 \end{array}\right] \end{aligned}