#### Please Solve RD Sharma Class 12 Chapter 6 Adjoint and Inverese of Matrices Exercise 6.1 Question 8 subquestion (vii) Maths Textbook Solution.

$A^{-1}=\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & \cos \alpha & \sin \alpha \\ 0 & \sin \alpha & -\cos \alpha \end{array}\right]$

Hint:

Here, we use basic concept of determinant and inverse of matrix

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

Given:

$A=\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & \cos \alpha & \sin \alpha \\ 0 & \sin \alpha & -\cos \alpha \end{array}\right]$

Solution:

Let’s find $\left | A \right |$

\begin{aligned} &|A|=1\left|\begin{array}{cc} \cos \alpha & \sin \alpha \\ \sin \alpha & -\cos \alpha \end{array}\right|-0+0 \\ &=-\cos ^{2} \alpha-\sin ^{2} \alpha \\ &|A|=-1 \end{aligned}

So  $A^{-1}$  exist

Cofactors of A are

\begin{aligned} &C_{11}=-1, C_{21}=0, C_{31}=0 \\ &C_{12}=0, C_{22}=-\cos \alpha, C_{32}=-\sin \alpha \\ &C_{13}=0, C_{23}=-\sin \alpha, C_{33}=\cos \alpha \\ &\operatorname{Adj}(A)=C_{{ij}}^{T} \end{aligned}

\begin{aligned} &\operatorname{Adj}(A)=\left[\begin{array}{ccc} -1 & 0 & 0 \\ 0 & -\cos \alpha & -\sin \alpha \\ 0 & -\sin \alpha & \cos \alpha \end{array}\right]^{T}=\left[\begin{array}{ccc} -1 & 0 & 0 \\ 0 & -\cos \alpha & -\sin \alpha \\ 0 & -\sin \alpha & \cos \alpha \end{array}\right] \\ &A^{-1}=\frac{1}{|A|} \times \operatorname{Adj}(A) \end{aligned}

\begin{aligned} &=\frac{1}{-1}\left[\begin{array}{ccc} -1 & 0 & 0 \\ 0 & -\cos \alpha & -\sin \alpha \\ 0 & -\sin \alpha & \cos \alpha \end{array}\right] \\ &A^{-1}=\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & \cos \alpha & \sin \alpha \\ 0 & \sin \alpha & -\cos \alpha \end{array}\right] \end{aligned}