#### Need solution for RD Sharma Maths Class 12 Chapter 6 Adjoint and Inverse of Matrices Excercise 6.1 Question 16 question (i)

Hence proved   $[F(\alpha)]^{-1}=F(-\alpha)$

Hint:

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

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

Given:

$F(\alpha)=\left[\begin{array}{ccc} \cos \alpha & -\sin \alpha & 0 \\ \sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1 \end{array}\right], G(\beta)=\left[\begin{array}{ccc} \cos \beta & 0 & \sin \beta \\ 0 & 1 & 0 \\ -\sin \beta & 0 & \cos \beta \end{array}\right]$

Solution:

\begin{aligned} &|F(\alpha)|=\left|\begin{array}{ccc} \cos \alpha & -\sin \alpha & 0 \\ \sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1 \end{array}\right| \\\\ &|F(\alpha)|=\cos ^{2} \alpha+\sin ^{2} \alpha=1 \end{aligned}

Cofactor of A are

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

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

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

$=\left[\begin{array}{ccc} \cos \alpha & \sin \alpha & 0 \\ -\sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1 \end{array}\right]$                                                (2)

From equation (1) and (2)

$[F(\alpha)]^{-1}=F(-\alpha)$