#### Please Solve RD Sharma Class 12 Chapter 6 Adjoint and Inverse of a Matrix Exercise 6.1 Question 1 Subquestion (iii) Maths Textbook Solution.

$\left[\begin{array}{cc} \cos \alpha & -\sin \alpha \\ -\sin \alpha & \cos \alpha \end{array}\right]$

Hint:

Here, we use basic concept of determinant.

Given:

$A= \left[\begin{array}{cc} \cos \alpha & \sin \alpha \\ \sin \alpha & \cos \alpha \end{array}\right]$

Solution:

Let’s find  $\left | A \right |$

$|A|=\cos ^{2} \alpha-\sin ^{2} \alpha=\cos 2 \alpha$

Let’s find cofactor

\begin{aligned} &C_{11}=\cos \alpha, C_{12}=-\sin \alpha, C_{21}=-\sin \alpha, C_{22}=\cos \alpha \\ &C_{i j}=\left[\begin{array}{cc} \cos \alpha & -\sin \alpha \\ -\sin \alpha & \cos \alpha \end{array}\right] \end{aligned}

Let’s transpose  $C_{ij}$

$Adj(A)= \left[\begin{array}{cc} \cos \alpha & -\sin \alpha \\ -\sin \alpha & \cos \alpha \end{array}\right]$

Let’s prove below

\begin{aligned} &\operatorname{Adj}(A) \times A=|A| \times A=A \times \operatorname{Adj}(A) \\ &\operatorname{Adj}(A) \times A=\left[\begin{array}{cc} \cos \alpha & -\sin \alpha \\ -\sin \alpha & \cos \alpha \end{array}\right] \times\left[\begin{array}{cc} \cos \alpha & \sin \alpha \\ \sin \alpha & \cos \alpha \end{array}\right] \end{aligned}
$=\left[\begin{array}{cc} \cos ^{2} \alpha-\sin ^{2} \alpha & \cos \alpha \sin \alpha-\cos \alpha \sin \alpha \\ -\sin \alpha \cos \alpha+\cos \alpha \sin \alpha & \cos ^{2} \alpha-\sin ^{2} \alpha \end{array}\right]$

$= \left[\begin{array}{cc} \cos 2\alpha &0 \\0 & \cos2 \alpha \end{array}\right]$                                    (1)

$\left | A \right |I= \cos2 \alpha\times \left[\begin{array}{cc} 1 &0 \\ 0 & 1\end{array}\right]$

$= \left[\begin{array}{cc} \cos 2\alpha &0 \\0 & \cos2 \alpha \end{array}\right]$                                    (2)

\begin{aligned} &A \times \operatorname{Adj}(A)=\left[\begin{array}{cc} \cos \alpha & \sin \alpha \\ \sin \alpha & \cos \alpha \end{array}\right] \times\left[\begin{array}{cc} \cos \alpha & -\sin \alpha \\ -\sin \alpha & \cos \alpha \end{array}\right] \\ &=\left[\begin{array}{cc} \cos ^{2} \alpha-\sin ^{2} \alpha & 0 \\ 0 & \cos ^{2} \alpha-\sin ^{2} \alpha \end{array}\right] \end{aligned}

$= \left[\begin{array}{cc} \cos 2\alpha &0 \\0 & \cos2 \alpha \end{array}\right]$                                    (3)

From equation (1), (2) and (3)

$A \times \operatorname{Adj}(A)=|A| I=\operatorname{Adj}(A) \times A$