#### Explain Solution for RD Sharma Maths Class 12 Chapter 6 Adjoint and inverse of a Matrix Exercise very short answer type Question 23 maths textbook solution

VSQ : 23

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

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

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

Hint : Find the cofactor and then transpose it

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

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

\begin{aligned} &c_{11}=\cos \theta, c_{12}=\sin \theta, c_{21}=-\sin \theta, c_{22}=\cos \theta \\ &|A|=\cos ^{2} \theta-\left(-\sin ^{2} \theta\right) \\ &=\cos ^{2} \theta+\sin ^{2} \theta \quad \end{aligned}                $\left[\because \cos ^{2} \theta+\sin ^{2} \theta=1\right] \\$

$=1$

$A^{-1}=\frac{A d j|A|}{|A|}=\frac{1}{1}$$\left[\begin{array}{cc} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right]$

Now transpose it, we get

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