#### Explain Solution for RD Sharma Maths Class 12 Chapter 6 Adjoint and inverse of a Matrix Exercise very short answer type Question 16 maths textbook solution

VSQ : 16

Given : $A=\left[\begin{array}{cc} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right] \& A(\operatorname{adj} A)=\left[\begin{array}{cc} K & 0 \\ 0 & K \end{array}\right]$

Hint : Find $A^{-1}$ and then multiply both side by A

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

\begin{aligned} &|A|=\cos ^{2} \theta+\sin ^{2} \theta \\ &=1 \\ &|A| \neq 0 \\ &A^{-1} \text { exists } \\ &A^{-1}=\frac{a d j A}{|A|} \\ &A^{-1}=\left[\begin{array}{cc} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array}\right] \\ &A^{-1}=a d j A \end{aligned}

multiply both side by A

\begin{aligned} &A A^{-1}=A \operatorname{adj} A \\ &I=\left[\begin{array}{cc} K & 0 \\ 0 & K \end{array}\right] \\ &{\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]=\left[\begin{array}{cc} K & 0 \\ 0 & K \end{array}\right]} \\ &K=1(\text { \text{On comparing }}) \end{aligned}