Find the inverse of the matrix

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

$\therefore|A|=\cos ^{2} \theta+\sin ^{2} \theta=1 \neq 0$

$A$ is a singular matrix. Therefore, it is invertible.

Let $C_{i j}$ be a cofactor of $a_{i j}$ in $A$.

The cofactors of element $A$ are given by

$C_{11}=\cos \theta$

$C_{12}=\sin \theta$

$C_{21}=-\sin \theta$

$C_{22}=\cos \theta$

Now,

$\operatorname{adj} A=\left[\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta\end{array}\right]^{T}=\left[\begin{array}{cc}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta\end{array}\right]$

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