If $A=\begin{bmatrix} acos\theta & bsin\theta \\-bsin\theta & acos\theta \end{bmatrix}$ then $A^{-1}$ is equal to :
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → $\frac{1}{a^2\cos^2\theta + b^2\sin^2\theta }\begin{bmatrix} a\cos\theta & -b\sin\theta \\b\sin\theta & a\cos\theta \end{bmatrix}$
$A=\begin{bmatrix} a\cos\theta & b\sin\theta \\-b\sin\theta & a\cos\theta \end{bmatrix}$
$|A|=a^2\cos^2θ+b^2\sin^2θ$
$Adj\,A=\begin{bmatrix} a\cos\theta & -b\sin\theta \\b\sin\theta & a\cos\theta \end{bmatrix}$
$A^{-1}=\frac{Adj\,A}{|A|}=\frac{1}{a^2\cos^2θ+b^2\sin^2θ}\begin{bmatrix} a\cos\theta & -b\sin\theta \\b\sin\theta & a\cos\theta \end{bmatrix}$