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# Inverse of the matrix $\begin{bmatrix}\cos 2\theta&-\sin 2\theta\\\sin 2\theta&\cos 2\theta\end{bmatrix}$ is

$\begin{array}{1 1}(a)\;\begin{bmatrix}\cos 2\theta&-\sin 2\theta\\\sin 2\theta&\cos 2\theta\end{bmatrix}&(b)\;\begin{bmatrix}\cos 2\theta&\sin 2\theta\\\sin 2\theta&-\cos 2\theta\end{bmatrix}\\(c)\;\begin{bmatrix}\cos 2\theta&\sin 2\theta\\\sin 2\theta&\cos 2\theta\end{bmatrix}&(d)\;\begin{bmatrix}\cos 2\theta&\sin 2\theta\\-\sin 2\theta&\cos 2\theta\end{bmatrix}\end{array}$

Can you answer this question?

Inverse of matrix $\begin{bmatrix}a&b\\c&d\end{bmatrix}$
$\Rightarrow \large\frac{1}{ad-bc}$$\begin{bmatrix}d&-b\\c&a\end{bmatrix} 4\Rightarrow ad-bc\neq 0 A^{-1}=\large\frac{1}{\cos^22\theta+\sin^22\theta}$$\begin{bmatrix}\cos 2\theta&\sin 2\theta\\-\sin 2\theta&\cos 2\theta\end{bmatrix}$
$\Rightarrow \begin{bmatrix}\cos 2\theta&\sin 2\theta\\-\sin 2\theta&\cos 2\theta\end{bmatrix}$
Hence (d) is the correct answer.
answered Nov 22, 2013