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Inverse of the matrix $\left[\begin{array}{cc}\cos 2 \theta & -\sin 2 \theta \\ \sin 2 \theta & \cos 2 \theta\end{array}\right]$ is
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$\left[\begin{array}{cc}\cos 2 \theta & \sin 2 \theta \\ -\sin 2 \theta & \cos 2 \theta\end{array}\right]$
Let $A=\left[\begin{array}{cc}\cos 2 \theta & -\sin 2 \theta \\ \sin 2 \theta & \cos 2 \theta\end{array}\right]$
Now, $|A|=\left|\begin{array}{cc}\cos 2 \theta & -\sin 2 \theta \\ \sin 2 \theta & \cos 2 \theta\end{array}\right|$
$=\cos ^{2} 2 \theta-\left(-\sin ^{2} 2 \theta\right)$
$=\cos ^{2} 2 \theta+\sin ^{2} 2 \theta=1$
Again, $A^{-1}=\frac{1}{|A|}\left|\begin{array}{cc}\cos 2 \theta & \sin 2 \theta \\ -\sin 2 \theta & \cos 2 \theta\end{array}\right|$
$\left.\because\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]^{-1}=\frac{1}{a d-b c}\left[\begin{array}{cc}d & -b \\ -c & a\end{array}\right]\right]$
$=\left[\begin{array}{cc}\cos 2 \theta & \sin 2 \theta \\ -\sin 2 \theta & \cos 2 \theta\end{array}\right]$
Now, $|A|=\left|\begin{array}{cc}\cos 2 \theta & -\sin 2 \theta \\ \sin 2 \theta & \cos 2 \theta\end{array}\right|$
$=\cos ^{2} 2 \theta-\left(-\sin ^{2} 2 \theta\right)$
$=\cos ^{2} 2 \theta+\sin ^{2} 2 \theta=1$
Again, $A^{-1}=\frac{1}{|A|}\left|\begin{array}{cc}\cos 2 \theta & \sin 2 \theta \\ -\sin 2 \theta & \cos 2 \theta\end{array}\right|$
$\left.\because\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]^{-1}=\frac{1}{a d-b c}\left[\begin{array}{cc}d & -b \\ -c & a\end{array}\right]\right]$
$=\left[\begin{array}{cc}\cos 2 \theta & \sin 2 \theta \\ -\sin 2 \theta & \cos 2 \theta\end{array}\right]$
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