A) \[\left[ \begin{matrix} -\cos 2x & \sin 2x \\ -\sin 2x & \cos 2x \\ \end{matrix} \right]\]
B) \[\left[ \begin{matrix} \cos 2x & -\sin 2x \\ \sin 2x & \cos 2x \\ \end{matrix} \right]\]
C) \[\left[ \begin{matrix} \cos 2x & \cos 2x \\ \sin 2x & \sin 2x \\ \end{matrix} \right]\]
D) none of these
Correct Answer: B
Solution :
[b] \[\left| A \right|=\left| \begin{matrix} 1 & \tan x \\ -\tan x & 1 \\ \end{matrix} \right|=1+{{\tan }^{2}}x\ne 0\] So A is invertible. Also, Adj \[A={{\left[ \begin{matrix} 1 & \tan x \\ -\tan x & 1 \\ \end{matrix} \right]}^{T}}=\left[ \begin{matrix} 1 & -\tan x \\ \tan x & 1 \\ \end{matrix} \right]\] Now, \[{{A}^{-1}}=\frac{1}{\left| A \right|}adjA\] \[=\frac{1}{(1+{{\tan }^{2}}x)}\left[ \begin{matrix} 1 & -\tan x \\ \tan x & 1 \\ \end{matrix} \right]\] \[=\left[ \begin{matrix} \frac{1}{1+{{\tan }^{2}}x} & \frac{-\tan x}{1+{{\tan }^{2}}x} \\ \frac{\tan x}{1+{{\tan }^{2}}} & \frac{1}{1+{{\tan }^{2}}x} \\ \end{matrix} \right]\] \[\therefore {{A}^{T}}{{A}^{-1}}=\left[ \begin{matrix} 1 & -\tan x \\ \tan x & 1 \\ \end{matrix} \right]\] \[\left[ \begin{matrix} \frac{1}{1+{{\tan }^{2}}x} & \frac{-\tan x}{1+{{\tan }^{2}}x} \\ \frac{\tan x}{1+{{\tan }^{2}}} & \frac{1}{1+{{\tan }^{2}}x} \\ \end{matrix} \right]\] =\[\left[ \begin{matrix} \frac{1-{{\tan }^{2}}x}{1+{{\tan }^{2}}x} & \frac{-2\tan x}{1+{{\tan }^{2}}x} \\ \frac{2\tan x}{1+{{\tan }^{2}}x} & \frac{1-{{\tan }^{2}}x}{1+{{\tan }^{2}}x} \\ \end{matrix} \right]\] \[=\left[ \begin{matrix} \cos 2x & -\sin 2x \\ \sin 2x & \cos 2x \\ \end{matrix} \right]\]
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