A) \[\left[ \begin{matrix} {{\cos }^{n}}x & {{\sin }^{n}}x \\ -{{\sin }^{n}}x & {{\cos }^{n}}x \\ \end{matrix} \right]\]
B) \[\left[ \begin{matrix} {{\cos }^{2}}nx & {{\sin }^{2}}nx \\ -{{\sin }^{2}}nx & {{\cos }^{2}}nx \\ \end{matrix} \right]\]
C) \[\left[ \begin{matrix} \cos nx & \sin nx \\ -\sin \,nx & \cos \,nx \\ \end{matrix} \right]\]
D) \[\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right]\]
Correct Answer: D
Solution :
We have \[A=\left[ \begin{matrix} \cos x & \sin x \\ -\sin x & \cos x \\ \end{matrix} \right]\] Now, \[{{A}^{2}}=\left[ \begin{matrix} \cos x & \sin x \\ -\sin x & \cos x \\ \end{matrix} \right]\left[ \begin{matrix} \cos x & \sin x \\ -\sin x & \cos x \\ \end{matrix} \right]\] \[=\left[ \begin{matrix} {{\cos }^{2}}x+{{\sin }^{2}}x-\sin x\cos x+\sin x\cos x \\ -\sin x\cos x+\cos x\sin x{{\sin }^{2}}x+{{\cos }^{2}}x \\ {} \\ \end{matrix} \right]\] \[=\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right]\] \[\therefore \] \[{{A}^{n}}={{\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right]}^{n-1}}\] \[=\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right]\]
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