A) \[\left[ \begin{matrix} \cos 3\theta & \sin 3\theta \\ \sin 3\theta & \cos 3\theta \\ \end{matrix} \right]\]
B) \[\left[ \begin{matrix} \cos 3\theta & -\sin 3\theta \\ \sin 3\theta & \cos 3\theta \\ \end{matrix} \right]\]
C) \[\left[ \begin{matrix} \cos 3\theta & \sin 3\theta \\ -\sin 3\theta & -\cos 3\theta \\ \end{matrix} \right]\]
D) \[\left[ \begin{matrix} -\cos 3\theta & -\sin 3\theta \\ -\sin 3\theta & -\cos 3\theta \\ \end{matrix} \right]\]
E) \[\left[ \begin{matrix} {{\cos }^{3}}\theta & {{\sin }^{3}}\theta \\ -{{\sin }^{3}}\theta & {{\cos }^{3}}\theta \\ \end{matrix} \right]\]
Correct Answer: A
Solution :
Given that, \[A=\left[ \begin{matrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \\ \end{matrix} \right]\] \[\therefore \] \[{{A}^{2}}=\left[ \begin{matrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \\ \end{matrix} \right]\left[ \begin{matrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \\ \end{matrix} \right]\] \[=\left[ \begin{matrix} {{\cos }^{2}}\theta -{{\sin }^{2}}\theta & 2\sin \theta \cos \theta \\ -2\sin \theta \cos \theta & {{\cos }^{2}}\theta -{{\sin }^{2}}\theta \\ \end{matrix} \right]\] \[=\left[ \begin{matrix} \cos 2\theta & \sin 2\theta \\ -\sin 2\theta & \cos 2\theta \\ \end{matrix} \right]\] \[{{A}^{3}}=\left[ \begin{matrix} \cos 2\theta & \sin 2\theta \\ -\sin 2\theta & \cos 2\theta \\ \end{matrix} \right]\left[ \begin{matrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \\ \end{matrix} \right]\] \[=\left[ \begin{matrix} \cos 2\theta \cos \theta -\sin \theta \sin 2\theta \\ -\sin 2\theta \cos \theta -\cos 2\theta \sin \theta \\ \end{matrix} \right.\] \[\left. \begin{matrix} \sin \theta \cos 2\theta +\sin 2\theta \cos \theta \\ -\sin 2\theta \sin \theta +\cos 2\theta \cos \theta \\ \end{matrix} \right]\] \[=\left[ \begin{matrix} \cos 3\theta & \sin 3\theta \\ \sin 3\theta & \cos 3\theta \\ \end{matrix} \right]\]
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