A) \[\left[ \begin{matrix} \cos 2\alpha & \sin 2\alpha \\ -\sin 2\alpha & \cos 2\alpha \\ \end{matrix} \right]\]
B) \[\left[ \begin{matrix} {{\cos }^{2}}\alpha & -{{\sin }^{2}}\alpha \\ -{{\sin }^{2}}\alpha & {{\cos }^{2}}\alpha \\ \end{matrix} \right]\]
C) \[\left[ \begin{matrix} {{\sin }^{2}}\alpha & {{\cos }^{2}}\alpha \\ -{{\cos }^{2}}\alpha & {{\sin }^{2}}\alpha \\ \end{matrix} \right]\]
D) \[\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right]\]
Correct Answer: A
Solution :
\[A=\left[ \begin{matrix} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \\ \end{matrix} \right]\] \[{{A}^{2}}=A.A\] \[=\left[ \begin{matrix} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \\ \end{matrix} \right]\left[ \begin{matrix} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \\ \end{matrix} \right]\] \[=\left[ \begin{matrix} {{\cos }^{2}}\alpha -{{\sin }^{2}}\alpha & \cos \alpha \sin \alpha +\sin \alpha \cos \alpha \\ -\sin \alpha \cos \alpha -\sin \alpha \cos \alpha & -{{\sin }^{2}}\alpha +{{\cos }^{2}}\alpha \\ \end{matrix} \right]\] \[=\left[ \begin{matrix} \cos 2\alpha & \sin 2\alpha \\ -\sin 2\alpha & \cos 2\alpha \\ \end{matrix} \right]\]
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