A) \[\left[ \begin{matrix} \cos \alpha & 2\sin \alpha \\ -2\sin \alpha & \cos \alpha \\ \end{matrix} \right]\]
B) \[\left[ \begin{matrix} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \\ \end{matrix} \right]\]
C) \[\left[ \begin{matrix} \cos \alpha & \sin \alpha \\ \sin \alpha & \cos \alpha \\ \end{matrix} \right]\]
D) \[\left[ \begin{matrix} 1 & 1 \\ 1 & 1 \\ \end{matrix} \right]\]
Correct Answer: B
Solution :
A square matrix is to be orthogonal matrix if \[{A}'A=I=A{A}'\] \[\Rightarrow \] \[A=\left[ \begin{matrix} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \\ \end{matrix} \right]\], \[{A}'=\left[ \begin{matrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \\ \end{matrix} \right]\] \[\Rightarrow \] \[A{A}'=\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right],\,{A}'A=\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right]\] \[\therefore \] \[A{A}'={A}'A=I\].
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