A) \[{{A}^{2}}={{B}^{2}}\]
B) \[A+B=B-A\]
C) \[AB=BA\]
D) None of these
Correct Answer: C
Solution :
Clearly, \[AB=\left[ \begin{matrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \\ \end{matrix} \right]\left[ \begin{matrix} \cos \beta & -\sin \beta \\ \sin \beta & \cos \beta \\ \end{matrix} \right]\] \[=\left[ \begin{matrix} \cos (\alpha +\beta ) & -\sin (\alpha +\beta ) \\ \sin (\alpha +\beta ) & \cos (\alpha +\beta ) \\ \end{matrix} \right]=BA\](verify).
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