A) \[{{A}^{T}}=A\]
B) \[{{A}^{T}}=-A\]
C) \[{{A}^{2}}=I\]
D) \[{{A}^{T}}={{A}^{-1}}\]
Correct Answer: D
Solution :
Given, \[A=\left[ \begin{matrix} \cos \,\theta & -\sin \theta \\ \sin \theta & \cos \theta \\ \end{matrix} \right]\] \[\therefore \] \[{{A}^{-1}}\left[ \begin{matrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \\ \end{matrix} \right]={{A}^{T}}\]
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