A) I
B) A
C) \[-A\]
D) \[{{A}^{2}}\]
Correct Answer: A
Solution :
Given, \[A=\left[ \begin{matrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \\ \end{matrix} \right]\] \[A'=\left[ \begin{matrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \\ \end{matrix} \right]\] \[AA'=\left[ \begin{matrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \\ \end{matrix} \right]\,\,\left[ \begin{matrix} \cos \theta & -\sin \theta \\ +\sin \theta & \cos \theta \\ \end{matrix} \right]\] \[=\left[ \begin{matrix} {{\cos }^{2}}\theta +{{\sin }^{2}}\theta & -\sin \theta .\cos \theta +\sin \theta .\cos \theta \\ -\sin \theta .\cos \theta +\cos \theta .\sin \theta & {{\sin }^{2}}\theta +{{\cos }^{2}}\theta \\ \end{matrix} \right]\] \[AA'=\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right]=I\] (unit matrix) Hence, \[AA'=I,\] which is called an orthogonal matrix.
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