Q103 / 180 · KCET 2018 · Mathematics
If \( \mathrm{A}=\left[\begin{array}{cc}\cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha\end{array}\right] \), then \( A A^{\prime}= \)
- A\( A \)
- Bzero matrix
- CA
- Correct answerD\( 11 \)
Solution
Given that \( A=\left[\begin{array}{cc}\cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha\end{array}\right] \)
So, \( A=\left[\begin{array}{cc}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha\end{array}\right] \)
Then,
\[ \begin{array}{l} A \cdot A=\left[\begin{array}{cc} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{array}\right]\left[\begin{array}{cc} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{array}\right] \\ =\left[\begin{array}{cc} \cos ^{2} \alpha+\sin ^{2} \alpha & -\cos \alpha \sin \alpha+\cos \alpha \sin \alpha \\ -\cos \alpha \sin \alpha+\cos \alpha \sin \alpha & \cos ^{2} \alpha+\sin ^{2} \alpha \end{array}\right] \\ =\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]=I \end{array} \]
So, \( A=\left[\begin{array}{cc}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha\end{array}\right] \)
Then,
\[ \begin{array}{l} A \cdot A=\left[\begin{array}{cc} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{array}\right]\left[\begin{array}{cc} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{array}\right] \\ =\left[\begin{array}{cc} \cos ^{2} \alpha+\sin ^{2} \alpha & -\cos \alpha \sin \alpha+\cos \alpha \sin \alpha \\ -\cos \alpha \sin \alpha+\cos \alpha \sin \alpha & \cos ^{2} \alpha+\sin ^{2} \alpha \end{array}\right] \\ =\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]=I \end{array} \]
Hence, the correct answer is option: 'D'.