If \(A = \begin{bmatrix}\cos\alpha&-\sin\alpha\\\sin\alpha&\cos\alpha\end{bmatrix}\), where \(0 \lt \alpha \lt \displaystyle\frac{\pi}{2}\), and \(A + A' = \sqrt{3}\,I\), where I is the identity matrix of order 2, then the value of \(\alpha\) is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\displaystyle\frac{\pi}{6}\)
The transpose of A is
\( \displaystyle A' = \begin{bmatrix}\cos\alpha&\sin\alpha\\-\sin\alpha&\cos\alpha\end{bmatrix} \)
Adding, the sine terms cancel:
\( \displaystyle A + A' = \begin{bmatrix}2\cos\alpha&0\\0&2\cos\alpha\end{bmatrix} \)
Comparing with \(\sqrt{3}\,I = \begin{bmatrix}\sqrt{3}&0\\0&\sqrt{3}\end{bmatrix}\):
\(2\cos\alpha = \sqrt{3}\)
\( \displaystyle \cos\alpha = \frac{\sqrt{3}}{2} \)
In \(0 \lt \alpha \lt \frac{\pi}{2}\) this gives
\( \displaystyle \alpha = \frac{\pi}{6} \)