If \(A = \begin{bmatrix}\cos\theta & -\sin\theta\\ \sin\theta & \cos\theta\end{bmatrix}\), where \(0 \lt \theta \lt \displaystyle\frac{\pi}{2}\), and \(A + A^{T} = \sqrt{3}\,I\), then \(\theta\) is equal to
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(\displaystyle\frac{\pi}{6}\)
The transpose interchanges rows and columns:
\( \displaystyle A^{T} = \begin{bmatrix}\cos\theta & \sin\theta\\ -\sin\theta & \cos\theta\end{bmatrix} \)
Adding, the sine terms cancel:
\( \displaystyle A + A^{T} = \begin{bmatrix}2\cos\theta & 0\\ 0 & 2\cos\theta\end{bmatrix} \)
By equality of matrices, comparing with \(\sqrt{3}\,I = \begin{bmatrix}\sqrt{3} & 0\\ 0 & \sqrt{3}\end{bmatrix}\):
\(2\cos\theta = \sqrt{3}\)
\( \displaystyle \cos\theta = \frac{\sqrt{3}}{2} \)
In \(0 \lt \theta \lt \frac{\pi}{2}\) this gives \(\theta = \frac{\pi}{6}\).