Question

If \[ A = \begin{bmatrix} \frac{1}{2}\cos x & -\sin x \\ \sin x & \frac{1}{2}\cos x \end{bmatrix} \] and \(A + A^T = I\), then the value of \(x \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\) is

• A. $\frac{\pi}{2}$
• B. $\frac{\pi}{3}$
• C. $0$
• D. $-\frac{\pi}{2}$

Hint:

Whenever $A+A^T$ appears, compute the transpose first and then add matrices element-wise.

Answer 1

The Correct Option is C

To find the value of \(x\) for which the given condition \(A + A^T = I\) holds true, let's start by analyzing the problem step-by-step.

1. We are provided with the matrix: \[ A = \begin{bmatrix} \frac{1}{2}\cos x & -\sin x \\ \sin x & \frac{1}{2}\cos x \end{bmatrix} \]
2. The condition given is \(A + A^T = I\), where \(A^T\) is the transpose of matrix \(A\), and \(I\) is the identity matrix. The transposed matrix \(A^T\) looks like: \[ A^T = \begin{bmatrix} \frac{1}{2}\cos x & \sin x \\ -\sin x & \frac{1}{2}\cos x \end{bmatrix} \]
3. Calculate \(A + A^T\): \[ A + A^T = \begin{bmatrix} \frac{1}{2}\cos x & -\sin x \\ \sin x & \frac{1}{2}\cos x \end{bmatrix} + \begin{bmatrix} \frac{1}{2}\cos x & \sin x \\ -\sin x & \frac{1}{2}\cos x \end{bmatrix} \]
4. This results in: \[ A + A^T = \begin{bmatrix} \cos x & 0 \\ 0 & \cos x \end{bmatrix} \]
5. The condition is \(A + A^T = I\), which means: \[ \begin{bmatrix} \cos x & 0 \\ 0 & \cos x \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \] This implies \(\cos x = 1\).
6. The value of \(x\) that satisfies \(\cos x = 1\) within the given range \(x \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\) is \(x = 0\).

Therefore, the correct answer is \(x = 0\).