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

Step 1: Write the transpose of matrix \( A \).
Given
\[ A = \begin{bmatrix} \frac{1}{2} \cos x & -\sin x \\ \sin x & \frac{1}{2} \cos x \end{bmatrix} \] The transpose becomes
\[ A^T = \begin{bmatrix} \frac{1}{2} \cos x & \sin x \\ -\sin x & \frac{1}{2} \cos x \end{bmatrix} \] Step 2: Compute \( A + A^T \).
\[ A + A^T = \begin{bmatrix} \frac{1}{2} \cos x & -\sin x \\ \sin x & \frac{1}{2} \cos x \end{bmatrix} + \begin{bmatrix} \frac{1}{2} \cos x & \sin x \\ -\sin x & \frac{1}{2} \cos x \end{bmatrix} \] Add corresponding elements
\[ = \begin{bmatrix} \cos x & 0 \\ 0 & \cos x \end{bmatrix} \] Step 3: Use the given condition.
Given
\[ A + A^T = I \] Identity matrix
\[ I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \] Thus
\[ \cos x = 1 \] Step 4: Solve the trigonometric equation.
\[ \cos x = 1 \] General solution
\[ x = 2n\pi \] But
\[ x \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \] Thus
\[ x = 0 \] Step 5: Final conclusion.
Hence the required value of \( x \) is
\[ x = 0 \] Final Answer: \( \boxed{0} \)