We are given the matrix: \[ A = \begin{bmatrix} \cos B & -\sin B \\ \sin B & \cos B \end{bmatrix} \] We need to find the value of \( B \) such that: \[ A + A^T = I \] where \( A^T \) is the transpose of \( A \), and \( I \) is the identity matrix. First, compute \( A^T \): \[ A^T = \begin{bmatrix} \cos B & \sin B \\ -\sin B & \cos B \end{bmatrix} \] Now, compute \( A + A^T \): \[ A + A^T = \begin{bmatrix} \cos B & -\sin B \\ \sin B & \cos B \end{bmatrix} + \begin{bmatrix} \cos B & \sin B \\ -\sin B & \cos B \end{bmatrix} = \begin{bmatrix} 2\cos B & 0 \\ 0 & 2\cos B \end{bmatrix} \] Set \( A + A^T \) equal to the identity matrix \( I \): \[ \begin{bmatrix} 2\cos B & 0 \\ 0 & 2\cos B \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \] This gives us the equation: \[ 2\cos B = 1 \] Solving for \( B \): \[ \cos B = \frac{1}{2} \] The general solution for \( B \) is: \[ B = 2n\pi \pm \frac{\pi}{3}, \quad n \in \mathbb{Z} \] From the given options, the value of \( B \) that satisfies \( A + A^T = I \) is: \[ B = \frac{\pi}{3} \] Correct Answer: \(\boxed{A}\)