Question

If \( A = \begin{bmatrix} \cos a & -\sin a \\ \sin a & \cos a \end{bmatrix} \) and \( A + A' = I \), then find the value of \( a \).

Hint:

For a matrix \( A \), if \( A + A' = I \), the diagonal elements of \( A \) must be equal and their value should satisfy the equation derived from the identity matrix.

Answer 1

We are given that \[ A = \begin{bmatrix} \cos a & -\sin a \\ \sin a & \cos a \end{bmatrix} \] and \[ A + A' = I, \] where \( A' \) is the transpose of matrix \( A \), and \( I \) is the identity matrix. First, find the transpose of \( A \): \[ A' = \begin{bmatrix} \cos a & \sin a \\ -\sin a & \cos a \end{bmatrix}. \] Now, add \( A \) and \( A' \): \[ A + A' = \begin{bmatrix} \cos a & -\sin a \\ \sin a & \cos a \end{bmatrix} + \begin{bmatrix} \cos a & \sin a \\ -\sin a & \cos a \end{bmatrix} = \begin{bmatrix} 2 \cos a & 0 \\ 0 & 2 \cos a \end{bmatrix}. \] We are told that \( A + A' = I \), so: \[ \begin{bmatrix} 2 \cos a & 0 \\ 0 & 2 \cos a \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}. \] This implies that: \[ 2 \cos a = 1. \] Solving for \( \cos a \): \[ \cos a = \frac{1}{2}. \] Thus, \( a = \cos^{-1} \left( \frac{1}{2} \right) \), which gives: \[ a = \frac{\pi}{3}. \]
Conclusion: The value of \( a \) is \( \boxed{\frac{\pi}{3}} \).