Question

If matrix \[ A = \begin{bmatrix} \cos\alpha & -\sin\alpha \\ \sin\alpha & \cos\alpha \end{bmatrix} \] and \[ A + A' = I, \] then the value of \( \alpha \) will be:

• A. \( \frac{\pi}{6} \)
• B. \( \frac{\pi}{3} \)
• C. \( \pi \)
• D. \( \frac{3\pi}{2} \)

Hint:

For matrix equations involving transpose, remember \( A' \) flips the off-diagonal elements while keeping the diagonal unchanged.

Answer 1

The Correct Option is B

The given matrix \( A \) is:

\[ A = \begin{bmatrix} \cos\alpha & -\sin\alpha \\ \sin\alpha & \cos\alpha \end{bmatrix}. \]

The transpose of \( A \) is:

\[ A' = \begin{bmatrix} \cos\alpha & \sin\alpha \\ -\sin\alpha & \cos\alpha \end{bmatrix}. \]

From the condition \( A + A' = I \), where \( I \) is the identity matrix:

\[ \begin{bmatrix} \cos\alpha & -\sin\alpha \\ \sin\alpha & \cos\alpha \end{bmatrix} + \begin{bmatrix} \cos\alpha & \sin\alpha \\ -\sin\alpha & \cos\alpha \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}. \]

Adding the matrices:

\[ \begin{bmatrix} 2\cos\alpha & 0 \\ 0 & 2\cos\alpha \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}. \]

Equating elements:

\[ 2\cos\alpha = 1 \quad \Rightarrow \quad \cos\alpha = \frac{1}{2}. \]

The value of \( \alpha \) satisfying \( \cos\alpha = \frac{1}{2} \) in the principal range is:

\[ \alpha = \frac{\pi}{3}. \]

Hence, the correct answer is B) \( \frac{\pi}{3} \).