Question

If \[ P = \begin{pmatrix} \cos\alpha & \sin\alpha \\ -\sin\alpha & \cos\alpha \end{pmatrix} \] and \(P + P^T = I\), the value of \(\alpha \; (0 \leq \alpha \leq \pi/2)\) is

• A. \(\dfrac{\pi}{2}\)
• B. \(\dfrac{\pi}{3}\)
• C. \(\dfrac{3\pi}{2}\)
• D. \(0\)

Hint:

For such matrix problems, use the property \(P + P^T\) often leads to a diagonal form when the off-diagonal terms cancel.

Answer 1

The Correct Option is B

Step 1: Write down \(P^T\).
\[ P^T = \begin{pmatrix} \cos\alpha & -\sin\alpha \\ \sin\alpha & \cos\alpha \end{pmatrix} \]
Step 2: Compute \(P + P^T\).
\[ P + P^T = \begin{pmatrix} \cos\alpha & \sin\alpha \\ -\sin\alpha & \cos\alpha \end{pmatrix} + \begin{pmatrix} \cos\alpha & -\sin\alpha \\ \sin\alpha & \cos\alpha \end{pmatrix} = \begin{pmatrix} 2\cos\alpha & 0 \\ 0 & 2\cos\alpha \end{pmatrix} \]
Step 3: Condition given.
We want \[ P + P^T = I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \] Thus, \[ 2\cos\alpha = 1 \quad \Rightarrow \quad \cos\alpha = \tfrac{1}{2} \]
Step 4: Solve for \(\alpha\).
\[ \alpha = \cos^{-1}\left(\tfrac{1}{2}\right) = \tfrac{\pi}{3} \]
Final Answer:
\[ \boxed{\alpha = \dfrac{\pi}{3}} \]