Question

If \( A = \begin{bmatrix} \cos x & -\sin x \\ \sin x & \cos x \end{bmatrix} \) and \( A + A^{\prime} = I \), then the value of \( x \in \left[ 0, \dfrac{\pi}{2} \right] \) is

• A. \( 0 \)
• B. \( \dfrac{\pi}{4} \)
• C. \( \dfrac{\pi}{3} \)
• D. \( \dfrac{\pi}{2} \)

Hint:

Add A and its transpose to get 2 cos x times the identity, then solve 2 cos x = 1 in the interval [0, pi/2].

Answer 1

The Correct Option is C

Step 1: Write the transpose:
The transpose \( A^{\prime} \) is formed by swapping rows and columns of \( A \).
\[ A^{\prime} = \begin{bmatrix} \cos x & \sin x \\ -\sin x & \cos x \end{bmatrix} \]

Step 2: Add the two matrices:
We add entry by entry.
\[ A + A^{\prime} = \begin{bmatrix} 2\cos x & -\sin x + \sin x \\ \sin x - \sin x & 2\cos x \end{bmatrix} = \begin{bmatrix} 2\cos x & 0 \\ 0 & 2\cos x \end{bmatrix} \]

Step 3: Compare with the identity matrix:
We are told \( A + A^{\prime} = I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \).
Equal matrices have equal entries, so \( 2\cos x = 1 \), that is \( \cos x = \dfrac{1}{2} \).

Step 4: Solve in the given interval:
In \( \left[ 0, \dfrac{\pi}{2} \right] \) the only angle with cosine \( \dfrac{1}{2} \) is \( x = \dfrac{\pi}{3} \).

Step 5: Why the other options are wrong:
(A) \( x = 0 \) gives \( 2\cos 0 = 2 \), so the sum is \( 2I \), not \( I \).
(B) \( x = \dfrac{\pi}{4} \) gives \( 2\cos\dfrac{\pi}{4} = \sqrt{2} \), so the sum is \( \sqrt{2}\,I \).
(D) \( x = \dfrac{\pi}{2} \) gives \( 2\cos\dfrac{\pi}{2} = 0 \), so the sum is the zero matrix.

Final Answer:
The value is \( x = \dfrac{\pi}{3} \), option (C). \[ \boxed{x = \dfrac{\pi}{3}} \]