Question

If $A = \begin{bmatrix} \sin \alpha & -\cos \alpha\\ \cos \alpha & \sin \alpha \end{bmatrix}$ and $\alpha \in (\frac{\pi}{2}, \frac{3\pi}{2})$. If $A + A^T = I$, then $\alpha =$}

• A. $\frac{2\pi}{3}$
• B. $\frac{5\pi}{6}$
• C. $\pi$
• D. $\frac{4\pi}{3}$

Hint:

Always check the given range for the angle after finding the trigonometric value to select the correct quadrant.

Answer 1

The Correct Option is B

Step 1: Concept

\(A^T\) is the transpose of matrix \(A\). The sum \(A + A^T\) is compared with the identity matrix \(I\).

Step 2: Matrix Representation

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

Therefore, \[ A + A^T = \begin{bmatrix} 2\sin\alpha & 0 \\ 0 & 2\sin\alpha \end{bmatrix} \]

Since \(A + A^T = I\), \[ \begin{bmatrix} 2\sin\alpha & 0 \\ 0 & 2\sin\alpha \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \]

Step 3: Analysis

Equating corresponding elements: \[ 2\sin\alpha = 1 \]

Therefore, \[ \sin\alpha = \frac{1}{2} \]

In the interval: \[ \left(\frac{\pi}{2}, \frac{3\pi}{2}\right) \] sine is positive only in the second quadrant.

Step 4: Conclusion

The required angle is: \[ \alpha = \pi - \frac{\pi}{6} = \frac{5\pi}{6} \]

Final Answer: \[ \boxed{\frac{5\pi}{6}} \] Hence, the correct option is (B).