Question

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

• A. \( \dfrac{2\pi}{3} \)
• B. \( \dfrac{5\pi}{6} \)
• C. \( \dfrac{\pi}{3} \)
• D. \( \dfrac{4\pi}{3} \)

Hint:

When two matrices are equal, compare corresponding entries directly to form equations.

Answer 1

The Correct Option is A

Concept: Transpose of a matrix is obtained by interchanging rows and columns. If: \[ A+A^T=I \] then corresponding elements of matrices must be equal.

Step 1: Find transpose of \(A\). Given: \[ A= \begin{bmatrix} \sin\alpha & -\cos\alpha \cos\alpha & \sin\alpha \end{bmatrix} \] Therefore: \[ A^T= \begin{bmatrix} \sin\alpha & \cos\alpha -\cos\alpha & \sin\alpha \end{bmatrix} \]

Step 2: Compute \(A+A^T\). \[ A+A^T = \begin{bmatrix} 2\sin\alpha & 0 0 & 2\sin\alpha \end{bmatrix} \] Given: \[ A+A^T=I \] Therefore: \[ \begin{bmatrix} 2\sin\alpha & 0 0 & 2\sin\alpha \end{bmatrix} = \begin{bmatrix} 1 & 0 0 & 1 \end{bmatrix} \] Hence: \[ 2\sin\alpha=1 \] \[ \sin\alpha=\frac12 \]

Step 3: Use the interval condition. General solutions: \[ \alpha=\frac{\pi}{6},\frac{5\pi}{6} \] But: \[ \alpha\in\left(\frac{\pi}{2},\frac{3\pi}{2}\right) \] Only: \[ \boxed{ \alpha=\frac{5\pi}{6} } \] satisfies the interval.

Step 4: Final conclusion. \[ \boxed{ \frac{5\pi}{6} } \] Hence correct option is: \[ \boxed{(2)} \]