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. $2\pi/3$
• B. $5\pi/6$
• C. $\pi$
• D. $4\pi/3$

Hint:

$\sin \theta = 1/2$ has two solutions in $(0, 2\pi)$. Use the interval $(\pi/2, 3\pi/2)$ to choose the correct one.

Answer 1

The Correct Option is B

Step 1: Concept

We add matrix \(A\) to its transpose \(A^T\) and equate the resulting matrix to the identity matrix \(I\).

Step 2: Matrix Representation

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

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

Adding \(A\) and \(A^T\): \[ A + A^T = \begin{bmatrix} \sin\alpha & -\cos\alpha \\ \cos\alpha & \sin\alpha \end{bmatrix} + \begin{bmatrix} \sin\alpha & \cos\alpha \\ -\cos\alpha & \sin\alpha \end{bmatrix} \]

\[ = \begin{bmatrix} 2\sin\alpha & 0 \\ 0 & 2\sin\alpha \end{bmatrix} \]

Step 3: Solving the Equation

Since: \[ A + A^T = I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \]

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

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

Step 4: Finding \(\alpha\)

In the interval: \[ \left(\frac{\pi}{2}, \frac{3\pi}{2}\right) \] the angle where \[ \sin\alpha = \frac{1}{2} \] is: \[ \alpha = \pi - \frac{\pi}{6} = \frac{5\pi}{6} \]

Final Answer:

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