Question

If \[ A= \begin{bmatrix} \sin\alpha & -\cos\alpha\\ \cos\alpha & \sin\alpha \end{bmatrix} \] and \(A+A^{-1}=I\), then \(\alpha=\)

• A. \(0\)
• B. \(\dfrac{\pi}{3}\)
• C. \(\dfrac{\pi}{6}\)
• D. \(\dfrac{\pi}{4}\)

Hint:

For a \(2\times 2\) matrix, first find \(A^{-1}\), then apply the given matrix equation and compare corresponding entries.

Answer 1

The Correct Option is C

Step 1: Find the determinant of \(A\).
Given, \[ A= \begin{bmatrix} \sin\alpha & -\cos\alpha\\ \cos\alpha & \sin\alpha \end{bmatrix} \] Now, \[ |A|=(\sin\alpha)(\sin\alpha)-(-\cos\alpha)(\cos\alpha) \] \[ |A|=\sin^2\alpha+\cos^2\alpha \] \[ |A|=1 \] So, \(A^{-1}\) exists.

Step 2: Find \(A^{-1}\).
For a matrix \[ \begin{bmatrix} a & b\\ c & d \end{bmatrix}, \] the inverse is \[ \frac{1}{ad-bc} \begin{bmatrix} d & -b\\ -c & a \end{bmatrix} \] Therefore, \[ A^{-1}= \begin{bmatrix} \sin\alpha & \cos\alpha\\ -\cos\alpha & \sin\alpha \end{bmatrix} \]

Step 3: Use the given condition \(A+A^{-1}=I\).
Now, \[ A+A^{-1} = \begin{bmatrix} \sin\alpha & -\cos\alpha\\ \cos\alpha & \sin\alpha \end{bmatrix} + \begin{bmatrix} \sin\alpha & \cos\alpha\\ -\cos\alpha & \sin\alpha \end{bmatrix} \] \[ A+A^{-1} = \begin{bmatrix} 2\sin\alpha & 0\\ 0 & 2\sin\alpha \end{bmatrix} \] Since, \[ A+A^{-1}=I \] we get \[ \begin{bmatrix} 2\sin\alpha & 0\\ 0 & 2\sin\alpha \end{bmatrix} = \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix} \]

Step 4: Compare corresponding elements.
Comparing diagonal elements, \[ 2\sin\alpha=1 \] \[ \sin\alpha=\frac{1}{2} \] Thus, \[ \alpha=\frac{\pi}{6} \]

Step 5: Final conclusion.
Therefore, \[ \boxed{\frac{\pi}{6}} \]