Question

If \(A=\begin{bmatrix} \sin \alpha & -\cos \alpha \\ \cos \alpha & \sin \alpha \end{bmatrix}\) and \(S=\left\{\alpha :\alpha \in \left(\frac{\pi }{2},\frac{3\pi }{2}\right),A+A^{-1}=I\right\}\), then \(\alpha =\) ...

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

Hint:

Find the inverse using the adjoint, then equate A plus its inverse to the identity matrix.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
For a \(2\times 2\) matrix, \(A^{-1}=\dfrac{\text{adj }A}{|A|}\).

Step 2: Find the inverse
\(|A|=\sin ^{2}\alpha +\cos ^{2}\alpha =1\).
So \(A^{-1}=\begin{bmatrix} \sin \alpha & \cos \alpha \\ -\cos \alpha & \sin \alpha \end{bmatrix}\).

Step 3: Add
\(A+A^{-1}=\begin{bmatrix} 2\sin \alpha & 0 \\ 0 & 2\sin \alpha \end{bmatrix}\).
Setting this equal to \(I\) gives \(2\sin \alpha =1\), so \(\sin \alpha =\dfrac{1}{2}\).

Step 4: Use the interval
For \(\alpha \in \left(\dfrac{\pi }{2},\dfrac{3\pi }{2}\right)\), the angle with sine \(\dfrac{1}{2}\) is \(\pi -\dfrac{\pi }{6}=\dfrac{5\pi }{6}\).

Final Answer:
The value of \(\alpha\) is \(\dfrac{5\pi }{6}\). \[ \boxed{\text{(C) }\dfrac{5\pi }{6}} \]