To solve this problem, we need to find the value of \(\alpha\) such that \(A + A^T = I\), where \(A\) is given as:
| \(A = \begin{bmatrix} \sin \alpha & -\cos \alpha\\ \cos \alpha & \sin \alpha \end{bmatrix}\) |
First, compute the transpose of \(A\), denoted as \(A^T\):
| \(A^T = \begin{bmatrix} \sin \alpha & \cos \alpha\\ -\cos \alpha & \sin \alpha \end{bmatrix}\) |
Now, calculate \(A + A^T\):
| \(A + A^T = \begin{bmatrix} \sin \alpha & -\cos \alpha\\ \cos \alpha & \sin \alpha \end{bmatrix} + \begin{bmatrix} \sin \alpha & \cos \alpha\\ -\cos \alpha & \sin \alpha \end{bmatrix}\) |
Perform the matrix addition:
| \(A + A^T = \begin{bmatrix} 2\sin \alpha & 0\\ 0 & 2\sin \alpha \end{bmatrix}\) |
We need \(A + A^T\) to equal the identity matrix \(I = \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}\). Thus, we require:
| \(2\sin \alpha = 1\) |
Solving for \(\sin \alpha\), we get:
| \(\sin \alpha = \frac{1}{2}\) |
We know from trigonometric identities that \(\sin \alpha = \frac{1}{2}\) when \(\alpha = \frac{\pi}{6}, \frac{5\pi}{6}\), etc. However, given the interval \(\left(\frac{\pi}{2}, \frac{3\pi}{2}\right)\), only \(\alpha = \frac{5\pi}{6}\) satisfies the condition.
Therefore, the correct value of \(\alpha\) is:
\(\frac{5\pi}{6}\)
This matches option: \(\frac{5\pi}{6}\).