If, then, if the value of α is
Question:

If $A=\left[\begin{array}{cc}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha\end{array}\right]$, then $A+A^{*}=I$, if the value of $\alpha$ is

A. $\frac{\pi}{6}$

B. $\frac{\pi}{3}$

C. π

D. $\frac{3 \pi}{2}$

Solution:

$A=\left[\begin{array}{lr}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha\end{array}\right]$

$\Rightarrow A^{\prime}=\left[\begin{array}{ll}\cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha\end{array}\right]$

Now, $A+A^{\prime}=I$

$\therefore\left[\begin{array}{cc}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha\end{array}\right]+\left[\begin{array}{ll}\cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha\end{array}\right]=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$

$\Rightarrow\left[\begin{array}{lc}2 \cos \alpha & 0 \\ 0 & 2 \cos \alpha\end{array}\right]=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$

Comparing the corresponding elements of the two matrices, we have:

$2 \cos \alpha=1$

$\Rightarrow \cos \alpha=\frac{1 \pi}{2}=\cos \frac{-}{3}$

$\therefore \alpha=\frac{\pi}{3}$