If $A = \begin{bmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{bmatrix}$, then $A + A' = I$, then the value of $\alpha$.

$\frac{\pi}{3}$

The correct answer is Option (2) → $\frac{\pi}{3}$ ##

Given that, $A = \begin{bmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{bmatrix}$

Also $A + A' = I$

$\Rightarrow \begin{bmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{bmatrix} + \begin{bmatrix} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$

$\Rightarrow \begin{bmatrix} 2\cos \alpha & 0 \\ 0 & 2\cos \alpha \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$

Equating corresponding entries, we have

$\Rightarrow 2\cos \alpha = 1$

$\Rightarrow \cos \alpha = \frac{1}{2}$

$\Rightarrow \cos \alpha = \cos \frac{\pi}{3}$

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