$\Rightarrow A^{\prime}=\left[\begin{array}{cc}\cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha\end{array}\right]$
Now $A+A^{\prime}=1$
$\therefore $ $\left[\begin{array}{cc}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha\end{array}\right]+\left[\begin{array}{cc}\cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha\end{array}\right]=\left[\begin{array}{cc}1 & 0 \\ 0 & 1\end{array}\right]$
$\Rightarrow $ $\left[\begin{array}{cc}2 \cos \alpha & 0 \\ 0 & 2 \cos \alpha\end{array}\right]=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$
Comparing the corresponding elements of the two matrices, we have :
$\cos \alpha=\frac{1}{2}$
$\alpha=\cos ^{-1}\left(\frac{1}{2}\right)$
$\therefore $ $\alpha=\frac{\pi}{3}$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
