Question

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

• A. $\frac{\pi }{6}$
• B. $\frac{\pi }{3}$
• C. $n$
• D. $\frac{3\pi }{2}$

Answer 1

The correct option is B $\frac{\pi }{3}$

$A=\left[\begin{array}{cc}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{array}\right]$
$\Rightarrow A^{\prime }=\left[\begin{array}{cc}\cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{array}\right]$
Now, $A+A^{\prime }=I$
$\therefore \left[\begin{array}{cc}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{array}\right]+\left[\begin{array}{cc}\cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{array}\right]=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$
$\Rightarrow \left[\begin{array}{cc}2\cos \alpha & 0\\ 0& 2\cos \alpha \end{array}\right]=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$
Comparing the corresponding elements of the two matrics, we have:
$2\cos \alpha =1$
$\Rightarrow \cos \alpha =\frac{1}{2}=\cos \frac{\pi }{3}$$\therefore \alpha =\frac{\pi }{3}$