Question

If $A=\left[\begin{array}{cc}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{array}\right]$, then number of values of $\alpha \in (0,6)$ satisfying $A+A^T=I$ is (Here $I$ is $2\times 2$ matrix and $A^T$ is transpose at matrix $A$)

• A. $4$
• B. $2$
• C. $8$
• D. $6$

Answer 1

The correct option is B $2$

Given matrix is

$A=\left(\begin{array}{cc}cos\alpha & -sin\alpha \\ sin\alpha & cos\alpha \end{array}\right)$

As per question

$A+A^T=I$

$\left(\begin{array}{cc}cos\alpha & -sin\alpha \\ sin\alpha & cos\alpha \end{array}\right)+\left(\begin{array}{cc}cos\alpha & sin\alpha \\ -sin\alpha & cos\alpha \end{array}\right)=\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)$

$\left(\begin{array}{cc}2cos\alpha & 0\\ 0& 2cos\alpha \end{array}\right)=\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)$

On comparing we get

$2cos\alpha =1⟹cos\alpha =\frac{1}{2}=cos\frac{\pi }{3},cos\frac{5\pi }{3}⟹\alpha =\frac{\pi }{3},\frac{5\pi }{3}$