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

.:

Option b is correct

Given that, $A = \left[ {\begin{array}{cccccccccccccccccccc}{\cos \alpha }&{ - \sin \alpha }\\{\sin \alpha }&{\cos \alpha }\end{array}} \right]$
$\Rightarrow$ $A' = \left[ {\begin{array}{cccccccccccccccccccc}{\cos \alpha }&{\sin \alpha }\\{ - \sin \alpha }&{\cos \alpha }\end{array}} \right]$

We know that, A + A’ = I
$\Rightarrow$ $\left[ {\begin{array}{cccccccccccccccccccc}{\cos \alpha }&{ - \sin \alpha }\\{\sin \alpha }&{\cos \alpha }\end{array}} \right]$+ $\left[ {\begin{array}{cccccccccccccccccccc}{\cos \alpha }&{\sin \alpha }\\{ - \sin \alpha }&{\cos \alpha }\end{array}} \right]$ $= \left[ {\begin{array}{cccccccccccccccccccc}1&0\\0&1\end{array}} \right]$

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

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

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