If

$(i)A = \left[ {\begin{array}{cccccccccccccccccccc}{\cos \alpha }&{\sin \alpha }\\{ - \sin \alpha }&{\cos \alpha }\end{array}} \right],$then verify that $A'A = I$

(ii) $A = \left[ {\begin{array}{cccccccccccccccccccc}{\sin \alpha }&{\cos \alpha }\\{ - \cos \alpha }&{\sin \alpha }\end{array}} \right],$ then verify that $A'A = I$

.:

(i) $A = \left[ {\begin{array}{cccccccccccccccccccc}{\cos \alpha }&{\sin \alpha }\\{ - \sin \alpha }&{\cos \alpha }\end{array}} \right]$ $\Rightarrow$ $A' = \left[ {\begin{array}{cccccccccccccccccccc}{\cos \alpha }&{ - \sin \alpha }\\{\sin \alpha }&{\cos \alpha }\end{array}} \right]$

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

$= \left[ {\begin{array}{cccccccccccccccccccc}{{{\cos }^2}\alpha + {{\sin }^2}\alpha }&{\cos \alpha \sin \alpha - \sin \alpha \cos \alpha }\\{\sin \alpha \cos \alpha - \cos \alpha \sin \alpha }&{{{\sin }^2}\alpha + {{\cos }^2}\alpha }\end{array}} \right]$

$= \left[ {\begin{array}{cccccccccccccccccccc}1&0\\0&1\end{array}} \right] = I$

Hence, $A'A = I$.

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

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

$= \left[ {\begin{array}{cccccccccccccccccccc}{{{\sin }^2}\alpha + {{\cos }^2}\alpha }&{\sin \alpha \cos \alpha - \cos \alpha \sin \alpha }\\{\cos \alpha \sin \alpha - \sin \alpha \cos \alpha }&{{{\cos }^2}\alpha + {{\sin }^2}\alpha }\end{array}} \right]$
$= \left[ {\begin{array}{cccccccccccccccccccc}1&0\\0&1\end{array}} \right] = I$

Hence, $A'A = I$.