If $A = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{\cos \alpha }&{\sin \alpha }\\{ - \sin \alpha }&{\cos \alpha }\end{array}} \right]$ and ${A^{ - 1}} = {A^\prime }$, then find the value of $\alpha$.

We have,
$A = \left[ {\begin{array}{cccccccccccccccccccc}{\cos a}&{\sin a}\\{ - \sin a}&{\cos a}\end{array}} \right]$ and

${A^\prime } = \left[ {\begin{array}{llllllllllllllllllll}{\cos a}&{ - \sin a}\\{\sin a}&{\cos a}\end{array}} \right]$

Hence, ${A^{ - 1}} = {A^\prime }$
$\Rightarrow$ $A{A^{ - 1}} = AA'$

$\Rightarrow$ $I = \left[ {\begin{array}{cccccccccccccccccccc}{\cos \alpha }&{\sin \alpha }\\{ - \sin \alpha }&{\cos \alpha }\end{array}} \right]\left[ {\begin{array}{cccccccccccccccccccc}{\cos \alpha }&{ - \sin \alpha }\\{\sin \alpha }&{\cos \alpha }\end{array}} \right]$

$\Rightarrow$ $\left[ {\begin{array}{llllllllllllllllllll}1&0\\0&1\end{array}} \right] = \left[ {\begin{array}{llllllllllllllllllll}{{{\cos }^2}\alpha + {{\sin }^2}\alpha }&{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0}\\0&{{{\sin }^2}\alpha + {{\cos }^2}\alpha }\end{array}} \right]$

By using equality of matrices,

we get
${\cos ^2}\alpha + {\sin ^2}\alpha = 1$

which is true for all real values of $\alpha$.