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

$\therefore  $  $A^{\prime}=\left[\begin{array}{ll}\sin \alpha & -\cos \alpha \\ \cos \alpha & \sin \alpha\end{array}\right]$

$A^{\prime} A=\left[\begin{array}{ll}\sin \alpha & -\cos \alpha \\ \cos \alpha & \sin \alpha\end{array}\right]\left[\begin{array}{cc}\sin \alpha & \cos \alpha \\ -\cos \alpha & \sin \alpha\end{array}\right]$

$\left[\begin{array}{cc}\sin \alpha & -\cos \alpha \\ \cos \alpha & \sin \alpha\end{array}\right]\left[\begin{array}{cc}\sin \alpha & \cos \alpha \\ -\cos \alpha & \sin \alpha\end{array}\right]$

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

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

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

Hence, we have verified that $A ^{\prime}A=1$.