If (i) , then verify that

Solution:

(i)

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

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

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

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

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

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

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

(ii)

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

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

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

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

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

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

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

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