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यदि $A =\left[\begin{array}{cc}\cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha\end{array}\right]$ हो तो सत्यापित कीजिए कि $A ^{\prime} A = I$
Solution
$A=\left[\begin{array}{ll}\cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha\end{array}\right]$
$\therefore A^{\prime}=\left[\begin{array}{cc}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha\end{array}\right]$
$A^{\prime} A=\left[\begin{array}{ll}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha\end{array}\right]\left[\begin{array}{cc}\cos \alpha & -\sin \alpha \\ -\sin \alpha & \cos \alpha\end{array}\right]$
$=\left[\begin{array}{cc}(\cos \alpha)(\cos \alpha)+(-\sin \alpha)(-\sin \alpha) & (\cos \alpha)(\sin \alpha)+(-\sin \alpha)(\cos \alpha) \\ (\sin \alpha)(\cos \alpha)+(\cos \alpha)(-\sin \alpha) & (\sin \alpha)(\sin \alpha)+(\cos \alpha)(\cos \alpha)\end{array}\right]$
$\left[\begin{array}{cc}\cos ^{2} \alpha+\sin ^{2} \alpha & \sin \alpha \cos \alpha-\sin \alpha \cos \alpha \\ \sin \alpha \cos \alpha-\sin \alpha \cos \alpha & \sin ^{2} \alpha+\cos ^{2} \alpha\end{array}\right]$
$=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]=I$
Hence, we have verified that $A ^{\prime}A=1$