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Question: Answered & Verified by Expert
If (i) $A=\left[\begin{array}{cc}\cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha\end{array}\right]$, then verify that $\mathbf{A}^{\prime} \mathbf{A}=\mathbf{I}$.
(ii) If $\mathrm{A}=\left[\begin{array}{cc}\sin \alpha & \cos \alpha \\ -\cos \alpha & \sin \alpha\end{array}\right]$, then verify that $\mathbf{A}^{\prime} \mathbf{A}=\mathbf{I}$.
MathematicsMatrices
Solution:
2689 Upvotes Verified Answer
(i) $\mathrm{A}=\left[\begin{array}{cc}\sin \alpha & \cos \alpha \\ -\sin \alpha & \cos \alpha\end{array}\right]$
$\Rightarrow \mathrm{A}^{\prime}=\left[\begin{array}{cc}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha\end{array}\right]$
L.H.S. $=\mathrm{A}^{\prime} \mathrm{A}$
$=\left[\begin{array}{cc}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha\end{array}\right]\left[\begin{array}{cc}\cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha\end{array}\right]$
$=\left[\begin{array}{cc}\cos ^2 \alpha+\sin ^2 \alpha & \cos \alpha \sin \alpha-\sin \alpha \cos \alpha \\ \sin \alpha \cos \alpha-\cos \alpha \sin \alpha & \sin ^2 \alpha+\cos ^2 \alpha\end{array}\right]$
$=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]=1$
Hence, A' A= I.
(ii) $A^{\prime}=\left[\begin{array}{cc}\sin \alpha & -\cos \alpha \\ \cos \alpha & \sin \alpha\end{array}\right]$
LHS $=A^{\prime} A$
$=\left[\begin{array}{cc}\sin ^2 \alpha+\cos ^2 \alpha & \sin \alpha \cos \alpha-\cos \alpha \sin \alpha \\ \cos \alpha \sin \alpha-\sin \alpha \cos \alpha & \cos ^2 \alpha+\sin ^2 \alpha\end{array}\right]$
$=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]=I$

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