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Question:

If $A=\left[\begin{array}{rr}\cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha\end{array}\right]$, then verify that $A^{\top} A=I_{2}$

Solution:

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

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

Now,

$A^{T} A=I_{2}$

Consider : LHS $=A^{T} A$

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

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

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

Hence proved.

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