If $A=\left[\begin{array}{rr}\sin \alpha & \cos \alpha \\ -\cos \alpha & \sin \alpha\end{array}\right]$, verify that $A^{\top} A=I_{2}$.
Given : $A=\left[\begin{array}{cc}\sin \alpha & \cos \alpha \\ -\cos \alpha & \sin \alpha\end{array}\right]$
$A^{T}=\left[\begin{array}{cc}\sin \alpha & -\cos \alpha \\ \cos \alpha & \sin \alpha\end{array}\right]$
Now,
$A^{T} A=\left[\begin{array}{cc}\sin \alpha & -\cos \alpha \\ \cos \alpha & \sin \alpha\end{array}\right]\left[\begin{array}{cc}\sin \alpha & \cos \alpha \\ -\cos \alpha & \sin \alpha\end{array}\right]$
$\Rightarrow A^{T} A=\left[\begin{array}{cc}(\sin \alpha)(\sin \alpha)+(-\cos \alpha)(-\cos \alpha) & (\sin \alpha)(\cos \alpha)+(-\cos \alpha)(\sin \alpha) \\ (\cos \alpha)(\sin \alpha)+(\sin \alpha)(-\cos \alpha) & (\cos \alpha)(\cos \alpha)+(\sin \alpha)(\sin \alpha)\end{array}\right]$
$\Rightarrow A^{T} A=\left[\begin{array}{cc}\sin ^{2} \alpha+\cos ^{2} \alpha & \sin \alpha \cos \alpha-\sin \alpha \cos \alpha \\ \sin \alpha \cos \alpha-\sin \alpha \cos \alpha & \cos ^{2} \alpha+\sin ^{2} \alpha\end{array}\right]$
$\Rightarrow A^{T} A=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]=I$
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.