Question

If $A=\left[\begin{array}{cc}\sin \mathrm{\alpha }& \cos \mathrm{\alpha }\\ -\cos \mathrm{\alpha }& \sin \mathrm{\alpha }\end{array}\right]$, verify that A^T A = I₂.

Answer 1

$$\begin{gathered} \mathrm{Given}:A=\left[\begin{array}{cc}\sin \alpha & \cos \alpha \\ -\cos \alpha & \sin \alpha \end{array}\right] \\ {A}^T=\left[\begin{array}{cc}\sin \alpha & -\cos \alpha \\ \cos \alpha & \sin \alpha \end{array}\right] \\ \mathrm{Now}, \\ {A}^{T}A=\left[\begin{array}{cc}\sin \alpha & -\cos \alpha \\ \cos \alpha & \sin \alpha \end{array}\right]\left[\begin{array}{cc}\sin \alpha & \cos \alpha \\ -\cos \alpha & \sin \alpha \end{array}\right] \\ \Rightarrow A^{T}A=\left[\begin{array}{cc}\left(\sin \alpha \right)\left(\sin \alpha \right)+\left(-\cos \alpha \right)\left(-\cos \alpha \right)& \left(\sin \alpha \right)\left(\cos \alpha \right)+\left(-\cos \alpha \right)\left(\sin \alpha \right)\\ \left(\cos \alpha \right)\left(\sin \alpha \right)+\left(\sin \alpha \right)\left(-\cos \alpha \right)& \left(\cos \alpha \right)\left(\cos \alpha \right)+\left(\sin \alpha \right)\left(\sin \alpha \right)\end{array}\right] \\ \Rightarrow A^{T}A=\left[\begin{array}{cc}{\sin }^2\alpha +{\cos }^2\alpha & \sin \alpha \cos \alpha -\sin \alpha \cos \alpha \\ \sin \alpha \cos \alpha -\sin \alpha \cos \alpha & {\cos }^2\alpha +{\sin }^2\alpha \end{array}\right] \\ \Rightarrow A^{T}A=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]=I \end{gathered}$$