Question

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

Answer 1

$$\begin{gathered} \mathrm{Given}:A=\left[\begin{array}{cc}\cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{array}\right] \\ {A}^T=\left[\begin{array}{cc}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{array}\right] \\ \mathrm{Now}, \\ {A}^{T}A=I_2 \\ \mathrm{Consider}:\mathrm{LHS}=A^{T}A \\ =\left[\begin{array}{cc}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{array}\right]\left[\begin{array}{cc}\cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{array}\right] \\ =\left[\begin{array}{cc}{\cos }^2\alpha +{\sin }^2\alpha & \cos \alpha \sin \alpha -\sin \alpha \cos \alpha \\ \sin \alpha \cos \alpha -\cos \alpha \sin \alpha & {\sin }^2\alpha +{\cos }^2\alpha \end{array}\right] \\ =\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]=\mathrm{RHS} \end{gathered}$$

Hence proved.