Question

If $A=\left[\begin{array}{cc}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{array}\right]$ then prove that $A.A^{\prime }=I_2$.

Answer 1

Given,

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

Now

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

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

or, $A.A^{\prime }=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$

or, $A.A^{\prime }=I_2$.