Question

If $A=\left(\begin{array}{cc}\cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{array}\right)$ and $A^{-1}=A^{\prime }$, find value of $\alpha$.

Answer 1

We have,

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

Also, $A^{-1}=A^{\prime }$

$\Rightarrow A{A}^{-1}=A{A}^{\prime }$

$\Rightarrow I=A{A}^{\prime }$

$\Rightarrow \left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]=$$\left[\begin{array}{cc}\cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{array}\right]$$\left[\begin{array}{cc}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{array}\right]$

$\Rightarrow \left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]=$$\left[\begin{array}{cc}{\cos }^2\alpha +{\sin }^2\alpha & 0\\ 0& {\sin }^2\alpha +{\cos }^2\alpha \end{array}\right]$

So, from above matrix we can say tha t${\cos }^2\alpha +{\sin }^2\alpha =1$ for all real values of $\alpha$