Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. If $ A=\left[ \begin{matrix} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \\ \end{matrix} \right], $ then $ {{A}^{2}} $ is equal to

Rajasthan PETRajasthan PET 2001

Solution:

$ A=\left[ \begin{matrix} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \\ \end{matrix} \right] $
$ {{A}^{2}}=A.A $
$ =\left[ \begin{matrix} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \\ \end{matrix} \right]\left[ \begin{matrix} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \\ \end{matrix} \right] $
$ =\left[ \begin{matrix} {{\cos }^{2}}\alpha -{{\sin }^{2}}\alpha & \cos \alpha \sin \alpha +\sin \alpha \cos \alpha \\ -\sin \alpha \cos \alpha -\sin \alpha \cos \alpha & -{{\sin }^{2}}\alpha +{{\cos }^{2}}\alpha \\ \end{matrix} \right] $ $ =\left[ \begin{matrix} \cos 2\alpha & \sin 2\alpha \\ -\sin 2\alpha & \cos 2\alpha \\ \end{matrix} \right] $