If $A = \left[ {\begin{array}{cccccccccccccccccccc}{\cos q}&{\sin q}\\{ - \sin q}&{\cos q}\end{array}} \right]$, then show that ${A^2} = \left[ {\begin{array}{cccccccccccccccccccc}{\cos 2q}&{\sin 2q}\\{ - \sin 2q}&{\cos 2q}\end{array}} \right]$.

We have, $A = \left[ {\begin{array}{cccccccccccccccccccc}{\cos q}&{\sin q}\\{ - \sin q}&{\cos q}\end{array}} \right]$

$\therefore$ ${A^2} = A \cdot A = \left[ {\begin{array}{cccccccccccccccccccc}{\cos q}&{\sin q}\\{ - \sin q}&{\cos q}\end{array}} \right] \cdot \left[ {\begin{array}{cccccccccccccccccccc}{\cos q}&{\sin q}\\{ - \sin q}&{\cos q}\end{array}} \right]$

$= \left[ {\begin{array}{cccccccccccccccccccc}{{{\cos }^2}q - {{\sin }^2}q}&{\cos q \cdot \sin q + \sin q\cos q}\\{ - \sin q\cos q - \cos q\sin q}&{ - {{\sin }^2}q + {{\cos }^2}q}\end{array}} \right]$

$= \left[ {\begin{array}{cccccccccccccccccccc}{\cos 2q}&{2\sin q\cos q}\\{ - 2\sin q\cos q}&{\cos 2q}\end{array}} \right]$

$= \left[ {\begin{array}{cccccccccccccccccccc}{\cos 2q}&{\sin 2q}\\{ - \sin 2q}&{\cos 2q}\end{array}} \right]$