Question

Let F $\left(\alpha \right)$ = $\left[\begin{array}{ccc}\cos \alpha & -\sin \alpha & 0\\ \sin \alpha & \cos \alpha & 0\\ 0& 0& 1\end{array}\right]$ where $\alpha ϵR$.
Then $F\left(\alpha \right).F(-\alpha )=I_3$ or $\left[F\right(\alpha ){]}^{-1}=F(-\alpha )$

• A. True
• B. False

Answer 1

The correct option is A True

$F\left(\alpha \right)=\left|\begin{array}{ccc}\cos \alpha & -\sin \alpha & 0\\ \sin \alpha & \cos \alpha & 0\\ 0& 0& 1\end{array}\right|$

$F\left(\alpha \right)F(-\alpha )=\left|\begin{array}{ccc}\cos \alpha & -\sin \alpha & 0\\ \sin \alpha & \cos \alpha & 0\\ 0& 0& 1\end{array}\right|\left|\begin{array}{ccc}\cos \alpha & \sin \alpha & 0\\ -\sin \alpha & \cos \alpha & 0\\ 0& 0& 1\end{array}\right|$

$=\left|\begin{array}{ccc}{\cos }^2\alpha +{\sin }^2\alpha +0& \cos \alpha \sin \alpha -\cos \alpha \sin \alpha & 0\\ \sin \alpha \cos \alpha -\sin \alpha \cos \alpha & {\sin }^2\alpha +{\cos }^2\alpha & 0+0+0\\ 0+0+0& 0+0+0& 0+0+1\end{array}\right|$

$=\left|\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right|=I_3$

$F\left(\alpha \right)F(-\alpha )=I$

$F(-\alpha )=\frac{1}{F\left(\alpha \right)}$

$F(-\alpha )=\left(F\right(\alpha ){)}^{-1}$