Question

Let $F(\alpha )=\left[\begin{array}{ccc}\cos \alpha & -\sin \alpha & 0\\ \sin \alpha & \cos \alpha & 0\\ 0& 0& 1\end{array}\right]$ where $\alpha \in R$. Then $[F(\alpha ){]}^{-1}$ is equal to

• A. $F(-\alpha )$
• B. $F\left({\alpha }^{-1}\right)$
• C. $F(2\alpha )$
• D. None of these

Answer 1

Answer: (A)

$F(\alpha )\cdot F(-\alpha )=\left[\begin{array}{ccc}\cos \alpha & -\sin \alpha & 0\\ \sin \alpha & \cos \alpha & 0\\ 0& 0& 1\end{array}\right]\left[\begin{array}{ccc}\cos (-\alpha )& -\sin (-\alpha )& 0\\ \sin (-\alpha )& \cos (-\alpha )& 0\\ 0& 0& 1\end{array}\right]$
$F(\alpha )\cdot F(-\alpha )=\left[\begin{array}{ccc}\cos \alpha & -\sin \alpha & 0\\ \sin \alpha & \cos \alpha & 0\\ 0& 0& 1\end{array}\right]\left[\begin{array}{ccc}\cos \alpha & \sin \alpha & 0\\ -\sin \alpha & \cos \alpha & 0\\ 0& 0& 1\end{array}\right]$
$=\left[\begin{array}{ccc}{\cos }^2\alpha +{\sin }^2\alpha +0& \cos \alpha \sin \alpha -\cos \alpha \sin \alpha +0& 0+0+0\\ \sin \alpha \cos \alpha -\sin \alpha \cos \alpha +0& {\sin }^2\alpha +{\cos }^2\alpha +0& 0+0+0\\ 0+0+0& 0+0+0& 0+0+1\end{array}\right]$
$=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]=I$
$\left[\because {\cos }^2\alpha +{\sin }^2\alpha =1\right]$
$F(\alpha )\cdot F(-\alpha )=1\therefore [F(\alpha ){]}^{-1}=F(-\alpha )$