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 $\left[F\right(\alpha ){]}^{-1}$ is equal to

• A. $F(-\alpha )$
• B. $F(-{\alpha }^{-1})$
• C. $F\left(2\alpha \right)$
• D. $Noneofthese$

Answer 1

The correct option is A $F(-\alpha )$

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