Question

If $F\left(x\right)=\left[\begin{array}{ccc}\cos x& -\sin x& 0\\ \sin x& \cos x& 0\\ 0& 0& 1\end{array}\right]$ then show that $F\left(x\right).F\left(y\right)=F(x+y)$. Hence prove that $\left[F\right(x){]}^{-1}=F(-x)$.

Answer 1

$F\left(x\right)=\left[\begin{array}{ccc}\cos x& -\sin x& 0\\ \sin x& \cos x& 0\\ 0& 0& 1\end{array}\right]=\cos x(\cos x-0)+\sin x(\sin x-0)={\cos }^{2}x+{\sin }^{2}x=1F\left(y\right)=1F\left(x\right).F\left(y\right)=F(x+y)$
Hence
$\left[F\right(x){]}^{-1}=\left[\begin{array}{ccc}\cos x& \sin x& 0\\ -\sin x& \cos x& 0\\ 0& 0& 1\end{array}\right]F(-x)=\left[\begin{array}{ccc}\cos (-x)& -\sin (-x)& 0\\ \sin (-x)& \cos (-x)& 0\\ 0& 0& 1\end{array}\right]=\left[\begin{array}{ccc}\cos x& \sin x& 0\\ -\sin x& \cos x& 0\\ 0& 0& 1\end{array}\right][F\left(x\right){]}^{-1}=F(-x)$