Question

If $F\left(x\right)=\left[\begin{array}{ccc}\cos x& -\sin x& 0\\ \sin x& \cos x& 0\\ 0& 0& 1\end{array}\right]$ and $G\left(x\right)=\left[\begin{array}{ccc}\cos x& 0& \sin x\\ 0& 1& 0\\ -\sin x& 0& \cos x\end{array}\right],$ then $\left[F\right(x).G(x){]}^{-1}$ is equal to:

• A. $F(-x)\cdot G(-x)$
• B. $G\left(x\right)\cdot F\left(x\right)$
• C. $G(-x)\cdot F(-x)$
• D. $F\left(x\right)\cdot G\left(x\right)$

Answer 1

The correct option is C $G(-x)\cdot F(-x)$

We can see, $|F\left(x\right)|=|G\left(x\right)|={\sin }^{2}x+{\cos }^{2}x=1$
$\Rightarrow \left[F\right(x){]}^{-1}={\left[\begin{array}{ccc}\cos x& -\sin x& 0\\ \sin x& \cos x& 0\\ 0& 0& 1\end{array}\right]}^T=\left[\begin{array}{ccc}\cos x& \sin x& 0\\ -\sin x& \cos x& 0\\ 0& 0& 1\end{array}\right]=F(-x)$ and
$\left[G\right(x){]}^{-1}={\left[\begin{array}{ccc}\cos x& 0& \sin x\\ 0& 1& 0\\ -\sin x& 0& \cos x\end{array}\right]}^T=\left[\begin{array}{ccc}\cos x& 0& -\sin x\\ 0& 1& 0\\ \sin x& 0& \cos x\end{array}\right]=G(-x)$
So, $\left[F\right(x).G(x){]}^{-1}=[G\left(x\right){]}^{-1}\cdot \left[F\right(x){]}^{-1}=G(-x)\cdot F(-x)$