Question

If $f\left(x\right)=\left[\begin{array}{ccc}cosx& -sinx& 0\\ sinx& cosx& 0\\ 0& 0& 1\end{array}\right]$ and $g\left(y\right)=\left[\begin{array}{ccc}cosy& 0& siny\\ 0& 1& 0\\ -siny& 0& cosy\end{array}\right]$, then $[{f\left(x\right)g\left(y\right)]}^{-1}$ is equal to

• A. $F\left(x\right)G(-y)$
• B. $G(-y)F(-x)$
• C. ${F\left(x\right)}^{-1}{G\left(y\right)}^{-1}$
• D. ${G(y}^{-1}\left){F(x}^{-1}\right)$

Answer 1

The correct option is B $G(-y)F(-x)$

Consider, $[{F\left(x\right)G\left(y\right)]}^{-1}=[{G\left(y\right)]}^{-1}[{F\left(x\right)]}^{-1}$ ....(1)
$F\left(x\right)=\left[\begin{array}{ccc}cosx& -sinx& 0\\ sinx& cosx& 0\\ 0& 0& 1\end{array}\right]$
$\Rightarrow |F\left(x\right)|={\cos }^{2}x+{\sin }^{2}x=1$
$adj\left(F\right(x\left)\right)=C^T={\left[\begin{array}{ccc}\cos x& -\sin x& 0\\ \sin x& \cos x& 0\\ 0& 0& 1\end{array}\right]}^T$
$\Rightarrow adj\left(F\right(x\left)\right)=\left[\begin{array}{ccc}\cos x& \sin x& 0\\ -\sin x& \cos x& 0\\ 0& 0& 1\end{array}\right]$
$\left[F\right(x){]}^{-1}=\left[\begin{array}{ccc}\cos x& \sin x& 0\\ -\sin x& \cos x& 0\\ 0& 0& 1\end{array}\right]$

or, $\left[F\right(x){]}^{-1}=\left[\begin{array}{ccc}\cos (-x)& -\sin (-x)& 0\\ \sin (-x)& \cos (-x)& 0\\ 0& 0& 1\end{array}\right]$
$\Rightarrow \left[F\right(x){]}^{-1}=F(-x)$ .....(2)
$G\left(y\right)=\left[\begin{array}{ccc}cosy& 0& siny\\ 0& 1& 0\\ -siny& 0& cosy\end{array}\right]$
$|G\left(y\right)|=1$
$adj\left(G\right(y\left)\right)=C^T={\left[\begin{array}{ccc}cosy& 0& siny\\ 0& 1& 0\\ -siny& 0& cosy\end{array}\right]}^T$
$adj\left(G\right(y\left)\right)=\left[\begin{array}{ccc}cosy& 0& -siny\\ 0& 1& 0\\ siny& 0& cosy\end{array}\right]$
$\Rightarrow \left[G\right(y){]}^{-1}=\left[\begin{array}{ccc}cosy& 0& -siny\\ 0& 1& 0\\ siny& 0& cosy\end{array}\right]$
$\Rightarrow \left[G\right(y){]}^{-1}=\left[\begin{array}{ccc}\cos (-y)& 0& \sin (-y)\\ 0& 1& 0\\ -\sin (-y)& 0& \cos (-y)\end{array}\right]$
$\Rightarrow \left[G\right(y){]}^{-1}=G(-y)$ ....(3)
Substitute the values from (2) and (3) in (1), we get
$[{F\left(x\right)G\left(y\right)]}^{-1}=G(-y\left)F\right(-x)$