Question

If f (x)= $\left[\begin{array}{ccc}\cos x& -\sin x& 0\\ \sin x& \cos x& 0\\ 0& 0& 1\end{array}\right]$ 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 ${\left[f\left(x\right)g\left(y\right)\right]}^{-1}$ is equal to

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

Answer 1

The correct option is C $g\left(-y\right)f\left(-x\right)$

$\left[f\right(x\left)g\right(y){]}^{-1}=(g\left(y\right){)}^{-1}\left(f\right(x){)}^{-1}$

$g\left(y\right)=\left[\begin{array}{ccc}\cos y& 0& \sin y\\ 0& 1& 0\\ -\sin y& 0& \cos y\end{array}\right]$

$|g\left(y\right)|={\cos }^{2}y+{\sin }^{2}y=1$

cofactors $=\left[\begin{array}{ccc}\cos y& 0& \sin y\\ 0& 1& 0\\ -\sin y& 0& \cos y\end{array}\right]$

$adj\left(g\right(y\left)\right)=\left[\begin{array}{ccc}\cos y& 0& \sin y\\ 0& 1& 0\\ \sin y& 0& \cos y\end{array}\right]$

$\left(g\right(y){)}^{-1}=\frac{adj\left(g\right(y\left)\right)}{|g\left(y\right)|}$

$=\left[\begin{array}{ccc}\cos (-y)& 0& \sin (-y)\\ 0& 1& 0\\ -\sin (-y)& 0& \cos (-y)\end{array}\right]$

$=g(-y)$

$f\left(x\right)=\left[\begin{array}{ccc}\cos x& -\sin x& 0\\ \sin x& \cos x& 0\\ 0& 0& 1\end{array}\right]$

$|f\left(x\right)|={\cos }^{2}x+{\sin }^{2}x=1$

Cofactors $=\left[\begin{array}{ccc}\cos x& -\sin x& 0\\ \sin x& \cos x& 0\\ 0& 0& 1\end{array}\right]$

$adj\left(f\right(x\left)\right)=\left[\begin{array}{ccc}\cos x& \sin x& 0\\ -\sin x& \cos x& 0\\ 0& 0& 1\end{array}\right]$

$\left[f\right(x){]}^{-1}=\left[\begin{array}{ccc}\cos (-x)& -\sin (-x)& 0\\ \sin (-x)& \cos (-x)& 0\\ 0& 0& 1\end{array}\right]$

$=f(-x)$

So, $\left[f\right(x\left)g\right(y){]}^{-1}$

$=\left[g\right(y\left){]}^{-1}\right[f\left(x\right){]}^{-1}$

$=g(-y)g(-x)$

So, the correct option is C