Mathematics - Matrices Question with Solution | TestHub

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

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

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

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

$\Rightarrow f\left(x\right)\times f\left(-x\right)=I$

$\Rightarrow f\left(-x\right)=f^{-1}\left(x\right)$

Hence statement-I is correct

Now, checking statement II

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

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

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

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

$\Rightarrow f\left(x\right)·f\left(y\right)=f(x+y)$

Hence statement-II is also correct.