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

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We are given that, $F(x) = \left[ {\begin{array}{cccccccccccccccccccc}{\cos x}&{ - \sin x}&0\\{\sin x}&{\cos x}&0\\0&0&1\end{array}} \right]$

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

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

$= \left[ {\begin{array}{cccccccccccccccccccc}{\cos x\cos y - \sin x\sin y}&{ - \sin y\cos x - \sin x\cos y}&0\\{\sin x\cos y + \cos x\sin y}&{ - \sin x\sin y + \cos x\cos y}&0\\0&0&1\end{array}} \right]$

$= \left[ {\begin{array}{cccccccccccccccccccc}{\cos (x + y)}&{ - \sin (x + y)}&0\\{\sin (x + y)}&{\cos (x + y)}&0\\0&0&1\end{array}} \right]$
Also, $F(x + y) = \left[ {\begin{array}{cccccccccccccccccccc}{\cos (x + y)}&{ - \sin (x + y)}&0\\{\sin (x + y)}&{\cos (x + y)}&0\\0&0&1\end{array}} \right]$
$\therefore$ F(x) $\cdot$F(y) = F(x + y).