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

We have,
$P(x) = \left[ {\begin{array}{llllllllllllllllllll}{\cos x}&{\sin x}\\{ - \sin x}&{\cos x}\end{array}} \right]$

$\therefore$ $P(y) = \left[ {\begin{array}{llllllllllllllllllll}{\cos y}&{\sin y}\\{ - \sin y}&{\cos y}\end{array}} \right]$

Now, $P(x) \cdot P(y) = \left[ {\begin{array}{llllllllllllllllllll}{\cos x}&{\sin x}\\{ - \sin x}&{\cos x}\end{array}} \right]\left[ {\begin{array}{cccccccccccccccccccc}{\cos y}&{\sin y}\\{ - \sin y}&{\cos y}\end{array}} \right]$

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

$= \left[ {\begin{array}{llllllllllllllllllll}{\cos (x + y)}&{\sin (x + y)}\\{ - \sin (x + y)}&{\cos (x + y)}\end{array}} \right]$ ……(i)

and $P(x + y) = \left[ {\begin{array}{cccccccccccccccccccc}{\cos (x + y)}&{\sin (x + y)}\\{ - \sin (x + y)}&{\cos (x + y)}\end{array}} \right]$ ….(ii)

Also we have, $P(y) \cdot P(x) = \left[ {\begin{array}{llllllllllllllllllll}{\cos y}&{\sin y}\\{ - \sin y}&{\cos y}\end{array}} \right]\left[ {\begin{array}{llllllllllllllllllll}{\cos x}&{\sin x}\\{ - \sin x}&{\cos x}\end{array}} \right]$

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

$= \left[ {\begin{array}{llllllllllllllllllll}{\cos (x + y)}&{\sin (x + y)}\\{ - \sin (x + y)}&{\cos (x + y)}\end{array}} \right]$ ……(iii)

Thus, we see from the equations (i), (ii) and (iii) that,
$P(x) \cdot P(y) = P(x + y) = P(y) \cdot P(x)$