Question

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

(ii) If $P=\left[\begin{array}{ccc}x& 0& 0\\ 0& y& 0\\ 0& 0& z\end{array}\right]\mathrm{and}Q=\left[\begin{array}{ccc}a& 0& 0\\ 0& b& 0\\ 0& 0& c\end{array}\right],\mathrm{prove}\mathrm{that}PQ=\left[\begin{array}{ccc}xa& 0& 0\\ 0& yb& 0\\ 0& 0& zc\end{array}\right]=QP$

Answer 1

(i) Given: $P\left(x\right)=\left[\begin{array}{cc}\cos x& \sin x\\ -\sin x& \cos x\end{array}\right]$

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

Now,
$$\begin{gathered} P\left(x\right)P\left(y\right)=\left[\begin{array}{cc}\cos x& \sin x\\ -\sin x& \cos x\end{array}\right]\left[\begin{array}{cc}\cos y& \sin y\\ -\sin y& \cos y\end{array}\right] \\ =\left[\begin{array}{cc}\cos x\cos y-\sin x\sin y& \cos x\sin y+\sin x\cos y\\ -\sin x\cos y-\cos x\sin y& -\sin x\sin y+\cos x\cos y\end{array}\right] \\ =\left[\begin{array}{cc}\cos \left(x+y\right)& \sin \left(x+y\right)\\ -\sin \left(x+y\right)& \cos \left(x+y\right)\end{array}\right]...\left(1\right) \end{gathered}$$

Also, $P(x+y)=\left[\begin{array}{cc}\cos \left(x+y\right)& \sin \left(x+y\right)\\ -\sin \left(x+y\right)& \cos \left(x+y\right)\end{array}\right]...\left(2\right)$

Now,
$$\begin{gathered} P\left(y\right)P\left(x\right)=\left[\begin{array}{cc}\cos y& \sin y\\ -\sin y& \cos y\end{array}\right]\left[\begin{array}{cc}\cos x& \sin x\\ -\sin x& \cos x\end{array}\right] \\ =\left[\begin{array}{cc}\cos y\cos x-\sin y\sin x& \cos y\sin x+\sin y\cos x\\ -\sin y\cos x-\cos y\sin x& -\sin y\sin x+\cos y\cos x\end{array}\right] \\ =\left[\begin{array}{cc}\cos \left(x+y\right)& \sin \left(x+y\right)\\ -\sin \left(x+y\right)& \cos \left(x+y\right)\end{array}\right]...\left(3\right) \end{gathered}$$

From (1), (2) and (3), we get

P(x) P(y) = P(x + y) = P(y) P(x)

(ii) Given: $P=\left[\begin{array}{ccc}x& 0& 0\\ 0& y& 0\\ 0& 0& z\end{array}\right]\mathrm{and}Q=\left[\begin{array}{ccc}a& 0& 0\\ 0& b& 0\\ 0& 0& c\end{array}\right]$

Now,
$$\begin{gathered} PQ=\left[\begin{array}{ccc}x& 0& 0\\ 0& y& 0\\ 0& 0& z\end{array}\right]\left[\begin{array}{ccc}a& 0& 0\\ 0& b& 0\\ 0& 0& c\end{array}\right] \\ =\left[\begin{array}{ccc}xa+0+0& 0+0+0& 0+0+0\\ 0+0+0& 0+yb+0& 0+0+0\\ 0+0+0& 0+0+0& 0+0+zc\end{array}\right] \\ =\left[\begin{array}{ccc}xa& 0& 0\\ 0& yb& 0\\ 0& 0& zc\end{array}\right]...\left(4\right) \end{gathered}$$

Also,
$$\begin{gathered} QP=\left[\begin{array}{ccc}a& 0& 0\\ 0& b& 0\\ 0& 0& c\end{array}\right]\left[\begin{array}{ccc}x& 0& 0\\ 0& y& 0\\ 0& 0& z\end{array}\right] \\ =\left[\begin{array}{ccc}ax+0+0& 0+0+0& 0+0+0\\ 0+0+0& 0+by+0& 0+0+0\\ 0+0+0& 0+0+0& 0+0+cz\end{array}\right] \\ =\left[\begin{array}{ccc}xa& 0& 0\\ 0& yb& 0\\ 0& 0& zc\end{array}\right]...\left(5\right) \end{gathered}$$

From (4) and (5), we get
$PQ=\left[\begin{array}{ccc}xa& 0& 0\\ 0& yb& 0\\ 0& 0& zc\end{array}\right]=QP$