# Solve this

Question:

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

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

Solution:

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

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

Now,

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

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

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

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

Now,

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

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

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

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

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

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

Now,

$P Q=\left[\begin{array}{lll}x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & z\end{array}\right]\left[\begin{array}{lll}a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c\end{array}\right]$

$=\left[\begin{array}{ccc}x a+0+0 & 0+0+0 & 0+0+0 \\ 0+0+0 & 0+y b+0 & 0+0+0 \\ 0+0+0 & 0+0+0 & 0+0+z c\end{array}\right]$

$=\left[\begin{array}{ccc}x a & 0 & 0 \\ 0 & y b & 0 \\ 0 & 0 & z c\end{array}\right]$     ...(4)

Also,

$Q P=\left[\begin{array}{lll}a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c\end{array}\right]\left[\begin{array}{lll}x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & z\end{array}\right]$

$=\left[\begin{array}{ccc}a x+0+0 & 0+0+0 & 0+0+0 \\ 0+0+0 & 0+b y+0 & 0+0+0 \\ 0+0+0 & 0+0+0 & 0+0+c z\end{array}\right]$

$=\left[\begin{array}{ccc}x a & 0 & 0 \\ 0 & y b & 0 \\ 0 & 0 & z c\end{array}\right]$            ...(5)

From (4) and (5), we get

$P Q=\left[\begin{array}{ccc}x a & 0 & 0 \\ 0 & y b & 0 \\ 0 & 0 & z c\end{array}\right]=Q P$