Question
If $\text{P}\big(\text{x}\big)=\begin{bmatrix}\cos\text{x}&\sin\text{x}\\-\sin\text{x}&\cos\text{x}\end{bmatrix},$ then show that P(x)P(y) = P(x + y) = P(y)P(x).

Answer

Given: $\text{P}\big(\text{x}\big)=\begin{bmatrix}\cos\text{x}&\sin\text{x}\\-\sin\text{x}&\cos\text{x}\end{bmatrix}$
Then, $\text{P}\big(\text{y}\big)=\begin{bmatrix}\cos\text{y}&\sin\text{y}\\-\sin\text{y}&\cos\text{y}\end{bmatrix}$
Now,
$\text{P}\big(\text{x}\big)\text{P}\big(\text{y}\big)=\begin{bmatrix}\cos\text{x}&\sin\text{x}\\-\sin\text{x}&\cos\text{x}\end{bmatrix}\begin{bmatrix}\cos\text{y}&\sin\text{y}\\-\sin\text{y}&\cos\text{y}\end{bmatrix}$
$=\begin{bmatrix}\cos\text{x}\cos\text{y}-\sin\text{x}\sin\text{y}&\cos\text{x}\sin\text{y}+\sin\text{x}\cos\text{y}\\-\sin\text{x}\cos\text{y}-\cos\text{x}\sin\text{y}&-\sin\text{x}\sin\text{y}+\cos\text{x}\cos\text{y}\end{bmatrix}$
$=\begin{bmatrix}\cos(\text{x}+\text{y})&\sin(\text{x}+\text{y})\\-\sin(\text{x}+\text{y})&\cos(\text{x}+\text{y})\end{bmatrix}\ \dots(1)$
Also, $\text{P}\big(\text{x}+\text{y}\big)=\begin{bmatrix}\cos(\text{x}+\text{y})&\sin(\text{x}+\text{y})\\-\sin(\text{x}+\text{y})&\cos(\text{x}+\text{y})\end{bmatrix}\ \dots(2)$
Now,
$=\text{P}\big(\text{y}\big)\text{P}\big(\text{x}\big)=\begin{bmatrix}\cos\text{y}&\sin\text{y}\\-\sin\text{y}&\cos\text{y}\end{bmatrix}\begin{bmatrix}\cos\text{x}&\sin\text{x}\\-\sin\text{x}&\cos\text{x}\end{bmatrix}$
$=\begin{bmatrix}\cos\text{y}\cos\text{x}-\sin\text{y}\sin\text{x}&\cos\text{y}\sin\text{x}+\sin\text{y}\cos\text{x}\\-\sin\text{y}\cos\text{x}\cos\text{y}\sin\text{x}&\sin\text{y}\sin\text{x}+\cos\text{y}\cos\text{x}\end{bmatrix}$
$=\begin{bmatrix}\cos(\text{x}+\text{y})&\sin(\text{x}+\text{y})\\-\sin(\text{x}+\text{y})&\cos(\text{x}+\text{y})\end{bmatrix}\ \dots(3)$
From (1), (2) and (3), we get
P(x)P(y) = P(x + y) = P(y)P(x)

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