Question

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

• A. $P\left(x\right).P\left(y\right)=P(x+y)$
• B. $P\left(x\right).P\left(y\right)=P\left(xy\right)$
• C. $P\left(x\right).P\left(y\right)=P\left(y\right).P\left(x\right)$
• D. $P\left(x\right).P\left(y\right)=P\left(y\right)+P\left(x\right)$

Answer 1

Answer: (A), (C)

The correct options are
A $P\left(x\right).P\left(y\right)=P(x+y)$
C $P\left(x\right).P\left(y\right)=P\left(y\right).P\left(x\right)$

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

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

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