#### Explain solution for rd sharma class class 12 chapter Algebra of matrices exercise 4.3 question 54 sub question (i) math

Hence proved

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

Hint: Matrix multiplication is only possible, when the number of columns of first matrix is equal to the number of rows of second matrix.

Given:

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

We have,

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

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

Since

$[\cos x \cos y-\sin x \sin y=\cos (x+y), \sin y \cos x+\sin x \cos y \\\\ =\sin (x+y),-\sin x \cos y- \cos x \sin y=-\sin (x+y)]$

We have,

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