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Mathematics Question on Matrices

If R(t)=[cos⁡tsin⁡t −sin⁡tcos⁡t]R\left(t\right) = \begin{bmatrix}\cos t&\sin t\\\ -\sin t&\cos t\end{bmatrix}, then R(s) R(t) equals

A

R (s + t)

B

R (s - t)

C

R(s) + R(t)

D

None of these

Answer

R (s + t)

Explanation

Solution

R(s)R(t)=[cos⁡ssin⁡s −sin⁡scos⁡s]×[cos⁡tsin⁡t −sin⁡tcos⁡t]R\left(s\right) R\left(t\right) = \begin{bmatrix}\cos s&\sin s\\\ -\sin s& \cos s\end{bmatrix} \times\begin{bmatrix}\cos t&\sin t\\\ -\sin t&\cos t\end{bmatrix}
=[cos⁡scos⁡t−sin⁡ssin⁡tcos⁡ssin⁡t+sin⁡scos⁡t −sin⁡scos⁡t−cos⁡ssin⁡t−sin⁡ssin⁡t+cos⁡scos⁡t]= \begin{bmatrix}\cos s \cos t -\sin s \sin t&\cos s \sin t + \sin s \cos t\\\ -\sin s \cos t -\cos s \sin t& -\sin s \sin t +\cos s \cos t\end{bmatrix}
=[cos⁡(s+t)sin⁡(s+t) −sin⁡(s+t)cos⁡(s+t)]=R(s+t)= \begin{bmatrix}\cos\left(s+t\right)&\sin\left(s+t\right)\\\ -\sin\left(s+t\right)&\cos\left(s+t\right)\end{bmatrix}=R\left(s+t\right)