Introduction: Here we shall find F(x)F(y) for the given matrix F(x).
Question : If \[F(x)\ =\left(\begin{matrix}cos(x)\ &-sin(x)\ &0\\sin(x)&cos(x)&0\\0&0&1\end{matrix}\right)\] then F(x)F(y) is
equal to
- \[F\left(x+y\right)\]
- \[F\left(x-y\right)\]
- \[F\left(x\right)+F\left(y\right)\]
- \[F\left(x\right)-F\left(y\right)\]
Solution :
Given matrix
\[F\left(x\right)=\left(\begin{matrix}\cos{\left(x\right)}&-\sin{\left(x\right)}&0\\\sin{\left(x\right)}&\cos{\left(x\right)}&0\\0&0&1\end{matrix}\right)\]
then
\[ F(y)\ =\left(\begin{matrix}cos(y)\ &-sin(y)\ &0\\sin(y)&cos(y)&0\\0&0&1\end{matrix}\right)\]
we get
\[F\left(x\right)F\left(y\right)=\left(\begin{matrix}\cos{\left(x\right)}&-\sin{\left(x\right)}&0 \\\sin{\left(x\right)}&\cos{\left(x\right)}&0\\0&0&1\end{matrix}\right)\left(\begin{matrix}\cos{\left(y\right)}&-\sin{\left(y\right)}&0\\\sin{\left(y\right)}&\cos{\left(y\right)}&0\\0&0&1\end{matrix}\right)\]
Hence the matrix multiplication
\[=\left(\begin{matrix}\cos{\left(x\right)}\cos{\left(y\right)}-\sin{\left(x\right)}\sin{\left(y\right)}&-\cos{\left(x\right)}\sin{\left(y\right)}-\sin{\left(x\right)}\cos{\left(y\right)}&0\\\cos{\left(y\right)}\sin{\left(x\right)}+\cos{\left(x\right)}\sin(y)&-\sin{\left(x\right)}\sin{\left(y\right)}+\cos{\left(x\right)}\cos(y)&0\\0&0&1\end{matrix}\right)\]
using formulas of sin(A+B), cos(A+B)
\[=\left(\begin{matrix}\cos{\left(x+y\right)}&-\sin{\left(x+y\right)}&0\\\sin{\left(x+y\right)}&\cos{\left(x+y\right)}&0\\0&0&1\\\end{matrix}\right)\]
\[F\left(x\right)F\left(y\right)=F\left(x+y\right)\]
correct option is (1)