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Mathematics Question on Matrices

If F(x)= [cos⁡xsin⁡x0sin⁡xcosx0\001]\begin{bmatrix}\cos x&\sin x&0\\\\\sin x&cos x&0\\\0&0&1\end{bmatrix}and F(y)=[cos⁡y−sin⁡y0sin⁡ycosy0\001]\begin{bmatrix}\cos y&-\sin y&0\\\\\sin y&cos y&0\\\0&0&1\end{bmatrix},show that F(x)+F(y)=F(x+y)

Answer

F(x)=[cos⁡x−sin⁡x0sin⁡xcosx0\001]\begin{bmatrix}\cos x&-\sin x&0\\\\\sin x&cos x&0\\\0&0&1\end{bmatrix},F(y)=[cos⁡y−sin⁡y0sin⁡ycosy0\001]\begin{bmatrix}\cos y&-\sin y&0\\\\\sin y&cos y&0\\\0&0&1\end{bmatrix}

F (x+y)=[cos⁡(x+y)−sin⁡(x+y)0sin⁡(x+y)cos(x+y)0\001]\begin{bmatrix}\cos (x+y)&-\sin (x+y)&0\\\\\sin (x+y)&cos (x+y)&0\\\0&0&1\end{bmatrix}

F(x)F(y)=\begin{bmatrix}\cos x&-\sin x&0\\\\\sin x&cos x&0\\\0&0&1\end{bmatrix}$$\begin{bmatrix}\cos y&-\sin y&0\\\\\sin y&cos y&0\\\0&0&1\end{bmatrix}

=[cos⁡xcos⁡y−sin⁡xsin⁡y+0−cos⁡xsin⁡y−sin⁡xcos⁡y+00sin⁡xcos⁡y+cos⁡xsin⁡y−sin⁡xsin⁡y+cos⁡xcos⁡y+00\000]\begin{bmatrix}\cos x\cos y-\sin x\sin y+0&-\cos x\sin y-\sin x\cos y+0&0\\\\\sin x\cos y+\cos x\sin y&-\sin x\sin y+\cos x\cos y+0&0\\\0&0&0\end{bmatrix}

=[cos⁡(x+y)−sin⁡(x+y)0sin⁡(x+y)cos⁡(x+y)0\001]\begin{bmatrix}\cos (x+y)&-\sin(x+y)&0\\\\\sin (x+y)&\cos(x+y)&0\\\0&0&1\end{bmatrix}
=F(x+y)

∴\therefore F(x)+F(y)=F(x+y)