Let G(x) = begin{bmatrix} cos x & -sin x & 0 | vedclass

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(C) Given $G(x) = \begin{bmatrix} \cos x & -\sin x & 0 \ \sin x & \cos x & 0 \ 0 & 0 & 1 \end{bmatrix}$.Since $x+y=0$,we have $y = -x$.Thus,$G(y) =

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identity Matrix

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(C) Given $G(x) = \begin{bmatrix} \cos x & -\sin x & 0 \ \sin x & \cos x & 0 \ 0 & 0 & 1 \end{bmatrix}$.Since $x+y=0$,we have $y = -x$.Thus,$G(y) = G(-x) = \begin{bmatrix} \cos(-x) & -\sin(-x) & 0 \ \sin(-x) & \cos(-x) & 0 \ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} \cos x & \sin x & 0 \ -\sin x & \cos x & 0 \ 0 & 0 & 1 \end{bmatrix}$.Now,$G(x)G(y) = \begin{bmatrix} \cos x & -\sin x & 0 \ \sin x & \cos x & 0 \ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} \cos x & \sin x & 0 \ -\sin x & \cos x & 0 \ 0 & 0 & 1 \end{bmatrix}$.Performing matrix multiplication:Row $1$,Col $1$: $(\cos x)(\cos x) + (-\sin x)(-\sin x) + (0)(0) = \cos^2 x + \sin^2 x = 1$.Row $1$,Col $2$: $(\cos x)(\sin x) + (-\sin x)(\cos x) + (0)(0) = 0$.Row $2$,Col $1$: $(\sin x)(\cos x) + (\cos x)(-\sin x) + (0)(0) = 0$.Row $2$,Col $2$: $(\sin x)(\sin x) + (\cos x)(\cos x) + (0)(0) = \sin^2 x + \cos^2 x = 1$.Thus,$G(x)G(y) = \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix} = I$,which is the identity matrix.

Let $G(x) = \begin{bmatrix} \cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1 \end{bmatrix}$. If $x+y=0$,then $G(x) G(y) =$

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