To solve this problem, we need to determine the product of two matrices [tex]\(A(x)\)[/tex] and [tex]\(A(y)\)[/tex], where each matrix [tex]\(A(z)\)[/tex] is defined as follows:
[tex]\[
A(z) = \begin{pmatrix}
\cos z & -\sin z & 0 \\
\sin z & \cos z & 0 \\
0 & 0 & 1
\end{pmatrix}
\][/tex]
Let's denote [tex]\(A(x)\)[/tex] and [tex]\(A(y)\)[/tex] as:
[tex]\[
A(x) = \begin{pmatrix}
\cos x & -\sin x & 0 \\
\sin x & \cos x & 0 \\
0 & 0 & 1
\end{pmatrix}
\][/tex]
[tex]\[
A(y) = \begin{pmatrix}
\cos y & -\sin y & 0 \\
\sin y & \cos y & 0 \\
0 & 0 & 1
\end{pmatrix}
\][/tex]
The product [tex]\(A(x) \cdot A(y)\)[/tex] is obtained by performing matrix multiplication:
[tex]\[
A(x) \cdot A(y) = \begin{pmatrix}
\cos x & -\sin x & 0 \\
\sin x & \cos x & 0 \\
0 & 0 & 1
\end{pmatrix}
\begin{pmatrix}
\cos y & -\sin y & 0 \\
\sin y & \cos y & 0 \\
0 & 0 & 1
\end{pmatrix}
\][/tex]
Carrying out the multiplication, we get:
[tex]\[
\begin{aligned}
&A(x) \cdot A(y) = \begin{pmatrix}
\cos x \cos y + (-\sin x \sin y) & \cos x (-\sin y) + (-\sin x \cos y) & 0 \\
\sin x \cos y + \cos x \sin y & \sin x (-\sin y) + \cos x \cos y & 0 \\
0 & 0 & 1
\end{pmatrix} \\
& = \begin{pmatrix}
\cos x \cos y - \sin x \sin y & -\cos x \sin y - \sin x \cos y & 0 \\
\sin x \cos y + \cos x \sin y & -\sin x \sin y + \cos x \cos y & 0 \\
0 & 0 & 1
\end{pmatrix} \\
& = \begin{pmatrix}
\cos(x + y) & -\sin(x + y) & 0 \\
\sin(x + y) & \cos(x + y) & 0 \\
0 & 0 & 1
\end{pmatrix} \\
&= A(x + y)
\end{aligned}
\][/tex]
Thus, we find that [tex]\(A(x) \cdot A(y) = A(x + y)\)[/tex].
Therefore, the correct option is:
B) [tex]\( A(x + y) \)[/tex]
The answer is [tex]\( \boxed{2} \)[/tex].
