= Solution
{c}
For <multivariable Newton iteration>, use the <Jacobian matrix>
$$
D\mathbf f(x,y,z)=\begin{pmatrix}
y&x&-3\\
2x&2y&2z\\
\cos x&e^y&-4
\end{pmatrix}.
$$
At $\mathbf x_0=(1/2,-1/2,1/2)$, direct evaluation gives
$$
\mathbf f(\mathbf x_0)=\begin{pmatrix}-3/4\\-1/4\\\sin(1/2)+e^{-1/2}-2\end{pmatrix}.
$$
With $\mathbf s_0=\mathbf x_1-\mathbf x_0$, the required <linear system> is therefore
$$
\boxed{\begin{pmatrix}
-1/2&1/2&-3\\
1&-1&1\\
\cos(1/2)&e^{-1/2}&-4
\end{pmatrix}\begin{pmatrix}s_x\\s_y\\s_z\end{pmatrix}
=\begin{pmatrix}3/4\\1/4\\2-\sin(1/2)-e^{-1/2}\end{pmatrix}.}
$$
After solving this <linear system>, form $\boxed{\mathbf x_1=\mathbf x_0+\mathbf s_0}$. To get the second iterate, evaluate both the <Jacobian matrix> and the residual anew at $\mathbf x_1$, solve
$$
D\mathbf f(\mathbf x_1)\mathbf s_1=-\mathbf f(\mathbf x_1),
$$
and set $\boxed{\mathbf x_2=\mathbf x_1+\mathbf s_1}$. Reusing the initial <Jacobian matrix> would define a different iteration. The requested <linear system> has deliberately been left unsolved.
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