= Solution
{c}
For <Newton's method for nonlinear systems>, solve $J(\mathbf z_0)\boldsymbol\delta=-\mathbf F(\mathbf z_0)$ and set $\mathbf z_1=\mathbf z_0+\boldsymbol\delta$. The <Jacobian matrix> here is
$$
J(x,y)=\begin{pmatrix}
3x^2+3&-2\\
3x^2&6y-1
\end{pmatrix}.
$$
At $(0,0)$,
$$
\mathbf F(0,0)=\begin{pmatrix}-1\\-1\end{pmatrix},\qquad
J(0,0)=\begin{pmatrix}3&-2\\0&-1\end{pmatrix}.
$$
The correction <linear system> is $3\delta_x-2\delta_y=1$, $-\delta_y=1$. Hence $\delta_y=-1$, $\delta_x=-1/3$, and one iteration gives
$$
\boxed{(x_1,y_1)=(-1/3,-1).}
$$
At the alternative starting value $(1,0)$,
$$
\mathbf F(1,0)=\begin{pmatrix}3\\0\end{pmatrix},\qquad
J(1,0)=\begin{pmatrix}6&-2\\3&-1\end{pmatrix}.
$$
This <Jacobian matrix> is a <singular matrix>: its first row is twice its second. In fact the correction equations would be
$$
6\delta_x-2\delta_y=-3,\qquad 3\delta_x-\delta_y=0,
$$
which are inconsistent, since twice the second left side equals the first while twice its right side is $0$, rather than $-3$. Thus \b[Newton's iteration cannot take its first step from $(1,0)$]. This is failure of the correction <linear system>, not merely nonuniqueness of a possible correction.
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