= Solution
{c}
The <gradient> and constant <Hessian matrix> are
$$
\nabla F=\begin{pmatrix}6x-4y+4\\4y-4x-4\end{pmatrix},\qquad
H_F=\begin{pmatrix}6&-4\\-4&4\end{pmatrix},\qquad
H_F^{-1}=\frac18\begin{pmatrix}4&4\\4&6\end{pmatrix}.
$$
At $\mathbf x_0=(1,0)^T$, $\nabla F=(10,-8)^T$ and $H_F^{-1}\nabla F=(1,-1)^T$. Hence one <multivariable Newton iteration> gives
$$
\boxed{\mathbf x_1=(1,0)^T-(1,-1)^T=(0,1)^T.}
$$
At this point the <gradient> vanishes, so every subsequent iterate is $(0,1)$. The one-step result follows because the <gradient> of a quadratic <polynomial> is affine and its linearization is exact. Since the leading principal minors of $H_F$ are $6>0$ and $8>0$, it is a <positive-definite matrix>. Alternatively,
$$
F(x,y)=x^2+2(y-x-1)^2-1
$$
directly proves that \b[$(0,1)$ is the unique <global minimum>, with value $-1$].
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