Answer: (b).

• (a) is nonsense: $J$ is a matrix, and you cannot "divide" a vector by a matrix. Division is replaced by solving a linear system.
• (c) has the wrong sign and multiplies by $J$ instead of by $J^{-1}$. We want the ramp to hit zero: $\mathbf{F}+J\Delta \mathbf{x}=0\Rightarrow J\Delta \mathbf{x}=-\mathbf{F}$.
• (b) is exactly Step 3–4 of the derivation: set the linear model to zero, solve for the step, walk there.

Row $i$ = derivatives of $f_i$ with respect to each variable, in order $(x,y)$. $J=\left(\begin{array}{cc}\frac{\partial f_1}{\partial x}& \frac{\partial f_1}{\partial y}\\ \frac{\partial f_2}{\partial x}& \frac{\partial f_2}{\partial y}\end{array}\right)=\left(\begin{array}{cc}2xy& x^2\\ 1& \cos y\end{array}\right).$ Check the shape: $2$ equations, $2$ unknowns $\Rightarrow$ a $2\times 2$ matrix. Every entry is a function, re-evaluated each iteration.

Jacobian: $J=\left(\begin{array}{cc}2x& 2y\\ y& x\end{array}\right)$. Evaluate at $(2,0.5)$:

• $\mathbf{F}=(4+0.25-4,1-1)=(0.25,0)$.
• $J=\left(\begin{array}{cc}4& 1\\ 0.5& 2\end{array}\right)$, $\det =4\cdot 2-1\cdot 0.5=7.5$. Solve $J\Delta \mathbf{x}=-\mathbf{F}=(-0.25,0)$: $\Delta \mathbf{x}=\frac{1}{7.5}\left(\begin{array}{cc}2& -1\\ -0.5& 4\end{array}\right)\left(\begin{array}{c}-0.25\\ 0\end{array}\right)=\frac{1}{7.5}\left(\begin{array}{c}-0.5\\ 0.125\end{array}\right)=(-0.0\overline{6},0.0166\overline{6}).$ Update: ${\mathbf{x}}_1=(2-0.0667,0.5+0.0167)=(1.9333,0.5167)$. Quick sanity: $\Vert \mathbf{F}\Vert$ went from $0.25$ toward $0$; one component was already exactly satisfied and the step barely disturbed it.

Here $\mathbf{F}=A\mathbf{x}-\mathbf{b}$, so $J=A$ constant — the linear model is exact, not an approximation. Therefore Newton converges in one step from anywhere.

• $\mathbf{F}(0)=A0-\mathbf{b}=(-3,-5)$.
• Solve $A\Delta \mathbf{x}=(3,5)$. $\det =2\cdot 3-1\cdot 1=5$. $\Delta \mathbf{x}=\frac{1}{5}\left(\begin{array}{cc}3& -1\\ -1& 2\end{array}\right)\left(\begin{array}{c}3\\ 5\end{array}\right)=\frac{1}{5}\left(\begin{array}{c}4\\ 7\end{array}\right)=(0.8,1.4).$
• ${\mathbf{x}}_1=(0.8,1.4)$. Check: $A{\mathbf{x}}_1=(1.6+1.4,0.8+4.2)=(3,5)=\mathbf{b}$. ✓ Exact.

Look at the figure: the circle and parabola cross in the first quadrant near the orange dot. Our start $(1,2)$ (blue) sits above the true crossing. Jacobian: $J=\left(\begin{array}{cc}2x& 2y\\ -2x& 1\end{array}\right)$. Step 1 — at $(1,2)$:

• $\mathbf{F}=(1+4-4,2-1)=(1,1)$.
• $J=\left(\begin{array}{cc}2& 4\\ -2& 1\end{array}\right)$, $\det =2\cdot 1-4\cdot (-2)=10$.
• $\Delta \mathbf{x}=\frac{1}{10}\left(\begin{array}{cc}1& -4\\ 2& 2\end{array}\right)\left(\begin{array}{c}-1\\ -1\end{array}\right)=\frac{1}{10}(3,-4)=(0.3,-0.4)$.
• ${\mathbf{x}}_1=(1.3,1.6)$. Step 2 — at $(1.3,1.6)$:
• $\mathbf{F}=(1.69+2.56-4,1.6-1.69)=(0.25,-0.09)$ — much smaller, good.
• $J=\left(\begin{array}{cc}2.6& 3.2\\ -2.6& 1\end{array}\right)$, $\det =2.6\cdot 1-3.2\cdot (-2.6)=2.6+8.32=10.92$.
• Solve $J\Delta \mathbf{x}=(-0.25,0.09)$: $\Delta \mathbf{x}=\frac{1}{10.92}\left(\begin{array}{cc}1& -3.2\\ 2.6& 2.6\end{array}\right)\left(\begin{array}{c}-0.25\\ 0.09\end{array}\right)=\frac{1}{10.92}(-0.538,-0.416)=(-0.04927,-0.03810).$
• ${\mathbf{x}}_2=(1.25073,1.56190)$. The exact root is ${\mathbf{x}}^{\ast }=(1.24962,1.56155)$ (from $x^2=\frac{-1+\sqrt{17}}{2}$, $y=x^2$). After two steps we already match to $3$ decimals — the error roughly squared.

Quadratic convergence means $\Vert {\mathbf{e}}_{k+1}\Vert \le C\Vert {\mathbf{e}}_k{\Vert }^2$.

• $\Vert {\mathbf{e}}_{k+1}\Vert \le 2\cdot (10^{-3}{)}^2=2\times 10^{-6}$.
• $\Vert {\mathbf{e}}_{k+2}\Vert \le 2\cdot (2\times 10^{-6}{)}^2=8\times 10^{-12}$. The count of correct digits roughly doubles each step ($3\to 6\to 12$). This is why Newton needs so few iterations once it is close — see Convergence order of iterative methods.

$J=\left(\begin{array}{cc}2x& 2y\\ -2x& 1\end{array}\right)$. At $(0,2)$: $J=\left(\begin{array}{cc}0& 4\\ 0& 1\end{array}\right),\det =0\cdot 1-4\cdot 0=0.$ The Jacobian is singular — its determinant is $0$, so $J\Delta \mathbf{x}=-\mathbf{F}$ has no unique solution and Newton cannot take a step. Geometric reason: both rows have a zero in the $x$-slot, so the linear model has no sensitivity to $x$ at that instant — both surfaces momentarily look flat in the $x$-direction. The ramp is degenerate. Any point on the $y$-axis ($x=0$) triggers this, since column $1$ of $J$ becomes $(0,0)$. Cure: perturb the start off the $y$-axis (e.g. $(0.1,2)$), or add damping / a small regularization. This is the $n$-D echo of scalar Newton stalling where $f^{\prime }=0$.

The full step ($t=1$) increased the residual, so the linear model was trusted too far. Backtracking line search: start with $t=1$; while $\Vert \mathbf{F}({\mathbf{x}}_k+t\Delta \mathbf{x})\Vert \ge \Vert \mathbf{F}({\mathbf{x}}_k)\Vert$, halve $t$ ($t\leftarrow t/2$). Why halving works: $\Delta \mathbf{x}$ is a descent direction for $\Vert \mathbf{F}{\Vert }^2$ (the linear model reduces the residual for small enough steps). So there is always some small $t>0$ that decreases $\Vert \mathbf{F}\Vert$; shrinking $t$ geometrically is guaranteed to find it. As $t\to 0$ you approach ${\mathbf{x}}_k$ where the residual is smaller, so eventually the test passes. Near the root you get $t=1$ back and full quadratic speed returns.

Jacobian (row $i$ = $\nabla f_i$): $J=\left(\begin{array}{ccc}2x& 1& 1\\ 1& 2y& 1\\ 1& 1& 2z\end{array}\right).$ At $(0.5,0.5,0.5)$: every variable equals $0.5$, so by symmetry each $f_i$ is equal:

• $f_1=0.25+0.5+0.5-3=-1.75$. Thus $\mathbf{F}=(-1.75,-1.75,-1.75)$.
• $J=\left(\begin{array}{ccc}1& 1& 1\\ 1& 1& 1\\ 1& 1& 1\end{array}\right)$. Problem! This $J$ is singular ($\det =0$, all rows identical). At the perfectly symmetric point the three surfaces present the same linear model — the step is not unique. This is a Mastery lesson: symmetry can create a singular Jacobian even when the root is simple. Cure — break symmetry. Start instead at ${\mathbf{x}}_0=(0.5,0.6,0.7)$:
• $f_1=0.25+0.6+0.7-3=-1.45$
• $f_2=0.5+0.36+0.7-3=-1.44$
• $f_3=0.5+0.6+0.49-3=-1.41$ $J=\left(\begin{array}{ccc}1& 1& 1\\ 1& 1.2& 1\\ 1& 1& 1.4\end{array}\right).$ Solving $J\Delta \mathbf{x}=-\mathbf{F}=(1.45,1.44,1.41)$ by elimination:
• Rows $2-1$: $0.2\Delta y=-0.01\Rightarrow \Delta y=-0.05$.
• Rows $3-1$: $0.4\Delta z=-0.04\Rightarrow \Delta z=-0.10$.
• Row $1$: $\Delta x+\Delta y+\Delta z=1.45\Rightarrow \Delta x=1.45+0.05+0.10=1.60$.
• ${\mathbf{x}}_1=(0.5+1.60,0.6-0.05,0.7-0.10)=(2.10,0.55,0.60)$. The step is large (we started far), but the loop now proceeds; iterating drives toward $(1,1,1)$. The mastery point: detect symmetry-induced singularity and perturb the start.

Do not stop. A tiny step with a large residual is the classic sign of a near-singular / ill-conditioned $J$: the solve of $J\Delta \mathbf{x}=-\mathbf{F}$ produced a small $\Delta \mathbf{x}$ not because we are near a root, but because $J$ barely moves the point in the needed direction. Correct stopping test requires BOTH: $\Vert \mathbf{F}({\mathbf{x}}_k)\Vert <{\text{tol}}_F\text{AND}\Vert \Delta \mathbf{x}\Vert <{\text{tol}}_x.$ Here $\Vert \mathbf{F}\Vert =0.4$ fails the first test — we are genuinely far from a root. Investigate the Jacobian (condition number), damp, or restart.

Build $\mathbf{F}$ ::: stack all equation values into one vector; root = all zero. Build $J$ ::: row $i$ is $\nabla f_i$; entry $J_{ij}=\partial f_i/\partial x_j$; re-evaluate each step. The update ::: solve $J\Delta \mathbf{x}=-\mathbf{F}$, then ${\mathbf{x}}_{k+1}={\mathbf{x}}_k+\Delta \mathbf{x}$. Never ::: form $J^{-1}$; solve instead (cheaper, stabler). Stop when ::: both $\Vert \mathbf{F}\Vert$ and $\Vert \Delta \mathbf{x}\Vert$ are below tolerance. If $\det J\approx 0$ ::: singular/ill-conditioned — perturb the start, damp, or regularize. Linear $\mathbf{F}=A\mathbf{x}-\mathbf{b}$ ::: converges in exactly one step ($J=A$ constant).