Answer: (b).

• (a) nonsense hai: $J$ ek matrix hai, aur aap ek vector ko matrix se "divide" nahi kar sakte. Division ki jagah linear system solve karna hota hai.
• (c) mein galat sign hai aur $J^{-1}$ ki jagah $J$ se multiply kiya gaya hai. Hum chahte hain ki ramp zero hit kare: $\mathbf{F}+J\Delta \mathbf{x}=0\Rightarrow J\Delta \mathbf{x}=-\mathbf{F}$.
• (b) bilkul derivation ke Step 3–4 jaisa hai: linear model ko zero set karo, step ke liye solve karo, wahan jao.

Row $i$ = $f_i$ ke derivatives har variable ke respect mein, order $(x,y)$ mein. $J=\left(\begin{array}{cc}\frac{\partial f_1}{\partial x}& \frac{\partial f_1}{\partial y}\\ \frac{\partial f_2}{\partial x}& \frac{\partial f_2}{\partial y}\end{array}\right)=\left(\begin{array}{cc}2xy& x^2\\ 1& \cos y\end{array}\right).$ Shape check karo: $2$ equations, $2$ unknowns $\Rightarrow$ ek $2\times 2$ matrix. Har entry ek function hai, jo har iteration mein re-evaluate hoti hai.

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

• $\mathbf{F}=(4+0.25-4,1-1)=(0.25,0)$.
• $J=\left(\begin{array}{cc}4& 1\\ 0.5& 2\end{array}\right)$, $\det =4\cdot 2-1\cdot 0.5=7.5$. $J\Delta \mathbf{x}=-\mathbf{F}=(-0.25,0)$ solve karo: $\Delta \mathbf{x}=\frac{1}{7.5}\left(\begin{array}{cc}2& -1\\ -0.5& 4\end{array}\right)\left(\begin{array}{c}-0.25\\ 0\end{array}\right)=\frac{1}{7.5}\left(\begin{array}{c}-0.5\\ 0.125\end{array}\right)=(-0.0\overline{6},0.0166\overline{6}).$ Update: ${\mathbf{x}}_1=(2-0.0667,0.5+0.0167)=(1.9333,0.5167)$. Quick sanity: $\Vert \mathbf{F}\Vert$ $0.25$ se $0$ ki taraf gaya; ek component pehle se exactly satisfy tha aur step ne use barely disturb kiya.

Yahan $\mathbf{F}=A\mathbf{x}-\mathbf{b}$ hai, isliye $J=A$ constant hai — linear model ek approximation nahi, exact hai. Isliye Newton kahin se bhi ek step mein converge karta hai.

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

Figure dekhein: circle aur parabola first quadrant mein orange dot ke paas cross karte hain. Hamara start $(1,2)$ (blue) true crossing ke upar hai. Jacobian: $J=\left(\begin{array}{cc}2x& 2y\\ -2x& 1\end{array}\right)$. Step 1 — $(1,2)$ par:

• $\mathbf{F}=(1+4-4,2-1)=(1,1)$.
• $J=\left(\begin{array}{cc}2& 4\\ -2& 1\end{array}\right)$, $\det =2\cdot 1-4\cdot (-2)=10$.
• $\Delta \mathbf{x}=\frac{1}{10}\left(\begin{array}{cc}1& -4\\ 2& 2\end{array}\right)\left(\begin{array}{c}-1\\ -1\end{array}\right)=\frac{1}{10}(3,-4)=(0.3,-0.4)$.
• ${\mathbf{x}}_1=(1.3,1.6)$. Step 2 — $(1.3,1.6)$ par:
• $\mathbf{F}=(1.69+2.56-4,1.6-1.69)=(0.25,-0.09)$ — bahut chhota, accha hai.
• $J=\left(\begin{array}{cc}2.6& 3.2\\ -2.6& 1\end{array}\right)$, $\det =2.6\cdot 1-3.2\cdot (-2.6)=2.6+8.32=10.92$.
• $J\Delta \mathbf{x}=(-0.25,0.09)$ solve karo: $\Delta \mathbf{x}=\frac{1}{10.92}\left(\begin{array}{cc}1& -3.2\\ 2.6& 2.6\end{array}\right)\left(\begin{array}{c}-0.25\\ 0.09\end{array}\right)=\frac{1}{10.92}(-0.538,-0.416)=(-0.04927,-0.03810).$
• ${\mathbf{x}}_2=(1.25073,1.56190)$. Exact root ${\mathbf{x}}^{\ast }=(1.24962,1.56155)$ hai (jo $x^2=\frac{-1+\sqrt{17}}{2}$, $y=x^2$ se aata hai). Do steps ke baad hum already $3$ decimals tak match kar rahe hain — error roughly square ho gayi.

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

• $\Vert {\mathbf{e}}_{k+1}\Vert \le 2\cdot (10^{-3}{)}^2=2\times 10^{-6}$.
• $\Vert {\mathbf{e}}_{k+2}\Vert \le 2\cdot (2\times 10^{-6}{)}^2=8\times 10^{-12}$. Correct digits ki count roughly har step mein double hoti hai ($3\to 6\to 12$). Isliye ek baar close aane par Newton ko itne kam iterations chahiye hote hain — dekho Convergence order of iterative methods.

$J=\left(\begin{array}{cc}2x& 2y\\ -2x& 1\end{array}\right)$. $(0,2)$ par: $J=\left(\begin{array}{cc}0& 4\\ 0& 1\end{array}\right),\det =0\cdot 1-4\cdot 0=0.$ Jacobian singular hai — determinant $0$ hai, isliye $J\Delta \mathbf{x}=-\mathbf{F}$ ka koi unique solution nahi aur Newton step nahi le sakta. Geometric reason: dono rows ke $x$-slot mein zero hai, isliye linear model ka $x$ ke liye koi sensitivity nahi us instant mein — dono surfaces momentarily $x$-direction mein flat dikhti hain. Ramp degenerate hai. $y$-axis ($x=0$) par koi bhi point yeh trigger karta hai, kyunki $J$ ka column $1$ $(0,0)$ ban jaata hai. Cure: start ko $y$-axis se hataao (jaise $(0.1,2)$), ya damping / thodi regularization add karo. Yeh scalar Newton ke $n$-D echo hai jahan $f^{\prime }=0$ par woh ruk jaata hai.

Full step ($t=1$) ne residual badha diya, isliye linear model par zyada trust kiya gaya. Backtracking line search: $t=1$ se shuru karo; jab tak $\Vert \mathbf{F}({\mathbf{x}}_k+t\Delta \mathbf{x})\Vert \ge \Vert \mathbf{F}({\mathbf{x}}_k)\Vert$, $t$ ko halve karo ($t\leftarrow t/2$). Halving kyun kaam karti hai: $\Delta \mathbf{x}$ $\Vert \mathbf{F}{\Vert }^2$ ke liye ek descent direction hai (linear model chhote enough steps ke liye residual reduce karta hai). Isliye hamesha koi chhota $t>0$ hota hai jo $\Vert \mathbf{F}\Vert$ reduce kare; $t$ ko geometrically shrink karna guaranteed hai ki woh mil jaayega. Jaise $t\to 0$, aap ${\mathbf{x}}_k$ ke paas pahunchte ho jahan residual chhota hai, isliye eventually test pass ho jaata hai. Root ke paas $t=1$ wapas milta hai aur full quadratic speed return hoti hai.

Jacobian (row $i$ = $\nabla f_i$): $J=\left(\begin{array}{ccc}2x& 1& 1\\ 1& 2y& 1\\ 1& 1& 2z\end{array}\right).$ $(0.5,0.5,0.5)$ par: har variable $0.5$ equals hai, isliye symmetry se har $f_i$ equal hai:

• $f_1=0.25+0.5+0.5-3=-1.75$. Isliye $\mathbf{F}=(-1.75,-1.75,-1.75)$.
• $J=\left(\begin{array}{ccc}1& 1& 1\\ 1& 1& 1\\ 1& 1& 1\end{array}\right)$. Problem! Yeh $J$ singular hai ($\det =0$, sab rows identical). Perfectly symmetric point par teeno surfaces same linear model present karti hain — step unique nahi hai. Yeh ek Mastery lesson hai: symmetry ek singular Jacobian create kar sakti hai chahe root simple ho. Cure — symmetry todo. Iske bajaay ${\mathbf{x}}_0=(0.5,0.6,0.7)$ se start karo:
• $f_1=0.25+0.6+0.7-3=-1.45$
• $f_2=0.5+0.36+0.7-3=-1.44$
• $f_3=0.5+0.6+0.49-3=-1.41$ $J=\left(\begin{array}{ccc}1& 1& 1\\ 1& 1.2& 1\\ 1& 1& 1.4\end{array}\right).$ Elimination se $J\Delta \mathbf{x}=-\mathbf{F}=(1.45,1.44,1.41)$ solve karo:
• 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 $1$: $\Delta x+\Delta y+\Delta z=1.45\Rightarrow \Delta x=1.45+0.05+0.10=1.60$.
• ${\mathbf{x}}_1=(0.5+1.60,0.6-0.05,0.7-0.10)=(2.10,0.55,0.60)$. Step bada hai (hum door se start kiye), lekin loop ab aage badhta hai; iterate karte karte $(1,1,1)$ ki taraf jaata hai. Mastery point: symmetry-induced singularity detect karo aur start perturb karo.

Mat ruko. Bade residual ke saath tiny step ek near-singular / ill-conditioned $J$ ka classic sign hai: $J\Delta \mathbf{x}=-\mathbf{F}$ ka solve chhota $\Delta \mathbf{x}$ diya, is liye nahi ki hum root ke paas hain, balki is liye ki $J$ zaruri direction mein point ko barely move kar paata hai. Correct stopping test ke liye DONO chahiye: $\Vert \mathbf{F}({\mathbf{x}}_k)\Vert <{\text{tol}}_F\text{AND}\Vert \Delta \mathbf{x}\Vert <{\text{tol}}_x.$ Yahan $\Vert \mathbf{F}\Vert =0.4$ pehla test fail karta hai — hum genuinely root se door hain. Jacobian investigate karo (condition number), damp karo, ya restart karo.

$\mathbf{F}$ banao ::: saari equation values ko ek vector mein stack karo; root = sab zero. $J$ banao ::: row $i$ hai $\nabla f_i$; entry $J_{ij}=\partial f_i/\partial x_j$; har step mein re-evaluate karo. Update ::: $J\Delta \mathbf{x}=-\mathbf{F}$ solve karo, phir ${\mathbf{x}}_{k+1}={\mathbf{x}}_k+\Delta \mathbf{x}$. Kabhi mat karo ::: $J^{-1}$ form karo; iske bajaay solve karo (sasta, stable). Tab ruko jab ::: $\Vert \mathbf{F}\Vert$ aur $\Vert \Delta \mathbf{x}\Vert$ dono tolerance se neeche hon. Agar $\det J\approx 0$ ::: singular/ill-conditioned — start perturb karo, damp karo, ya regularize karo. Linear $\mathbf{F}=A\mathbf{x}-\mathbf{b}$ ::: exactly ek step mein converge hota hai ($J=A$ constant).