Question bank — Solving linear systems — Gaussian elimination with partial pivoting

False. Ye tiny pivots se divide karne se bhi bachata hai, jo bade multipliers paida karte hain jo round-off amplify karte hain; point sirf legality nahi, stability hai.

True. Hum largest-magnitude column entry ko diagonal par rakhte hain, isliye $|a_{ik}|\le |a_{kk}|$ aur $|m_{ik}|=|a_{ik}/a_{kk}|\le 1$.

False. Do equations ko swap karna sirf unhe reorder karta hai; solution set unchanged rehta hai — swapping ek legal elementary row operation hai.

False. Ek nonzero-but-tiny pivot floating point mein numerically disastrous hota hai; partial pivoting tab bhi bahut help karta hai.

False. Kuch non-singular matrices mid-way ek zero pivot deti hain (jisko swap chahiye), aur process exact zero hit kiye bina bhi accuracy bahut kharab kar sakta hai.

False. Complete pivoting poore remaining submatrix ko search karta hai, isliye wo alag pivot choose kar sakta hai aur zyada cost par (usually zyada) accurate result de sakta hai.

False. Sirf last equation mein ek hi unknown hota hai; har $x_i$ ko already-known $x_j$ chahiye jahan $j>i$, isliye tum bottom-up jaana hi padega.

False. Aise matrices pivoting ke bina stable hote hain (Cholesky ko koi nahi chahiye); wahan pivoting symmetry bhi destroy kar sakta hai. Dekho Condition Number and Numerical Stability.

True. Koi swap usable pivot supply nahi kar sakta, isliye column $k$ (rows $k\dots n$) sab zero hai — rows dependent hain aur $\det A=0$.

False. Ye sirf $O(n^2)$ comparisons add karta hai $\frac{2}{3}{n}^3$ elimination cost ke upar — ye sasta insurance hai, doubling nahi.

Wrong axis. Partial pivoting current column ke neeche entries compare karta hai (rows $k\dots n$), kisi row ke along nahi; row search alag variables ko mix up kar deta.

Swap mein right-hand side bhi shamil hona chahiye. Augmented $[A|\mathbf{b}]$ par kaam karo; half-swap galat RHS values ko galat equations ke saath pair karta hai aur answer corrupt ho jaata hai.

Ratio ulta hai. Hume chahiye $a_{ik}-m_{ik}{a}_{kk}=0$, jo deta hai $m_{ik}=a_{ik}/a_{kk}$ — denominator mein pivot.

Row operation ${\text{Row}}_i\leftarrow {\text{Row}}_i-m_{ik}{\text{Row}}_k$ har column par act karta hai jisme RHS bhi shamil hai: $b_i\leftarrow b_i-m_{ik}{b}_k$. $\mathbf{b}$ skip karna galat system solve karta hai.

Pivot se divide karna zaroori hai: $x_i=\frac{1}{u_{ii}}(c_i-{\sum }_{j>i}{u}_{ij}{x}_j)$. $u_{ii}$ chhorna har unknown ko galat scale kar deta hai siwaye jab $u_{ii}=1$.

Zero ko scale karna zero hi rehta hai; nonzero pivot ko scale karna kuch essential nahi badalta. Sahi move ek row swap hai us row ke saath jisme nonzero column entry ho.

Pivoting absolute value use karta hai: $\arg {\max }_{i\ge k}|a_{ik}|$. Ek large-magnitude negative entry (jaise $-9$ over $+2$) better, zyada stable pivot hai.

Update $a_{ij}\leftarrow a_{ij}-m_{ik}{a}_{kj}$ pivot row ko $m_{ik}$ se scale karta hai; ek bada $m_{ik}$ $a_{ij}$ ko swamp kar sakta hai isliye chhoti term ke significant digits round-off mein kho jaate hain. Dekho Round-off Error in Floating Point.

$k$ se upar wali rows already finalized hain (unke pivots set hain), aur doosre columns un variables ke hain jinhe hum abhi tak nahi pahunche; sirf current column ki sub-entries legally is step ka pivot supply kar sakti hain.

Elimination $n$ columns mein se har ek ke liye ek $O(n)\times O(n)$ block touch karta hai ($\sim \frac{2}{3}{n}^3$), jabki back-substitution $n$ unknowns ke liye har unknown ke liye ek $O(n)$-length sum karta hai ($\sim n^2$).

$\det A$ pivots ke product ke barabar hai times $(-1{)}^s$ $s$ row swaps ke liye; elimination already wo pivots produce karta hai, $n!$ cofactor route se bahut sasta. Dekho Determinants aur Cramer's Rule.

Bounded multipliers $U$ ki entries ko wildly badhne se rokti hain, jo round-off ki amplification ko bound karti hai — exactly yahi "numerical stability" hume deti hai.

Pivoted elimination hi ek $LU$ factorization $PA=LU$ hai; $L$, $U$, $P$ store karna hume $A\mathbf{x}=\mathbf{b}$ kai alag $\mathbf{b}$ ke liye re-eliminating ke bina solve karne deta hai. Dekho LU Decomposition aur Back-substitution and Forward-substitution.

Huge sparse systems ke liye, $\frac{2}{3}{n}^3$ direct work aur fill-in prohibitive ho jaate hain, jabki har iteration $O(\text{nonzeros})$ hai. Dekho Iterative Methods (Jacobi, Gauss-Seidel).

Koi swap nonzero pivot yield nahi karta, isliye $A$ singular hai; system ka ya koi solution nahi ya infinitely many hain, kabhi bhi unique nahi.

Wo row ko khud se swap karta hai — ek no-op. Pivoting kabhi hurt nahi karta; best case mein ye sirf current pivot confirm karta hai.

Dono valid hain; ek common tie-break lowest-index row hai. Choice intermediate numbers affect karta hai lekin exact solution nahi.

Saari below-diagonal entries zero hain, isliye har multiplier zero hai aur koi swap kuch improve nahi karta; tum seedha back-substitution par ja sakte ho.

Haan — us lower row ko nonzero pivot supply karne ke liye upar swap karo. Ye precisely wahi case hai jisko partial pivoting rescue karne ke liye design kiya gaya tha.

Pivoting phir bhi run hoga aur ek answer dega, lekin bada condition number matlab hai ki chhoti data errors badi solution errors cause kar sakti hain; soch-samajh ke report karo. Dekho Condition Number and Numerical Stability.

Eliminate karne ke liye kuch nahi hai; back-substitution deta hai $x=b/a$, aur pivoting sirf check karta hai ki $a\ne 0$ (warna singular).