Worked examples — Solving linear systems — Gaussian elimination with partial pivoting

Before anything, three words you must own (each built in the parent, repeated here so line one stands alone):

• Pivot — the number sitting on the diagonal that we divide by. Picture the top-left corner of the block we are currently clearing.
• Multiplier $m_{ik}=a_{ik}/a_{kk}$ — how many copies of the pivot row we subtract from row $i$ to punch a zero into position $(i,k)$.
• Augmented matrix $[A|\mathbf{b}]$ — the coefficient grid with the right-hand side stapled on as one extra column, so every row operation touches the answer column too.

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The scenario matrix

Every linear-system problem this topic throws lands in exactly one of these cells. Our job below is to hit all of them.

Cell	What makes it special	Danger it probes	Hit by
C1 Clean unique	biggest pivot already on top, nice numbers	none — the baseline	Ex 1
C2 Swap needed	biggest entry is below the diagonal	forgetting to pivot	Ex 2
C3 Tiny pivot	a near-zero would-be pivot	round-off blow-up	Ex 3
C4 Negative pivots & signs	negative diagonal, negative multipliers	sign slips in $m_{ik}$	Ex 4
C5 Zero pivot, but fixable	a literal $0$ on the diagonal that a swap rescues	"divide by zero" panic	Ex 5
C6 Singular — no solution	whole row collapses to $0=c$, $c\ne 0$	recognising failure	Ex 6
C6′ Singular — infinitely many	whole row collapses to $0=0$	free variables	Ex 6′
C7 Word problem (real units)	mixture / circuit modelled as $A\mathbf{x}=\mathbf{b}$	translating words → matrix	Ex 7
C8 Exam twist (parameter)	an unknown $t$ decides solvability	limiting behaviour	Ex 8

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Ex 1 — Cell C1: the clean baseline

Forecast: column 1 has $|4|>|2|$, so no swap. Guess the answer is small positive numbers.

1. Check the pivot. Why this step? Partial pivoting first looks at column 1, rows 1–2: $|4|$ vs $|2|$. The biggest is already on top, so no swap.
2. Multiplier. Why this step? We need a $0$ in position $(2,1)$. $m_{21}=a_{21}/a_{11}=2/4=0.5$.
3. Eliminate. Why this step? Subtracting $0.5\times$row 1 kills $a_{21}$ and updates the rest of the row (including the RHS): $a_{22}=5-0.5(3)=3.5,b_2=11-0.5(10)=6.$ $\left[\begin{array}{cccc}4& 3& |& 10\\ 0& 3.5& |& 6\end{array}\right].$
4. Back-substitute. Why bottom-up? The last row now has one unknown: $x_2=\frac{6}{3.5}=\frac{12}{7}\approx 1.7143,x_1=\frac{10-3x_2}{4}=\frac{10-36/7}{4}=\frac{34/7}{4}=\frac{17}{14}\approx \mathrm{1.2143.}$

Verify: eq 2: $2(1.2143)+5(1.7143)=2.4286+8.5714=11✓$.

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Ex 2 — Cell C2: a swap is forced

Forecast: $|3|>|1|$ down column 1, so we must swap. Guess: swapping first, then a clean elimination.

1. Scan column 1. Why this step? Among rows 1–2, $|3|>|1|$; the biggest sits in row 2. Swap rows 1 and 2 — the entire rows, RHS included. $\left[\begin{array}{cccc}3& 2& |& 7\\ 1& 4& |& 9\end{array}\right].$
2. Multiplier. Why this step? Now the pivot is a safe $3$. $m_{21}=1/3\approx 0.3333$.
3. Eliminate. Why: punch the zero and update: $a_{22}=4-\frac{1}{3}(2)=\frac{10}{3}\approx 3.3333,b_2=9-\frac{1}{3}(7)=\frac{20}{3}\approx \mathrm{6.6667.}$
4. Back-substitute. Why bottom-up? The eliminated row 2 now holds only $x_2$, so we solve it first, then feed that value up into row 1: $x_2=\frac{20/3}{10/3}=2,x_1=\frac{7-2(2)}{3}=1.$

Verify: eq 1 (original): $1(1)+4(2)=9✓$; eq 2: $3(1)+2(2)=7✓$.

The figure shows why pivoting is a geometric idea: each equation is a line, the solution is their crossing point, and swapping rows never moves that crossing — it just relabels which line we clear first.

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Ex 3 — Cell C3: the tiny-pivot trap (with vs without)

Forecast: the exact answer (solving the two equations) is $x_1\approx 1.6668,x_2\approx 1.9992$. Guess the no-pivot route corrupts $x_1$.

Route A — WITHOUT pivoting (the cautionary tale).

1. Keep the tiny pivot $0.002$. Why show this? To expose the failure the parent warned about. $m_{21}=3/0.002=1500$. A monster multiplier magnifies every rounding error.
2. Round each result to 4 sig figs. Why: our imaginary calculator only holds 4 digits. $a_{22}=1-1500(4)=1-6000=-5999\to -5999,$ $b_2=7-1500(8)=7-12000=-11993\to -11990(\text{4 sig figs}).$
3. $x_2=-11990/-5999=1.999$. Then $x_1=(8-4(1.999))/0.002$. Why it breaks: $4(1.999)=7.996$, and $8-7.996=0.004$ — a subtraction of near-equal 4-digit numbers keeps only one good digit. So $x_1=0.004/0.002=2.000$, badly wrong (true value $1.667$).

Route B — WITH pivoting.

1. Swap first. Why this step? Partial pivoting scans column 1, rows 1–2: $|3|>|0.002|$, so the biggest goes on top. This is the whole point — never divide by the tiny entry. $\left[\begin{array}{cccc}3& 1& |& 7\\ 0.002& 4& |& 8\end{array}\right].$
2. Multiplier. Why: to zero the $(2,1)$ entry. $m_{21}=0.002/3=0.0006667$ (tiny magnitude ⇒ tiny error amplification).
3. Eliminate, rounding to 4 sig figs. $a_{22}=4-0.0006667(1)=3.999$; $b_2=8-0.0006667(7)=7.995$.
4. Back-substitute. Why bottom-up: row 2 now has one unknown. $x_2=7.995/3.999=1.999$; $x_1=(7-1(1.999))/3=5.001/3=1.667$.

Verify: the exact solution has $x_1\approx 1.6668$. Numerically, eq 1 with $(1.667,1.999)$: $0.002(1.667)+4(1.999)=0.003334+7.996=7.999\approx 8✓$. Pivoting kept $x_1$; no-pivoting destroyed it. See Round-off Error in Floating Point and Condition Number and Numerical Stability.

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Ex 4 — Cell C4: negative pivots and negative multipliers

Forecast: $|-4|>|3|$, so no swap; the pivot is negative — watch the signs.

1. Scan. Why: $|-4|=4>3=|3|$, biggest already on top. No swap. The pivot is $-4$; negative is fine — only small magnitude is bad.
2. Multiplier. $m_{21}=a_{21}/a_{11}=3/(-4)=-0.75$. Why the minus matters: subtracting a negative multiple means we effectively add. Keep the sign attached.
3. Eliminate. Row${}_2\leftarrow$Row${}_2-(-0.75)\cdot$Row${}_1$: $a_{22}=-5-(-0.75)(2)=-5+1.5=-3.5,b_2=-1-(-0.75)(2)=-1+1.5=\mathrm{0.5.}$
4. Back-substitute. $x_2=\frac{0.5}{-3.5}=-\frac{1}{7}\approx -0.1429,x_1=\frac{2-2x_2}{-4}=\frac{2-2(-1/7)}{-4}=\frac{2+2/7}{-4}=\frac{16/7}{-4}=-\frac{4}{7}\approx -\mathrm{0.5714.}$

Verify: eq 2: $3(-4/7)-5(-1/7)=-12/7+5/7=-7/7=-1✓$.

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Ex 5 — Cell C5: a literal zero pivot that a swap rescues

Forecast: $a_{11}=0$ — you cannot divide by it. But there is a nonzero entry below, so a swap saves us. Pivoting handles this automatically.

1. Scan column 1. Why this step? Rows 1–2 give $|0|$ vs $|3|$. Largest is $3$ in row 2 — swap. The rule that always picks the biggest magnitude never leaves a $0$ on the diagonal when a nonzero exists below. $\left[\begin{array}{cccc}3& 1& |& 5\\ 0& 2& |& 4\end{array}\right].$
2. Lucky elimination. $m_{21}=0/3=0$: row 2 is already clear. Why: the swap did all the work.
3. Back-substitute. $x_2=\frac{4}{2}=2,x_1=\frac{5-1(2)}{3}=1.$

Verify: original eq 1: $0(1)+2(2)=4✓$; eq 2: $3(1)+1(2)=5✓$.

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Ex 6 — Cell C6: genuinely singular (no solution)

Forecast: row 2 is almost half of row 1 — the left sides are proportional but the right sides are not. Guess: no solution (parallel lines).

1. Scan / no swap. $|2|>|1|$, pivot $=2$.
2. Eliminate. $m_{21}=1/2=0.5$: $a_{22}=2-0.5(4)=0,b_2=5-0.5(6)=2.$ $\left[\begin{array}{cccc}2& 4& |& 6\\ 0& 0& |& 2\end{array}\right].$
3. Read the bottom row. Why this step? The last row says $0\cdot x_1+0\cdot x_2=2$, i.e. $0=2$ — impossible. The whole column-2 sub-block collapsed to zero, so there is no pivot and the RHS is nonzero ⇒ no solution at all.

Verify: $\det A=2\cdot 2-4\cdot 1=0$, confirming singularity — see Determinants and why Cramer's Rule also fails here (it divides by $\det A=0$).

The two lines are parallel and distinct — they never cross, which is the geometric face of "$0=2$".

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Ex 6′ — Cell C6′: singular with infinitely many solutions

Forecast: now row 2 is exactly half of row 1 — same line twice. Guess: infinitely many solutions along one line.

1. Scan / no swap. Why: $|2|>|1|$ down column 1, pivot $=2$ already on top.
2. Eliminate. Why this step? Punch the zero in $(2,1)$. $m_{21}=1/2=0.5$: $a_{22}=2-0.5(4)=0,b_2=3-0.5(6)=0.$ $\left[\begin{array}{cccc}2& 4& |& 6\\ 0& 0& |& 0\end{array}\right].$
3. Read the bottom row. Why this step? It now says $0=0$ — always true, no information. So $x_2$ is a free variable: pick any value $x_2=s$.
4. Back-substitute with the free variable. Why: row 1 still constrains $x_1$ in terms of $s$. From $2x_1+4s=6$: $x_1=\frac{6-4s}{2}=3-2s,x_2=s,s\in \mathbb{R}.$ Infinitely many solutions, one for each $s$.

Verify: pick $s=0$: $(3,0)$ gives $2(3)+4(0)=6✓$ and $1(3)+2(0)=3✓$. Pick $s=1$: $(1,1)$ gives $2(1)+4(1)=6✓$ and $1(1)+2(1)=3✓$. Both lie on the single shared line.

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Ex 7 — Cell C7: a real word problem (mixtures, with units)

Forecast: total volume 10 L, so guess A is the larger share (32% is closer to 20%).

1. Translate to a system. Why this step? Let $x_1$ = litres of A, $x_2$ = litres of B. Volume balance and acid balance give $\{\begin{array}{l}{x}_1+x_2=10\\ 0.20x_1+0.50x_2=0.32\times 10=3.2\end{array}\Rightarrow \left[\begin{array}{cccc}1& 1& |& 10\\ 0.2& 0.5& |& 3.2\end{array}\right].$
2. Pivot check. Why this step? Scan column 1, rows 1–2: $|1|>|0.2|$, so the biggest is already on top — no swap, pivot $=1$.
3. Eliminate. Why this step? We want a $0$ in position $(2,1)$ so row 2 holds only one unknown. $m_{21}=0.2/1=0.2$; then $a_{22}=0.5-0.2(1)=0.3$, $b_2=3.2-0.2(10)=1.2$.
4. Back-substitute. Why bottom-up: row 2 now has just $x_2$. $x_2=\frac{1.2}{0.3}=4\text{ L (B)},x_1=10-4=6\text{ L (A)}.$

Verify (units + acid): total $6+4=10$ L $✓$. Acid: $0.2(6)+0.5(4)=1.2+2.0=3.2$ L, and $3.2/10=32\%✓$. As forecast, A is the larger share.

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Ex 8 — Cell C8: exam twist with a parameter (limiting behaviour)

Forecast: a zero pivot will appear after elimination for one special $t$. Guess $t=4$.

1. Pivot honestly — a swap is forced. Why this step? Scan column 1, rows 1–2: $|2|>|1|$ for every $t$ (column 1 does not contain $t$). Partial pivoting therefore swaps rows 1 and 2 before doing anything — we must not assume a pivot of $1$. $\left[\begin{array}{cccc}2& t& |& 7\\ 1& 2& |& 3\end{array}\right].$
2. Eliminate symbolically. Why this step? Punch a $0$ in $(2,1)$; the resulting $(2,2)$ entry tells us when the method stalls. Pivot $=2$, so $m_{21}=1/2$: $a_{22}=2-\frac{1}{2}t=2-\frac{t}{2},b_2=3-\frac{1}{2}(7)=-\frac{1}{2}.$ $\left[\begin{array}{cccc}2& t& |& 7\\ 0& 2-\frac{t}{2}& |& -\frac{1}{2}\end{array}\right].$
3. Find the failure. Why this step? Back-substitution divides by $a_{22}=2-\frac{t}{2}$. This is zero when $2-\frac{t}{2}=0$, i.e. $\boxed{t=4}$. Then the bottom row reads $0=-\frac{1}{2}$ ⇒ no unique solution. This is the limiting value of the parameter (and indeed $\det A=1\cdot t-2\cdot 2=t-4$ vanishes there too).
4. Solve for $t=6$. Why this step? Plug the concrete value into the eliminated system so both unknowns follow. Here $a_{22}=2-\frac{6}{2}=-1$: $x_2=\frac{-1/2}{-1}=\frac{1}{2}=0.5,x_1=\frac{7-6x_2}{2}=\frac{7-3}{2}=2.$

Verify: at $t=6$, original eq 2: $2(2)+6(0.5)=4+3=7✓$; eq 1: $1(2)+2(0.5)=3✓$. At $t=4$, $\det A=t-4=0$ — singular exactly as predicted.

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Recall The whole matrix in one breath

Baseline (C1) ::: biggest pivot already on top, just eliminate and climb back. Forced swap (C2) ::: biggest entry is below — swap whole rows first. Tiny pivot (C3) ::: pivoting avoids a monster multiplier that would wreck round-off. Negative pivot (C4) ::: negative is fine; only small magnitude is the enemy — keep the multiplier's sign. Zero pivot, fixable (C5) ::: a nonzero below rescues you; pivoting does it automatically. Singular, no solution (C6) ::: last row becomes $0=c$ with $c\ne 0$ ($\det A=0$). Singular, infinitely many (C6′) ::: last row becomes $0=0$ ⇒ free variable. Word problem (C7) ::: name the unknowns, write one equation per balance law, then solve. Parameter twist (C8) ::: after the forced swap the pivot-2 entry is $2-t/2$; it vanishes at $t=4$.

See also LU Decomposition (elimination remembered as a factorisation), Back-substitution and Forward-substitution, and Iterative Methods (Jacobi, Gauss-Seidel) for when $n$ is huge.