Worked examples — Gaussian elimination — forward elimination, back substitution

This is the workbench child of the parent note. The parent told you the rules; here we play every hand the game can deal. Before a single example, we lay out a matrix of all the situations elimination can throw at you — then we hunt down each one.

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The scenario matrix

Every linear system you feed to elimination lands in one of these cells. The last column names the example that nails it.

#	Case class	What's special	Outcome you'll see	Example
C1	Clean square system	every pivot nonzero as-is	unique solution	Ex 1
C2	Zero pivot, swap fixes it	diagonal pivot $a_{kk}=0$ but a lower row has a nonzero	unique solution after swap	Ex 2
C3	Tiny pivot (stability)	pivot $\approx 0$, division blows up round-off	swap for accuracy	Ex 3
C4	Inconsistent	a row becomes $0\dots 0\mid d,d\ne 0$	no solution	Ex 4
C5	Free variable / infinite	a row becomes $0\dots 0\mid 0$	infinitely many	Ex 5
C6	Negative & fractional pivots	signs and fractions everywhere	unique, sign discipline	Ex 6
C7	Rectangular ($m\ne n$: rows $\ne$ columns)	over- ($m>n$) and under-determined ($m<n$)	rank tells the story	Ex 7a, 7b
C8	Word problem (real units)	translate → solve → interpret	numeric answer with units	Ex 8
C9	Exam twist (unknown parameter)	a letter $k$ in the matrix	condition on $k$ for each outcome	Ex 9

We'll cover every cell. Cells C4/C5/C9 connect to Solving Linear Systems — Consistency; C7 to Rank of a Matrix; C3 to Pivoting and Numerical Stability.

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Ex 1 — Clean square system (cell C1)

Forecast: every pivot is already nonzero, so we should march straight down with no swaps and land on one unique triple. Guess whether the numbers are "nice" (they are).

Augmented matrix: $\left[\begin{array}{cccc}1& 1& 1& 6\\ 2& 3& 7& 25\\ 1& 4& -1& 4\end{array}\right]$

1. Pivot $a_{11}=1$; clear column 1. $m_{21}=2/1=2$ so $R_2\to R_2-2R_1$; $m_{31}=1/1=1$ so $R_3\to R_3-R_1$. Why this step? We want the $2$ and the $1$ under the pivot to become $0$ — the multiplier is exactly "how many pivot-rows cancel them." $\left[\begin{array}{cccc}1& 1& 1& 6\\ 0& 1& 5& 13\\ 0& 3& -2& -2\end{array}\right]$
2. Pivot $a_{22}=1$; clear below. $m_{32}=3/1=3$ so $R_3\to R_3-3R_2$. Why this step? Kill the $3$ in column 2 so only the last variable survives in row 3. $\left[\begin{array}{cccc}1& 1& 1& 6\\ 0& 1& 5& 13\\ 0& 0& -17& -41\end{array}\right]$
3. Back-substitute (bottom-up). Why bottom-up? Only the last row has a single unknown.
   • Row 3: $-17z=-41\Rightarrow z=\frac{41}{17}$.
   • Row 2: $y+5z=13\Rightarrow y=13-5\cdot \frac{41}{17}=\frac{221-205}{17}=\frac{16}{17}$.
   • Row 1: $x+y+z=6\Rightarrow x=6-\frac{16}{17}-\frac{41}{17}=\frac{102-57}{17}=\frac{45}{17}$.

Answer: $x=\frac{45}{17},y=\frac{16}{17},z=\frac{41}{17}$.

Worked example Verify

Row 3 original: $x+4y-z=\frac{45}{17}+\frac{64}{17}-\frac{41}{17}=\frac{68}{17}=4$ ✓.

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Ex 2 — Zero pivot, swap saves the day (cell C2)

Forecast: position $(1,1)$ is $0$ — you cannot divide by it. Predict that a row swap is forced (not optional), and after it we get a unique solution.

1. Swap $R_1\leftrightarrow R_2$. Why this step? A pivot must be nonzero; row 2 offers a clean $1$ in column 1. $\left[\begin{array}{cccc}1& 2& 1& 6\\ 0& 1& 2& 4\\ 3& 1& 0& 7\end{array}\right]$
2. Pivot $a_{11}=1$; clear column 1. Only $R_3$ needs work: $m_{31}=3$, $R_3\to R_3-3R_1$. Why this step? Zero out the $3$ below the pivot. $\left[\begin{array}{cccc}1& 2& 1& 6\\ 0& 1& 2& 4\\ 0& -5& -3& -11\end{array}\right]$
3. Pivot $a_{22}=1$; clear below. $m_{32}=-5/1=-5$ so $R_3\to R_3-(-5)R_2=R_3+5R_2$. Why this step? A negative multiplier means we add the pivot row — the "$-m$" template handles the sign automatically. $\left[\begin{array}{cccc}1& 2& 1& 6\\ 0& 1& 2& 4\\ 0& 0& 7& 9\end{array}\right]$
4. Back-substitute. Row 3: $7z=9\Rightarrow z=\frac{9}{7}$. Row 2: $y+2z=4\Rightarrow y=4-\frac{18}{7}=\frac{10}{7}$. Row 1: $x+2y+z=6\Rightarrow x=6-\frac{20}{7}-\frac{9}{7}=6-\frac{29}{7}=\frac{13}{7}$.

Answer: $x=\frac{13}{7},y=\frac{10}{7},z=\frac{9}{7}$.

Worked example Verify

Row 1 original ($0\cdot x+y+2z=4$): $\frac{10}{7}+\frac{18}{7}=\frac{28}{7}=4$ ✓. Row 3 original: $3\cdot \frac{13}{7}+\frac{10}{7}=\frac{49}{7}=7$ ✓.

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Ex 3 — Tiny pivot, swap for accuracy (cell C3)

Forecast: the pivot $0.001$ is legal (nonzero) but tiny. Dividing by it makes the multiplier huge and round-off eats the answer. Predict: partial pivoting (swap in the bigger entry) is the fix — see Pivoting and Numerical Stability.

1. Naive way (no swap). $m_{21}=1/0.001=1000$, $R_2\to R_2-1000R_1$:
   • new $a_{22}=1-1000\cdot 1=-999$
   • new RHS $=2-1000\cdot 1=-998$. Round to 3 sig figs: $-999y=-998\Rightarrow y\approx 0.999$. Then $0.001x+0.999=1\Rightarrow x=\frac{0.001}{0.001}=1.00$. Here it's still ok, but with more digits lost the small pivot amplifies error.
2. Stable way (swap first). Why this step? Put the largest column-1 entry on top so the multiplier is $\le 1$ and round-off can't blow up. $\left[\begin{array}{ccc}1& 1& 2\\ 0.001& 1& 1\end{array}\right]$ $m_{21}=0.001/1=0.001$, $R_2\to R_2-0.001R_1$: $a_{22}=1-0.001=0.999$, RHS $=1-0.002=0.998$. Back-sub: $0.999y=0.998\Rightarrow y\approx 0.999$; $x=2-y\approx 1.00$.

Answer (true value): $x=\frac{1000}{999},y=\frac{998}{999}$. Exact system: $0.001x+y=1,x+y=2$. Subtract: $(1-0.001)x=1\Rightarrow x=\frac{1}{0.999}=\frac{1000}{999}$, and $y=2-x=\frac{998}{999}$.

Worked example Verify

$x=\frac{1000}{999},y=\frac{998}{999}$: eqn 1 $0.001x+y=\frac{1}{999}+\frac{998}{999}=\frac{999}{999}=1$ ✓; eqn 2 $x+y=\frac{1998}{999}=2$ ✓.

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Ex 4 — Inconsistent system, no solution (cell C4)

Forecast: the left column of row 2 is exactly $-2$ times row 1's — the coefficients are proportional but the RHS is not. Predict a row of the form $00\mid d$ with $d\ne 0$: no solution.

1. Clear column 1. $m_{21}=-2/1=-2$, $R_2\to R_2-(-2)R_1=R_2+2R_1$. Why this step? Standard elimination; the negative multiplier means add. $\left[\begin{array}{ccc}1& -2& 3\\ 0& 0& 7\end{array}\right]$
2. Read the bottom row. It says $0x+0y=7$, i.e. $0=7$ — impossible. Why does this reveal contradiction? Elimination only makes $0=$ nonzero when the equations genuinely disagree.

Answer: Inconsistent — no solution. (Two parallel lines, never meeting.)

Worked example Verify

The RHS after elimination: $1+2\cdot 3=7\ne 0$, confirming inconsistency. See Solving Linear Systems — Consistency.

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Ex 5 — Free variable, infinitely many solutions (cell C5)

Forecast: same left side as Ex 4, but now the RHS is also $-2\times$ row 1's. Predict a full zero row $00\mid 0$: infinitely many solutions with one free variable.

1. Clear column 1. $m_{21}=-2$, $R_2\to R_2+2R_1$. $\left[\begin{array}{ccc}1& -2& 3\\ 0& 0& 0\end{array}\right]$
2. Interpret. Row 2 is $0=0$ — always true, carries no info. Why free variable? Only one real equation $x-2y=3$ for two unknowns; one variable is free.
3. Parametrise. Let $y=t$ (free). Then $x=3+2t$.

Answer: infinitely many: $(x,y)=(3+2t,t)$ for any real $t$.

Worked example Verify

Take $t=1$: $(x,y)=(5,1)$. Eqn 1: $5-2=3$ ✓. Eqn 2: $-10+4=-6$ ✓. Take $t=0$: $(3,0)$, eqn 1 $=3$ ✓.

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Ex 6 — Negative and fractional pivots (cell C6)

Forecast: the first pivot is negative. Predict fractions appearing but a clean unique answer if you keep the "$R_i\to R_i-m{R}_k$" template religiously.

1. Pivot $a_{11}=-2$; clear column 1. $m_{21}=4/(-2)=-2$, $R_2\to R_2+2R_1$; $m_{31}=-6/(-2)=3$, $R_3\to R_3-3R_1$. Why negative pivot is fine? The multiplier formula never assumed positivity — it just cancels. $\left[\begin{array}{cccc}-2& 1& 3& 9\\ 0& 1& 8& 20\\ 0& 2& -10& -30\end{array}\right]$
2. Pivot $a_{22}=1$; clear below. $m_{32}=2/1=2$, $R_3\to R_3-2R_2$. $\left[\begin{array}{cccc}-2& 1& 3& 9\\ 0& 1& 8& 20\\ 0& 0& -26& -70\end{array}\right]$
3. Back-substitute. Row 3: $-26z=-70\Rightarrow z=\frac{70}{26}=\frac{35}{13}$. Row 2: $y+8z=20\Rightarrow y=20-\frac{280}{13}=\frac{260-280}{13}=-\frac{20}{13}$. Row 1: $-2x+y+3z=9\Rightarrow -2x=9+\frac{20}{13}-\frac{105}{13}=9-\frac{85}{13}=\frac{32}{13}\Rightarrow x=-\frac{16}{13}$.

Answer: $x=-\frac{16}{13},y=-\frac{20}{13},z=\frac{35}{13}$.

Worked example Verify

Row 2 original: $4x-y+2z=-\frac{64}{13}+\frac{20}{13}+\frac{70}{13}=\frac{26}{13}=2$ ✓.

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Ex 7a — Rectangular: over-determined, $m>n$ (cell C7)

Forecast: here $m=3$ equations and $n=2$ unknowns, so $m>n$. Usually over-determined systems have no solution — unless the third equation is a hidden combination of the first two. Predict we must finish eliminating, then inspect the last row (per the strategy box above): if it collapses to $0=0$, the extra equation was redundant and a unique solution survives. (This is Rank of a Matrix in action.)

1. Clear column 1. $R_2\to R_2-R_1$, $R_3\to R_3-2R_1$. $\left[\begin{array}{ccc}1& 1& 2\\ 0& -2& -2\\ 0& -1& -1\end{array}\right]$
2. Clear column 2. Pivot $a_{22}=-2$, $m_{32}=-1/-2=\frac{1}{2}$, $R_3\to R_3-\frac{1}{2}{R}_2$. Why this step? Test whether row 3 adds new information or collapses. $\left[\begin{array}{ccc}1& 1& 2\\ 0& -2& -2\\ 0& 0& 0\end{array}\right]$
3. Inspect, then interpret. Bottom row $0=0$: consistent and carries no info, so the third equation was dependent. Two pivots for two unknowns → rank $=n=2$ → unique solution. Back-sub: $-2y=-2\Rightarrow y=1$; $x+y=2\Rightarrow x=1$.

Answer: $x=1,y=1$ (and it satisfies all three originals).

Worked example Verify

Eqn 2: $x-y=1-1=0$ ✓. Eqn 3: $2x+y=2+1=3$ ✓. Rank $=2=$ number of unknowns → unique.

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Ex 7b — Rectangular: under-determined, $m<n$ (cell C7)

Forecast: here $m=2$ equations and $n=3$ unknowns, so $m<n$. There are more variables than pivots can pin down, so predict at least one free variable and hence infinitely many solutions (assuming no contradiction). Rank is the tell: see Rank of a Matrix.

1. Clear column 1. $m_{21}=2/1=2$, $R_2\to R_2-2R_1$. Why this step? Zero the $2$ below the first pivot, as always. $\left[\begin{array}{cccc}1& 2& 1& 4\\ 0& 1& 1& 3\end{array}\right]$
2. Count pivots (the stop-and-inspect step). Pivots sit in columns 1 and 2 (values $1$ and $1$). Column 3 has no pivot. No row is $0\dots 0\mid d$, so the system is consistent; rank $=2<n=3$. Why does this matter? A column without a pivot corresponds to a free variable — nothing forces its value, so we expect a whole family, not a single point.
3. Parametrise the free variable. The pivot-less column is column 3, whose unknown is $z$, so we set $z=t$ (free) and let it range over all real numbers. Why $z$ and not $x$ or $y$? $x$ and $y$ each own a pivot (columns 1 and 2), so their values are forced once $z$ is chosen; only the pivot-less variable is genuinely free.
   • Row 2: $y+z=3\Rightarrow y=3-t$. Why this substitution? With $z=t$ known, row 2 has a single remaining unknown $y$.
   • Row 1: $x+2y+z=4\Rightarrow x=4-2(3-t)-t=4-6+2t-t=-2+t$. Why last? Climbing up, row 1 now has only $x$ unknown once $y$ and $z$ are in hand.

Answer: infinitely many: $(x,y,z)=(-2+t,3-t,t)$ for any real $t$. (Rank $=2<3=$ unknowns, so exactly one free variable — a line of solutions in 3-D space.)

Worked example Verify

Take $t=2$: $(0,1,2)$. Eqn 1: $0+2+2=4$ ✓. Eqn 2: $0+5+6=11$ ✓. Take $t=0$: $(-2,3,0)$. Eqn 1: $-2+6+0=4$ ✓; eqn 2: $-4+15+0=11$ ✓.

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Ex 8 — Word problem with real units (cell C8)

Forecast: translate three sentences into three equations, then it's a plain $3\times 3$. Guess whether the counts come out as whole numbers (they must — you cannot have half a coin!).

Let $n,d,q$ be the counts. $\begin{array}{rlrl}n+d+q& =12& & \text{(count)}\\ 5n+10d+25q& =170& & \text{(cents)}\\ d& =n+q\Rightarrow -n+d-q=0& & \text{(dimes rule)}\end{array}$

$\left[\begin{array}{cccc}1& 1& 1& 12\\ 5& 10& 25& 170\\ -1& 1& -1& 0\end{array}\right]$

1. Clear column 1. $m_{21}=5/1=5$, $R_2\to R_2-5R_1$; $m_{31}=-1/1=-1$, $R_3\to R_3+R_1$. Why this step? Zero the $5$ and the $-1$ under the first pivot so rows 2–3 lose the variable $n$. $\left[\begin{array}{cccc}1& 1& 1& 12\\ 0& 5& 20& 110\\ 0& 2& 0& 12\end{array}\right]$
2. Clear column 2. Pivot $a_{22}=5$, $m_{32}=2/5$, $R_3\to R_3-\frac{2}{5}{R}_2$. Why this step? Kill the $2$ in column 2 so row 3 holds only the last variable $q$. $\left[\begin{array}{cccc}1& 1& 1& 12\\ 0& 5& 20& 110\\ 0& 0& -8& -32\end{array}\right]$
3. Back-substitute (bottom-up). Why bottom-up? The last row has a single unknown $q$. $-8q=-32\Rightarrow q=4$. Then $5d+20\cdot 4=110\Rightarrow 5d=30\Rightarrow d=6$. Then $n+6+4=12\Rightarrow n=2$.

Answer: $n=2$ nickels, $d=6$ dimes, $q=4$ quarters.

Worked example Verify (with units)

Count: $2+6+4=12$ coins ✓. Value: $2(5)+6(10)+4(25)=10+60+100=170$¢ =\1.70$✓.Dimesrule:$d=6,\ n+q=6$ ✓.

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Ex 9 — Exam twist: a parameter $k$ (cell C9)

Forecast: the pivot in column 2 will contain $k$. Predict a special value of $k$ that makes that pivot $0$; then the RHS decides between "no solution" and "infinitely many." This is a Solving Linear Systems — Consistency classic.

1. Clear column 1. $m_{21}=2/1=2$, $R_2\to R_2-2R_1$. Why this step? Isolate all the $k$-dependence into one entry so the fork is easy to see. $\left[\begin{array}{ccc}1& 2& 3\\ 0& k-4& 2\end{array}\right]$
2. Case-split on the pivot $k-4$. Why split here? The next step divides by $k-4$; the behaviour changes exactly when that pivot is zero.
   • If $k\ne 4$: pivot $\ne 0$, so $y=\frac{2}{k-4}$, then $x=3-2y=3-\frac{4}{k-4}$. Unique solution.
   • If $k=4$: row 2 becomes $00\mid 2$, i.e. $0=2$. Inconsistent — no solution.
   • Infinitely many? Would need row 2 $=00\mid 0$. That requires $k-4=0$ and RHS $=0$; but the RHS is $2\ne 0$, so it never happens here.

Answer: unique for all $k\ne 4$; no solution at $k=4$; infinitely many for no value of $k$.

Worked example Verify

$k=6$: $y=\frac{2}{2}=1,x=3-2=1$. Check eqn 2: $2(1)+6(1)=8$ ✓. $k=4$: row gives $0=2$, inconsistent ✓.

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Recall checkpoint

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Connections

• Solving Linear Systems — Consistency — cells C4, C5, C9 are exactly its three outcomes.
• Rank of a Matrix — number of pivots (Ex 7a, 7b) = rank; rank vs. unknowns decides uniqueness.
• Pivoting and Numerical Stability — the "why swap on the biggest" from Ex 3.
• Row Echelon Form / Reduced Row Echelon Form — the staircase shapes these examples produce.
• LU Decomposition — the multipliers we computed are the entries of $L$.
• Determinant — a full zero row (Ex 5, Ex 7a) signals a zero determinant.