Worked examples — LU decomposition (numerical)

Before anything, the vocabulary we reuse (all built in the parent): $A=LU$ means we split a square number-grid $A$ into a lower-triangular $L$ (zeros above the diagonal, 1's on it) times an upper-triangular $U$ (zeros below the diagonal). We solve $A\mathbf{x}=\mathbf{b}$ by first walking down ($L\mathbf{y}=\mathbf{b}$, forward substitution) then up ($U\mathbf{x}=\mathbf{y}$, back substitution). If you're shaky on any of these words, re-read Gaussian Elimination and Forward and Back Substitution first.

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The scenario matrix

Every LU problem you'll meet falls into one of these cells. We will hit each one with a worked example.

Cell	Case class	What's special	Example
C1	Clean $2\times 2$, nonzero pivots	The textbook happy path	Ex 1
C2	Clean $3\times 3$	Watch the sums grow	Ex 2
C3	Zero pivot, matrix still invertible	$u_{11}=0$, must pivot ($PA=LU$)	Ex 3
C4	Tiny pivot vs. big pivot	Rounding blows up without pivoting	Ex 4
C5	Multiple right-hand sides	The whole point: factor once, solve twice	Ex 5
C6	Singular matrix	A pivot becomes $0$ with nothing to swap	Ex 6
C7	Determinant via LU (with a swap)	Sign flip from the permutation	Ex 7
C8	Word problem (real world)	Set up $A\mathbf{x}=\mathbf{b}$ from a story	Ex 8

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C1 — Clean $2\times 2$: the happy path

Forecast: guess the sign of both answers before computing. (Hint: the second equation is "heavier".)

1. $u_{11}=4,u_{12}=3$. Why this step? Row 1 of $L$ is $(1,0)$, so multiplying gives the top row of $U$ = top row of $A$ untouched.
2. ${\ell }_{21}=a_{21}/u_{11}=8/4=2$. Why? This is the Gaussian multiplier — "what times row 1 do I subtract from row 2 to kill the $8$?"
3. $u_{22}=a_{22}-{\ell }_{21}{u}_{12}=11-2\cdot 3=5$. Why? Subtract what the column-1 elimination dumped into position $(2,2)$.

$L=\left(\begin{array}{cc}1& 0\\ 2& 1\end{array}\right),U=\left(\begin{array}{cc}4& 3\\ 0& 5\end{array}\right).$

4. Forward solve $L\mathbf{y}=\mathbf{b}$: $y_1=10$; then $2y_1+y_2=30\Rightarrow y_2=30-20=10$. Why top-first? The first row of $L$ has only one unknown.
5. Back solve $U\mathbf{x}=\mathbf{y}$: $5x_2=10\Rightarrow x_2=2$; then $4x_1+3\cdot 2=10\Rightarrow 4x_1=4\Rightarrow x_1=1$. Why bottom-first? The last row of $U$ has only one unknown.

$\boxed{\mathbf{x}=(1,2{)}^{\top }}$

Verify: $A\mathbf{x}=(4\cdot 1+3\cdot 2,8\cdot 1+11\cdot 2)=(10,30)=\mathbf{b}$ ✓. And $LU=A$ ✓.

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C2 — Clean $3\times 3$: watch the sums grow

Forecast: will all three pivots $u_{11},u_{22},u_{33}$ come out positive? Guess now.

1. Row 1 of $U$: $u_{11}=2,u_{12}=4,u_{13}=-2$. Why? Same reason as C1 — top row copies over.
2. Column 1 of $L$: ${\ell }_{21}=4/2=2$, ${\ell }_{31}=-2/2=-1$. Why? Multipliers to zero the two entries below the first pivot.
3. Row 2 of $U$: $u_{22}=9-{\ell }_{21}{u}_{12}=9-2\cdot 4=1$; $u_{23}=-3-{\ell }_{21}{u}_{13}=-3-2\cdot (-2)=1$. Why? Subtract row-1's contribution using the Doolittle sum ${\sum }_k{\ell }_{ik}{u}_{kj}$.
4. Column 2 of $L$: ${\ell }_{32}=\frac{a_{32}-{\ell }_{31}{u}_{12}}{u_{22}}=\frac{-3-(-1)\cdot 4}{1}=\frac{-3+4}{1}=1$. Why? The multiplier to kill entry $(3,2)$ after column-1 was cleared.
5. Row 3 of $U$: $u_{33}=a_{33}-{\ell }_{31}{u}_{13}-{\ell }_{32}{u}_{23}=7-(-1)(-2)-1\cdot 1=7-2-1=4$. Why? The last pivot is what's left after both earlier columns paid their share.

$L=\left(\begin{array}{ccc}1& 0& 0\\ 2& 1& 0\\ -1& 1& 1\end{array}\right),U=\left(\begin{array}{ccc}2& 4& -2\\ 0& 1& 1\\ 0& 0& 4\end{array}\right).$

Verify: multiply $LU$ — row 2 gives $(2,9,-3)$ ✓, row 3 gives $(-2,-3,7)$ ✓. Pivots $2,1,4$ are all positive, as forecast for this nice symmetric-looking matrix.

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C3 — Zero pivot, but $A$ is invertible: you MUST pivot

Forecast: does LU simply fail here, or is there a rescue?

1. Naive attempt: $u_{11}=a_{11}=0$, then ${\ell }_{21}=3/0$ — division by zero. Why this step matters? It shows a perfectly invertible matrix ($\det A=-6\ne 0$) can still break plain Doolittle.
2. Partial pivoting: swap the two rows so the biggest first-column entry ($3>0$) sits on the pivot. Record it in a permutation matrix $P=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$. Why? Swapping rows of $A$ = multiplying by $P$; we now factor $PA=LU$.
3. Factor $PA=\left(\begin{array}{cc}3& 4\\ 0& 2\end{array}\right)$: it's already upper-triangular, so $u_{11}=3,u_{12}=4$, ${\ell }_{21}=0/3=0$, $u_{22}=2$.

$P=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right),L=\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right),U=\left(\begin{array}{cc}3& 4\\ 0& 2\end{array}\right).$

4. Solve — remember to permute $\mathbf{b}$! Solve $LU\mathbf{x}=P\mathbf{b}$, where $P\mathbf{b}=(11,8{)}^{\top }$. Forward: $L$ is identity so $\mathbf{y}=P\mathbf{b}=(11,8)$. Back: $2x_2=8\Rightarrow x_2=4$; $3x_1+4\cdot 4=11\Rightarrow 3x_1=-5\Rightarrow x_1=-5/3$.

$\boxed{\mathbf{x}=(-5/3,4{)}^{\top }}$

Verify: $A\mathbf{x}=(0\cdot (-5/3)+2\cdot 4,3\cdot (-5/3)+4\cdot 4)=(8,-5+16)=(8,11)=\mathbf{b}$ ✓. See Partial Pivoting for the full swapping rule.

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C4 — Tiny pivot vs. big pivot: accuracy matters

Forecast: both routes reach the same exact answer. Which route survives rounding?

1. No-pivot factor: $u_{11}=0.001$, ${\ell }_{21}=1/0.001=1000$, $u_{22}=1-1000\cdot 1=-999$. Why note this? The multiplier $1000$ is huge — that's the danger sign.
2. No-pivot solve (rounded to 3 sig figs): $y_1=1$, $y_2=2-1000\cdot 1=-998\approx -1.00\times 10^3$. Back: $x_2=(-998)/(-999)\approx 0.999$; then $0.001x_1=1-0.999=0.001\Rightarrow x_1\approx 1.00$. Rounding $2-1000=-998$ to 3 figures gives $-1.00\times 10^3$, so $x_2\to 1.00$, and then $x_1\to (1-1)/0.001=0$ — catastrophically wrong ($x_1$ should be $\approx 1$). Why? Subtracting a huge number swamped the small $1$; the true information was lost.
3. With pivot: swap rows (biggest first-column entry is $1$). $PA=\left(\begin{array}{cc}1& 1\\ 0.001& 1\end{array}\right)$, so ${\ell }_{21}=0.001$, $u_{22}=1-0.001=0.999\approx 1.00$. Multiplier is tiny — safe.
4. Pivoted solve: $P\mathbf{b}=(2,1)$. $y_1=2$, $y_2=1-0.001\cdot 2=0.998$. Back: $x_2=0.998/0.999\approx 0.999$; $x_1=2-x_2\approx 1.00$. Correct.

$\text{Exact answer: }\mathbf{x}=\left(\frac{1000}{999},\frac{998}{999}\right)\approx (1.001,0.999).$

Verify: with exact arithmetic, $A\mathbf{x}=(0.001\cdot \frac{1000}{999}+\frac{998}{999},\frac{1000}{999}+\frac{998}{999})=(\frac{1+998}{999},\frac{1998}{999})=(1,2)=\mathbf{b}$ ✓. Moral: big pivot, small error — see Condition Number and Numerical Stability.

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C5 — Multiple right-hand sides: the reason LU exists

Forecast: how much new work per right-hand side? (Answer: no factoring, just two substitutions.)

Recall $L=\left(\begin{array}{cc}1& 0\\ 2& 1\end{array}\right),U=\left(\begin{array}{cc}4& 3\\ 0& 5\end{array}\right)$.

1. ${\mathbf{b}}_1$ — forward: $y_1=7$, $y_2=14-2\cdot 7=0$. Back: $5x_2=0\Rightarrow x_2=0$; $4x_1+0=7\Rightarrow x_1=7/4$. Why no re-factoring? $A$ didn't change, only $\mathbf{b}$; the elimination steps are frozen in $L,U$.
2. ${\mathbf{b}}_2$ — forward: $y_1=0$, $y_2=20-2\cdot 0=20$. Back: $5x_2=20\Rightarrow x_2=4$; $4x_1+3\cdot 4=0\Rightarrow 4x_1=-12\Rightarrow x_1=-3$.

$\boxed{{\mathbf{x}}_1=(7/4,0{)}^{\top },{\mathbf{x}}_2=(-3,4{)}^{\top }}$

Verify: $A{\mathbf{x}}_1=(4\cdot \frac{7}{4}+0,8\cdot \frac{7}{4}+0)=(7,14)$ ✓; $A{\mathbf{x}}_2=(-12+12,-24+44)=(0,20)$ ✓. Two solves, zero re-factoring — that's the payoff.

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C6 — Singular matrix: a pivot dies with no rescue

Forecast: the second row is exactly half the first. What should the second pivot be?

1. Row 1 of $U$: $u_{11}=2,u_{12}=4$. Why? Top row copies over as always.
2. ${\ell }_{21}=a_{21}/u_{11}=1/2$. Why? Standard multiplier.
3. $u_{22}=a_{22}-{\ell }_{21}{u}_{12}=2-\frac{1}{2}\cdot 4=0$. Why this is fatal? $u_{22}=0$ is the last pivot; there is no row below to swap in. Back substitution would need $0\cdot x_2=y_2$, unsolvable (or infinitely many).

$L=\left(\begin{array}{cc}1& 0\\ 1/2& 1\end{array}\right),U=\left(\begin{array}{cc}2& 4\\ 0& 0\end{array}\right).$

Verify: the factorization is still valid ($LU=A$ ✓), but $\det A=u_{11}{u}_{22}=2\cdot 0=0$, confirming $A$ is singular. A zero pivot at the last position means no unique solution exists — pivoting cannot save you here. This is the honest signal that $A\mathbf{x}=\mathbf{b}$ has no unique answer.

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C7 — Determinant via LU (with a swap)

Forecast: we needed a row swap in row 1 (since $a_{11}=0$). Will that flip the sign of $\det$?

1. Swap needed: $a_{11}=0$, so pivot on the largest first-column entry, which is $2$ in row 3. One swap → $(-1{)}^1$ factor. After swapping rows 1 and 3: $PA=\left(\begin{array}{ccc}2& 1& 0\\ 1& 2& 1\\ 0& 1& 2\end{array}\right),P=\left(\begin{array}{ccc}0& 0& 1\\ 0& 1& 0\\ 1& 0& 0\end{array}\right).$ Why? Determinants flip sign once per row swap; we must count swaps.
2. Factor $PA$. $u_{11}=2$. ${\ell }_{21}=1/2$, ${\ell }_{31}=0$. $u_{22}=2-\frac{1}{2}\cdot 1=\frac{3}{2}$; $u_{23}=1-\frac{1}{2}\cdot 0=1$. ${\ell }_{32}=\frac{1-0}{3/2}=\frac{2}{3}$. $u_{33}=2-0-\frac{2}{3}\cdot 1=\frac{4}{3}$. Why? Straight Doolittle sums on the pivoted matrix.
3. Assemble determinant: $\det A=(-1{)}^{\#\text{swaps}}\prod u_{ii}=(-1{)}^1\cdot 2\cdot \frac{3}{2}\cdot \frac{4}{3}=(-1)\cdot 4=-4.$ Why the minus? One swap = one sign flip.

$\boxed{\det A=-4}$

Verify: expand $A$ directly along the top row: $0\cdot (\dots )-1\cdot (1\cdot 0-1\cdot 2)+2\cdot (1\cdot 1-2\cdot 2)=-1\cdot (-2)+2\cdot (1-4)=2+2\cdot (-3)=2-6=-4$ ✓.

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C8 — Word problem: a real setup

Forecast: both counts must be positive (you can't make negative batches) — check this at the end.

1. Factor $A$. $u_{11}=2,u_{12}=1$; ${\ell }_{21}=1/2$; $u_{22}=3-\frac{1}{2}\cdot 1=\frac{5}{2}$. Why? Standard Doolittle; the numbers are the recipe grid.
2. Forward $L\mathbf{y}=\mathbf{b}$: $y_1=8$; $y_2=14-\frac{1}{2}\cdot 8=14-4=10$. Why? $L$ lower-triangular, solve top-down.
3. Back $U\mathbf{x}=\mathbf{y}$: $\frac{5}{2}{x}_B=10\Rightarrow x_B=4$; $2x_A+1\cdot 4=8\Rightarrow 2x_A=4\Rightarrow x_A=2$. Why? $U$ upper-triangular, solve bottom-up.

$\boxed{x_A=2\text{ units},x_B=4\text{ units}}$

Verify (units!): apple $=2\cdot 2+1\cdot 4=8$ L ✓; mango $=1\cdot 2+3\cdot 4=14$ L ✓. Both counts positive — physically sensible.

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The figure above places every cell of our scenario matrix on one map: green cells are "LU just works", yellow cells "works but pivot first", pink cells "LU tells you something is wrong or special". Keep this mental map when a new matrix lands on your desk.

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Recall Answers

A zero pivot means the multiplier below it divides by $0$; skipping isn't allowed, you must swap in a nonzero row ::: — but only if one exists. Pivoting kept the multiplier small so finite-precision rounding didn't wipe out real digits ::: (stability, not exactness). None — $L$ and $U$ were reused; only two substitutions ran ::: (the whole point of LU). The zero was the last pivot with no row beneath to swap — the matrix is singular ::: ($\det A=0$). One row swap contributed a factor $(-1{)}^1$ ::: to $\det A=(-1{)}^{\#\text{swaps}}\prod u_{ii}$.

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Connections

• Parent: LU decomposition — the machinery these examples exercise.
• Gaussian Elimination · Forward and Back Substitution — the two engines used in every solve.
• Partial Pivoting · Permutation Matrices — the rescue in C3, C4, C7.
• Determinants — the sign-counting in C7.
• Condition Number and Numerical Stability — why C4 matters.
• Crout Decomposition · Cholesky Decomposition — sibling factorizations for other matrix shapes.