Recipe

Factor a matrix into L and U

Use when: you have a square matrix $A$ and need $L$ (lower triangular, $1$ s on the diagonal) and $U$ (upper triangular) with $A=LU$

Goal: a complete $L$ and $U$ that multiply back out to $A$ exactly, with every multiplier ordinary elimination used recorded in $L$

Taught in LU factorization · a later section extends this recipe with more cases

Quick recipe

Confirm elimination won't need a row swap
Clear each pivot column below the pivot, recording each multiplier in $L$ as you go
Repeat column by column until $U$ is upper triangular — a zero multiplier still counts
Verify by multiplying $L$ and $U$ back together

1
Run through the pivot columns of $A$ in your head and check that each one already has a nonzero entry sitting in the pivot row once earlier columns are cleared — no swap required to keep going.
- Not yet: elimination needs a row swap to reach a nonzero pivot — covered in Permutation matrices, PA = LU, and pivoting
Why: the plain factorization $A=LU$ (no permutation) only exists when ordinary elimination never has to swap rows; running the rest of this recipe on a matrix that does need one produces an $L$ and $U$ that don't multiply back to $A$ .
2
Start with $U=A$ and $L=I_n$ . For the pivot in column $k$ , and each row $i$ below it, compute $m_{ik}=U_{ik}/U_{kk}$ , apply $R_i\to R_i-m_{ik}R_k$ to $U$ , and write $m_{ik}$ into row $i$ , column $k$ of $L$ — every other entry of $L$ stays at its starting value ( $1$ on the diagonal, $0$ elsewhere) until its own turn comes.

Example

Start with $U=N$ and $L=I_3$ for $N=\begin{bmatrix}4&3&2\\8&9&7\\4&3&5\end{bmatrix}$ . Clear column 1 below the pivot $U_{11}=4$ : $m_{21}=\frac{8}{4}=2,\ R_2\to R_2-2R_1 \qquad m_{31}=\frac{4}{4}=1,\ R_3\to R_3-1R_1$ $U=\begin{bmatrix}4&3&2\\0&3&3\\0&0&3\end{bmatrix},\qquad L=\begin{bmatrix}1&0&0\\2&1&0\\1&0&1\end{bmatrix}$

Why: this is exactly the multiplier ordinary elimination already uses to zero out $U_{ik}$ — recording it instead of throwing it away is the only new bookkeeping $LU$ factorization adds to elimination you'd be doing anyway.
3
Move to the next pivot column and repeat. If the entry below a pivot is already $0$ , its multiplier is $0$ and $L$ 's entry there stays $0$ — that's a completed part of the process, not a step you skip.

Example

Continuing with the $U$ and $L$ above, column 2's pivot is $U_{22}=3$ , and the entry below it, $U_{32}$ , is already $0$ : $m_{32}=\frac{0}{3}=0,\qquad R_3\to R_3-0R_2\ \text{(no change)}$ $U$ was already upper triangular after column 1, so elimination is finished — $L$ 's $(3,2)$ entry stays the $0$ it started at: $U=\begin{bmatrix}4&3&2\\0&3&3\\0&0&3\end{bmatrix},\qquad L=\begin{bmatrix}1&0&0\\2&1&0\\1&0&1\end{bmatrix}$

Why: elimination isn't finished — and neither is $L$ — until every column left of the diagonal has had its multipliers computed, even when some of them turn out to be zero.
4
Compute $LU$ entry by entry and confirm it reproduces $A$ exactly.

Example

$LU=\begin{bmatrix}1&0&0\\2&1&0\\1&0&1\end{bmatrix}\begin{bmatrix}4&3&2\\0&3&3\\0&0&3\end{bmatrix}=\begin{bmatrix}4&3&2\\8&9&7\\4&3&5\end{bmatrix}=N$

Why: this is the direct check that $L$ and $U$ actually multiply back to the original matrix, catching an arithmetic slip in any multiplier before it propagates into a later solve.

Common mistakes

Recording a multiplier with the wrong sign — $L$ stores $m_{ik}$ itself, the number you multiplied row $k$ by before subtracting, not its negative.
Skipping a multiplier that happens to be $0$ , leaving $L$ incomplete even though no visible elimination step occurred there.
Writing a multiplier into the wrong slot of $L$ — $m_{ik}$ (row $i$ used pivot row $k$ ) belongs at row $i$ , column $k$ , never the transposed position.
Forcing plain $LU$ on a matrix that actually needs a row swap partway through elimination, instead of recognizing it needs $PA=LU$ .