LINEAR-ALGEBRA · Unit 4 · Video 2 · Interactive Practice

A = LU Factorization

IKey Formulas

Formula	Name	Description
$A = LU$	LU Factorization	Lower triangular $L$ times upper triangular $U$
$\ell_{ij} = \dfrac{a_{ij}}{a_{jj}}$	Multiplier	Entry to eliminate divided by the pivot
$L$ : ones on diagonal, multipliers below	Structure of $L$	$L$ comes free from elimination
$A = LDU'$	LDU form	$D$ holds pivots; both $L$ and $U'$ have ones on diagonal

IIVisualization 1 — Elimination builds U, multipliers build L

Each elimination step hollows out $U$ below the diagonal, and the multiplier it uses drops straight into $L$ .

Step 0

Step 0 · Start

U = A

and

L = I

. Elimination zeros

U

below the diagonal; each multiplier used lands in

L

Step 1 · Clear (2,1)

\ell_{21} = \tfrac{4}{2} = 2

, so

R_2 \leftarrow R_2 - 2R_1

. Store

2

L_{21}

Step 2 · Clear (3,1)

\ell_{31} = \tfrac{8}{2} = 4

, so

R_3 \leftarrow R_3 - 4R_1

. Store

4

L_{31}

Step 3 · Clear (3,2)

\ell_{32} = \tfrac{3}{1} = 3

, so

R_3 \leftarrow R_3 - 3R_2

. Store

3

L_{32}

U

is upper triangular ✓

💡 Because $L$ stores exactly the row operations, $LU$ replays them and rebuilds $A$ — the factorization is elimination run backward.

IIIVisualization 2 — Why the inverse product stays clean

Multiplying the $E$ 's forward breeds a cross-term at $(3,1)$ ; multiplying their inverses back gives a clean $L$ .

m₂₁ 2 m₃₂ 5

IVVisualization 3 — LDU: factoring the pivots into D

Pull the pivots out of $U$ into a diagonal $D$ , leaving $L$ and $U'$ both with ones on the diagonal.

a₁₁ 2 a₁₂ 1 a₂₁ 4 a₂₂ 5

💡 When $A$ is symmetric, $U' = L^{\mathsf{T}}$ and the factorization collapses to the elegant $A = LDL^{\mathsf{T}}$ .

VQuiz Questions

Question 1

Given the matrix $A = \begin{bmatrix} 3 & 1 \\ 6 & 5 \end{bmatrix}$ , what is the elimination multiplier $\ell_{21}$ ?

✅ Correct! The multiplier is $a_{21}/a_{11} = 6/3 = 2$ .

❌ Not quite. Remember: the multiplier is the entry to eliminate divided by the pivot above it: $\ell_{21} = a_{21} / a_{11}$ .

Show solution

Solution:

The multiplier is the entry to eliminate divided by the pivot:

$\ell_{21} = \frac{a_{21}}{a_{11}} = \frac{6}{3} = 2$

The multiplier is 2. We subtract 2 times row 1 from row 2 to produce a zero below the pivot.

Question 2

In the LU factorization $A = LU$ , the diagonal entries of $L$ are always:

✅ Correct! $L$ always has ones on its diagonal — the multipliers sit below the diagonal, and the pivots live in $U$ .

❌ Not quite. Think about the structure of $L$ : it's the product of inverse elimination matrices, each of which has ones on its diagonal.

Show solution

Solution:

$L$ is lower triangular with ones on the diagonal and the elimination multipliers below:

$L = \begin{bmatrix} 1 & 0 & 0 \\ \ell_{21} & 1 & 0 \\ \ell_{31} & \ell_{32} & 1 \end{bmatrix}$

The pivots appear on the diagonal of $U$ (or in $D$ in the LDU form), not $L$ .
The multipliers go below the diagonal of $L$ , not on it.

Question 3

True or False: When multiplying elimination matrices $E_{32} \cdot E_{21}$ , a parasitic cross-term can appear in position $(3,1)$ . But multiplying their inverses in reverse order $E_{21}^{-1} \cdot E_{32}^{-1}$ avoids this cross-term and produces a clean $L$ matrix.

✅ Correct! The inverse product avoids cross-terms — that's exactly why we define $L$ as the product of inverse elimination matrices, and why the multipliers slot directly into $L$ .

❌ Not quite. Try Visualization 2 above with non-zero multipliers and compare the (3,1) entry in both products.

Show solution

Solution: True

When computing $E_{32} \cdot E_{21}$ , the $(-m_{32})$ entry in $E_{32}$ multiplies the $(-m_{21})$ entry in $E_{21}$ , creating a cross-term $m_{32} \cdot m_{21}$ in position $(3,1)$ :

$E_{32} \cdot E_{21} = \begin{bmatrix} 1 & 0 & 0 \\ -m_{21} & 1 & 0 \\ m_{32}m_{21} & -m_{32} & 1 \end{bmatrix}$

But multiplying the inverses in the correct order avoids this entirely:

$L = E_{21}^{-1} \cdot E_{32}^{-1} = \begin{bmatrix} 1 & 0 & 0 \\ m_{21} & 1 & 0 \\ 0 & m_{32} & 1 \end{bmatrix}$

No cross-term — each multiplier sits cleanly in its natural position. This is because each inverse only modifies a row that later inverses don't touch.

Question 4

Given $L = \begin{bmatrix} 1 & 0 \\ 3 & 1 \end{bmatrix}$ and $U = \begin{bmatrix} 4 & 2 \\ 0 & 5 \end{bmatrix}$ , what is $A = LU$ ?

✅ Correct! Row 2 of $A$ : $[3 \cdot 4 + 1 \cdot 0, \; 3 \cdot 2 + 1 \cdot 5] = [12, \; 11]$ .

❌ Not quite. Remember: each entry of $A$ is a dot product of a row of $L$ with a column of $U$ . Check position $(2,2)$ : $3 \cdot 2 + 1 \cdot 5 = ?$

Show solution

Solution:

Multiply $L \times U$ row by row:

Row 1: $[1, 0] \cdot \begin{bmatrix} 4 & 2 \\ 0 & 5 \end{bmatrix} = [1 \cdot 4 + 0 \cdot 0, \; 1 \cdot 2 + 0 \cdot 5] = [4, \; 2]$

Row 2: $[3, 1] \cdot \begin{bmatrix} 4 & 2 \\ 0 & 5 \end{bmatrix} = [3 \cdot 4 + 1 \cdot 0, \; 3 \cdot 2 + 1 \cdot 5] = [12, \; 11]$

$A = LU = \begin{bmatrix} 4 & 2 \\ 12 & 11 \end{bmatrix}$

Common mistakes:

$\begin{bmatrix} 4 & 2 \\ 12 & 6 \end{bmatrix}$ : forgot the $1 \cdot 5$ term in position $(2,2)$ — only computed $3 \cdot 2 = 6$ .
$\begin{bmatrix} 4 & 2 \\ 3 & 6 \end{bmatrix}$ : added rows element-wise instead of computing dot products.
$\begin{bmatrix} 4 & 2 \\ 0 & 5 \end{bmatrix}$ : this is just $U$ — forgot to multiply by $L$ .

Solved: 0 / 4