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

Elimination Matrices: How Matrix Multiplication Encodes Row Operations

IKey Formulas

Concept	Formula / Rule	Description
Column view	$A\mathbf{x} = x_1\mathbf{c}_1 + x_2\mathbf{c}_2 + x_3\mathbf{c}_3$	Matrix × column = linear combination of columns
Row view	$\mathbf{x}^T A = x_1\mathbf{r}_1 + x_2\mathbf{r}_2 + x_3\mathbf{r}_3$	Row × matrix = linear combination of rows
Elimination matrix	Start from $I$ , place $-\ell$ at position $(i,j)$	Subtracts $\ell$ times row $j$ from row $i$
Full elimination	$E_{32}\,E_{21}\,A = U$	Product of elimination matrices $\to$ upper triangular

IIVisualization 1 — Rows Combine on the Left

A row vector times $A$ mixes the rows of $A$ — so what does $[c_1, c_2, c_3]\,A$ produce?

c₁ 0 c₂ 1 c₃ 0

IIIVisualization 2 — Building E from the Identity

Drop a single $-\ell$ into $I$ at position $(i,j)$ : which row operation does $E$ then perform on $A$ ?

Target row Pivot row Multiplier ℓ 3

IVVisualization 3 — Order Matters

Does $E_{32}E_{21}$ give the same result as $E_{21}E_{32}$ ? Watch position $(3,1)$ and the final shape.

Product order

💡 Associativity lets you regroup — $(E_{32}E_{21})A = E_{32}(E_{21}A)$ — but commutativity fails, so the order of elimination matrices matters.

VQuiz Questions

Question 1

The elimination matrix $E_{21}$ has the entry $-3$ in position $(2, 1)$ :

$E_{21} = \begin{bmatrix} 1 & 0 & 0 \\ -3 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$

What row operation does left-multiplying by $E_{21}$ perform?

✅ Correct! E₂₁ targets row 2 using pivot row 1, and the negative entry means subtraction.

❌ Not quite. The subscript (2,1) tells you: target row 2, pivot row 1. The negative entry means subtraction, not addition.

Show solution

Solution: Replace row 2 with row 2 − 3·(row 1).

The subscript $E_{21}$ : target = row 2, pivot = row 1. The entry $-3$ sits in position $(2,1)$ .

Row 2 of $E_{21}$ is $[-3, 1, 0]$ . Left-multiplying gives: $(-3)\cdot\text{row}_1 + (1)\cdot\text{row}_2 + (0)\cdot\text{row}_3 = \text{row}_2 - 3\cdot\text{row}_1$

The multiplier is $3$ , so we placed $-3$ (always negative) in the matrix. The operation subtracts.

Question 2 · True or False

Matrix multiplication is associative but not commutative. This means $(E_{32}\,E_{21})\,A = E_{32}\,(E_{21}\,A)$ always holds, but $E_{32}\,E_{21} \neq E_{21}\,E_{32}$ in general.

✅ Correct! Associativity lets you regroup freely. Commutativity fails — order matters in matrix multiplication.

❌ Not quite. Read carefully: the statement claims associativity holds AND commutativity fails. Both of these claims are true!

Show solution

Solution: True.

Associativity means we can regroup parentheses freely: $(E_{32}\,E_{21})\,A = E_{32}\,(E_{21}\,A)$ Both produce the same result $U$ . ✅

Commutativity would mean we could swap the order. But: $E_{32}E_{21} = \begin{bmatrix} 1 & 0 & 0 \\ -3 & 1 & 0 \\ 6 & -2 & 1 \end{bmatrix} \neq \begin{bmatrix} 1 & 0 & 0 \\ -3 & 1 & 0 \\ 0 & -2 & 1 \end{bmatrix} = E_{21}E_{32}$

Position $(3,1)$ : $6 \neq 0$ . Order matters! ❌

Question 3

You need to subtract $5$ times row 1 from row 3. Which elimination matrix accomplishes this when left-multiplied?

✅ Correct! The negative of the multiplier goes in the $(i, j)$ position of the identity matrix.

❌ Not quite. Remember: the negative of the multiplier goes in the position $(\text{target row}, \text{source row})$ of the identity matrix.

Show solution

Solution: The correct matrix has $-5$ in position $(3, 1)$ :

$E_{31} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ -5 & 0 & 1 \end{bmatrix}$

The negative of the multiplier ( $-5$ ) goes in position $(3, 1)$ because we are subtracting $5$ times row 1 from row 3. The rest of the matrix is the identity.

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