Matrix multiplication — definition, associativity, non-commutativity: Worked examples

Symbols we will use (built from zero)

Definition The vocabulary, in plain words

A matrix is a rectangular grid of numbers. We say it is $m \times n$ meaning $m$ rows tall, $n$ columns wide (rows first, always).
A vector, written $x$ (bold letter), is just a list of numbers stacked in a single column — that is, a matrix with one column. For example $x = (4 - 2)$ is a $2 \times 1$ matrix; its entries are $x_{1} = 4$ , $x_{2} = - 2$ . Picture it as an arrow from the origin to the point $(4, - 2)$ .
A matrix acting on a vector, written $A x$ , is just the product rule with $B$ replaced by the one-column matrix $x$ : entry $i$ of the answer is (row $i$ of $A$ ) dotted with $x$ , i.e. $(A x)_{i} = \sum_{k} A_{ik} x_{k}$ . Geometrically $A$ moves the arrow $x$ to a new arrow — this is the transformation picture.
$A_{ij}$ means "the number in row $i$ , column $j$ of matrix $A$ ". Row index first, column index second.
$\sum_{k = 1}^{n}$ (a sum) is shorthand for "add up the following as $k$ runs $1, 2, \dots, n$ ". So $\sum_{k = 1}^{3} x_{k} = x_{1} + x_{2} + x_{3}$ . It is just an adding machine with a counter $k$ .
The product $C = A B$ has entries $C_{ij} = \sum_{k = 1}^{n} A_{ik} B_{k j}$ — slide row $i$ of $A$ across column $j$ of $B$ , multiply matching numbers, add. This is a Dot Product of that row and that column.
$I$ (the identity) is the "do-nothing" matrix: $1$ 's down the diagonal, $0$ 's elsewhere. $I A = A I = A$ . See Identity and Inverse Matrices.
$0$ (the zero matrix) is the grid whose every entry is $0$ . Written boldface to distinguish it from the plain number $0$ . Adding it changes nothing; as a map it sends every arrow to the origin.
$A^{⊤}$ (the transpose, read "A-transpose") is the matrix you get by flipping $A$ across its diagonal — row $i$ becomes column $i$ . So the entry in row $i$ , column $j$ of $A^{⊤}$ is $A_{j i}$ . See Transpose.
$[A, B] := A B - B A$ (the commutator) measures how much order matters: it is the zero matrix exactly when $A B = B A$ .

Everything else is earned as it appears.

The scenario matrix

Every product falls into one of these cells. The examples below are labelled with the cell(s) they cover, so you can check the whole grid is filled.

Cell	Situation	What could trip you up	Example
A	Square × square (both $2 \times 2$ )	mechanics of row·column	E1
B	Rectangular, shapes align	3-term sum; result shape	E2
C	Shapes don't align	product is undefined	E3
D	Multiply by identity $I$	$I$ changes nothing	E4
E	Zero matrix / zero divisors	$A B = 0$ with $A, B \neq = 0$	E5
F	Order matters: $A B \neq = B A$	geometry of composition	E6
G	Associativity $(A B) C = A (B C)$	bracketing is free; cost differs	E7
H	Inverse of a product	order reverses	E8
I	Word problem (real world)	reading a chain as a matrix	E9
J	Exam twist: $(A + B)^{2}$	binomial trap	E10

Sign cases (positive, negative, zero entries) appear throughout — E1, E2 and E6 all carry negatives so no sign is left unshown.

Worked examples

E1 — Cell A: square × square, the plain mechanics

Worked example Multiply two

$2 \times 2$ matrices $A = (20 - 1 3), B = (4 - 2 51) . Find A B .$

Forecast: guess before computing — will the top-left entry be positive or negative? (Row $(2, - 1)$ against column $(4, - 2)$ .)

Step 1 — check shapes. Why this step? A product only exists if the inner dimensions match. $A$ is $2 \times 2$ , $B$ is $2 \times 2$ : inner dims $2 = 2$ ✓, result is $2 \times 2$ .

Step 2 — top-left $(A B)_{11}$ : row 1 of $A$ · column 1 of $B$ . Why? $(A B)_{ij}$ is always row $i$ dotted with column $j$ . $(A B)_{11} = 2 \cdot 4 + (- 1) \cdot (- 2) = 8 + 2 = 10.$ Positive — the two negatives multiplied to $+ 2$ . (Forecast check!)

Step 3 — remaining entries, same recipe. $(A B)_{12} = 2 \cdot 5 + (- 1) \cdot 1 = 9, (A B)_{21} = 0 \cdot 4 + 3 \cdot (- 2) = - 6, (A B)_{22} = 0 \cdot 5 + 3 \cdot 1 = 3.$ $A B = (10 - 6 93) .$

Verify: column view — column 1 of $A B$ should equal $A$ acting on the vector $v = (4 - 2)$ (column 1 of $B$ ): $A v = (2 \cdot 4 - 1 \cdot (- 2) 0 \cdot 4 + 3 \cdot (- 2)) = (10 - 6)$ ✓ matches column 1.

E2 — Cell B: rectangular, three-term dot products, signs

Worked example A

$2 \times 3$ times a $3 \times 2$ $A = (13 - 2 1 0 - 1), B = 2 - 1 5 041 .$

Forecast: what shape is $A B$ ? (Outer dimensions.)

Step 1 — shapes. Why? Inner dims: $A$ is $2 \times 3$ , $B$ is $3 \times 2$ , $3 = 3$ ✓. Outer dims $2$ and $2$ ⟹ result is $2 \times 2$ . (Forecast: $2 \times 2$ .)

Step 2 — each entry now sums over $k = 1, 2, 3$ . Why three terms? Because the shared dimension is $3$ ; the sum $\sum_{k = 1}^{3}$ runs over all three. $(A B)_{11} = 1 \cdot 2 + (- 2) (- 1) + 0 \cdot 5 = 2 + 2 + 0 = 4.$ $(A B)_{12} = 1 \cdot 0 + (- 2) \cdot 4 + 0 \cdot 1 = - 8.$ $(A B)_{21} = 3 \cdot 2 + 1 \cdot (- 1) + (- 1) \cdot 5 = 6 - 1 - 5 = 0.$ $(A B)_{22} = 3 \cdot 0 + 1 \cdot 4 + (- 1) \cdot 1 = 3.$ $A B = (40 - 8 3) .$

Verify: here $B A$ does exist (it is $3 \times 3$ ), but it has a different shape from $A B$ (which is $2 \times 2$ ), so the two can never be equal — a shape-mismatch flavour of non-commutativity. (Contrast E3, where the product is genuinely undefined; here $B A$ is defined, just a different size.)

E3 — Cell C: shapes refuse to align (the undefined case)

E4 — Cell D: the identity changes nothing

E5 — Cell E: zero divisors, $A B = 0$ with neither matrix zero

E6 — Cell F: order matters (geometry)

Worked example Rotate-then-flip vs flip-then-rotate

$R = (01 - 1 0)$ (rotate $9 0^{\circ}$ anticlockwise), $F = (10 0 - 1)$ (flip the $y$ -axis). Find $R F$ and $F R$ , where we track the vector $x = (10)$ (the arrow to the point $(1, 0)$ ).

Forecast: where does the point $(1, 0)$ land under " $F$ then $R$ " versus " $R$ then $F$ "? Predict, then read on.

Figure — what you're looking at. The gray arrow is the starting point

(1, 0)

. The blue arrow shows where it lands under

R F

(flip first, then rotate): straight up to

(0, 1)

. The orange arrow shows

F R

(rotate first, then flip): straight down to

(0, - 1)

. Same start, two different endpoints — that gap between blue and orange is exactly the non-commutativity we compute below.

Step 1 — compute $R F$ (this is " $F$ first, then $R$ ", because acting on any vector $x$ gives $(R F) x = R (F x)$ — the matrix nearest $x$ acts first). $R F = (01 - 1 0) (10 0 - 1) = (0110) .$ Why the rightmost matrix acts first: $x$ meets $F$ first, its output meets $R$ .

Step 2 — compute $F R$ (" $R$ first, then $F$ "). $F R = (10 0 - 1) (01 - 1 0) = (0 - 1 - 1 0) .$

Step 3 — track $x = (10)$ (look at the figure). Under $R F$ : $(0110) (10) = (01)$ (up, the blue arrow). Under $F R$ : $(0 - 1 - 1 0) (10) = (0 - 1)$ (down, the orange arrow). Different landing points.

Verify: $R F - F R = (0220) \neq = 0$ , so the commutator $[R, F] \neq = 0$ : order genuinely counts. See Linear Transformations.

E7 — Cell G: associativity is free (but the cost isn't)

Worked example Check

$(A B) C = A (B C)$ $A = (1021), B = (1301), C = (2110) .$

Forecast: the two bracketings must give the same matrix. Guess whether they also cost the same effort.

Step 1 — left grouping. Why start here: to compute $(A B) C$ we must first form the inner product $A B$ , then multiply the result by $C$ . $A B = (1 \cdot 1 + 2 \cdot 3 0 \cdot 1 + 1 \cdot 3 1 \cdot 0 + 2 \cdot 1 0 \cdot 0 + 1 \cdot 1) = (7321)$ . Then $(A B) C = (7321) (2110) = (16773) .$

Step 2 — right grouping. Why redo it grouped the other way: the whole point is to test whether the bracket position changes the answer, so we now form the inner product $B C$ first. $B C = (1 \cdot 2 + 0 \cdot 1 3 \cdot 2 + 1 \cdot 1 1 \cdot 1 + 0 \cdot 0 3 \cdot 1 + 1 \cdot 0) = (2713)$ . Then $A (B C) = (1021) (2713) = (16773) .$

Step 3 — compare. Identical. Why guaranteed: each side applies maps in the same order $C, B, A$ to any vector. Entrywise, both sides equal the same triple sum $((A B) C)_{i ℓ} = (A (B C))_{i ℓ} = \sum_{k} \sum_{j} A_{ik} B_{k j} C_{j ℓ},$ where — recalling our index convention (row first, column second) — $i$ is the output row, $ℓ$ is the output column of the final matrix, while $k$ and $j$ are the two inner indices being summed away.

Verify: both equal $(16773)$ ✓. (Note: for tall/thin matrices one bracketing can be far cheaper — associativity lets you choose.)

E8 — Cell H: inverse of a product reverses order

Worked example

$(A B)^{- 1} = B^{- 1} A^{- 1}$ $A = (1011)$ , $B = (1201)$ . Verify $(A B)^{- 1} = B^{- 1} A^{- 1}$ , and show $A^{- 1} B^{- 1}$ is wrong.

Forecast: "socks then shoes" — to undo, which comes off first?

Step 1 — inverses of each. Why these: $A$ shears by $+ 1$ , its inverse shears by $- 1$ : $A^{- 1} = (10 - 1 1)$ . Similarly $B^{- 1} = (1 - 2 01)$ .

Step 2 — form $A B$ . Why this step: to talk about $(A B)^{- 1}$ we first need the actual matrix $A B$ so we can later confirm the candidate inverse multiplies it to $I$ . $A B = (1011) (1201) = (3211) .$

Step 3 — test $B^{- 1} A^{- 1}$ against $A B$ . Why this order: to undo " $B$ then $A$ " you must undo $A$ first. $B^{- 1} A^{- 1} = (1 - 2 01) (10 - 1 1) = (1 - 2 - 1 3) .$ Multiply: $(A B) (B^{- 1} A^{- 1}) = (3211) (1 - 2 - 1 3) = (1001) = I$ ✓.

Step 4 — the wrong order. $A^{- 1} B^{- 1} = (10 - 1 1) (1 - 2 01) = (3 - 2 - 1 1)$ , and $(A B) (A^{- 1} B^{- 1}) = (74 - 2 - 1) \neq = I$ .

Verify: only $B^{- 1} A^{- 1}$ gives $I$ — order reverses. The transpose $A^{⊤}$ (flip across the diagonal, defined in the symbols list) reverses the same way: $(A B)^{⊤} = B^{⊤} A^{⊤}$ . See Identity and Inverse Matrices and Transpose.

E9 — Cell I: a real-world word problem

Worked example Two-stage supply chain

A factory turns raw materials into parts, and parts into products. Rows are outputs, columns are inputs. Parts-per-product: $P = (2013)$ (rows = part A, part B; cols = product 1, product 2). Materials-per-part: $M = (1540)$ (rows = wood, steel; cols = part A, part B). Question: how much wood and steel does each product need?

Forecast: which product, $M P$ or $P M$ , answers "materials per product"?

Step 1 — chain the maps. Why $M P$ : we want materials←parts←products, i.e. do "products→parts" ( $P$ ) first, then "parts→materials" ( $M$ ). Composition order puts the later stage on the left, giving $M P$ . $M P = (1540) (2013) = (1 \cdot 2 + 4 \cdot 0 5 \cdot 2 + 0 \cdot 0 1 \cdot 1 + 4 \cdot 3 5 \cdot 1 + 0 \cdot 3) = (210135) .$

Step 2 — read it off (with units!). Column $j$ is product $j$ ; row $1$ is wood, row $2$ is steel. So product 1 needs $2$ wood and $10$ steel; product 2 needs $13$ wood and $5$ steel.

Verify (sanity + units). Product 1 uses $2$ of part A and $0$ of part B. Part A alone needs $1$ wood and $5$ steel, so $2$ of it needs $2$ wood and $10$ steel — matches column 1 ✓. Every number carries units (material units)/(product), dimensionally consistent. (Note $P M$ would answer a different question and generally differ — order counts again.)

E10 — Cell J: the exam twist, $(A + B)^{2}$

Worked example Expand a matrix square correctly

$A = (1011)$ , $B = (1101)$ . Compute $(A + B)^{2}$ and compare to the (wrong) $A^{2} + 2 A B + B^{2}$ .

Forecast: for numbers $(a + b)^{2} = a^{2} + 2 ab + b^{2}$ . Will the middle term be $2 A B$ here?

Step 1 — the honest expansion. Why not $2 A B$ : $(A + B)^{2} = (A + B) (A + B) = A^{2} + A B + B A + B^{2}$ ; you may only merge $A B + B A \to 2 A B$ if $A B = B A$ .

Step 2 — direct compute. $A + B = (2112)$ , so $(A + B)^{2} = (2112) (2112) = (5445) .$

Step 3 — the pieces. $A^{2} = (1021)$ , $B^{2} = (1201)$ , $A B = (2111)$ , $B A = (1112)$ . Correct sum: $A^{2} + A B + B A + B^{2} = (5445)$ ✓. Wrong formula: $A^{2} + 2 A B + B^{2} = (6444) \neq = (A + B)^{2}$ .

Verify: the honest expansion matches; the binomial shortcut is off because the commutator is nonzero: $[A, B] = A B - B A = (2111) - (1112) = (10 0 - 1) \neq = 0 .$ Since $[A, B] \neq = 0$ , the matrices do not commute, so $A B + B A \neq = 2 A B$ and the binomial shortcut must fail — exactly as we saw.

Active recall

Row

i

A

against column

j

B

gives which entry?

(A B)_{ij}

, the dot product

\sum_{k} A_{ik} B_{k j}

A

2 \times 3

B

3 \times 3

— is

A B

defined, and what shape?

Yes, inner dims

3 = 3

; result is

2 \times 3

A

2 \times 2

B

3 \times 3

— is

A B

defined?

No; columns of

A

(

2

) ≠ rows of

B

(

3

Can a nonzero matrix square to the zero matrix

0

Yes — e.g.

(0010)^{2} = 0

(zero divisors).

To undo "apply

B

then

A

", which inverse acts first?

A^{- 1}

(A B)^{- 1} = B^{- 1} A^{- 1}

Why is

(A + B)^{2} \neq = A^{2} + 2 A B + B^{2}

in general?

The cross terms are

A B + B A

, which only merge to

2 A B

A, B

commute.

In the supply-chain problem, which product gives materials-per-product?

M P

— do products→parts first (

P

), then parts→materials (

M

Connections

Matrix multiplication — definition, associativity, non-commutativity — the parent rule these examples exercise.
Linear Transformations — E6 is composition of a rotation and a flip.
Identity and Inverse Matrices — E4 (identity) and E8 (order-reversing inverse).
Dot Product — every entry $(A B)_{ij}$ is one.
Determinants — a fast $2 \times 2$ invertibility check for E8.
Transpose — $(A B)^{⊤} = B^{⊤} A^{⊤}$ reverses order like the inverse does.
Matrix Powers and Diagonalization — E10's commutator controls when powers behave nicely.

Worked examples — Matrix multiplication — definition, associativity, non-commutativity