Matrix multiplication — definition, associativity, non-commutativity: Exercises

Before we start, one picture of the whole "row crosses column" motion — keep it in your head for every single problem below.

Figure — Matrix multiplication — definition, associativity, non-commutativity

Read the figure first. On the left, the pale-blue arrow sweeps across row $i$ of $A$ (the entries $a_{11}, a_{12}, a_{13}$ ). On the right, the pink arrow runs down column $j$ of $B$ (the entries $b_{1 j}, b_{2 j}, b_{3 j}$ ). The yellow curved arrow shows those two streams meeting at a single yellow dot — that dot is the output entry $(A B)_{ij}$ . The lesson the picture teaches: one output number costs one full row-times-column pass. Every solution below is just this figure repeated for each $(i, j)$ .

Level 1 — Recognition

(Can you read shapes and grab the right row and column?)

Recall Solution L1.1

First, what those two words mean. Write each matrix as (rows) $\times$ (columns): $A$ is $rows 2 \times cols 3$ . When you line up a product $X Y$ as $(rows_{X} \times cols_{X}) (rows_{Y} \times cols_{Y})$ , the inner dimensions are the two numbers that sit next to each other in the middle — $cols_{X}$ and $rows_{Y}$ — and the outer dimensions are the two on the outside — $rows_{X}$ and $cols_{Y}$ . The rule: to form $X Y$ the inner dimensions must match (so each row of $X$ has exactly as many entries as each column of $Y$ , letting the dot product line up), and the result takes the outer dimensions as its (rows) $\times$ (columns).

$A B$ : $(2 \times 3) (3 \times 4)$ → inner $3 = 3$ ✓ → size (outer) $2 \times 4$ . Defined.
$B A$ : $(3 \times 4) (2 \times 3)$ → inner $4 \neq = 2$ ✗. Not defined.
$B C$ : $(3 \times 4) (4 \times 2)$ → inner $4 = 4$ ✓ → size $3 \times 2$ . Defined.
$C A$ : $(4 \times 2) (2 \times 3)$ → inner $2 = 2$ ✓ → size $4 \times 3$ . Defined.
$A C$ : $(2 \times 3) (4 \times 2)$ → inner $3 \neq = 4$ ✗. Not defined. What we did: just matched the two inner numbers and copied the two outer numbers. Nothing computed — this is pure bookkeeping.

Recall Solution L1.2

$(A B)_{21}$ = row $2$ of $A$ dotted with column $1$ of $B$ . Row 2 of $A$ is $(0, 3)$ ; column 1 of $B$ is $(4, 6)$ : $(A B)_{21} = 0 \cdot 4 + 3 \cdot 6 = 18.$ $(A B)_{12}$ = row $1$ of $A$ dotted with column $2$ of $B$ . Row 1 of $A$ is $(2, - 1)$ ; column 2 of $B$ is $(5, 7)$ : $(A B)_{12} = 2 \cdot 5 + (- 1) \cdot 7 = 10 - 7 = 3.$ Why the index reads $ij$ = (row $i$ )(column $j$ ): the first index is which output row, the second is which output column. The output row is decided by which row of $A$ you slide; the output column by which column of $B$ you drop it onto.

Level 2 — Application

(Turn the crank on full products.)

Recall Solution L2.1

Four dot products, each row of $A$ against each column of $B$ .

$(A B)_{11} = (2, - 1) \cdot (4, 6) = 8 - 6 = 2$
$(A B)_{12} = (2, - 1) \cdot (5, 7) = 10 - 7 = 3$ (from L1.2)
$(A B)_{21} = (0, 3) \cdot (4, 6) = 0 + 18 = 18$ (from L1.2)
$(A B)_{22} = (0, 3) \cdot (5, 7) = 0 + 21 = 21$ $A B = (218321) .$

Recall Solution L2.2

Inner dims $3 = 3$ ✓, so $A B$ is $2 \times 2$ . Each entry is a three-term dot product over the shared index $k = 1, 2, 3$ .

$(A B)_{11} = (1, 2, - 1) \cdot (2, 0, 4) = 2 + 0 - 4 = - 2$
$(A B)_{12} = (1, 2, - 1) \cdot (1, - 1, 2) = 1 - 2 - 2 = - 3$
$(A B)_{21} = (0, 1, 3) \cdot (2, 0, 4) = 0 + 0 + 12 = 12$
$(A B)_{22} = (0, 1, 3) \cdot (1, - 1, 2) = 0 - 1 + 6 = 5$ $A B = (- 2 12 - 3 5) .$

Recall Solution L2.3

(a) Row·column. Each output entry is a row of $M$ dotted with $x = (2, - 1)$ : $M x = 1 \cdot 2 + 2 \cdot (- 1) 3 \cdot 2 + 4 \cdot (- 1) 5 \cdot 2 + 6 \cdot (- 1) = 024 .$ (b) Column combination. $M x$ is $2$ copies of column 1 minus $1$ copy of column 2: $2135 - 1246 = 2 - 2 6 - 4 10 - 6 = 024 .$ Why both work: the row·column rule and "weight the columns by the vector's entries" are the same arithmetic re-grouped. View (b) is the one that reveals matrices as linear transformations — the output is always a weighted mix of the columns.

Level 3 — Analysis

(Explain and exploit why, not just compute.)

Recall Solution L3.1

FR=\begin{pmatrix}1&0\\0&-1\end{pmatrix}\begin{pmatrix}0&-1\\1&0\end{pmatrix}=\begin{pmatrix}0&-1\\-1&0\end{pmatrix}.$$ They differ, so $RF\neq FR$. **Reading the notation:** $RF$ means "$F$ first, then $R$" — remember the *rightmost matrix touches the vector first*. Track the point $\mathbf{p}=(1,0)$ (the white "start" arrow in **both panels of the figure below**): - $RF\,\mathbf{p}$ (**left panel**): flip $y$ first — $(1,0)$ stays $(1,0)$ (it's on the $x$-axis) — then rotate $90^\circ$ → $(0,1)$, the blue "end" arrow. Check: $\begin{pmatrix}0&1\\1&0\end{pmatrix}\begin{pmatrix}1\\0\end{pmatrix}=\begin{pmatrix}0\\1\end{pmatrix}$ ✓. - $FR\,\mathbf{p}$ (**right panel**): rotate first → $(0,1)$, then flip $y$ → $(0,-1)$, the pink "end" arrow. Check: $\begin{pmatrix}0&-1\\-1&0\end{pmatrix}\begin{pmatrix}1\\0\end{pmatrix}=\begin{pmatrix}0\\-1\end{pmatrix}$ ✓. Same white start arrow, two *different* coloured end arrows → order genuinely matters. That contrast is the whole point of the figure.

The two panels above sit side by side precisely so you can see one identical white starting arrow $(1, 0)$ produce a blue ending arrow on the left ( $R F$ , pointing up) and a pink ending arrow on the right ( $F R$ , pointing down). If matrix multiplication were commutative, the two coloured arrows would coincide — they don't, and that visible gap is non-commutativity.

Recall Solution L3.2

BA=\begin{pmatrix}1&0\\1&1\end{pmatrix}\begin{pmatrix}1&1\\0&1\end{pmatrix}=\begin{pmatrix}1&1\\1&2\end{pmatrix}.$$ $$[A,B]=AB-BA=\begin{pmatrix}2-1&1-1\\1-1&1-2\end{pmatrix}=\begin{pmatrix}1&0\\0&-1\end{pmatrix}.$$ **Meaning:** the commutator is $0$ *if and only if* $A$ and $B$ commute. Here it is nonzero, so $AB\neq BA$ — and the *entries* of $[A,B]$ literally measure how far apart the two products land (they disagree on the diagonal by $\pm1$).

Recall Solution L3.3

First $A B = (1201) (1021) = (1225)$ . Then $(A B) C = (1225) (0110) = (2512)$ . Now the other bracketing: $B C = (1021) (0110) = (2110)$ . Then $A (B C) = (1201) (2110) = (2512)$ . Equal ✓. Why guaranteed: both spell out "do $C$ , then $B$ , then $A$ " — the same pipeline; only the order you paused to multiply changed, never the order maps act.

Level 4 — Synthesis

(Combine several rules in one problem.)

Recall Solution L4.1

The rule from the parent note: to undo " $B$ then $A$ " you undo $A$ first, so $(A B)^{- 1} = B^{- 1} A^{- 1} .$ Compute: $B^{- 1} A^{- 1} = (1 - 3 01) (2111) = (2 - 5 1 - 2) .$ Proof of ordering (why not $A^{- 1} B^{- 1}$ ): using the identity and associativity, $(A B) (B^{- 1} A^{- 1}) = A (B B^{- 1}) A^{- 1} = A I A^{- 1} = A A^{- 1} = I . ✓$ The inner $B B^{- 1}$ cancels because it's adjacent; the reversed order is what makes the cancellation line up. The wrong order $A^{- 1} B^{- 1}$ would leave a mismatched $B A^{- 1}$ in the middle that does not collapse.

Recall Solution L4.2

$A B = (1023) (4102) = (6346)$ , so $(A B)^{⊤} = (6436)$ . Now the transpose side. $B^{⊤} = (4012)$ , $A^{⊤} = (1203)$ : $B^{⊤} A^{⊤} = (4012) (1203) = (6436) .$ Match ✓. Why order reverses again: transpose swaps rows↔columns, and since $(A B)_{ij}$ pairs row $i$ of $A$ with column $j$ of $B$ , transposing turns that into column $i$ of $A$ (= row $i$ of $A^{⊤}$ )... the only consistent reassembly is $B^{⊤} A^{⊤}$ .

Recall Solution L4.3

$A + B = (1011) + (1101) = (2112)$ . $(A + B)^{2} = (2112) (2112) = (5445) .$ The correct expansion keeps order: $(A + B)^{2} = A^{2} + A B + B A + B^{2}$ . From L3.2, $A B = (2111)$ and $B A = (1112)$ . Also $A^{2} = (1021)$ , $B^{2} = (1201)$ . Sum: $A^{2} + A B + B A + B^{2} = (1021) + (2111) + (1112) + (1201) = (5445) ✓$ The wrong $A^{2} + 2 A B + B^{2}$ would use $2 A B = (4222)$ instead of $A B + B A = (3223)$ , giving $(6244)$ — wrong, because here $A B \neq = B A$ .

Level 5 — Mastery

(Prove / discover.)

Recall Solution L5.1

Take $N = (0010) \neq = 0$ . Compute $N^{2} = N \cdot N$ entry by entry, each entry a row·column dot product — no shortcuts.

$(N^{2})_{11} =$ row 1 $(0, 1) \cdot$ col 1 $(0, 0) = 0 \cdot 0 + 1 \cdot 0 = 0$ .
$(N^{2})_{12} =$ row 1 $(0, 1) \cdot$ col 2 $(1, 0) = 0 \cdot 1 + 1 \cdot 0 = 0$ .
$(N^{2})_{21} =$ row 2 $(0, 0) \cdot$ col 1 $(0, 0) = 0 \cdot 0 + 0 \cdot 0 = 0$ .
$(N^{2})_{22} =$ row 2 $(0, 0) \cdot$ col 2 $(1, 0) = 0 \cdot 1 + 0 \cdot 0 = 0$ . Notice the bottom row of $N$ is all zeros, so any dot product using it is automatically $0$ — that kills $(N^{2})_{21}$ and $(N^{2})_{22}$ . The top row survives, but the only nonzero piece ( $1$ in position 2) always meets a $0$ from the columns of $N$ , so it too dies. Hence $N^{2} = (0000) .$ Why numbers can't: for reals, $x^{2} = 0 \Rightarrow x = 0$ (no zero divisors). Matrices do have zero divisors, so " $X Y = 0 \Rightarrow X = 0$ or $Y = 0$ " is false. Geometrically $N$ collapses the plane onto the $x$ -axis, then the leftover pushes that line to the origin — two squashes kill everything.

Recall Solution L5.2

Write $A = (a c b d)$ . The hypothesis says $A M = M A$ for every $M$ ; in particular it must hold for specially chosen test matrices, and each choice squeezes out a constraint on $a, b, c, d$ . Test 1: $E = (1000)$ . $A E = (a c b d) (1000) = (a c 00), E A = (1000) (a c b d) = (a 0 b 0) .$ Setting $A E = E A$ entrywise: the $(2, 1)$ entries give $c = 0$ ; the $(1, 2)$ entries give $b = 0$ . So $A$ must be diagonal, $A = (a 0 0 d)$ . Test 2: $S = (0110)$ (the swap matrix). Using the now-diagonal $A$ : $A S = (a 0 0 d) (0110) = (0 d a 0), S A = (0110) (a 0 0 d) = (0 a d 0) .$ Setting $A S = S A$ : the $(1, 2)$ entries force $a = d$ . Hence $A = (a 0 0 a) = a I$ , i.e. $A = c I$ with $c = a$ . Why this finishes the proof. Commuting with all $M$ is an extremely strong demand — yet we only needed two cleverly chosen $M$ 's to pin down all four entries: Test 1 killed the off-diagonal entries, Test 2 tied the two diagonal entries together. Conversely, $c I$ obviously commutes with everything, since $(c I) M = c M = M (c I)$ for any $M$ . So the set of "commutes with everything" matrices is exactly ${c I}$ — nothing more, nothing less. $■$

Recall Solution L5.3

For a $2 \times 2$ , $det (p r q s) = p s - q r$ (see Determinants). $det A = 1 \cdot 4 - 2 \cdot 3 = - 2$ . $det B = 0 \cdot 2 - 1 \cdot (- 1) = 1$ . Product $= - 2$ . Now $A B = (1324) (0 - 1 12) = (- 2 - 4 511)$ , so $det (A B) = (- 2) (11) - (5) (- 4) = - 22 + 20 = - 2$ . ✓ Why remarkable: determinants measure area-scaling. Doing two maps multiplies the scalings — and multiplying scalars doesn't care about order, so $det (A B) = det (B A) = det A det B$ always, even when the matrices themselves don't commute. Order-sensitivity lives in the matrix, not in its determinant.

Active recall

Which product is defined:

(2 \times 3) (3 \times 4)

(3 \times 4) (2 \times 3)

The first, giving

2 \times 4

; the second has mismatched inner dims

4 \neq = 2

A B x

, which matrix touches

x

first?

B

— read products right-to-left.

What does

[A, B] = 0

mean?

A

and

B

commute,

A B = B A

Correct expansion of

(A + B)^{2}

for matrices?

A^{2} + A B + B A + B^{2}

(A B)^{- 1} = ?

and

(A B)^{⊤} = ?

B^{- 1} A^{- 1}

and

B^{⊤} A^{⊤}

— order reverses.

Give a nonzero

N

with

N^{2} = 0

(0010)

Which matrices commute with every matrix?

Only scalar multiples of the identity,

c I

Connections

Matrix multiplication — definition, associativity, non-commutativity — the parent these exercises drill.
Dot Product — every entry of a product is one.
Linear Transformations — the column view (L2.3) and order-of-composition (L3.1).
Identity and Inverse Matrices — L4.1 and L5.2.
Transpose — L4.2 order reversal.
Determinants — L5.3 multiplicativity.
Matrix Powers and Diagonalization — commuting matrices (L3.2, L5.2) share eigenvectors.