Matrix multiplication — definition, associativity, non-commutativity: Question bank

True or false — justify

State true/false, then give the reason. A bare verdict scores nothing.

A B

and

B A

always have the same size.

False. If

A

m \times n

and

B

n \times p

then

A B

m \times p

but

B A

needs

p = m

to even exist; e.g. a

2 \times 3

times

3 \times 2

gives

2 \times 2

one way,

3 \times 3

the other.

If both

A B

and

B A

exist and are the same size, then

A B = B A

False. Same size is necessary but not sufficient;

(1011)

and

(1101)

are both

2 \times 2

yet their products differ on the diagonal.

Matrix multiplication is associative.

True. It represents composing maps "

C

then

B

then

A

", one unambiguous pipeline; entrywise both bracketings equal the same double sum

\sum_{j, k} A_{ik} B_{k j} C_{j ℓ}

Matrix multiplication distributes over addition.

True.

A (B + C) = A B + A C

and

(A + B) C = A C + B C

— but note you must keep the left/right side fixed because order still matters.

A B = A C

then

B = C

False. Cancellation needs

A

to be invertible. With

A = (1000)

you can change the second row of

B

freely without touching

A B

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

for all square

A, B

False. Expanding gives

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

; you may only merge

A B + B A

into

2 A B

when

A

and

B

commute.

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

whenever the inverses exist.

False. Order reverses:

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

, because to undo "

B

then

A

" you must undo

A

first. See Identity and Inverse Matrices.

(A B)^{⊤} = A^{⊤} B^{⊤}

False. Transpose also reverses order:

(A B)^{⊤} = B^{⊤} A^{⊤}

. See Transpose.

A B = 0

forces

A = 0

B = 0

False. Matrices have zero divisors;

(0010)^{2} = 0

though the matrix is nonzero.

A^{2} = 0

implies

A = 0

False. Same nilpotent example:

(0010)

squares to zero but is not itself zero.

The identity

I

commutes with every square matrix of matching size.

True.

A I = I A = A

always, because "do nothing then

A

" equals "do

A

then nothing" — a rare pair that truly commutes.

Any matrix commutes with its own powers:

A A^{2} = A^{2} A

True. Both equal

A^{3}

; a matrix always commutes with itself, and associativity lets us drop brackets. See Matrix Powers and Diagonalization.

det (A B) = det (A) det (B)

even though

A B \neq = B A

True. The determinant is multiplicative regardless of order, so

det (A B) = det (B A)

always, even when the products themselves differ. See Determinants.

Spot the error

Each line contains a plausible-but-wrong claim or step. Name the flaw.

(A - B) (A + B) = A^{2} - B^{2}

, by difference of squares."

Error: it actually equals

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

; the middle terms cancel only if

A, B

commute.

"To compute

C = A B

I multiplied entry by entry:

C_{ij} = A_{ij} B_{ij}

Error: that is the Hadamard (entrywise) product, which breaks

(A B) x = A (B x)

. The real rule is row

i

A

dotted with column

j

B

A

3 \times 2

and

B

3 \times 2

, so

A B

has size

3 \times 2

Error: inner dimensions

2

and

3

do not match, so

A B

is undefined; only outer dims survive after the inner ones agree.

"Since

A B = B A

for these two matrices I found, multiplication must be commutative in general."

Error: a single commuting pair proves nothing about the general rule; special pairs (like

A

with

I

or with

A^{- 1}

) commute, most do not.

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

Error: the whole chain reverses —

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

, undoing the last-applied map first.

"Row

i

A B

equals row

i

A

times row

i

B

Error: row

i

A B

equals (row

i

A

) times the whole matrix

B

; you never multiply a row of

A

by a single row of

B

A^{2} B^{2} = (A B)^{2}

Error:

(A B)^{2} = A B A B

, which equals

A^{2} B^{2}

only if

B A = A B

; in general the inner

B A

can't be swapped.

Why questions

$::$ prompts for the underlying reason, not a formula.

Why is the row-times-column rule forced, rather than chosen for convenience?

It is the only rule making

(A B) x = A (B x)

hold for every

x

; demanding "

A B

= compose the two maps" pins the coefficient of

x_{j}

to exactly

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

Why must the inner dimensions match?

Each entry is a dot product of a length-

n

row of

A

with a length-

n

column of

B

; if those lengths differ there is no way to pair terms for the sum. See Dot Product.

Why is associativity "free" while commutativity is not?

Composing functions in a fixed order is always associative — "

C

then

B

then

A

" means one thing however you bracket — but the order itself changes the pipeline, so swapping is generally not allowed.

Why does inverting a product reverse the order?

To undo "put on socks then shoes" you take off shoes first:

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

, so the undo of the last-applied map comes first.

Why can a nonzero matrix square to zero when a nonzero number never can?

A matrix can map everything into a subspace that a second application collapses entirely (e.g. shear then re-shear onto zero); numbers have no such "hidden direction" to hide inside.

Why does non-commutativity not break associativity or distributivity?

Those laws never ask us to swap

A

and

B

; they only regroup or split sums, operations that respect the fixed order of application.

Why does

det (A B) = det (B A)

hold even though

A B \neq = B A

Determinant measures the total volume-scaling of a map, and scaling factors multiply regardless of order, so both products scale volume by

det A \cdot det B

. See Determinants.

Edge cases

Degenerate and boundary inputs the topic quietly assumes you can handle.

What is

A B

when

A

1 \times n

and

B

n \times 1

1 \times 1

matrix — a single number, exactly the dot product of the row and the column; this is the atomic case every entry of a general product reduces to.

What is

A B

when

A

n \times 1

and

B

1 \times n

n \times n

matrix (an outer product) of rank at most

1

; swapping to

B A

gives instead a

1 \times 1

scalar, a stark shape-based non-commutation.

Does the zero matrix commute with everything?

Yes:

0 \cdot A = A \cdot 0 = 0

whenever the shapes allow, since composing with the map that sends all vectors to

0

gives

0

either way.

A

m \times n

with

m \neq = n

, can

A^{2}

exist?

No:

A^{2} = AA

needs the inner dimensions

n

and

m

to match, so only square matrices have powers. See Matrix Powers and Diagonalization.

Is the commutator

[A, B] = A B - B A

ever guaranteed to be zero without

A, B

being special?

No — it is zero precisely when they commute; for generic square matrices it is a nonzero matrix measuring exactly how much order matters.

A

commutes with

B

and

B

commutes with

C

, must

A

commute with

C

No — commuting is not transitive; the identity commutes with everything yet the matrices it links need not commute with each other.

What does multiplying any square matrix

A

I

on either side reveal about

I

's role?

A I = I A = A

shows

I

is the "do-nothing" map, the multiplicative identity, and one of the few things guaranteed to commute with

A

. See Identity and Inverse Matrices.

Connections

Matrix Multiplication — definition, associativity, non-commutativity — the parent this bank stress-tests.
Linear Transformations — the composition view that resolves every trap.
Identity and Inverse Matrices — inverse-of-product order reversal.
Transpose — transpose-of-product order reversal.
Dot Product — each entry as a row·column sum.
Determinants — multiplicative despite non-commutativity.
Matrix Powers and Diagonalization — when matrices do commute.