Matrix multiplication — conditions, process, non-commutativity: Question bank

This page is a workout for your understanding, not your arithmetic. Every item below is a place where students confidently write the wrong thing. Read the question, cover the answer, commit to a reason out loud, then reveal. If your reason doesn't match the reasoning given, that's a trap you just escaped.

Prerequisites you should already own: the parent note (dot-product rule and the dimension-matching condition), plus a feel for Linear transformations as the meaning behind the numbers.

True or false — justify

A B

is defined, then

B A

is also defined.

False.

A B

needs columns-of-

A

= rows-of-

B

;

B A

needs columns-of-

B

= rows-of-

A

— a completely different condition. E.g.

A

2 \times 3

B

3 \times 4

A B

works,

B A

does not.

If both

A B

and

B A

are defined, they must have the same shape.

False in general.

A_{2 \times 3}

and

B_{3 \times 2}

give

A B

of shape

2 \times 2

but

B A

of shape

3 \times 3

. Same shape is only forced when both are square of the same size.

For square matrices of the same size,

A B = B A

always.

False. Both products exist and share a shape, but they are usually different — see the swap-matrix example in the parent where

A B = [2413]

but

B A = [3142]

. Multiplication being defined both ways is not the same as it commuting.

Matrix multiplication is associative:

(A B) C = A (B C)

True. It mirrors function composition, which is associative — applying

C

then

B

then

A

groups either way to the same overall transformation.

Matrix multiplication is entry-by-entry, like addition.

False. Addition is entry-wise, but multiplication is a row-column dot product. The entry-wise version is a different operation (the Hadamard product

A \circ B

A B = O

(the zero matrix), then at least one of

A

B

must be zero.

False. Non-zero matrices can multiply to zero, e.g.

[1000] [0001] = [0000]

. Matrices have "zero divisors" — numbers do not.

A B = A C

, then

B = C

False. You cannot "cancel"

A

unless

A

is invertible. If

A

collapses space (has no inverse), different

B

and

C

can land on the same output.

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

False — the order reverses:

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

. See Matrix transpose properties; transposing flips the composition order, just like undoing socks-then-shoes.

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

True, and notably

det (A B) = det (B A)

even when

A B \neq = B A

. The determinant measures area/volume scaling, and total scaling doesn't care about order.

Multiplying by the Identity matrix can change a matrix.

False.

A I = I A = A

; the identity is the "do-nothing" transformation, so it leaves every matrix untouched from either side.

Spot the error

A

2 \times 2

B

3 \times 1

, so

A B

2 \times 1

." Find the mistake.

The inner dimensions must match first:

A

2

columns vs

B

3

rows (

2 \neq = 3

), so

A B

is undefined — you never reach a shape at all.

"To compute

C_{21}

, dot column 2 of

A

with row 1 of

B

." Find the mistake.

Reversed.

C_{ij}

uses ==row

i

A

with column

j

B

==, so

C_{21}

is row 2 of

A

with column 1 of

B

A B

exists and

B A

exists, so multiplication is commutative here." Find the mistake.

Existence of both products is a dimension fact; commutativity is a value fact. You must actually compute both and compare — they rarely agree.

"Since

2 \times 3

times

3 \times 2

gives

2 \times 2

, and

3 \times 2

times

2 \times 3

gives

3 \times 3

, both are the identity-sized answer." Find the mistake.

Neither is automatically an identity; they're just valid products of different shapes. Shape ≠ content, and being square doesn't make you

I

"I'll factor:

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

." Find the mistake.

The correct distributive law is

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

; you factor out the common

A

, you do not add the

A

's.

(A + A) (B + C) = 2 A (B + C)

is a totally different quantity.

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

." Find the mistake.

Expanding gives

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

, and since

A B \neq = B A

in general you cannot merge them into

2 A B

. The binomial shortcut assumes commutativity.

Why questions

Why is the

(i, j)

entry a dot product of a row and a column, not a product of two numbers?

Because column

j

B

says where a basis vector goes, and row

i

A

says how to read off one coordinate of the result — summing

A_{ik} B_{k j}

over

k

blends all intermediate dimensions into one output coordinate.

Why must the inner dimensions match?

A

's rows expect an input with

n

components;

B

's columns must supply exactly

n

components. If

B

's columns have a different height, there's nothing consistent to dot against.

Why does order matter — why is

A B

generally

\neq = B A

Because

A B

means "do

B

first, then

A

", and composing transformations in the opposite order lands vectors elsewhere — rotate-then-stretch differs from stretch-then-rotate.

Why is

det (A B) = det (B A)

even though

A B \neq = B A

The determinant is a scalar recording how much area/volume is scaled. Both orders apply the same two scalings, and

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

because numbers commute — the geometry of "total stretch" is order-free.

Why does

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

reverse the order?

Transposing swaps the roles of rows and columns, which is exactly the "undo in reverse" pattern — the last operation applied becomes the first to be transposed.

Why can't we cancel

A

from

A B = A C

in general?

Cancelling secretly multiplies by

A^{- 1}

, but if

A

squashes a dimension it has no inverse. A collapsing map can send different

B, C

to the same image, so cancellation is illegal.

Edge cases

What is

A \cdot O

where

O

is the zero matrix of compatible shape?

The zero matrix. Every entry is a dot product with an all-zero column, so every sum is

0

— the transformation sends everything to the origin.

What happens when you multiply a

1 \times n

row by an

n \times 1

column?

You get a

1 \times 1

matrix — a single number, the plain dot product of the two vectors. This is the smallest non-trivial matrix product.

What happens when you multiply an

n \times 1

column by a

1 \times n

row (the reverse order)?

You get an

n \times n

matrix (an "outer product"), not a scalar. Same two vectors, opposite order, wildly different shape — a vivid reminder that order controls even the shape.

Is a

1 \times 1

matrix times a

1 \times 1

matrix just ordinary number multiplication?

Yes — the dot-product rule collapses to a single term

A_{11} B_{11}

, so matrix multiplication contains ordinary multiplication as its smallest case.

Can

A^{2}

be undefined for a valid matrix

A

Yes, if

A

is non-square:

A^{2} = A \cdot A

needs columns-of-

A

= rows-of-

A

, which only holds when

A

is square. Powers are a square-matrix privilege.

D

is diagonal, does

D A

scale

A

's rows or columns?

D A

scales the rows of

A

(each row

i

times

D_{ii}

), while

A D

scales the columns — a clean case where you can see why left- and right-multiplication differ.

Recall One-line summary of every trap here

The shape rules (inner match, order changes shape), the value rules ( $A B \neq = B A$ , no cancelling, zero divisors), and the algebra rules ( $(A B)^{T} = B^{T} A^{T}$ , $A (B + C) = A B + A C$ ) all trace back to one idea: matrices are ordered compositions of transformations, not numbers.