Matrix multiplication — conditions, process, non-commutativity: Exercises

This page is a ladder. Each rung asks slightly more of you than the last. Do the problem before opening the solution — the solutions are collapsed on purpose so the page tests you.

Everything here rests on one single rule from the parent note:

Figure — Matrix multiplication — conditions, process, non-commutativity

Here the red path is row $i$ of $A$ sliding across, meeting the red column $j$ of $B$ sliding down. Where they cross, one number of the answer is born.

Level 1 — Recognition

(Can you even tell whether a product is legal, and what shape it has?)

Exercise 1.1

For each pair, say whether $A B$ is defined, and if so give the size of $A B$ .

(a) $A$ is $2 \times 3$ , $B$ is $3 \times 5$ .
(b) $A$ is $4 \times 2$ , $B$ is $4 \times 2$ .
(c) $A$ is $1 \times 3$ (a row), $B$ is $3 \times 1$ (a column).
(d) $A$ is $3 \times 1$ , $B$ is $1 \times 3$ .

Recall Solution 1.1

Rule: for $A_{m \times n}$ times $B_{n^{'} \times p}$ , we need $n = n^{'}$ ; the answer is $m \times p$ .

(a) inner numbers $3$ and $3$ match → defined, size $2 \times 5$ . ✓
(b) inner numbers $2$ and $4$ do not match → undefined. ✗
(c) inner numbers $3$ and $3$ match → defined, size $1 \times 1$ (a single number — this is exactly a dot product).
(d) inner numbers $1$ and $1$ match → defined, size $3 \times 3$ (a whole grid from a column times a row!).

Notice (c) and (d) use the same two matrices but in swapped order and give completely different shapes. Order matters even for the shape.

Level 2 — Application

(Run the rule by hand, all entries, no shortcuts.)

Exercise 2.1

Compute $A B$ where $A = [20 - 1 3], B = [15 4 - 2] .$

Recall Solution 2.1

Both are $2 \times 2$ , inner numbers match, answer is $2 \times 2$ . Each entry = (row of $A$ )·(column of $B$ ).

$C_{11} = (2) (1) + (- 1) (5) = 2 - 5 = - 3$ $C_{12} = (2) (4) + (- 1) (- 2) = 8 + 2 = 10$ $C_{21} = (0) (1) + (3) (5) = 0 + 15 = 15$ $C_{22} = (0) (4) + (3) (- 2) = 0 - 6 = - 6$ $A B = [- 3 15 10 - 6]$

Exercise 2.2

Compute the product of the row and column $[25 - 1] 3 - 2 4 .$

Recall Solution 2.2

$1 \times 3$ times $3 \times 1$ → answer is $1 \times 1$ . Just one dot product: $(2) (3) + (5) (- 2) + (- 1) (4) = 6 - 10 - 4 = - 8.$ Answer: $[- 8]$ . This is the smallest possible matrix product — the atom every bigger product is built from.

Exercise 2.3

$A$ is $2 \times 3$ , $B$ is $3 \times 2$ : $A = [1 - 1 0321], B = 401 1 - 2 1 .$ Find $A B$ .

Recall Solution 2.3

Answer is $2 \times 2$ . $C_{11} = (1) (4) + (0) (0) + (2) (1) = 4 + 0 + 2 = 6$ $C_{12} = (1) (1) + (0) (- 2) + (2) (1) = 1 + 0 + 2 = 3$ $C_{21} = (- 1) (4) + (3) (0) + (1) (1) = - 4 + 0 + 1 = - 3$ $C_{22} = (- 1) (1) + (3) (- 2) + (1) (1) = - 1 - 6 + 1 = - 6$ $A B = [6 - 3 3 - 6]$

Level 3 — Analysis

(Now you must reason about why, not just crank.)

Exercise 3.1

Using $A = [1324]$ and $B = [2102]$ , compute both $A B$ and $B A$ and confirm they differ. Then state in one sentence what geometric fact this demonstrates (Linear transformations).

Recall Solution 3.1

$A B = [(1) (2) + (2) (1) (3) (2) + (4) (1) (1) (0) + (2) (2) (3) (0) + (4) (2)] = [41048]$ $B A = [(2) (1) + (0) (3) (1) (1) + (2) (3) (2) (2) + (0) (4) (1) (2) + (2) (4)] = [27410]$ $A B \neq = B A$ . Meaning: matrices are transformations, and "do $B$ then $A$ " is generally a different combined transformation than "do $A$ then $B$ " — order of composition matters.

Exercise 3.2

Let $D = [0010]$ . Compute $D^{2} = D D$ . What does the answer tell you about "zero divisors" — can a non-zero matrix square to the zero matrix?

Recall Solution 3.2

$D^{2} = [(0) (0) + (1) (0) (0) (0) + (0) (0) (0) (1) + (1) (0) (0) (1) + (0) (0)] = [0000] .$ So a non-zero matrix can square to the zero matrix. With ordinary numbers, $x^{2} = 0 \Rightarrow x = 0$ ; that guarantee fails for matrices. Such $D$ is called nilpotent, and its existence is exactly why you cannot "cancel" matrices freely (unlike with invertible ones).

Exercise 3.3

Find a $2 \times 2$ matrix $A$ (not the zero matrix, not the identity) such that $A^{2} = A$ . Verify.

Recall Solution 3.3

A matrix with $A^{2} = A$ is a projection. Try $A = [1000] .$ $A^{2} = [(1) (1) + (0) (0) (0) (1) + (0) (0) (1) (0) + (0) (0) (0) (0) + (0) (0)] = [1000] = A . ✓$ Geometrically it flattens every vector onto the $x$ -axis; flattening an already-flat thing changes nothing, so applying it twice equals applying it once.

Level 4 — Synthesis

(Combine multiplication with other tools.)

Exercise 4.1

Verify $(A B)^{T} = B^{T} A^{T}$ (see Matrix transpose properties) for $A = [1021], B = [3410] .$ Here $M^{T}$ (transpose) means "flip rows and columns": the entry in row $i$ , col $j$ moves to row $j$ , col $i$ .

Recall Solution 4.1

First $A B$ : $A B = [(1) (3) + (2) (4) (0) (3) + (1) (4) (1) (1) + (2) (0) (0) (1) + (1) (0)] = [11410] .$ Transpose (flip across the diagonal): $(A B)^{T} = [11140] .$ Now the other side. $B^{T} = [3140]$ , $A^{T} = [1201]$ . $B^{T} A^{T} = [(3) (1) + (4) (2) (1) (1) + (0) (2) (3) (0) + (4) (1) (1) (0) + (0) (1)] = [11140] .$ They match. Notice the order flips: it is $B^{T} A^{T}$ , not $A^{T} B^{T}$ .

Exercise 4.2

Check that $det (A B) = det (A) det (B)$ (Determinants) for $A = [2113], B = [0 - 1 12],$ where for a $2 \times 2$ matrix $det [a c b d] = a d - b c$ .

Recall Solution 4.2

$det A = (2) (3) - (1) (1) = 5$ . $det B = (0) (2) - (1) (- 1) = 1$ . So the product of determinants is $5 \cdot 1 = 5$ . Now $A B$ : $A B = [(2) (0) + (1) (- 1) (1) (0) + (3) (- 1) (2) (1) + (1) (2) (1) (1) + (3) (2)] = [- 1 - 3 47] .$ $det (A B) = (- 1) (7) - (4) (- 3) = - 7 + 12 = 5.$ ✓ Equal, as promised. (Determinant is multiplicative — a rare place where order doesn't sting, because scalars commute.)

Exercise 4.3

Solve the system using matrix-vector multiplication (System of linear equations). Given $A = [21 1 - 1], x = [x y], b = [51],$ write out the two scalar equations $A x = b$ represents, then solve for $x, y$ .

Recall Solution 4.3

$A x$ means: row 1 of $A$ dotted with $x$ , then row 2 of $A$ dotted with $x$ . $[2 x + y x - y] = [51] \Rightarrow {2 x + y = 5 x - y = 1$ Add the two equations: $3 x = 6 \Rightarrow x = 2$ . Back-substitute: $2 - y = 1 \Rightarrow y = 1$ . Check: $2 (2) + 1 = 5$ ✓ and $2 - 1 = 1$ ✓. So $x = [21]$ .

Level 5 — Mastery

(One problem that leans on everything above.)

Exercise 5.1

Let $R = [01 - 1 0] (a 9 0^{\circ} rotation), S = [2001] (a horizontal stretch) .$ The unit vector $e = [10]$ points along the positive $x$ -axis.

(a) Compute $R S$ and $S R$ . Confirm $R S \neq = S R$ .
(b) Apply the transformation "stretch then rotate" to $e$ . Which matrix product represents this — $R S$ or $S R$ ? Compute the resulting vector.
(c) Apply "rotate then stretch" to $e$ and compute the result.
(d) In one sentence, explain the difference you see geometrically.

Recall Solution 5.1

(a) $R S = [(0) (2) + (- 1) (0) (1) (2) + (0) (0) (0) (0) + (- 1) (1) (1) (0) + (0) (1)] = [02 - 1 0]$ $S R = [(2) (0) + (0) (1) (0) (0) + (1) (1) (2) (- 1) + (0) (0) (0) (- 1) + (1) (0)] = [01 - 2 0]$ $R S \neq = S R$ . ✓

(b) "First $S$ , then $R$ " means apply $S$ nearest the vector: $R (S e) = (R S) e$ . So use $RS$ . Step 1 — stretch: $S e = [2001] [10] = [20]$ (arrow doubles along $x$ ). Step 2 — rotate that: $R [20] = [02]$ . Result: $[02]$ (points up, length 2). Same as $(R S) e$ : $[02 - 1 0] [10] = [02]$ . ✓

(c) "First $R$ , then $S$ " = $(S R) e$ . Step 1 — rotate: $R e = [01]$ (arrow now points up, still length 1). Step 2 — stretch horizontally: $S [01] = [01]$ (the horizontal stretch does nothing to a vertical arrow!). Result: $[01]$ . Same as $(S R) e = [01 - 2 0] [10] = [01]$ . ✓

(d) Stretching first lengthens the arrow while it still lies along $x$ , so the stretch bites (final length 2); rotating first turns the arrow off the $x$ -axis before the horizontal stretch, so the stretch has nothing to grab and the length stays 1. Same two operations, opposite order, different vector — that is non-commutativity, made visible.

[!recall]- One-line self-check before you leave

Which acts on

x

first in the product

A B C x

C

(rightmost, nearest the vector); then

B

; then

A

For

A_{4 \times 3}

and

B_{3 \times 6}

, size of

A B

4 \times 6

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

No — the order reverses:

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

Connections

Matrix multiplication — conditions, process, non-commutativity — the parent rules these drills exercise
Determinants — used in Exercise 4.2
Matrix transpose properties — used in Exercise 4.1
System of linear equations — used in Exercise 4.3
Linear transformations — the geometry behind Level 5
Inverse of a matrix — same order-reversal idea as transpose
Identity matrix — the "do nothing" transformation lurking behind $A^{2} = A$