Linear-Algebra-Done-Right · Unit 6 · Video 2 · Interactive Practice

Matrix Operations and Why Multiplication Is Row Times Column

IKey Formulas

Formula	Name	Meaning
$(A + C)_{j,k} = A_{j,k} + C_{j,k}$ , $\ (\lambda A)_{j,k} = \lambda A_{j,k}$	Entrywise sum and scalar multiple	Same size only; they give $\mathcal{M}(S + T) = \mathcal{M}(S) + \mathcal{M}(T)$ and $\mathcal{M}(\lambda T) = \lambda \mathcal{M}(T)$
$\dim \mathbf{F}^{m,n} = mn$	The space of $m \times n$ matrices	Basis: the $mn$ matrices $E_{j,k}$ with a single $1$ in row $j$ , column $k$ ; $A = \sum_{j,k} A_{j,k} E_{j,k}$
$(AB)_{j,k} = \sum_{r=1}^{n} A_{j,r} B_{r,k}$	Matrix product	$A$ is $m \times n$ , $B$ is $n \times p$ , $AB$ is $m \times p$ : row $j$ of $A$ times column $k$ of $B$
$\mathcal{M}(ST) = \mathcal{M}(S)\,\mathcal{M}(T)$	Composition becomes multiplication	$T \in \mathcal{L}(U, V)$ , $S \in \mathcal{L}(V, W)$ , one basis for each of $U$ , $V$ , $W$

Key Insight: The row-times-column rule is forced, not chosen: it is the only product that makes $\mathcal{M}(ST) = \mathcal{M}(S)\mathcal{M}(T)$ true, and the inner size $n$ that must match is $\dim V$ , the space in the middle.

IIAdding and Scaling Entry by Entry

Each entry of $\lambda A + C$ uses only the entries of $A$ and $C$ in the same position.

λ 2 Entry

IIIRow Times Column

Entry $(j, k)$ of $AB$ pairs row $j$ of $A$ with column $k$ of $B$ , so the inner sizes must agree.

Order Entries computed 1

IVComposition Becomes Multiplication

Applying $T$ then $S$ moves $u$ exactly as $AB = \mathcal{M}(S)\mathcal{M}(T)$ does; the other order uses $BA$ .

Pair

💡 Order aside, matrix multiplication keeps the laws that products of linear maps have: $(AB)C = A(BC)$ , $A(B + C) = AB + AC$ and $(A + B)C = AC + BC$ .

VQuiz Questions

Problem 1 · A Linear Combination of Matrices

Given: $A = \begin{pmatrix} 1 & -2 \\ 3 & 0 \end{pmatrix}$ and $C = \begin{pmatrix} 4 & 1 \\ -1 & 5 \end{pmatrix}$ . Find $3A - C$ .

✅ Correct! Each entry is

3A_{j,k} - C_{j,k}

; for example the bottom-left entry is

3 \cdot 3 - (-1) = 10

❌ Not quite. That is

3A + C

. Subtracting

C

means adding

(-1)C

, so every entry of

C

changes sign.

❌ Not quite. That is

3(A - C) = 3A - 3C

. The scalar

3

multiplies only

A

❌ Not quite. That is

A - 3C

: the

3

was applied to the wrong matrix.

Show solution

Scalar multiplication multiplies every entry, and $3A - C = 3A + (-1)C$ is computed entry by entry:

3A = \begin{pmatrix} 3 & -6 \\ 9 & 0 \end{pmatrix}, \qquad 3A - C = \begin{pmatrix} 3 - 4 & -6 - 1 \\ 9 - (-1) & 0 - 5 \end{pmatrix} = \begin{pmatrix} -1 & -7 \\ 10 & -5 \end{pmatrix}

Problem 2 · Sizes Decide What Exists

Given: $A \in \mathbf{F}^{2,5}$ , $B \in \mathbf{F}^{5,3}$ and $C \in \mathbf{F}^{3,2}$ .

Which product is NOT defined?

What is the size of $ABC$ ?

What is $\dim \mathbf{F}^{2,5}$ , the space containing $A$ ?

✅ Correct!

BA

fails because

B

has

3

columns and

A

has

2

rows;

ABC

2 \times 2

; and

\dim \mathbf{F}^{2,5} = 2 \cdot 5 = 10

❌ Not quite.

C

3 \times 2

and

A

2 \times 5

: the inner sizes are both

2

, so

CA

exists and is

3 \times 5

❌ Not quite.

B

5 \times 3

and

C

3 \times 2

: the inner sizes are both

3

, so

BC

exists and is

5 \times 2

❌ Not quite.

A

2 \times 5

and

B

5 \times 3

: the inner sizes are both

5

, so

AB

exists and is

2 \times 3

❌ Not quite.

2 \times 3

is the size of

AB

; multiplying by the

3 \times 2

matrix

C

changes the column count to

2

❌ Not quite.

5 \times 3

is the size of

B

. A product keeps the OUTER sizes: rows from the first factor, columns from the last.

❌ Not quite.

3 \times 2

is the size of

C

alone; the row count of

ABC

comes from

A

❌ Not quite.

2 + 5

counts rows plus columns, but the basis has one matrix

E_{j,k}

per position, and there are

2 \cdot 5

positions.

❌ Not quite.

5

is the number of columns. Each of the

2 \cdot 5

positions carries its own basis matrix

E_{j,k}

❌ Not quite.

25 = 5 \cdot 5

would be

\dim \mathbf{F}^{5,5}

; a

2 \times 5

matrix has

2 \cdot 5

positions.

Show solution

A product of an $m \times n$ and a $q \times p$ matrix is defined exactly when $n = q$ , because the index $r$ in $\sum_{r} A_{j,r}B_{r,k}$ runs over the columns of the first and the rows of the second at once.

$AB$ : $(2 \times 5)(5 \times 3)$ , defined, $2 \times 3$ . $\ BC$ : $(5 \times 3)(3 \times 2)$ , defined, $5 \times 2$ . $\ CA$ : $(3 \times 2)(2 \times 5)$ , defined, $3 \times 5$ .
$BA$ : $(5 \times 3)(2 \times 5)$ , and $3 \ne 2$ , so $BA$ is not defined.
$ABC = (AB)C$ : $(2 \times 3)(3 \times 2)$ , which is $2 \times 2$ .
$\mathbf{F}^{2,5}$ has the basis $E_{j,k}$ , one for each of the $2 \cdot 5$ positions, so its dimension is $10$ .

Problem 3 · Two Orders, Two Entries

Given: $A = \begin{pmatrix} 1 & 2 & 0 \\ -1 & 3 & 1 \end{pmatrix}$ and $B = \begin{pmatrix} 2 & 1 \\ 0 & -1 \\ 4 & 2 \end{pmatrix}$ . Find $(AB)_{2,1}$ and $(BA)_{3,2}$ .

What is $(AB)_{2,1}$ ?

What is $(BA)_{3,2}$ ?

✅ Correct!

(AB)_{2,1} = (-1)(2) + 3 \cdot 0 + 1 \cdot 4 = 2

and

(BA)_{3,2} = 4 \cdot 2 + 2 \cdot 3 = 14

❌ Not quite. Check the sign of the first product:

A_{2,1}B_{1,1} = (-1)(2) = -2

, not

2

❌ Not quite.

-1

(AB)_{1,2}

, row

1

against column

2

. Entry

(2, 1)

uses row

2

A

and column

1

B

❌ Not quite.

1

(BA)_{2,1}

: it uses row

2

B

and column

1

A

. In

AB

the rows come from

A

❌ Not quite.

2

(BA)_{3,3}

; column

2

A

(2, 3)

, not

(0, 1)

❌ Not quite.

AB

2 \times 2

, but

BA

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

, a

3 \times 3

matrix, so its

(3, 2)

entry exists.

❌ Not quite.

-1

(BA)_{2,3}

, with the indices swapped. Entry

(3, 2)

uses row

3

B

and column

2

A

Show solution

$A$ is $2 \times 3$ and $B$ is $3 \times 2$ , so $AB$ is $2 \times 2$ and $BA$ is $3 \times 3$ . Entry $(j, k)$ is row $j$ of the first factor times column $k$ of the second:

(AB)_{2,1} = \sum_{r=1}^{3} A_{2,r}B_{r,1} = (-1)(2) + (3)(0) + (1)(4) = 2

(BA)_{3,2} = \sum_{r=1}^{2} B_{3,r}A_{r,2} = (4)(2) + (2)(3) = 14

In full, $AB = \begin{pmatrix} 2 & -1 \\ 2 & -2 \end{pmatrix}$ and $BA = \begin{pmatrix} 1 & 7 & 1 \\ 1 & -3 & -1 \\ 2 & 14 & 2 \end{pmatrix}$ : the two orders do not even have the same size.

Problem 4 · The Matrix of a Composition

Given: $T \in \mathcal{L}(\mathbf{F}^2, \mathbf{F}^3)$ and $S \in \mathcal{L}(\mathbf{F}^3, \mathbf{F}^2)$ with $T(x, y) = (x + y,\ 2y,\ x)$ and $S(x, y, z) = (z,\ x - y)$ , standard bases throughout. Find $\mathcal{M}(ST)$ .

✅ Correct!

\mathcal{M}(S)\mathcal{M}(T)

(2 \times 3)(3 \times 2) = 2 \times 2

, and it matches

(ST)(x, y) = (x,\ x - y)

computed directly.

❌ Not quite. Column

k

of a matrix holds the coordinates of the image of the

k

-th basis vector:

(ST)(1, 0) = (1, 1)

is the FIRST COLUMN, not the first row.

❌ Not quite. That

3 \times 3

matrix is

\mathcal{M}(T)\mathcal{M}(S) = \mathcal{M}(TS)

. In

ST

the map

T

acts first, and

ST

maps

\mathbf{F}^2

\mathbf{F}^2

, so its matrix is

2 \times 2

❌ Not quite. The second coordinate of

S(T(x, y))

(x + y) - 2y = x - y

; the

y

inside

x + y

was dropped.

Show solution

Column $k$ of each matrix records the image of the $k$ -th standard basis vector:

\mathcal{M}(S) = \begin{pmatrix} 0 & 0 & 1 \\ 1 & -1 & 0 \end{pmatrix}, \qquad \mathcal{M}(T) = \begin{pmatrix} 1 & 1 \\ 0 & 2 \\ 1 & 0 \end{pmatrix}

Row times column:

\mathcal{M}(S)\mathcal{M}(T) = \begin{pmatrix} 0 + 0 + 1 & 0 + 0 + 0 \\ 1 - 0 + 0 & 1 - 2 + 0 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 1 & -1 \end{pmatrix}

Check directly: $(ST)(x, y) = S(x + y,\ 2y,\ x) = (x,\ x - y)$ , so $(ST)(1, 0) = (1, 1)$ and $(ST)(0, 1) = (0, -1)$ are the two columns, as $\mathcal{M}(ST) = \mathcal{M}(S)\mathcal{M}(T)$ promises.

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