Multivariable-Calculus · Unit 3 · Video 3 · Interactive Practice

Rows Dot Columns: Packaging a Change of Coordinates as AX = U

IKey Formulas

Formula	Name	What you need
$u_i = a_{i1}x_1 + a_{i2}x_2 + a_{i3}x_3$	One line of the system	The coefficients of the $i$ th equation
$AX = U$	Matrix form	Coefficient table $A$ , input column $X$
$(AB)_{ij} = a_{i1}b_{1j} + a_{i2}b_{2j} + \cdots + a_{in}b_{nj}$	Entry of a product	Row $i$ of $A$ , column $j$ of $B$
$(m \times n)(n \times p) = m \times p$	Shape rule	Width of $A$ must equal height of $B$

Key Insight: One row, one column, one dot product, one entry — every other rule about matrix products follows from that one.

IIVisualization 1 — One Row Makes One Output

Each line of the system is one row of $A$ dotted with the column $X$ .

x₁ 1 x₂ 2 x₃ 1 Output

IIIVisualization 2 — Which Row, Which Column?

An entry's position names its ingredients: the row to its left, the column above it.

💡 Order is not negotiable: $BA$ would pair $B$ 's rows with $A$ 's columns, and with these shapes it does not even exist.

IVVisualization 3 — When Does $AB$ Exist?

A dot product needs two lists of the same length, and that single fact decides everything.

rows of A 3 cols of A 4 rows of B 4 cols of B 2

VQuiz Questions

Problem 1 · One Entry of a Product

Given: the first row of $A$ is $\begin{pmatrix} 2 & -1 & 3 \end{pmatrix}$ and the second column of $B$ is $\begin{pmatrix} 4 \\ 6 \\ 5 \end{pmatrix}$ — find the entry $(AB)_{12}$ .

✅ Correct!

2(4) + (-1)(6) + 3(5) = 8 - 6 + 15 = 17

, and it sits in row 1, column 2 because it used row 1 of

A

and column 2 of

B

❌ Close, but check the sign. The second entry of the row is

-1

, so that term is

-6

, not

+6

❌ Not quite. The two lists pair up in order — first with first, second with second, third with third — not front to back.

❌ Not quite. A dot product multiplies the matched entries first and adds the products; it is not the sum of the six numbers.

Show solution

The entry in row $1$ , column $2$ of $AB$ is row $1$ of $A$ dotted with column $2$ of $B$ :

(AB)_{12} = 2(4) + (-1)(6) + 3(5) = 8 - 6 + 15 = 17

Both lists have three entries, so every term of the row has a partner in the column — which is exactly the condition that lets the product exist.

Problem 2 · Packing a System into $AX = U$

Given: the system $u_1 = 4x_2 + 2x_1$ , $\ u_2 = x_3 - 3x_1$ , $\ u_3 = x_1 + 5x_3$ — find the matrix $A$ for which $AX = U$ .

✅ Correct! Row

i

holds the coefficients of

x_1, x_2, x_3

in equation

i

, in that order, with

0

wherever a variable is absent.

❌ Not quite. A column of

A

belongs to one variable, so the coefficients go in the order

x_1, x_2, x_3

— not in the order the terms happen to be written.

❌ Not quite. That is the table with rows and columns swapped: row

i

must reproduce equation

i

when dotted with

X

, so equation 2 belongs in row 2, not column 2.

❌ Close, but check a sign. In

u_2 = x_3 - 3x_1

the coefficient of

x_1

-3

Show solution

Rewrite each equation in the order $x_1, x_2, x_3$ with every coefficient written out — including the $0$ s and $1$ s that are usually left implicit:

u_1 = 2x_1 + 4x_2 + 0x_3, \qquad u_2 = -3x_1 + 0x_2 + 1x_3, \qquad u_3 = 1x_1 + 0x_2 + 5x_3

Lifting the coefficients out, keeping their positions:

A = \begin{pmatrix} 2 & 4 & 0 \\ -3 & 0 & 1 \\ 1 & 0 & 5 \end{pmatrix}

Check row 2: $(-3)x_1 + 0x_2 + 1x_3 = x_3 - 3x_1 = u_2$ ✓

Problem 3 · Shapes First

Given: $A$ is a $2 \times 5$ matrix and $B$ is a $5 \times 3$ matrix — find the size of each product, or say it does not exist.

Size of $AB$

Size of $BA$

✅ Correct!

(2 \times 5)(5 \times 3)

has matching inner numbers and gives

2 \times 3

;

(5 \times 3)(2 \times 5)

does not, so

BA

is undefined.

❌ The two sizes are the wrong way round. The answer inherits the height of the left matrix and the width of the right one.

❌ Check the inner numbers. A row of

A

has

5

entries and a column of

B

has

5

entries, so the dot products do exist.

❌ Check the inner numbers. In

BA

a row of

B

has

3

entries while a column of

A

has only

2

Show solution

$AB$ : line up the shapes as $(2 \times 5)(5 \times 3)$ . The inner numbers are $5$ and $5$ , so every row of $A$ and every column of $B$ are lists of the same length and the product exists. Its size is the outer pair:

(2 \times 5)(5 \times 3) = 2 \times 3

$BA$ : now the shapes line up as $(5 \times 3)(2 \times 5)$ . A row of $B$ has $3$ entries, a column of $A$ has $2$ ; the third entry of the row has nothing to pair with, so $BA$ does not exist.

Reversing the factors changes the question entirely — matrix multiplication is not commutative, and the two products need not even have the same shape.

Problem 4 · The Change of Coordinates (Video Example)

Given: the video's matrix $A = \begin{pmatrix} 2 & 3 & 3 \\ 2 & 4 & 5 \\ 1 & 1 & 2 \end{pmatrix}$ and the point with old coordinates $X = \begin{pmatrix} 2 \\ 1 \\ -1 \end{pmatrix}$ — find the new coordinates $U = AX$ .

✅ Correct! Three rows, three dot products, three new coordinates — all three equations at once.

❌ Close, but check the sign. The third old coordinate is

-1

, so each third term subtracts:

3(-1) = -3

5(-1) = -5

2(-1) = -2

❌ Not quite. Those are the columns of

A

dotted with

X

. Each entry of

AX

uses a row of

A

— the coefficients of one equation.

❌ Not quite. Each output uses the entire row and the entire column: all three terms, not just the

x_1

term.

Show solution

Dot each row of $A$ with the column $X = (2,\, 1,\, -1)$ :

u_1 = 2(2) + 3(1) + 3(-1) = 4 + 3 - 3 = 4

u_2 = 2(2) + 4(1) + 5(-1) = 4 + 4 - 5 = 3

u_3 = 1(2) + 1(1) + 2(-1) = 2 + 1 - 2 = 1

So $U = \begin{pmatrix} 4 \\ 3 \\ 1 \end{pmatrix}$ : the same point, now measured in the new axes.

The shapes agree with the rule: $(3 \times 3)(3 \times 1) = 3 \times 1$ , a column of three new coordinates.

Solved: 0 / 4

Rows Dot Columns: Packaging a Change of Coordinates as AX = U

IKey Formulas

IIVisualization 1 — One Row Makes One Output

IIIVisualization 2 — Which Row, Which Column?

IVVisualization 3 — When Does ABABAB Exist?

VQuiz Questions

IVVisualization 3 — When Does $AB$ Exist?