LINEAR-ALGEBRA · Unit 1 · Video 4 · Interactive Practice

Matrix-Vector Multiplication: Two Ways to Compute Ax

IKey Formulas

Formula	Name	Description
$(Ax)_i = \text{row}_i(A) \cdot x$	Row (dot product) method	Each entry of $Ax$ is a dot product of a row of $A$ with $x$
$Ax = x_1 \mathbf{a}_1 + x_2 \mathbf{a}_2 + \cdots$	Column (linear combination) method	$Ax$ is a weighted sum of the columns of $A$
$\mathbf{a} \cdot \mathbf{b} = a_1 b_1 + a_2 b_2$	Dot product	Multiply corresponding entries and sum
Column space of $A$ $= \{Ax : x \in \mathbb{R}^n\}$	Column space	The set of all possible outputs $Ax$

IIVisualization 1 — Two Groupings, One Answer

The row method groups the same arithmetic horizontally, the column method vertically — the result is identical.

a₁₁ 2 a₁₂ 5 a₂₁ 1 a₂₂ 3 x₁ 1 x₂ 2

Row method · dot products

Column method · linear combination

IIIVisualization 2 — Building Ax from Columns

$Ax$ is the head-to-tail sum of each column of $A$ scaled by its weight $x_1, x_2$ .

x₁ (weight on col₁) 1 x₂ (weight on col₂) 2

💡 Bridge: because these two columns aren't parallel, the right weights can reach every point in the plane — that reachable set is the column space explored next.

IVVisualization 3 — The Column Space

The outputs $Ax$ fill whatever the columns of $A$ span — the whole plane when they're independent, a line when they're parallel.

col₁ x 2 col₁ y 1 col₂ x 5 col₂ y 3

VQuiz Questions

Question 1 · Row Method Calculation

Using the row (dot product) method, compute the first entry of $Ax$ where:

$A = \begin{bmatrix} 3 & -2 \\ 4 & 1 \end{bmatrix}, \quad x = \begin{bmatrix} 2 \\ 5 \end{bmatrix}$

✅ Correct! Row 1 · x = 3(2) + (−2)(5) = 6 − 10 = −4.

❌ Not quite. Use row 1 of A: multiply 3 by 2 and (−2) by 5, then add the products.

Show solution

Solution:

The first entry of $Ax$ is the dot product of row 1 of $A$ with $x$ :

$\text{Row 1} \cdot x = (3)(2) + (-2)(5) = 6 + (-10) = 6 - 10 = -4$

The answer is $\mathbf{-4}$ .

Common mistakes:

Getting 16 by computing $3(2) + 2(5) = 16$ (forgetting the negative sign on $-2$ )
Getting 11 by computing $3(2) + 1(5) = 11$ (mixing entries from different rows)

Question 2 · Column Method Expression

Using the column (linear combination) method, which expression correctly represents $Ax$ for:

$A = \begin{bmatrix} 3 & -2 \\ 4 & 1 \end{bmatrix}, \quad x = \begin{bmatrix} 2 \\ 5 \end{bmatrix}$

✅ Correct! We weight each column of A by the corresponding entry of x.

❌ Not quite. Remember: extract the columns (vertical slices) of A, then weight by entries of x.

Show solution

Solution:

The column method says:

$Ax = x_1 \cdot (\text{col 1}) + x_2 \cdot (\text{col 2})$

The columns of $A$ are vertical slices:

Column 1 = $\begin{bmatrix} 3 \\ 4 \end{bmatrix}$
Column 2 = $\begin{bmatrix} -2 \\ 1 \end{bmatrix}$

With $x_1 = 2$ and $x_2 = 5$ :

$Ax = 2 \begin{bmatrix} 3 \\ 4 \end{bmatrix} + 5 \begin{bmatrix} -2 \\ 1 \end{bmatrix}$

Written compactly: 2 · [3, 4] + 5 · [−2, 1]

Why the others are wrong:

"2 · [3, −2] + 5 · [4, 1]" uses rows instead of columns
"3 · [2, 5] + (−2) · [2, 5]" doesn't extract columns of A at all
"[3(2)+(−2)(5), 4(2)+1(5)]" computes the result entry-by-entry — that's the row method

Question 3 · Column Space (True or False)

True or False: If the two columns of a $2 \times 2$ matrix $A$ point in the same direction (one is a scalar multiple of the other), then the column space of $A$ is all of $\mathbb{R}^2$ .

✅ Correct! Parallel (dependent) columns can only produce vectors along a single line, not all of R².

❌ Not quite. Think about what happens when you combine two parallel vectors — can you reach every direction?

Show solution

Solution: False.

If one column is a scalar multiple of the other (e.g., $\text{col}_2 = k \cdot \text{col}_1$ ), then every linear combination:

$x_1 \cdot \text{col}_1 + x_2 \cdot \text{col}_2 = x_1 \cdot \text{col}_1 + x_2 k \cdot \text{col}_1 = (x_1 + k x_2) \cdot \text{col}_1$

is just a scalar multiple of $\text{col}_1$ . The column space is a line through the origin, not all of $\mathbb{R}^2$ .

To fill all of $\mathbb{R}^2$ , the two columns must be linearly independent — pointing in genuinely different directions. You can verify this in Visualization 3 above by making the columns parallel!

Question 4 · Interpreting Ax = b

The equation $Ax = b$ can be reinterpreted using the column method. Which statement best describes what $Ax = b$ is really asking?

✅ Correct! The equation Ax = b asks whether b can be built from A's columns — the central idea of the column method.

❌ Not quite. Remember: the column method says Ax is always a linear combination of A's columns. So Ax = b asks whether b is one of those combinations.

Show solution

Solution: "Is b a linear combination of the columns of A?"

The column method rewrites $Ax = b$ as:

$x_1 \mathbf{a}_1 + x_2 \mathbf{a}_2 + \cdots + x_n \mathbf{a}_n = b$

This asks: can we find weights $x_1, x_2, \ldots, x_n$ such that a linear combination of $A$ 's columns equals $b$ ?

Equivalently: is $b$ in the column space of $A$ ?

This reframe is the key insight from the video — it powers everything ahead in linear algebra:

Solving systems: finding the right weights $x$
Existence of solutions: does $b$ lie in the column space?
Linear transformations: $A$ maps $x$ to a combination of its columns

Every time you see $Ax$ , think: which combination of A's columns am I making?

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