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

Three Ways to See Linear Equations: Row Picture, Column Picture & Matrix Form

IKey Formulas

Formula	Name	Description
$A\mathbf{x} = \mathbf{b}$	Matrix form	Coefficient matrix times unknowns equals right-hand side
$x\,\mathbf{col}_1 + y\,\mathbf{col}_2 = \mathbf{b}$	Column picture	Linear combination of columns of $A$ equals target $\mathbf{b}$
$c_1\mathbf{v}_1 + c_2\mathbf{v}_2 + \cdots + c_n\mathbf{v}_n$	Linear combination	Scalars times vectors, added together — the central operation

IIVisualization 1 — The Row Picture

Each equation is a line; the solution is the single point where $2x - y = 0$ and $-x + 2y = 3$ cross.

x 0 y 0

IIIVisualization 2 — The Column Picture

Which multiples of the columns $(2, -1)$ and $(-1, 2)$ add up tip-to-tail to the target $\mathbf{b} = (0, 3)$ ?

x (col 1) 0 y (col 2) 0

💡 Every target is reachable this way — unless the two columns point along the same line, which Visualization 3 explores.

IVVisualization 3 — Reachability & Span

When do combinations of the two columns fill the whole plane, and when do they collapse onto a line?

col 2 angle 115°

💡 Challenge: rotate column 2 until the sample points collapse onto a single line — then $A\mathbf{x} = \mathbf{b}$ has no solution for most targets.

VQuiz Questions

Question 1

In the row picture of a system of two linear equations in two unknowns ( $x$ and $y$ ), what does each equation represent geometrically in 2D?

✅ Correct! Each linear equation in two unknowns traces out a line. The solution is where the lines intersect.

❌ Not quite. Think about all the $(x, y)$ pairs that satisfy a single equation like $2x - y = 0$ . What shape do they form?

Show solution

Solution:

In 2D (two unknowns), each linear equation defines a line in the $xy$ -plane.

For our system:

$2x - y = 0$ is the line $y = 2x$
$-x + 2y = 3$ is the line $y = \frac{x + 3}{2}$

The solution to the system is the intersection point of these lines: $(1, 2)$ .

Note: In 3D (three unknowns), each equation would instead define a plane.

Question 2

The column picture rewrites our system as a linear combination:

$x \begin{bmatrix} 2 \\ -1 \end{bmatrix} + y \begin{bmatrix} -1 \\ 2 \end{bmatrix} = \begin{bmatrix} 0 \\ 3 \end{bmatrix}$

What values of $x$ and $y$ solve this equation?

✅ Correct! One copy of column 1 plus two copies of column 2 hits the target exactly.

❌ Not quite. Try plugging your values in: compute $x \cdot [2, -1]^T + y \cdot [-1, 2]^T$ and check if you get $[0, 3]^T$ .

Show solution

Solution:

Check $x = 1, \; y = 2$ :

$1 \begin{bmatrix} 2 \\ -1 \end{bmatrix} + 2 \begin{bmatrix} -1 \\ 2 \end{bmatrix} = \begin{bmatrix} 2 \\ -1 \end{bmatrix} + \begin{bmatrix} -2 \\ 4 \end{bmatrix} = \begin{bmatrix} 0 \\ 3 \end{bmatrix} \; \checkmark$

Why not the others?

$x=2, y=1$ : gives $\begin{bmatrix}4\\-2\end{bmatrix} + \begin{bmatrix}-1\\2\end{bmatrix} = \begin{bmatrix}3\\0\end{bmatrix} \neq \begin{bmatrix}0\\3\end{bmatrix}$
$x=0, y=\tfrac{3}{2}$ : gives $\begin{bmatrix}0\\0\end{bmatrix} + \begin{bmatrix}-3/2\\3\end{bmatrix} = \begin{bmatrix}-3/2\\3\end{bmatrix} \neq \begin{bmatrix}0\\3\end{bmatrix}$
$x=3, y=0$ : gives $\begin{bmatrix}6\\-3\end{bmatrix} + \begin{bmatrix}0\\0\end{bmatrix} = \begin{bmatrix}6\\-3\end{bmatrix} \neq \begin{bmatrix}0\\3\end{bmatrix}$

Question 3

True or False: The matrix-vector product $A\mathbf{x}$ computes a linear combination of the rows of $A$ , weighted by the entries of $\mathbf{x}$ .

✅ Correct! $A\mathbf{x}$ is a linear combination of the columns of $A$ , not the rows. Each entry of $\mathbf{x}$ scales the corresponding column.

❌ Not quite. Look carefully at how $A\mathbf{x}$ splits apart: each column of $A$ gets multiplied by the corresponding entry of $\mathbf{x}$ .

Show solution

Solution: False

$A\mathbf{x}$ is a linear combination of the columns of $A$ (not the rows), weighted by the entries of $\mathbf{x}$ :

$\begin{bmatrix} 2 & -1 \\ -1 & 2 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = x \underbrace{\begin{bmatrix} 2 \\ -1 \end{bmatrix}}_{\text{column 1}} + \; y \underbrace{\begin{bmatrix} -1 \\ 2 \end{bmatrix}}_{\text{column 2}}$

This is one of the central insights of the video: "When you multiply a matrix by a vector, you're combining columns." The matrix form $A\mathbf{x} = \mathbf{b}$ is the column picture written compactly.

Question 4

Suppose the two columns of a $2 \times 2$ matrix $A$ point in the same direction (they are parallel). What can you say about the system $A\mathbf{x} = \mathbf{b}$ ?

✅ Correct! Parallel (dependent) columns only span a line, so most targets in $\mathbb{R}^2$ are unreachable.

❌ Not quite. Think about Visualization 3: when the columns are parallel, their combinations only cover a line — not the whole plane.

Show solution

Solution: Some vectors $\mathbf{b}$ in $\mathbb{R}^2$ cannot be reached.

When both columns point in the same direction, every linear combination

$x\,\mathbf{col}_1 + y\,\mathbf{col}_2$

lies on a single line through the origin (since both columns are parallel, any weighted sum stays in that direction).

A line is only a 1-dimensional subset of $\mathbb{R}^2$ . Most target vectors $\mathbf{b}$ are not on that line and therefore cannot be reached — the system has no solution for those $\mathbf{b}$ .

"Always exactly one solution" is wrong because for reachable $\mathbf{b}$ there are infinitely many solutions.
"Always infinitely many solutions" is wrong because it ignores the unreachable targets.

This is the concept of linear dependence: dependent columns cannot span all of $\mathbb{R}^2$ .

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