LINEAR-ALGEBRA · Unit 1 · Video 2 · Interactive Practice

Systems of Equations in 3D: Row Picture vs Column Picture

IKey Formulas

Concept	Formula	Description
Matrix equation	$A\mathbf{x} = \mathbf{b}$	Compact form of a linear system
Row picture (3D)	Each equation $\rightarrow$ a plane	Solution = intersection of all planes
Column picture	$x\mathbf{c}_1 + y\mathbf{c}_2 + z\mathbf{c}_3 = \mathbf{b}$	Find scalars to combine columns into $\mathbf{b}$
Linear combination	$\alpha_1\mathbf{v}_1 + \alpha_2\mathbf{v}_2 + \cdots + \alpha_n\mathbf{v}_n$	Sum of scaled vectors

IIVisualization 1 — The Row Picture

Each equation is a plane; stacking all three narrows the solution from a plane, to a line, to one point.

Equations shown

💡 The row picture stops scaling past three unknowns — in four variables each equation is a 3-D hyperplane nobody can draw, which is exactly the gap the column picture closes.

IIIVisualization 2 — The Column Picture

Which scalars $x, y, z$ scale the columns so they add tip-to-tail onto the target $\mathbf{b} = [0, -1, 4]$ ?

x · c₁ 1/2 y · c₂ 1/2 z · c₃ 1/2

💡 Challenge: reach $\mathbf{b}$ with a single nonzero scalar — the column picture makes that shortcut obvious.

IVVisualization 3 — Column Picture, Any Target

Swap the target $\mathbf{b}$ and the same tip-to-tail process lands on a different solution $(x, y, z)$ .

Target b

💡 Whether the columns have 3 entries or 3000, the question never changes — can they combine to reach $\mathbf{b}$ ? — and that dimension-independence is why the column picture underlies modern linear algebra.

VQuiz Questions

Question 1

In the row picture of a system of 3 equations with 3 unknowns, each equation represents what geometric object in 3D space?

✅ Correct! Each linear equation in 3 unknowns defines a plane in 3D.

❌ Not quite. Think about how many degrees of freedom one equation in three unknowns leaves.

Show solution

Solution:

One linear equation in three unknowns has two free variables, so its solution set is a 2-dimensional surface in 3D — a plane.

In 2D: 1 equation with 2 unknowns → a line (1D)
In 3D: 1 equation with 3 unknowns → a plane (2D)
In nD: 1 equation with n unknowns → a hyperplane ((n-1)D)

Question 2

For $A = \begin{bmatrix}2 & -1 & 0\\-1 & 2 & -1\\0 & -3 & 4\end{bmatrix}$ and $\mathbf{b} = \begin{bmatrix}0\\-1\\4\end{bmatrix}$ , the column picture reveals the solution immediately because $\mathbf{b}$ is identical to which column of $A$ ?

✅ Correct! b = [0, -1, 4] matches column 3, giving the solution (0, 0, 1).

❌ Not quite. The column sums are [2-1+0, -1+2-1, 0-3+4] = [1, 0, 1], which does not equal b. Compare b to each column individually.

❌ Not quite. Compare b = [0, -1, 4] to each column component by component.

Show solution

Solution:

Compare $\mathbf{b} = [0, -1, 4]$ to each column of $A$ :

Column 1: $[2, -1, 0]$ — doesn't match
Column 2: $[-1, 2, -3]$ — doesn't match
Column 3: $[0, -1, 4]$ — perfect match! ✓

Since $\mathbf{b}$ equals column 3, the solution is: $0 \cdot \mathbf{c}_1 + 0 \cdot \mathbf{c}_2 + 1 \cdot \mathbf{c}_3 = \mathbf{b} \implies (x, y, z) = (0, 0, 1)$

Question 3

True or False: The row picture (visualizing intersecting geometric objects) can be used effectively for systems with 10 unknowns.

✅ Correct! We can only draw in 2D and 3D. The row picture is impossible to visualize in higher dimensions.

❌ Not quite. Can you draw a 9-dimensional hyperplane? The row picture requires exactly that in 10D.

Show solution

Answer: False

The row picture requires drawing geometric objects in $n$ -dimensional space:

Dimension	Row Picture Object	Visualizable?
2D	Lines	Yes
3D	Planes	Yes
4D	3D hyperplanes	No
10D	9D hyperplanes	No

Humans can only visualize up to 3 dimensions. The column picture, by contrast, always asks: "Can I combine these vectors to reach b?" — that concept works identically in any dimension.

Question 4

Using the same matrix $A = \begin{bmatrix}2 & -1 & 0\\-1 & 2 & -1\\0 & -3 & 4\end{bmatrix}$ , if the target changes to $\mathbf{b} = [1, 1, -3]$ , what is the solution $(x, y, z)$ ?

Hint: try adding column 1 and column 2 of $A$ component by component.

✅ Correct! Column 1 + Column 2 = [1, 1, -3] = b, so (x, y, z) = (1, 1, 0).

❌ That was the solution for b = [0, -1, 4]. This question uses b = [1, 1, -3]. Try adding columns 1 and 2.

❌ Not quite. Try adding column 1 and column 2 component by component and compare to b.

Show solution

Solution:

Add column 1 and column 2 of $A$ : $\begin{bmatrix}2\\-1\\0\end{bmatrix} + \begin{bmatrix}-1\\2\\-3\end{bmatrix} = \begin{bmatrix}2+(-1)\\-1+2\\0+(-3)\end{bmatrix} = \begin{bmatrix}1\\1\\-3\end{bmatrix} = \mathbf{b} \;\checkmark$

So $x = 1$ , $y = 1$ , $z = 0$ : one copy of column 1 plus one copy of column 2, zero copies of column 3.

Verification: $A\begin{bmatrix}1\\1\\0\end{bmatrix} = \begin{bmatrix}2(1)+(-1)(1)+0(0)\\(-1)(1)+2(1)+(-1)(0)\\0(1)+(-3)(1)+4(0)\end{bmatrix} = \begin{bmatrix}1\\1\\-3\end{bmatrix}$ ✓

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