LINEAR-ALGEBRA · Unit 8 · Video 1 · Interactive Practice

When Does Ax = b Have a Solution?

IKey Formulas

Concept	Formula / Rule	Meaning
Column space	$C(A) = \{A\mathbf{x} : \mathbf{x} \in \mathbb{R}^n\}$	All vectors $\mathbf{b}$ reachable as $A\mathbf{x}$
Solvability	$A\mathbf{x} = \mathbf{b}$ solvable $\iff \mathbf{b} \in C(A)$	$\mathbf{b}$ must lie in the column space
Row test	If $c_1 R_1 + c_2 R_2 + \cdots = \mathbf{0}$ , then $c_1 b_1 + c_2 b_2 + \cdots = 0$	Row dependencies constrain $\mathbf{b}$
Contradiction	$[0 \; 0 \; \cdots \; 0 \mid d]$ with $d \neq 0$	Zero row with nonzero RHS → no solution

IIVisualization 1 — Landing in the Column Space

$A\mathbf{x} = \mathbf{b}$ is solvable exactly when $\mathbf{b}$ lands on the column space — here the line $b_3 = b_1 + b_2$ .

b₁+b₂ 6 b₃ 6

💡 The condition $b_1 + b_2 = b_3$ comes from $A = \begin{bmatrix} 1 & 2 & 2 & 2 \\ 2 & 4 & 6 & 8 \\ 3 & 6 & 8 & 10 \end{bmatrix}$ , whose rows satisfy $R_1 + R_2 = R_3$ — so $C(A)$ is a rank-2 plane, not all of $\mathbb{R}^3$ .

IIIVisualization 2 — Elimination Exposes the Condition

Feed a right-hand side into $[A \mid \mathbf{b}]$ and eliminate: the last row collapses to $0 = b_3 - b_1 - b_2$ .

b₁ 1 b₂ 5 b₃ 6

[ A | b ] — original

[ U | b′ ] — eliminated

💡 That last-row residual is $R_1 + R_2 - R_3$ applied to $\mathbf{b}$ : a row dependency in $A$ becomes a constraint $\mathbf{b}$ must satisfy.

IVVisualization 3 — How Rare Is Solvability?

Random right-hand sides almost never hit $b_3 = b_1 + b_2$ exactly — the solvable set is a plane of zero volume.

Tolerance 0.50 Sample

💡 With 4 unknowns but only 3 equations, most $\mathbf{b}$ still give no solution: the dependency $R_1 + R_2 = R_3$ pins $C(A)$ to a 2-D plane inside $\mathbb{R}^3$ .

VQuiz Questions

Question 1

For $A = \begin{bmatrix} 1 & 2 & 2 & 2 \\ 2 & 4 & 6 & 8 \\ 3 & 6 & 8 & 10 \end{bmatrix}$ , which right-hand side $\mathbf{b}$ makes $A\mathbf{x} = \mathbf{b}$ solvable?

✅ Correct! Since $2 + 3 = 5$ , the condition $b_1 + b_2 = b_3$ is satisfied.

❌ Not quite. Check whether $b_1 + b_2 = b_3$ holds for each option.

Show solution

Solution:

The solvability condition is $b_1 + b_2 = b_3$ (because Row 1 + Row 2 = Row 3 in $A$ ).

Check each option:

$\mathbf{b} = (1, 2, 4)$ : $1 + 2 = 3 \neq 4$ ✗
$\mathbf{b} = (2, 3, 5)$ : $2 + 3 = 5$ ✓ Solvable!
$\mathbf{b} = (1, 1, 1)$ : $1 + 1 = 2 \neq 1$ ✗
$\mathbf{b} = (0, 0, 1)$ : $0 + 0 = 0 \neq 1$ ✗

The answer is $\mathbf{b} = (2, 3, 5)$ .

Question 2

During Gaussian elimination on an augmented matrix $[A|\mathbf{b}]$ , you obtain a row $[0 \;\; 0 \;\; 0 \;\; 0 \;\mid\; 0]$ (all zeros on both sides). What does this tell you?

✅ Correct! $0 = 0$ is always true — it's a redundant equation, not a contradiction.

❌ Not quite. The equation says $0 = 0$ . Is that ever false? A contradiction only arises when the right side is nonzero.

Show solution

Solution:

A row $[0\; 0\; 0\; 0 \mid 0]$ states $0 = 0$ , which is always true. This is a redundant equation — it provides no new constraint.

Key distinctions:

Zero row + zero RHS → $0 = 0$ → no contradiction → system may still be solvable
Zero row + nonzero RHS → $0 = d \neq 0$ → contradiction → no solution

Whether the system has one or infinitely many solutions depends on the number of pivot columns vs. unknowns, not on the zero row itself.

Question 3

True or False: If a system $A\mathbf{x} = \mathbf{b}$ has more unknowns than equations, it is guaranteed to have at least one solution for every $\mathbf{b}$ .

✅ Correct! Row dependencies can make the system unsolvable for certain $\mathbf{b}$ , regardless of how many unknowns there are.

❌ Not quite. Think about the $3 \times 4$ example from the video — does every $\mathbf{b}$ give a solution?

Show solution

Solution: False.

The video's example has $A$ of size $3 \times 4$ (3 equations, 4 unknowns). Yet $A\mathbf{x} = \mathbf{b}$ has no solution whenever $b_1 + b_2 \neq b_3$ .

Why? Row dependencies in $A$ force constraints on $\mathbf{b}$ . When rows are dependent, some combinations produce the zero row, and $\mathbf{b}$ must satisfy the same combinations.

What more unknowns than equations does guarantee is: if a solution exists, it is not unique (there will be free variables). But existence itself still depends on $\mathbf{b}$ .

Question 4

A $3 \times 3$ matrix $B$ has the row relationship $2 \cdot (\text{Row 1}) - 1 \cdot (\text{Row 2}) + 1 \cdot (\text{Row 3}) = \mathbf{0}$ .

What condition must $\mathbf{b} = (b_1, b_2, b_3)$ satisfy for $B\mathbf{x} = \mathbf{b}$ to be solvable?

✅ Correct! Apply the exact same coefficients $(2, -1, 1)$ to the entries of $\mathbf{b}$ .

❌ Not quite. The coefficients in the row combination transfer directly: $2 \cdot b_1 + (-1) \cdot b_2 + 1 \cdot b_3 = 0$ .

Show solution

Solution:

The key rule from the video: whatever linear combination of rows of $A$ gives the zero row, the same combination applied to $\mathbf{b}$ must give zero.

Given: $2 \cdot R_1 - 1 \cdot R_2 + 1 \cdot R_3 = \mathbf{0}$

Apply the same coefficients $(2,\, -1,\, 1)$ to $\mathbf{b}$ :

$2 b_1 - b_2 + b_3 = 0$

This is the required solvability condition. The vector $(2, -1, 1)$ belongs to the left null space of $B$ .

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