LINEAR-ALGEBRA · Unit 8 · Video 2 · Interactive Practice

Every Solution at Once: The Complete Solution to Ax = b

IKey Formulas

Formula	Name	Description
$\mathbf{x} = \mathbf{x}_p + \mathbf{x}_n$	Complete solution	Particular solution + null space vector
$A\mathbf{x}_p = \mathbf{b}$	Particular solution	Found by setting all free variables to 0
$\mathbf{x}_n = c_1\mathbf{x}_{s1} + c_2\mathbf{x}_{s2} + \cdots$	Null space component	Linear combination of null space basis vectors
$\text{\# free variables} = n - \text{rank}(A)$	Dimension of null space	$n$ = number of columns

IIVisualization 1 — Assembling the Complete Solution

As $c_1$ and $c_2$ sweep every value, does $A\mathbf{x}$ ever leave $\mathbf{b}$ ?

c₁ 0 c₂ 0

💡 With $\mathbf{x}_p = (-2, 0, \tfrac{3}{2}, 0)$ and null basis $\mathbf{s}_1 = (-2, 1, 0, 0),\ \mathbf{s}_2 = (2, 0, -2, 1)$ , each $A\mathbf{s}_i = \mathbf{0}$ , so adding any $c_1\mathbf{s}_1 + c_2\mathbf{s}_2$ moves $\mathbf{x}$ but never $A\mathbf{x}$ .

IIIVisualization 2 — Shifting the Null Space

The solution set of $x_1 + 2x_2 = 4$ is the null-space line lifted off the origin — by how much?

Shift 100%

💡 The fully shifted line is an affine subspace (a "flat"): same direction and dimension as $N(A)$ , but because it misses the origin when $\mathbf{b} \neq \mathbf{0}$ it is not a subspace.

IVVisualization 3 — Existence vs Uniqueness

Two independent switches — does a solution exist, and is it unique — decide the entire solution structure.

Existence Uniqueness

💡 The column space governs existence; the null space governs uniqueness — two independent tests, and only when both pass is the solution a single point.

VQuiz Questions

Question 1

The video's system row-reduces to echelon form: $\begin{bmatrix} 1 & 2 & 2 & 2 \\ 0 & 0 & 2 & 4 \\ 0 & 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix} = \begin{bmatrix} 1 \\ 3 \\ 0 \end{bmatrix}$

Using the "set free variables to zero" recipe, what is the particular solution $\mathbf{x}_p$ ?

✅ Correct! Set free variables ( $x_2, x_4$ ) to zero, then back-substitute to get $x_3 = 3/2$ and $x_1 = -2$ .

❌ Not quite. Remember: the recipe sets all free variables (columns without pivots) to zero first, then solves for pivot variables by back-substitution.

Show solution

Solution:

Step 1: Identify free variables. Pivots are in columns 1 and 3, so $x_2$ and $x_4$ are free.

Step 2: Set free variables to zero: $x_2 = 0$ , $x_4 = 0$ .

Step 3: Back-substitute from the bottom row up.

From row 2: $\;2x_3 + 4(0) = 3 \implies 2x_3 = 3 \implies x_3 = \frac{3}{2}$

From row 1: $\;x_1 + 2(0) + 2\!\left(\frac{3}{2}\right) + 2(0) = 1 \implies x_1 + 3 = 1 \implies x_1 = -2$

$\mathbf{x}_p = \begin{pmatrix} -2 \\ 0 \\ 3/2 \\ 0 \end{pmatrix}$

Question 2

True or False: When $\mathbf{b} \neq \mathbf{0}$ , the solution set of $A\mathbf{x} = \mathbf{b}$ is a subspace of $\mathbb{R}^n$ .

✅ Correct! The zero vector doesn't satisfy $A\mathbf{x} = \mathbf{b}$ when $\mathbf{b} \neq \mathbf{0}$ , so the solution set is NOT a subspace — it's an affine subspace.

❌ Not quite. Does the zero vector satisfy $A\mathbf{x} = \mathbf{b}$ when $\mathbf{b} \neq \mathbf{0}$ ? Every subspace must contain $\mathbf{0}$ .

Show solution

Solution: False

A subspace must contain the zero vector $\mathbf{0}$ . But when $\mathbf{b} \neq \mathbf{0}$ :

$A\mathbf{0} = \mathbf{0} \neq \mathbf{b}$

So $\mathbf{0}$ is not in the solution set — it fails the most basic subspace test.

The solution set is an affine subspace (or "flat"): it has the same shape and dimension as the null space, but it is translated away from the origin by $\mathbf{x}_p$ . A flat is only a subspace when it passes through the origin (i.e., when $\mathbf{b} = \mathbf{0}$ ).

Question 3

A $4 \times 7$ matrix $A$ has rank 3 (three pivot columns). Assuming the system $A\mathbf{x} = \mathbf{b}$ is consistent, what is the dimension of the solution set?

✅ Correct! With 7 columns and rank 3, there are $7 - 3 = 4$ free variables, so the solution set is 4-dimensional.

❌ Not quite. The dimension equals the number of free variables: (number of columns) − (rank). Try again!

Show solution

Solution: 4

Step 1: Count unknowns. The matrix has 7 columns, so there are 7 unknowns.

Step 2: Count free variables. $\text{free variables} = n - \text{rank}(A) = 7 - 3 = 4$

Step 3: Each free variable contributes one basis vector to the null space. $\dim(N(A)) = 4$

Step 4: The solution set is $\mathbf{x}_p$ + (4-dimensional null space), which is a 4-dimensional flat (affine subspace) in $\mathbb{R}^7$ .

The dimension of the solution set always equals the number of free variables.

Question 4

Which pair of conditions guarantees that $A\mathbf{x} = \mathbf{b}$ has exactly one solution?

✅ Correct! Existence needs $\mathbf{b} \in C(A)$ ; uniqueness needs a trivial null space. Together they guarantee exactly one solution.

❌ Not quite. Remember: the column space governs existence and the null space governs uniqueness. You need both conditions for exactly one solution.

Show solution

Solution: b in C(A) and N(A) = {0}

For exactly one solution, both conditions must hold simultaneously:

	Column Space Check	Null Space Check	Result
$\mathbf{b} \in C(A)$ and $N(A) = \{\mathbf{0}\}$	✅ Exists	✅ Unique	Exactly one
$\mathbf{b} \in C(A)$ and $N(A) \neq \{\mathbf{0}\}$	✅ Exists	❌ Not unique	Infinitely many
$\mathbf{b} \notin C(A)$	❌ Doesn't exist	Irrelevant	No solution

Existence requires $\mathbf{b} \in C(A)$ — otherwise there's no $\mathbf{x}_p$ at all.
Uniqueness requires $N(A) = \{\mathbf{0}\}$ — so the complete solution $\mathbf{x} = \mathbf{x}_p + \mathbf{x}_n$ reduces to just $\mathbf{x} = \mathbf{x}_p$ .

Both conditions are needed; neither alone is sufficient.

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