Solution Sets of Linear Systems and Free Variables

11 min read

Before Free variables come from row reduction, and the solution set is written with vectors, as in Ax = b. Builds on: The Matrix Equation Ax = b · Systems of Linear Equations and Row Reduction

Next The solutions of Ax = 0 lead to linear independence and to the null space. Leads to: Linear Independence · Null Space, Column Space, and Basis

Reference: Lay, Linear Algebra and Its Applications, Section 1.5: Solution Sets of Linear Systems

Video: Dr Lynch’s MATH 231 lectures (YouTube playlists)

Before You Start: Prerequisite Check

Can you do these? (Click to reveal self-test)

Test yourself on these prerequisite skills:

RREF and free variables: In the RREF below, identify the pivot columns and free columns. \[ \left[\begin{array}{rrrr|r}1&2&0&3&5\\0&0&1&-1&2\end{array}\right] \]

Check

Pivot columns: 1 and 3 (columns containing the leading 1s). Free columns: 2 and 4 (no leading 1). Variables $x_2$ and $x_4$ are free.
Parametric reading: From the RREF above, write $x_1$ and $x_3$ in terms of the free variables.

Check

Row 1: $x_1 + 2x_2 + 3x_4 = 5 \implies x_1 = 5 - 2x_2 - 3x_4$.

Row 2: $x_3 - x_4 = 2 \implies x_3 = 2 + x_4$.
Reading the equation: For which vectors $\mathbf{x}$ does $A\mathbf{x} = \mathbf{0}$, and what is that solution set called?

Check

It asks for which vectors $\mathbf{x}$ the column combination $A\mathbf{x}$ equals the zero vector. The solution set is called the null space of $A$.

If you struggled: Review The Matrix Equation Ax = b and Systems and Row Reduction.

Try This First

Before any new vocabulary, work through this.

Find all solutions to the system $x_1 + x_2 = 0$.

There is only one equation in two unknowns. Write every solution. Can you express them as a single formula involving one free parameter?

Show Solution

From $x_1 + x_2 = 0$ we get $x_1 = -x_2$. Let $x_2 = t$ for any real number $t$.

\[ \begin{pmatrix}x_1\\x_2\end{pmatrix} = \begin{pmatrix}-t\\t\end{pmatrix} = t\begin{pmatrix}-1\\1\end{pmatrix}, \quad t \in \mathbb{R} \]

Every solution is a scalar multiple of $\begin{pmatrix}-1\\1\end{pmatrix}$. Geometrically, the solution set is a line through the origin in $\mathbb{R}^2$, in the direction $\begin{pmatrix}-1\\1\end{pmatrix}$.

What you just did: You expressed the solution set in parametric vector form. When free variables appear, the solution set is always described this way: a combination of vectors, each scaled by a free parameter. The structure of those vectors depends only on $A$, not on $\mathbf{b}$.

The Two Types of Systems

Understanding solution sets requires distinguishing two kinds of systems:

Homogeneous: $A\mathbf{x} = \mathbf{0}$ (right-hand side is the zero vector).

Always consistent: $\mathbf{x} = \mathbf{0}$ is always a solution.
May have additional nontrivial solutions when free variables exist.

Nonhomogeneous: $A\mathbf{x} = \mathbf{b}$ with $\mathbf{b} \neq \mathbf{0}$.

May be consistent or inconsistent.
When consistent, the solution set is a translate of the homogeneous solution set.

The homogeneous system is the key to understanding the structure of all solutions.

Prerequisite Map

Quick Reference

Property	Value
Course	MATH 231
Difficulty	Intermediate
Time	~30 minutes
Source	Lay, Linear Algebra and Its Applications, Section 1.5

Solution Set Structure

System Type	Solution Set
$A\mathbf{x} = \mathbf{0}$ with no free variables	Unique solution: $\mathbf{x} = \mathbf{0}$
$A\mathbf{x} = \mathbf{0}$ with $k$ free variables	Span of $k$ vectors through origin
$A\mathbf{x} = \mathbf{b}$ consistent, no free variables	Unique solution: one point
$A\mathbf{x} = \mathbf{b}$ consistent, $k$ free variables	Particular solution + span of $k$ vectors
$A\mathbf{x} = \mathbf{b}$ inconsistent	Empty set

Key Concepts

Homogeneous Systems

A homogeneous system has the form $A\mathbf{x} = \mathbf{0}$. It is always consistent because $\mathbf{x} = \mathbf{0}$ satisfies it; this is the trivial solution.

The system has a nontrivial solution (a solution with at least one nonzero entry) exactly when there is at least one free variable after row reduction.

Test for free variables: If $A$ has more columns than pivot columns, free variables exist.

Source: Lay, Linear Algebra and Its Applications, Section 1.5: Solution Sets of Linear Systems.

Parametric Vector Form

When free variables exist, express the solution set by assigning a parameter to each free variable, then collecting the results into vectors.

Worked example (homogeneous):

Row-reduce $A$ and suppose the RREF of $[A \mid \mathbf{0}]$ is: \[ \left[\begin{array}{rrrr|r} 1 & 3 & 0 & -2 & 0 \\ 0 & 0 & 1 & 4 & 0 \end{array}\right] \]

Pivot columns: 1, 3. Free columns: 2, 4. Let $x_2 = s$ and $x_4 = t$.

From the rows:

Row 1: $x_1 + 3s - 2t = 0 \implies x_1 = -3s + 2t$
Row 2: $x_3 + 4t = 0 \implies x_3 = -4t$

Assemble: \[ \mathbf{x} = \begin{pmatrix}-3s + 2t \\ s \\ -4t \\ t\end{pmatrix} = s\begin{pmatrix}-3\\1\\0\\0\end{pmatrix} + t\begin{pmatrix}2\\0\\-4\\1\end{pmatrix} \]

The null space is $\text{Nul}(A) = \text{Span}\left\{\begin{pmatrix}-3\\1\\0\\0\end{pmatrix}, \begin{pmatrix}2\\0\\-4\\1\end{pmatrix}\right\}$.

The Null Space

The null space of $A$, written $\text{Nul}(A)$, is the solution set of the homogeneous equation $A\mathbf{x} = \mathbf{0}$. It is always a subspace of $\mathbb{R}^n$:

Contains $\mathbf{0}$ (trivial solution).
Closed under addition: if $A\mathbf{u} = \mathbf{0}$ and $A\mathbf{v} = \mathbf{0}$, then $A(\mathbf{u}+\mathbf{v}) = \mathbf{0}$.
Closed under scalar multiplication: $A(c\mathbf{u}) = c(A\mathbf{u}) = \mathbf{0}$.

Geometrically, the null space is a subspace of $\mathbb{R}^n$ whose dimension equals the number of free variables.

The General Solution: Particular + Homogeneous

Theorem: Suppose $A\mathbf{x} = \mathbf{b}$ is consistent with particular solution $\mathbf{x}_p$ (that is, $A\mathbf{x}_p = \mathbf{b}$). Then the general solution is:

\[ \mathbf{x} = \mathbf{x}_p + \mathbf{v}_h \]

where $\mathbf{v}_h$ ranges over all solutions of the homogeneous equation $A\mathbf{x} = \mathbf{0}$.

So the solution set of $A\mathbf{x} = \mathbf{b}$ is a translate of the null space $\text{Nul}(A)$, shifted by $\mathbf{x}_p$.

Worked example (nonhomogeneous):

Suppose the RREF of $[A \mid \mathbf{b}]$ is: \[ \left[\begin{array}{rrrr|r} 1 & 3 & 0 & -2 & 3 \\ 0 & 0 & 1 & 4 & 1 \end{array}\right] \]

Particular solution (set free variables to 0: $s = t = 0$): $\mathbf{x}_p = \begin{pmatrix}3\\0\\1\\0\end{pmatrix}$

Homogeneous part (same as before): $s\begin{pmatrix}-3\\1\\0\\0\end{pmatrix} + t\begin{pmatrix}2\\0\\-4\\1\end{pmatrix}$

General solution: \[ \mathbf{x} = \begin{pmatrix}3\\0\\1\\0\end{pmatrix} + s\begin{pmatrix}-3\\1\\0\\0\end{pmatrix} + t\begin{pmatrix}2\\0\\-4\\1\end{pmatrix}, \quad s, t \in \mathbb{R} \]

The solution set is a plane in $\mathbb{R}^4$, passing through $\mathbf{x}_p$, parallel to the null space.

Common Pitfalls

Mistake	Why It Is Wrong	Correct Approach
Concluding a homogeneous system has only the trivial solution	Free variables always produce nontrivial solutions	Count free variables: if any exist, nontrivial solutions exist
Thinking the particular solution is unique	Any solution to $A\mathbf{x} = \mathbf{b}$ is a particular solution	Choose the most convenient one: set all free variables to 0
Concluding the solution set of $A\mathbf{x} = \mathbf{b}$ (with $\mathbf{b} \neq \mathbf{0}$) is a subspace	It does not contain $\mathbf{0}$ when $\mathbf{b} \neq \mathbf{0}$, so it cannot be a subspace	The solution set is an affine subspace (a translate of the null space), not a subspace itself
Assigning free variable parameters but forgetting to substitute back	The basic (pivot) variables must be expressed in terms of the parameters	Always rewrite each pivot variable’s equation using the substitution

The Solution Set of $A\mathbf{x} = \mathbf{b}$ Is Not a Subspace

A subspace must contain $\mathbf{0}$. When $\mathbf{b} \neq \mathbf{0}$, any $\mathbf{x}$ satisfying $A\mathbf{x} = \mathbf{b}$ gives $A\mathbf{0} = \mathbf{0} \neq \mathbf{b}$, so $\mathbf{0}$ is not in the solution set.

Instead, the solution set is an affine subspace: it looks like the null space, but shifted to pass through the particular solution $\mathbf{x}_p$.

In $\mathbb{R}^2$: the solution set of $x_1 + x_2 = 3$ is the line $x_1 + x_2 = 3$, not a line through the origin. The null space of $A = [1 \; 1]$ is the line $x_1 + x_2 = 0$. The solution set is the null space shifted by $\mathbf{x}_p = \begin{pmatrix}3\\0\end{pmatrix}$.

Practice Problems

Level 1 Parametric Form from a Single Equation

Write all solutions to $2x_1 - 4x_2 + x_3 = 0$ in parametric vector form. Identify the free variables and the direction vectors.

Thought Process

This is one equation in three unknowns. Column 1 has the only pivot (coefficient of $x_1$ is $2$). So $x_2$ and $x_3$ are free. Solve for $x_1$ in terms of $x_2$ and $x_3$, then assemble the parametric form.

Show Answer

Solve for $x_1$: $x_1 = 2x_2 - \frac{1}{2}x_3$.

Let $x_2 = s$ and $x_3 = t$.

\[ \begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix} = \begin{pmatrix}2s - \frac{1}{2}t\\s\\t\end{pmatrix} = s\begin{pmatrix}2\\1\\0\end{pmatrix} + t\begin{pmatrix}-\frac{1}{2}\\0\\1\end{pmatrix} \]

Free variables: $x_2 = s$ and $x_3 = t$. Direction vectors: $\begin{pmatrix}2\\1\\0\end{pmatrix}$ and $\begin{pmatrix}-\frac{1}{2}\\0\\1\end{pmatrix}$.

The null space is a plane through the origin in $\mathbb{R}^3$.

Level 2 Homogeneous System from Row Reduction

Find all solutions to $A\mathbf{x} = \mathbf{0}$ where

\[ A = \begin{pmatrix}1&2&-3\\2&4&-6\end{pmatrix} \]

in parametric vector form. Describe $\text{Nul}(A)$ geometrically.

Thought Process

Row-reduce $[A \mid \mathbf{0}]$. Look at how many pivots there are and how many free variables appear. Then write the parametric solution.

Show Answer

$[A \mid \mathbf{0}]$: \[ \left[\begin{array}{rrr|r}1&2&-3&0\\2&4&-6&0\end{array}\right] \]

$R_2 \to R_2 - 2R_1$: \[ \left[\begin{array}{rrr|r}1&2&-3&0\\0&0&0&0\end{array}\right] \]

One pivot, in column 1. Free variables: $x_2 = s$, $x_3 = t$.

From row 1: $x_1 = -2s + 3t$.

\[ \mathbf{x} = s\begin{pmatrix}-2\\1\\0\end{pmatrix} + t\begin{pmatrix}3\\0\\1\end{pmatrix} \]

Geometric description: $\text{Nul}(A)$ is a plane through the origin in $\mathbb{R}^3$, spanned by $\begin{pmatrix}-2\\1\\0\end{pmatrix}$ and $\begin{pmatrix}3\\0\\1\end{pmatrix}$.

Notice that the two rows of $A$ are identical (row 2 is $2 \times$ row 1), so the two equations encode the same constraint. One constraint in $\mathbb{R}^3$ defines a plane.

Level 3 General Solution to a Nonhomogeneous System

Row-reduce the following augmented matrix to RREF. Write the general solution in parametric vector form as $\mathbf{x} = \mathbf{x}_p + s\mathbf{v}_1 + t\mathbf{v}_2$.

\[ \left[\begin{array}{rrrr|r} 1 & -2 & 0 & 3 & 2 \\ 0 & 0 & 1 & -1 & -1 \\ 0 & 0 & 0 & 0 & 0 \end{array}\right] \]

(This matrix is already in RREF.)

Thought Process

Identify the pivot columns (1 and 3) and free columns (2 and 4). Assign parameters to $x_2$ and $x_4$. Read each basic variable from its pivot row. Collect the particular solution (all parameters zero) and the direction vectors (one per parameter).

Show Answer

Pivot columns: 1, 3. Free columns: 2, 4. Let $x_2 = s$, $x_4 = t$.

Row 1: $x_1 - 2s + 3t = 2 \implies x_1 = 2 + 2s - 3t$
Row 2: $x_3 - t = -1 \implies x_3 = -1 + t$

Assemble: \[ \mathbf{x} = \begin{pmatrix}2 + 2s - 3t \\ s \\ -1 + t \\ t\end{pmatrix} = \underbrace{\begin{pmatrix}2\\0\\-1\\0\end{pmatrix}}_{\mathbf{x}_p} + s\underbrace{\begin{pmatrix}2\\1\\0\\0\end{pmatrix}}_{\mathbf{v}_1} + t\underbrace{\begin{pmatrix}-3\\0\\1\\1\end{pmatrix}}_{\mathbf{v}_2} \]

Verify $\mathbf{x}_p$: Row 1: $1(2) - 2(0) + 0(-1) + 3(0) = 2$ (correct); Row 2: $0(2) + 0(0) + 1(-1) + (-1)(0) = -1$ (correct).

Level 4 Identify Which Vectors Are in the Null Space

Let $A = \begin{pmatrix}1&-2&1\\0&1&-3\\2&-4&2\end{pmatrix}$.

(a) Row-reduce $A$ and write $\text{Nul}(A)$ in parametric vector form.

(b) Which of the following vectors is in $\text{Nul}(A)$?

$\mathbf{u} = \begin{pmatrix}5\\3\\1\end{pmatrix}$
$\mathbf{v} = \begin{pmatrix}7\\3\\1\end{pmatrix}$

Thought Process

For part (a): row-reduce $[A \mid \mathbf{0}]$. Find free variables.

For part (b): a vector $\mathbf{w}$ is in $\text{Nul}(A)$ exactly when $A\mathbf{w} = \mathbf{0}$. Compute $A\mathbf{u}$ and $A\mathbf{v}$ directly.

Show Answer

(a) Row-reduce $[A \mid \mathbf{0}]$:

$R_3 \to R_3 - 2R_1$: \[ \left[\begin{array}{rrr|r}1&-2&1&0\\0&1&-3&0\\0&0&0&0\end{array}\right] \]

This is already in REF (and RREF after back-substituting column 2):

$R_1 \to R_1 + 2R_2$: \[ \left[\begin{array}{rrr|r}1&0&-5&0\\0&1&-3&0\\0&0&0&0\end{array}\right] \]

Pivot columns: 1, 2. Free column: 3. Let $x_3 = t$.

$x_2 = 3t$
$x_1 = 5t$

\[ \text{Nul}(A) = \text{Span}\left\{\begin{pmatrix}5\\3\\1\end{pmatrix}\right\} \]

(b) Check $\mathbf{u} = \begin{pmatrix}5\\3\\1\end{pmatrix}$: this is the spanning vector with $t=1$, so $A\mathbf{u} = \mathbf{0}$. $\mathbf{u}$ IS in $\text{Nul}(A)$.

Check $\mathbf{v} = \begin{pmatrix}7\\3\\1\end{pmatrix}$: if $\mathbf{v} = t\begin{pmatrix}5\\3\\1\end{pmatrix}$, then from the third entry $t=1$, but from the first entry $t = 7/5$. Inconsistent; $\mathbf{v}$ is NOT in $\text{Nul}(A)$.

Quick check: $A\mathbf{v}$, row 1: $1(7)-2(3)+1(1) = 7-6+1 = 2 \neq 0$. Confirmed not in $\text{Nul}(A)$.

Level 5 Translate the Null Space

Let $A = \begin{pmatrix}1&0&-5\\0&1&3\end{pmatrix}$ and $\mathbf{b} = \begin{pmatrix}4\\-2\end{pmatrix}$.

(a) Find $\text{Nul}(A)$ in parametric vector form.

(b) Find one particular solution $\mathbf{x}_p$ to $A\mathbf{x} = \mathbf{b}$.

(d) If $\mathbf{x}_1 = \begin{pmatrix}4\\-2\\0\end{pmatrix}$ and $\mathbf{x}_2 = \begin{pmatrix}9\\-5\\1\end{pmatrix}$ are both solutions to $A\mathbf{x} = \mathbf{b}$, show that their difference $\mathbf{x}_1 - \mathbf{x}_2$ is in $\text{Nul}(A)$.

Thought Process

The matrix $A$ is already in RREF. Identify pivot and free variables.

For $\text{Nul}(A)$: use $\mathbf{b} = \mathbf{0}$, let the free variable be a parameter.

For a particular solution to $A\mathbf{x} = \mathbf{b}$: set the free variable to 0.

For part (d): if $A\mathbf{x}_1 = \mathbf{b}$ and $A\mathbf{x}_2 = \mathbf{b}$, compute $A(\mathbf{x}_1 - \mathbf{x}_2)$ using linearity.

Show Answer

(a) $A$ is in RREF. Pivot columns: 1, 2. Free column: 3. For $A\mathbf{x} = \mathbf{0}$:

$x_1 - 5x_3 = 0 \implies x_1 = 5t$
$x_2 + 3x_3 = 0 \implies x_2 = -3t$

\[ \text{Nul}(A) = \text{Span}\left\{\begin{pmatrix}5\\-3\\1\end{pmatrix}\right\} \]

Geometrically, $\text{Nul}(A)$ is a line through the origin in $\mathbb{R}^3$.

(b) Set $x_3 = 0$ in $A\mathbf{x} = \mathbf{b}$:

$x_1 = 4$, $x_2 = -2$, $x_3 = 0$

Particular solution: $\mathbf{x}_p = \begin{pmatrix}4\\-2\\0\end{pmatrix}$.

(c) General solution: \[ \mathbf{x} = \begin{pmatrix}4\\-2\\0\end{pmatrix} + t\begin{pmatrix}5\\-3\\1\end{pmatrix}, \quad t \in \mathbb{R} \]

Geometrically, the solution set is a line in $\mathbb{R}^3$ passing through $\begin{pmatrix}4\\-2\\0\end{pmatrix}$, parallel to the null space direction $\begin{pmatrix}5\\-3\\1\end{pmatrix}$.

(d) Using linearity of $A$: \[ A(\mathbf{x}_1 - \mathbf{x}_2) = A\mathbf{x}_1 - A\mathbf{x}_2 = \mathbf{b} - \mathbf{b} = \mathbf{0} \]

So $\mathbf{x}_1 - \mathbf{x}_2 = \begin{pmatrix}4\\-2\\0\end{pmatrix} - \begin{pmatrix}9\\-5\\1\end{pmatrix} = \begin{pmatrix}-5\\3\\-1\end{pmatrix} = -\begin{pmatrix}5\\-3\\1\end{pmatrix}$, which is in $\text{Nul}(A)$.

This confirms the theorem: the difference of any two solutions to $A\mathbf{x} = \mathbf{b}$ is always in the null space.

Common Misconceptions

Common misconception

the homogeneous system $A\mathbf{x} = \mathbf{0}$ has only the trivial solution. The trivial solution $\mathbf{x} = \mathbf{0}$ is always present, but whenever $A$ has a free variable (a non-pivot column), infinitely many nontrivial solutions exist as well. Specifically, any vector in the span of the null space direction vectors is a solution.

Common misconception

the particular solution $\mathbf{x}_p$ is uniquely determined. Any vector satisfying $A\mathbf{x}_p = \mathbf{b}$ is a valid particular solution. The most convenient choice is to set all free variables to zero in the RREF solution, but choosing a different value for the free variables gives a different particular solution that is equally valid. The general solution looks different if you choose a different $\mathbf{x}_p$, but describes the same set of vectors.

Common misconception

the solution set of $A\mathbf{x} = \mathbf{b}$ (with $\mathbf{b} \neq \mathbf{0}$) is a subspace of $\mathbb{R}^n$. A subspace must contain the zero vector. When $\mathbf{b} \neq \mathbf{0}$, checking $\mathbf{x} = \mathbf{0}$ gives $A\mathbf{0} = \mathbf{0} \neq \mathbf{b}$, so $\mathbf{0}$ is not in the solution set. The solution set is an affine subspace, a geometric object that looks like a subspace shifted away from the origin.

Mastery Checklist

Novice (Level 1-2):

I can solve a single equation with multiple unknowns in parametric form.
I can identify free variables from a RREF matrix.
I can describe what the null space of a matrix is.

Competent (Level 3-4):

I can write the general solution to $A\mathbf{x} = \mathbf{b}$ as a particular solution plus the null space.
I can determine which vectors lie in $\text{Nul}(A)$ by direct computation.
I can describe the solution set geometrically (line, plane, point, or empty).

Proficient (Level 5):

I can explain why the difference of two solutions to $A\mathbf{x} = \mathbf{b}$ lies in $\text{Nul}(A)$.
I can explain why the solution set of $A\mathbf{x} = \mathbf{b}$ is an affine subspace, not a subspace.
I can extract both the null space and a particular solution from a single row reduction.

Mental Model

The Shift Principle

Imagine the null space as a shape through the origin: a line, a plane, or higher-dimensional flat surface. The particular solution $\mathbf{x}_p$ is a single anchor point off the origin that satisfies $A\mathbf{x} = \mathbf{b}$.

The full solution set is the null space shape, rigidly moved (shifted) so it passes through $\mathbf{x}_p$ instead of through the origin. The shape does not rotate or deform; it just translates.

This is why the choice of $\mathbf{x}_p$ does not matter: any particular solution gives a valid anchor, and shifting the same null space to any anchor produces the same full solution set.

Connections

Looking ahead in MATH 231:

Linear Independence: whether the null space contains only the zero vector is exactly the question of whether the columns of $A$ are linearly independent
Null Space, Column Space, and Basis: the null space developed here becomes a fundamental subspace whose dimension is the rank-nullity gap
The Invertible Matrix Theorem: a square matrix is invertible exactly when its null space contains only $\mathbf{0}$

Real-world connections:

In network flow, the solutions to a flow conservation system have exactly the structure studied here: a particular flow plus combinations of circulation vectors.
In signal processing, the null space of a filter matrix describes all inputs the filter cannot distinguish from zero.

Source: Lay, Linear Algebra and Its Applications, Section 1.5: Solution Sets of Linear Systems.

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Last updated: 2026-06-16

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