Evanalysis
2.4Source-backedEmbedded interactionEstimated reading time: 3 min

2.4 Solution-set types

Use RREF to classify a linear system rigorously, identify pivot and free variables, and explain why a system has exactly one solution, infinitely many solutions, or no solution.

Note collections

MATH1030: Linear algebra I

Rigorous linear algebra notes on systems, matrices, structure, and proof, with interaction used only where it clarifies the mathematics.

Language

ENEnglish繁Traditional Chinese简Simplified Chinese
All note collectionsSeries overview

Chapter 1

Systems of equations

Learn to read equations as full solution sets.

1.1

1.1 Equations and solution sets

Chapter 2

Matrices and elimination

Build matrix intuition and use row reduction with purpose.

2.1

2.1 Matrix basics

2.2

2.2 Augmented matrices and row operations

2.3

2.3 Gaussian elimination and RREF

2.4

2.4 Solution-set types

Chapter 3

Matrix algebra

Matrix multiplication, transpose, and structural matrix notation.

3.1

3.1 Matrix multiplication and identity matrices

3.2

3.2 Transpose and special matrices

Chapter 4

Solution structure

Homogeneous systems, null spaces, and the shape of full solution sets.

4.1

4.1 Homogeneous systems and null space

Chapter 5

Invertibility

Understand when a matrix can be undone and why that matters.

5.1

5.1 Invertible matrices

Member

Chapter 6

Vector spaces

Move from matrix procedures to the structure of spaces, span, independence, and basis.

6.1

6.1 Vector spaces

Member

6.2

6.2 Subspaces

Member

6.3

6.3 Linear combinations and span

Member

6.4

6.4 Linear dependence and independence

Member

6.5

6.5 Basis and dimension

Member

6.6

6.6 Column space, row space, and rank

Member

Gaussian elimination is not finished once the matrix looks simpler. The point of row reduction is that the reduced form lets you read the structure of the solution set. This note makes that reading process precise.

There are only three possible outcomes for a linear system:

  1. exactly one solution,
  2. infinitely many solutions,
  3. no solution.

The classification is not based on guesswork. It follows from the position of the pivot columns in the reduced augmented matrix.

Dependent variables and free variables

When a consistent linear system is reduced to RREF, the pivot columns tell you which variables are determined by the others. Columns without pivots correspond to variables that may be chosen freely.

Definition

Dependent variables and free variables

Suppose the augmented matrix of a consistent linear system is row-equivalent to a matrix BB in RREF. If column j of BB is a pivot column, then the variable xjx_j is called a dependent variable or leading variable. A variable whose column is not a pivot column is called a free variable.

This language matters because it tells you how the final answer should be written. Free variables become parameters. Dependent variables are then solved in terms of those parameters.

The consistency test

The first question is not whether the system has one solution or many. The first question is whether it has any solution at all.

Theorem

Consistency criterion from RREF

Let AA be the augmented matrix of an m×nm \times n linear system, and let BB be the RREF row-equivalent to AA. Then the system is inconsistent if and only if the last column of BB is a pivot column.

Equivalently, the system is inconsistent if and only if BB contains a row of the form

[00⋯0d]with d≠0.\left[ \begin{array}{cccc|c} 0 & 0 & \cdots & 0 & d \end{array} \right] \qquad\text{with } d \neq 0.

Such a row represents the equation

0=d,0 = d,

which is impossible when d≠0d \neq 0. That is why contradiction rows settle the question immediately.

Why only three cases can occur

Once the system is known to be consistent, the remaining issue is the number of free variables.

Theorem

Classification by the number of pivots

Suppose an m×nm \times n linear system is consistent, and suppose its augmented matrix is row-equivalent to an RREF matrix with r pivot columns. Then r≤nr \le n. Moreover:

  1. if r=nr = n, the system has a unique solution;
  2. if r<nr < n, the system has infinitely many solutions.

The reason is structural.

  • If r=nr = n, every variable column contains a pivot, so there are no free variables.
  • If r<nr < n, then n−r>0n - r > 0, so at least one variable is free.
  • Every choice of the free variables gives a solution, so consistency plus a free variable automatically creates infinitely many solutions.

Putting the consistency theorem and the pivot-count theorem together gives the standard trichotomy.

Theorem

Possible solution sets for a linear system

A linear system has exactly one of the following three solution-set types:

  1. a unique solution;
  2. infinitely many solutions;
  3. no solution.

Three reduced forms, three different readings

The reduced matrix already contains the whole story. The only question is whether you know what to look for.

Worked example

Read the matrix before solving

Consider the following three RREF augmented matrices.

First,

[10301−2].\left[ \begin{array}{cc|c} 1 & 0 & 3 \\ 0 & 1 & -2 \end{array} \right].

Both variable columns are pivot columns, so there are no free variables. The system has the unique solution

x1=3,x2=−2.x_1 = 3,\qquad x_2 = -2.

Second,

[12050000].\left[ \begin{array}{ccc|c} 1 & 2 & 0 & 5 \\ 0 & 0 & 0 & 0 \end{array} \right].

Column 1 is a pivot column, but columns 2 and 3 are not. So x2x_2 and x3x_3 are free variables. The equation is

x1+2x2=5,x_1 + 2x_2 = 5,

which gives

x1=5−2x2.x_1 = 5 - 2x_2.

Hence the solution set is

{(5−2s,s,t)∣s,t∈R}.\{(5 - 2s, s, t) \mid s, t \in R\}.

Third,

[103001].\left[ \begin{array}{cc|c} 1 & 0 & 3 \\ 0 & 0 & 1 \end{array} \right].

The last row says 0=10 = 1, so the system is inconsistent and has no solution.

The essential habit is this:

  1. check for contradiction rows;
  2. identify pivot columns;
  3. count free variables;
  4. write the dependent variables in terms of the free ones.

A larger parameterized example

The source notes emphasize that a free-variable answer should be written as a full set, not as a vague phrase such as "there are many solutions."

Worked example

Write the solution set explicitly

Suppose a system in variables x1,x2,x3,x4,x5x_1, x_2, x_3, x_4, x_5 has RREF

[1−100360010−21000149000000].\left[ \begin{array}{ccccc|c} 1 & -1 & 0 & 0 & 3 & 6 \\ 0 & 0 & 1 & 0 & -2 & 1 \\ 0 & 0 & 0 & 1 & 4 & 9 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{array} \right].

Columns 1, 3, and 4 are pivot columns, so x1x_1, x3x_3, and x4x_4 are dependent variables. Columns 2 and 5 are non-pivot columns, so x2x_2 and x5x_5 are free.

Reading the rows gives

x1−x2+3x5=6,x3−2x5=1,x4+4x5=9.\begin{aligned} x_1 - x_2 + 3x_5 &= 6, \\ x_3 - 2x_5 &= 1, \\ x_4 + 4x_5 &= 9. \end{aligned}

Let

x2=s,x5=t.x_2 = s,\qquad x_5 = t.

Then

x1=6+s−3t,x3=1+2t,x4=9−4t.\begin{aligned} x_1 &= 6 + s - 3t, \\ x_3 &= 1 + 2t, \\ x_4 &= 9 - 4t. \end{aligned}

Therefore the solution set is

{(6+s−3t, s, 1+2t, 9−4t, t)∣s,t∈R}.\{(6 + s - 3t,\ s,\ 1 + 2t,\ 9 - 4t,\ t) \mid s, t \in R\}.

Two free variables produce a two-parameter family of solutions.

A useful corollary when there are more variables than equations

One common classroom situation is a consistent system with fewer equations than variables.

Theorem

If n>mn > m, consistency forces infinitely many solutions

Suppose a consistent linear system has m equations in n variables, and suppose n>mn > m. Then the system has infinitely many solutions.

Why? In RREF there can be at most m nonzero rows, so there can be at most m pivot columns. If n>mn > m, then at least one of the n variable columns must be non-pivot. Hence there is at least one free variable, and consistency then forces infinitely many solutions.

This corollary does not say that every underdetermined system is consistent. It says that once such a system is consistent, uniqueness is impossible.

Compare the three cases interactively

The classifier below is useful because it makes the reading process explicit: you check contradiction rows, look for pivots, and then decide how many free variables remain.

Read and try

Read the shape of a solution set

The live classifier compares three representative reduced matrices and explains what each structure means.

1002
010-1
0013

Why it works

Every variable is a pivot variable, so the system has one solution.

Common mistakes

Common mistake

A free variable is not an unfinished calculation

A free variable is a genuine feature of the system. It means the solution set is parameterized. The algebra is complete, even though the answer is a family of vectors rather than one point.

Common mistake

A zero row is not the same as a contradiction row

The row

[00000]\left[ \begin{array}{cccc|c} 0 & 0 & 0 & 0 & 0 \end{array} \right]

adds no new condition and is perfectly compatible with consistency. By contrast,

[00001]\left[ \begin{array}{cccc|c} 0 & 0 & 0 & 0 & 1 \end{array} \right]

is a contradiction and destroys consistency.

Quick checks

Quick check

What does the row [0001]\left[\begin{array}{ccc|c}0 & 0 & 0 & 1\end{array}\right] tell you about the system?

Interpret the row as an equation.

Solution

Answer

Quick check

If a consistent system in four variables has three pivot columns, how many free variables does it have?

Count the non-pivot variable columns.

Solution

Answer

Exercises

Quick check

Classify the solution set of the system whose RREF is [102401−3−10000]\left[\begin{array}{ccc|c}1 & 0 & 2 & 4 \\ 0 & 1 & -3 & -1 \\ 0 & 0 & 0 & 0\end{array}\right], and write the solution set explicitly.

Start by naming the free variable, then solve for the pivot variables.

Solution

Guided solution

Quick check

Why can a consistent system never have exactly two solutions?

Answer from the free-variable theorem, not from a picture.

Solution

Guided solution

Related notes

Read 2.3 Gaussian elimination and RREF for the mechanics of row reduction, and 2.2 Augmented matrices and row operations for the meaning of augmented matrices and elementary row operations.

Prerequisites

2.3

2.3 Gaussian elimination and RREF

Key terms in this unit

More notes in this series

Previous section

2.3 Gaussian elimination and RREF

Next section

3.1 Matrix multiplication and identity matrices