[Linear Algebra Series Part 11] When a System Has One Solution, No Solution, or Infinitely Many

What this post covers

This post classifies the solution structure of a linear system.

When a system has exactly one solution
When a system has no solution
When a system has infinitely many solutions
How pivots and free variables reveal the difference

Key terms

system of equations: several equations solved together
pivot: the first nonzero entry in a nonzero row after elimination
free variable: a variable from a non-pivot column that remains unconstrained
solution set: the full collection of solutions to the system

Core idea

In Part 10, we used Gaussian elimination to turn a system into echelon form. Now we use that simplified form to classify the solution itself: unique, none, or infinitely many.

Once we turn a system into echelon form, we can stop thinking only about “the answer” and start asking a better question:

does a solution exist?
if it exists, is it unique?
if it is not unique, where does the extra freedom come from?

That is what solution structure is about.

A system is called consistent if it has at least one solution. It is inconsistent if elimination produces a contradiction.

For example, a row like

0x + 0y + 0z = 1

means the system is inconsistent, because the left side is always zero while the right side is not.

If the system is consistent and every variable column of the coefficient matrix A contains a pivot, then the solution is unique.

If the system is consistent but some variable column has no pivot, then a free variable remains. That free parameter leaves room for infinitely many solutions.

Here is the decision rule in one place:

Condition	Rank comparison	Free variables	Solution count
Consistent and every variable column has a pivot	`rank([A	b]) = rank(A) = number of variables`	none
Contradiction appears	`rank([A	b]) > rank(A)`	not the issue
Consistent but some variable column has no pivot	`rank([A	b]) = rank(A) < number of variables`	at least one

Step-by-step examples

Example 1) One solution

x + y = 3
x - y = 1

These two equations describe two different lines that meet at exactly one point, so the system has one solution. Elimination gives 2y = 2, so y = 1, and then x = 2. There is a pivot in every variable column.

Example 2) No solution

x + y = 1
x + y = 3

The left sides match but the right sides disagree. Elimination exposes a contradiction, so the system has no solution. In echelon form, this becomes a row like 0 = 2, which cannot be satisfied.

Example 3) Infinitely many solutions

x + y = 2
2x + 2y = 4

The second equation is just twice the first. After elimination we get

x + y = 2
0 = 0

That means one genuine constraint remains, but not enough to determine both variables uniquely. If we set y = t, then x = 2 - t, so the solutions are

(x, y) = (2 - t, t)

for any real number t. Here the y column has no pivot, so y is a free variable.

Example 4) The general shape of a solvable system

If a nonhomogeneous system Ax = b has at least one solution, then every solution can be written as

x = x_particular + x_null

where x_particular is one specific solution and x_null belongs to the null space of A.

That is why the solution set can become a point, a line, a plane, or a higher-dimensional shifted subspace. When b != 0, the solution set is typically an affine subspace, meaning a translated copy of the null space rather than a linear subspace through the origin.

Math notes

A unique solution appears when the system is consistent and no free variables remain.
No solution appears when the augmented system has a contradiction.
Infinitely many solutions appear when the system is consistent and at least one free variable remains.
In rank language, if rank([A|b]) > rank(A), the system is inconsistent.
If rank([A|b]) = rank(A) but that rank is smaller than the number of variables, free variables remain and infinitely many solutions appear.
The system is consistent if and only if rank([A|b]) = rank(A).

So the structure of the solution set depends on both consistency and how many independent constraints survive elimination.

Common mistakes

Thinking the number of equations and variables is enough

It is not. Redundancy and contradiction matter more than raw counts.

Treating a free variable as just a temporary letter

A free variable tells you that the solution set has genuine geometric size, not just one isolated point.

Thinking infinitely many solutions mean something went wrong

Often they reveal an underconstrained system, a parameterized model, or hidden structure that matters later in null space and dimension arguments.

Practice or extension

Classify each system as having one solution, no solution, or infinitely many solutions.

x + y = 2, x - y = 0
x + y = 1, 2x + 2y = 3
x + y = 2, 2x + 2y = 4

Then explain what the pivots and free variables are telling you in each case.

Wrap-up

This post classified the three main solution structures.

A consistent system with no free variables has one solution.
A contradictory row means no solution.
A consistent system with at least one free variable has infinitely many solutions.
This viewpoint prepares us for span, basis, and null space.

Next, we will ask a new question: given some vectors, what can they produce together?