Solution Sets of Linear Systems

Quick answer

Once a system is in reduced echelon form, its pivots describe the whole solution set. A pivot in the augmented column means no solution. Otherwise every variable with a pivot is determined by the free variables, the ones whose columns have no pivot. With no free variables the solution is unique; with free variables there are infinitely many, and they are written in parametric vector form as one particular solution plus multiples of the solutions of the matching homogeneous system.

What you'll learn

Classify a system as inconsistent, uniquely solvable or underdetermined
Identify basic and free variables from pivot positions
Write a solution set in parametric vector form
Relate the solutions of Ax = b to those of Ax = 0

Three possibilities, no others

Two lines in the plane miss each other, cross once, or coincide. That is the whole story for every linear system, in any number of unknowns:

Reduced form	Solutions
a pivot in the augmented column	none: the system is inconsistent
a pivot in every variable column	exactly one
consistent, with a column lacking a pivot	infinitely many

Nothing else can happen. Two different solutions of the same system can be averaged, or extended past each other, to produce more solutions, so the moment there are two there are infinitely many.

A variable whose column holds a pivot is a basic variable. A variable whose column holds none is a free variable: it may take any value, and the basic variables adjust to match.

Why free variables give a line, a plane, or more

Suppose a consistent system reduces to

\left[\begin{array}{ccc|c} 1 & 2 & 0 & 3\\ 0 & 0 & 1 & 1 \end{array}\right]

Columns $1$ and $3$ hold pivots, so $x_1$ and $x_3$ are basic and $x_2$ is free. The rows say $x_1 = 3 - 2x_2$ and $x_3 = 1$ . Writing $x_2 = t$ and collecting the three components into one vector:

\mathbf{x} = \begin{bmatrix} 3 - 2t \\ t \\ 1 \end{bmatrix} = \begin{bmatrix} 3 \\ 0 \\ 1 \end{bmatrix} + t\begin{bmatrix} -2 \\ 1 \\ 0 \end{bmatrix}

This is parametric vector form: a fixed vector plus $t$ times a direction. As $t$ runs over the real numbers the solutions trace a line through $(3, 0, 1)$ . Each free variable contributes one direction vector, so the solution set is a point, a line, a plane, or a higher-dimensional flat, one dimension per free variable.

Example 1 — Reading a solution set

Solve $x_1 + 2x_2 + 3x_3 = 6$ and $2x_1 + 4x_2 + 9x_3 = 15$ .

Row reduce:

\left[\begin{array}{ccc|c} 1 & 2 & 3 & 6\\ 2 & 4 & 9 & 15 \end{array}\right] \xrightarrow{\;R_2 - 2R_1\;} \left[\begin{array}{ccc|c} 1 & 2 & 3 & 6\\ 0 & 0 & 3 & 3 \end{array}\right] \longrightarrow \left[\begin{array}{ccc|c} 1 & 2 & 0 & 3\\ 0 & 0 & 1 & 1 \end{array}\right]

So $x_3 = 1$ , $x_2$ is free, and $x_1 = 3 - 2x_2$ :

\mathbf{x} = \begin{bmatrix} 3 \\ 0 \\ 1 \end{bmatrix} + t\begin{bmatrix} -2 \\ 1 \\ 0 \end{bmatrix}

Checking $t = 4$ : the vector $(-5, 4, 1)$ gives $-5 + 8 + 3 = 6$ ✓ and $-10 + 16 + 9 = 15$ ✓.

The homogeneous system

Setting every right-hand side to zero gives $A\mathbf{x} = \mathbf{0}$ , a homogeneous system. It is always consistent, since $\mathbf{x} = \mathbf{0}$ works; that is the trivial solution. The interesting question is whether there are others, and the answer is yes exactly when there is a free variable.

The two problems are linked. If $\mathbf{p}$ solves $A\mathbf{x} = \mathbf{b}$ and $\mathbf{v}$ solves $A\mathbf{x} = \mathbf{0}$ , then $\mathbf{p} + \mathbf{v}$ solves $A\mathbf{x} = \mathbf{b}$ as well, and every solution arises this way.

The solution set of $A\mathbf{x} = \mathbf{b}$ , when it is not empty, is a translate of the solution set of $A\mathbf{x} = \mathbf{0}$ .

In the example above, $(3, 0, 1)$ is the translation and $t(-2, 1, 0)$ is the line of homogeneous solutions.

x₁ + 2x₂ = 3

Solutions run from (3, 0) in the direction (−2, 1)

Worked examples

Common mistakes

Practice problems

A system in $4$ unknowns reduces to a consistent form with pivots in columns 1, 2 and 4. Which variables are free?

Answer

$x_3$ only

Full solution

Every column without a pivot belongs to a free variable, and here that is the third.
How many solutions does that system have?

Answer

Infinitely many

Full solution

One free variable gives a line of solutions, one solution for each value of $x_3$ .
Write the solution set of $x_1 + 3x_2 = 6$ in parametric vector form.

Answer

$\mathbf{x} = \begin{bmatrix} 6 \\ 0 \end{bmatrix} + t\begin{bmatrix} -3 \\ 1 \end{bmatrix}$

Full solution

$x_2 = t$ is free and $x_1 = 6 - 3t$ . Collecting the components gives a particular solution plus $t$ times a direction.
Write the solution set of $x_1 + 3x_2 = 0$ in parametric vector form.

Answer

$\mathbf{x} = t\begin{bmatrix} -3 \\ 1 \end{bmatrix}$

Full solution

The homogeneous version has the same direction and no particular solution: it is the same line moved to pass through the origin.
Does $A\mathbf{x} = \mathbf{0}$ always have a solution?

Answer

Yes, $\mathbf{x} = \mathbf{0}$

Full solution

The zero vector satisfies every homogeneous equation, so such a system is never inconsistent. It is the trivial solution.
A system of $2$ equations in $4$ unknowns is consistent. What can you say about its solutions?

Answer

It has infinitely many, with at least two free variables.

Full solution

At most two pivots fit in two rows, leaving at least two of the four columns without one.
Solve $x_1 + x_2 - x_3 = 2$ , $2x_1 - x_2 + x_3 = 1$ , $x_1 + 2x_2 - 2x_3 = 3$ .

Hint

Row reduce, and watch for a zero row.

Answer

$\mathbf{x} = \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix} + t\begin{bmatrix} 0 \\ 1 \\ 1 \end{bmatrix}$

Full solution

Row reduction gives $\left[\begin{array}{ccc|c} 1 & 0 & 0 & 1\\ 0 & 1 & -1 & 1\\ 0 & 0 & 0 & 0 \end{array}\right]$ . So $x_1 = 1$ , $x_3 = t$ is free, and $x_2 = 1 + t$ .
Check that $t = 2$ in exercise 7 gives a solution.

Answer

$(1, 3, 2)$ satisfies all three equations.

Full solution

$1 + 3 - 2 = 2$ ✓, $2 - 3 + 2 = 1$ ✓ and $1 + 6 - 4 = 3$ ✓.
If $\mathbf{p}$ and $\mathbf{q}$ both solve $A\mathbf{x} = \mathbf{b}$ , what does $\mathbf{p} - \mathbf{q}$ solve?

Answer

$A\mathbf{x} = \mathbf{0}$

Full solution

Subtracting the two systems cancels $\mathbf{b}$ , so the difference of two solutions solves the homogeneous system. This is why every solution set is one solution plus the homogeneous set.
A student solves a consistent system with one free variable and reports the single solution found by setting the free variable to $0$ . What went wrong?

Hint

What do other values of the free variable give?

Answer

That is one solution among infinitely many. The answer is the whole set $\mathbf{p} + t\mathbf{v}$ .

Full solution

Setting the free variable to $0$ picks out a particular solution, which is a fine starting point. Every other value of the parameter gives another solution, so reporting one hides the line that the solution set actually is.

Frequently asked questions

How many solutions can a linear system have?

None, exactly one, or infinitely many. A linear system never has, say, exactly two solutions.

What is a free variable?

A variable whose column has no pivot. It can take any value, and the variables with pivots are then determined by it.

What is parametric vector form?

The solution set written as p + t₁d₁ + ⋯ + t_kd_k: one particular solution p plus multiples of direction vectors, one per free variable.

What is a homogeneous system?

One of the form Ax = 0. It is always consistent, since x = 0 is a solution, called the trivial solution.

How are the solutions of Ax = b and Ax = 0 related?

If Ax = b is consistent with one solution p, its whole solution set is p plus the solution set of Ax = 0.

What to learn next

For teachersPrint problems onlyTick to print a blank worksheet; leave it clear to print the lesson with the full answer key.