Math Core

When a system has infinitely many solutions, "list them all" is impossible, but "describe them all" is not. This lesson shows that every solution set of a consistent linear system has a clean geometric shape: a span of a few vectors, possibly shifted away from the origin. Writing solutions in parametric vector form makes that shape visible.

Homogeneous systems

Definition

Homogeneous system

A linear system is homogeneous if it can be written as $A\mathbf{x} = \mathbf{0}$ , where $A$ is $m \times n$ and $\mathbf{0}$ is the zero vector in $\mathbb{R}^m$ .

A homogeneous system always has at least one solution, namely $\mathbf{x} = \mathbf{0}$ , called the trivial solution. So a homogeneous system is never inconsistent, and the interesting question is whether it has a nontrivial solution, some $\mathbf{x} \ne \mathbf{0}$ with $A\mathbf{x} = \mathbf{0}$ . By the existence and uniqueness theorem, a consistent system has more than one solution exactly when it has a free variable.

Nontrivial solutions

The homogeneous equation $A\mathbf{x} = \mathbf{0}$ has a nontrivial solution if and only if the equation has at least one free variable.

In particular, if $A$ has more columns than rows (more unknowns than equations), $A\mathbf{x} = \mathbf{0}$ always has a nontrivial solution.

The second sentence holds because an $m \times n$ matrix has at most $m$ pivots, one per row; if $n$ is larger than $m$ , some column is left without a pivot.

Parametric vector form

Worked example: Solving a homogeneous system

Describe all solutions of $A\mathbf{x} = \mathbf{0}$ for $A = \begin{bmatrix} 1 & 2 & -1 \\ 2 & 5 & 1 \end{bmatrix}$ .

Row reduce $[\,A \mid \mathbf{0}\,]$ . $R_2 - 2R_1$ gives $[\,0 \;\; 1 \;\; 3 \mid 0\,]$ , then $R_1 - 2R_2$ gives the RREF

\left[\begin{array}{ccc|c} 1 & 0 & -7 & 0 \\ 0 & 1 & 3 & 0 \end{array}\right].

So $x_1 = 7x_3$ , $x_2 = -3x_3$ , and $x_3$ is free. Write the general solution as a vector and factor out the free variable:

\mathbf{x} = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} 7x_3 \\ -3x_3 \\ x_3 \end{bmatrix} = x_3\begin{bmatrix} 7 \\ -3 \\ 1 \end{bmatrix}.

The solution set is $\operatorname{Span}\{\mathbf{v}\}$ with $\mathbf{v} = (7, -3, 1)$ : a line through the origin in $\mathbb{R}^3$ . Geometrically, it is the line where the two planes $x_1 + 2x_2 - x_3 = 0$ and $2x_1 + 5x_2 + x_3 = 0$ intersect.

The final expression is the parametric vector form of the solution: the solution vector written as a combination of fixed vectors, with free variables as the weights. When there are $k$ free variables, the solution set of $A\mathbf{x} = \mathbf{0}$ is the span of $k$ vectors, one per free variable.

Worked example: Two free variables

Solve the single equation $x_1 - 3x_2 + 2x_3 = 0$ in $\mathbb{R}^3$ .

The coefficient matrix $[\,1 \;\; {-3} \;\; 2\,]$ is already in RREF with one pivot, so $x_2$ and $x_3$ are free and $x_1 = 3x_2 - 2x_3$ :

\mathbf{x} = \begin{bmatrix} 3x_2 - 2x_3 \\ x_2 \\ x_3 \end{bmatrix} = x_2\begin{bmatrix} 3 \\ 1 \\ 0 \end{bmatrix} + x_3\begin{bmatrix} -2 \\ 0 \\ 1 \end{bmatrix}.

The solution set is $\operatorname{Span}\{\mathbf{u}, \mathbf{v}\}$ with $\mathbf{u} = (3, 1, 0)$ and $\mathbf{v} = (-2, 0, 1)$ . Neither is a multiple of the other, so it is a plane through the origin, which is exactly the plane described by the original equation.

Nonhomogeneous systems

Now let $\mathbf{b} \ne \mathbf{0}$ . Solve $A\mathbf{x} = \mathbf{b}$ for the same $A$ as in the first example, with $\mathbf{b} = \begin{bmatrix} 1 \\ 3 \end{bmatrix}$ . The same row operations give

\left[\begin{array}{ccc|c} 1 & 2 & -1 & 1 \\ 2 & 5 & 1 & 3 \end{array}\right] \to \left[\begin{array}{ccc|c} 1 & 0 & -7 & -1 \\ 0 & 1 & 3 & 1 \end{array}\right],

so $x_1 = -1 + 7x_3$ , $x_2 = 1 - 3x_3$ , $x_3$ free, and

\mathbf{x} = \begin{bmatrix} -1 \\ 1 \\ 0 \end{bmatrix} + x_3\begin{bmatrix} 7 \\ -3 \\ 1 \end{bmatrix} = \mathbf{p} + t\mathbf{v}.

The vector $\mathbf{v}$ is the same one that spans the homogeneous solutions, and $\mathbf{p} = (-1, 1, 0)$ is one particular solution of $A\mathbf{x} = \mathbf{b}$ (the one with $x_3 = 0$ ). The solution set is the line through $\mathbf{p}$ parallel to the homogeneous solution line.

This is no accident. If $A\mathbf{p} = \mathbf{b}$ and $A\mathbf{v}_h = \mathbf{0}$ , linearity gives $A(\mathbf{p} + \mathbf{v}_h) = \mathbf{b} + \mathbf{0} = \mathbf{b}$ . Conversely, if $\mathbf{w}$ is any solution of $A\mathbf{x} = \mathbf{b}$ , then $A(\mathbf{w} - \mathbf{p}) = \mathbf{b} - \mathbf{b} = \mathbf{0}$ , so $\mathbf{w} - \mathbf{p}$ is a homogeneous solution.

Structure of solution sets

Suppose $A\mathbf{x} = \mathbf{b}$ is consistent and $\mathbf{p}$ is one solution. Then the solution set of $A\mathbf{x} = \mathbf{b}$ is the set of all vectors

\mathbf{w} = \mathbf{p} + \mathbf{v}_h,

where $\mathbf{v}_h$ is any solution of the homogeneous equation $A\mathbf{x} = \mathbf{0}$ . Geometrically, it is the homogeneous solution set translated by $\mathbf{p}$ .

In two variables the picture is easy to draw. The solutions of $x_1 - 2x_2 = 0$ form a line through the origin spanned by $(2, 1)$ ; the solutions of $x_1 - 2x_2 = 4$ are that same line shifted by the particular solution $\mathbf{p} = (4, 0)$ .

Ax = 0 is the line through the origin (direction v = (2, 1)); Ax = b is the parallel line through p = (4, 0).Open in grapher →

Common mistake

A particular solution $\mathbf{p}$ is not unique; any single solution of $A\mathbf{x} = \mathbf{b}$ works, and different choices describe the same set. Also, the translated set $\mathbf{p} + \operatorname{Span}\{\mathbf{v}\}$ generally does not contain $\mathbf{0}$ , so it is not a span. Only homogeneous solution sets pass through the origin.

A recipe for parametric vector form

Row reduce the augmented matrix to RREF.
Write each basic variable in terms of the free variables.
Write a typical solution $\mathbf{x}$ as a vector whose entries depend on the free variables.
Split $\mathbf{x}$ into a constant vector $\mathbf{p}$ plus one vector for each free variable, with that free variable as its weight.

Tip

Check a parametric answer in two parts: $A\mathbf{p}$ should equal $\mathbf{b}$ , and $A\mathbf{v}$ should equal $\mathbf{0}$ for each direction vector $\mathbf{v}$ . If both hold, every vector $\mathbf{p} + t\mathbf{v}$ is a solution.

Practice

Practice 1

A homogeneous system has $3$ equations and $5$ unknowns. Which statement must be true?

Practice 2

How many free variables does the homogeneous equation $x_1 - 3x_2 + 2x_3 = 0$ (in $\mathbb{R}^3$ ) have?

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 3

For $A = \begin{bmatrix} 1 & 2 & -1 \\ 2 & 5 & 1 \end{bmatrix}$ , every solution of $A\mathbf{x} = \mathbf{0}$ has the form $x_3\mathbf{v}$ . If $x_3 = 2$ , what is $x_1$ ?

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 4

For the same $A$ and $\mathbf{b} = \begin{bmatrix} 1 \\ 3 \end{bmatrix}$ , the solutions of $A\mathbf{x} = \mathbf{b}$ are $\mathbf{x} = \mathbf{p} + x_3\mathbf{v}$ . Find the solution with $x_3 = 1$ , as $(x_1, x_2, x_3)$ .

Enter a point like (2, -3)

Practice 5

Suppose $A\mathbf{p} = \mathbf{b}$ and $A\mathbf{v} = \mathbf{0}$ . Which vector is also a solution of $A\mathbf{x} = \mathbf{b}$ ?

Practice 6

The solution set of the system $x_1 + x_2 + x_3 = 3$ , $\;x_1 - x_2 = 1$ in $\mathbb{R}^3$ is

Practice 7

Let $A = \begin{bmatrix} 1 & -1 & 2 & 1 \\ 2 & -1 & 5 & 4 \\ 1 & 0 & 3 & 3 \end{bmatrix}$ . Write the solutions of $A\mathbf{x} = \mathbf{0}$ in parametric vector form. For the solution with $x_3 = 1$ and $x_4 = -1$ , what is $(x_1, x_2)$ ?

Enter a point like (2, -3)