Homogeneity of Linear Systems

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Table of Contents

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Definition of homogeneous linear system

Homogeneous Linear System

A linear system is said to be homogeneous if it can be written in the form \(\mathbf{Ax = 0}\) where \(\mathbf{A}\) is an \(m \times n\) matrix, \(\mathbf{x} \in \mathbb{R}^n\) is an unknown vector, and \(\mathbf{0} \in \mathbb{R}^m\) is the zero vector of \(\mathbb{R}^m\).

Note

The equation \(\mathbf{Ax = 0}\) always admits the trivial solution \(\mathbf{x = 0}\).

Besides the trivial solution \(\mathbf{x = 0}\), the equation \(\mathbf{Ax = 0}\) may admit non trivial solutions:

• If \(\mathbf{Ax = 0}\) admits a non trivial solution then it admits infinitely many solutions: simply multiply the non trivial solution by a scalar other than zero or one to obtain another non trivial solution.

• The linear system \(\mathbf{Ax = 0}\) has non trivial solutions iff the system has at least one free variable (i.e., the system has fewer pivots than there are columns in the coefficient matrix \(\mathbf{A}\)).

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Example

The following linear system is homogeneous.

\( \underset{\mathbf{A}}{ \begin{bmatrix*}[r] 1 & 0 & 2 \\ 0 & 2 & 0 \\ 2 & 0 & 3 \\ \end {bmatrix*} } \underset{\mathbf{x}}{ \begin{bmatrix*}[r] x_1 \\ x_2 \\ x_3 \\ \end {bmatrix*} } = \underset{\mathbf{0}}{ \begin{bmatrix*}[r] 0 \\ 0 \\ 0 \\ \end {bmatrix*} } \)

\( \begin{bmatrix} \boxed{1} & 0 & 2 \\ 0 & 2 & 0 \\ 2 & 0 & 3 \\ \end {bmatrix} \underset{r_3 \leftarrow\,\,\, r_3 + (-2)r_1}{\rightarrow} \underbrace{ \begin{bmatrix} \boxed{1} & 0 & 2 \\ 0 & \boxed{2} & 0 \\ 0 & 0 & \boxed{-1} \\ \end {bmatrix} }_{\textbf{row echelon}} \underset{r_3 \leftarrow\,\,\, (-1)r_3}{\rightarrow} \begin{bmatrix} \boxed{1} & 0 & 2 \\ 0 & \boxed{2} & 0 \\ 0 & 0 & \boxed{1} \\ \end {bmatrix} \)

\( \begin{bmatrix} \boxed{1} & 0 & 2 \\ 0 & \boxed{2} & 0 \\ 0 & 0 & \boxed{1} \\ \end {bmatrix} \underset{r_1 \leftarrow\,\,\, r_1 + (-2)r_3}{\rightarrow} \begin{bmatrix} \boxed{1} & 0 & 0 \\ 0 & \boxed{2} & 0 \\ 0 & 0 & \boxed{1} \\ \end {bmatrix} \underset{r_2 \leftarrow\,\,\, \frac{1}{2} r_2}{\rightarrow} \underbrace{ \begin{bmatrix} \boxed{1} & 0 & 0 \\ 0 & \boxed{1} & 0 \\ 0 & 0 & \boxed{1} \\ \end {bmatrix} }_{\textbf{reduced}} \)

Since the linear system has no free variable the only solution to the system is the trivial solution \(\mathbf{x = 0}\).

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Example

The following linear system is homogeneous.

\( \underset{\mathbf{A}}{ \begin{bmatrix*}[r] 3 & 4 & 0 \\ 6 & 8 & 5 \\ 0 & 0 & 7 \\ \end {bmatrix*} } \underset{\mathbf{x}}{ \begin{bmatrix*}[r] x_1 \\ x_2 \\ x_3 \\ \end {bmatrix*} } = \underset{\mathbf{0}}{ \begin{bmatrix*}[r] 0 \\ 0 \\ 0 \\ \end {bmatrix*} } \)

\( \begin{bmatrix} \boxed{3} & 4 & 0 \\ 6 & 8 & 5 \\ 0 & 0 & 7 \\ \end {bmatrix} \underset{r_2 \leftarrow\,\,\, r_2 + (-2)r_1}{\rightarrow} \begin{bmatrix} \boxed{3} & 4 & 0 \\ 0 & 0 & 5 \\ 0 & 0 & 7 \\ \end {bmatrix} \underset{r_3 \leftarrow\,\,\, r_3 + (-\frac{7}{5})r_2}{\rightarrow} \underbrace{ \begin{bmatrix} \boxed{3} & 4 & 0 \\ 0 & 0 & \boxed{5} \\ 0 & 0 & 0 \\ \end {bmatrix} }_{\textbf{row echelon}} \)

\( \begin{bmatrix} \boxed{3} & 4 & 0 \\ 0 & 0 & \boxed{5} \\ 0 & 0 & 0 \\ \end {bmatrix} \underset{r_2 \leftarrow\,\,\, \frac{1}{5} r_2}{\rightarrow} \begin{bmatrix} \boxed{3} & 4 & 0 \\ 0 & 0 & \boxed{1} \\ 0 & 0 & 0 \\ \end {bmatrix} \underset{r_1 \leftarrow\,\,\, \frac{1}{3} r_1}{\rightarrow} \underbrace{ \begin{bmatrix} \boxed{1} & \frac{4}{3} & 0 \\ 0 & 0 & \boxed{1} \\ 0 & 0 & 0 \\ \end {bmatrix} }_{\textbf{reduced}} \)

The linear system has infinitely many non trivial solutions–multiples of the vector

\( \begin{bmatrix*}[r] -4 \\ 3 \\ 0 \\ \end {bmatrix*} \)

as well as the trivial solution \(\mathbf{x = 0}\)–since the system has at least one free variable.

\( \begin{aligned} x_1 + \frac{4}{3} x_2 &= 0 && \iff && x_1 = -\frac{4}{3} x_2 \\ & && && x_2 \,\,\,\text{free} \\ & && && x_3 = 0 \\ \end {aligned} \)

\( \mathbf{x} = \begin{bmatrix*}[r] -\frac{4}{3} t \\ t \\ 0 \\ \end{bmatrix*} = t \begin{bmatrix*}[r] -\frac{4}{3} \\ 1 \\ 0 \\ \end{bmatrix*} \)

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Recipe for determining whether a homogeneous linear system has non trivial solutions

Recipe for determining whether a homogeneous linear system has non trivial solutions

STEP 1 - Put the homogeneous linear system into its augmented matrix representation (the last column is a zero column).

STEP 2 - Row reduce the augmented matrix until a row echelon matrix is obtained (the last column is still a zero column).

STEP 3 - Identify free variables–pivot-free columns in the coefficient matrix–if any.

STEP 4 - If there exists at least one free variable then there are infinitely many non trivial solutions. Otherwise, the trivial solution is the only solution.

(STEP 5) - Row reduce further to obtain a reduced row echelon matrix in order to read off the non trivial solutions.

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Example

PROBLEM STATEMENT

Does the following linear system admit non trivial solutions?

\( \begin{aligned} x_1 && - && 2x_2 && + && 3x_3 && = && 0 \\ -2x_1 && - && 3x_2 && - && 4x_3 && = && 0 \\ 2x_1 && - && 4x_2 && + && 9x_3 && = && 0 \\ \end {aligned} \)

ROW REDUCE

\( \left [ \begin{matrix*}[r] \boxed{1} & -2 & 3 \\ -2 & -3 & -4 \\ 2 & -4 & 9 \\ \end {matrix*} \left | \begin{matrix*}[r] 0 \\ 0 \\ 0 \\ \end {matrix*} \right. \right] \underset{r_2 \leftarrow\,\,\, r_2 + 2r_1}{\rightarrow} \left [ \begin{matrix*}[r] \boxed{1} & -2 & 3 \\ 0 & -7 & 2 \\ 2 & -4 & 9 \\ \end {matrix*} \left | \begin{matrix*}[r] 0 \\ 0 \\ 0 \\ \end {matrix*} \right. \right] \underset{r_3 \leftarrow\,\,\, r_3 + (-2)r_1}{\rightarrow} \underbrace{ \left [ \begin{matrix*}[r] \boxed{1} & -2 & 3 \\ 0 & \boxed{-7} & 2 \\ 0 & 0 & \boxed{3} \\ \end {matrix*} \left | \begin{matrix*}[r] 0 \\ 0 \\ 0 \\ \end {matrix*} \right. \right] }_{\textbf{row echelon}} \)

The linear system admits only the trivial solution \(\mathbf{x = 0}\) because the row-equivalent row echelon matrix contains no pivot-free column.

\( \left [ \begin{matrix*}[r] \boxed{1} & -2 & 3 \\ 0 & \boxed{-7} & 2 \\ 0 & 0 & \boxed{3} \\ \end {matrix*} \left | \begin{matrix*}[r] 0 \\ 0 \\ 0 \\ \end {matrix*} \right. \right] \underset{r_3 \leftarrow\,\,\, \frac{1}{3} r_3}{\rightarrow} \left [ \begin{matrix*}[r] \boxed{1} & -2 & 3 \\ 0 & \boxed{-7} & 2 \\ 0 & 0 & \boxed{1} \\ \end {matrix*} \left | \begin{matrix*}[r] 0 \\ 0 \\ 0 \\ \end {matrix*} \right. \right] \underset{r_2 \leftarrow\,\,\, r_2 + (-2)r_3}{\rightarrow} \left [ \begin{matrix*}[r] \boxed{1} & -2 & 3 \\ 0 & \boxed{-7} & 0 \\ 0 & 0 & \boxed{1} \\ \end {matrix*} \left | \begin{matrix*}[r] 0 \\ 0 \\ 0 \\ \end {matrix*} \right. \right] \)

\( \left [ \begin{matrix*}[r] \boxed{1} & -2 & 3 \\ 0 & \boxed{-7} & 0 \\ 0 & 0 & \boxed{1} \\ \end {matrix*} \left | \begin{matrix*}[r] 0 \\ 0 \\ 0 \\ \end {matrix*} \right. \right] \underset{r_1 \leftarrow\,\,\, r_1 + (-3)r_3}{\rightarrow} \left [ \begin{matrix*}[r] \boxed{1} & -2 & 0 \\ 0 & \boxed{-7} & 0 \\ 0 & 0 & \boxed{1} \\ \end {matrix*} \left | \begin{matrix*}[r] 0 \\ 0 \\ 0 \\ \end {matrix*} \right. \right] \underset{r_2 \leftarrow\,\,\, (-\frac{1}{7})r_2}{\rightarrow} \left [ \begin{matrix*}[r] \boxed{1} & -2 & 0 \\ 0 & \boxed{1} & 0 \\ 0 & 0 & \boxed{1} \\ \end {matrix*} \left | \begin{matrix*}[r] 0 \\ 0 \\ 0 \\ \end {matrix*} \right. \right] \)

\( \left [ \begin{matrix*}[r] \boxed{1} & -2 & 0 \\ 0 & \boxed{1} & 0 \\ 0 & 0 & \boxed{1} \\ \end {matrix*} \left | \begin{matrix*}[r] 0 \\ 0 \\ 0 \\ \end {matrix*} \right. \right] \underset{r_1 \leftarrow\,\,\, r_1 + 2r_2}{\rightarrow} \underbrace{ \left [ \begin{matrix*}[r] \boxed{1} & 0 & 0 \\ 0 & \boxed{1} & 0 \\ 0 & 0 & \boxed{1} \\ \end {matrix*} \left | \begin{matrix*}[r] 0 \\ 0 \\ 0 \\ \end {matrix*} \right. \right] }_{\textbf{reduced}} \)

In geometric terms, the solution set consists of a single point: the origin of \(\mathbb{R}^3\).

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Example

PROBLEM STATEMENT

Does the following linear system admit non trivial solutions?

\( \begin{aligned} 2x_1 && + && 4x_2 && - && 6x_3 && = && 0 \\ 4x_1 && + && 8x_2 && - && 10x_3 && = && 0 \\ \end {aligned} \)

ROW REDUCE

\( \left [ \begin{matrix*}[r] \boxed{2} & 4 & -6 \\ 4 & 8 & -10 \\ \end {matrix*} \left | \begin{matrix*}[r] 0 \\ 0 \\ \end {matrix*} \right. \right] \underset{r_2 \leftarrow\,\,\, r_2 + (-2)r_1}{\rightarrow} \underbrace{ \left [ \begin{matrix*}[r] \boxed{2} & 4 & -6 \\ 0 & 0 & \boxed{2} \\ \end {matrix*} \left | \begin{matrix*}[r] 0 \\ 0 \\ \end {matrix*} \right. \right] }_{\textbf{row echelon}} \underset{r_2 \leftarrow\,\,\, \frac{1}{2} r_2}{\rightarrow} \left [ \begin{matrix*}[r] \boxed{2} & 4 & -6 \\ 0 & 0 & \boxed{1} \\ \end {matrix*} \left | \begin{matrix*}[r] 0 \\ 0 \\ \end {matrix*} \right. \right] \)

The linear system admits infinitely many non trivial solutions because the row-equivalent row echelon matrix contains at least one pivot-free column.

\( \left [ \begin{matrix*}[r] \boxed{2} & 4 & -6 \\ 0 & 0 & \boxed{1} \\ \end {matrix*} \left | \begin{matrix*}[r] 0 \\ 0 \\ \end {matrix*} \right. \right] \underset{r_1 \leftarrow\,\,\, r_1 + 6r_2}{\rightarrow} \left [ \begin{matrix*}[r] \boxed{2} & 4 & 0 \\ 0 & 0 & \boxed{1} \\ \end {matrix*} \left | \begin{matrix*}[r] 0 \\ 0 \\ \end {matrix*} \right. \right] \underset{r_1 \leftarrow\,\,\, \frac{1}{2} r_1}{\rightarrow} \underbrace{ \left [ \begin{matrix*}[r] \boxed{1} & 2 & 0 \\ 0 & 0 & \boxed{1} \\ \end {matrix*} \left | \begin{matrix*}[r] 0 \\ 0 \\ \end {matrix*} \right. \right] }_{\textbf{reduced}} \)

\( \begin{aligned} x_1 + 2x_2 &= 0 && \iff && x_1 = -2x_2 \\ & && && x_2 \,\,\,\text{free} \\ & && && x_3 = 0 \\ \end {aligned} \)

\( \mathbf{x} = \begin{bmatrix*}[r] -2t \\ t \\ 0 \\ \end{bmatrix*} = t\begin{bmatrix*}[r] -2 \\ 1 \\ 0 \\ \end{bmatrix*} \)

In geometric terms, the solution set is a line through the origin

\( \text{span} \left \{ \begin{bmatrix*}[r] -2 \\ 1 \\ 0 \\ \end {bmatrix*} \right\} \)

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Example

PROBLEM STATEMENT

Does the following linear system admit non trivial solutions?

\( \begin{aligned} x_1 && - && 4x_2 && - && x_3 && = && 0 \\ \end {aligned} \)

ROW REDUCE

\( \underbrace{ \left [ \begin{matrix*}[r] \boxed{1} & -4 & -1 \\ \end {matrix*} \left | \begin{matrix*}[r] 0 \\ \end {matrix*} \right. \right] }_{\textbf{reduced row echelon}} \)

The linear system admits infinitely many non trivial solutions because the row-equivalent row echelon matrix contains at least one pivot-free column.

\( \begin{aligned} x_1 - 4x_2 - x_3 &= 0 && \iff && x_1 = 4x_2 + x_3 \\ & && && x_2 \,\,\,\text{free} \\ & && && x_3 \,\,\,\text{free} \\ \end {aligned} \)

\( \mathbf{x} = \begin{bmatrix*}[r] 4s + t \\ s \\ t \\ \end{bmatrix*} = \begin{bmatrix*}[r] 4s \\ s \\ 0 \\ \end{bmatrix*} + \begin{bmatrix*}[r] t \\ 0 \\ t \\ \end{bmatrix*} = s \begin{bmatrix*}[r] 4 \\ 1 \\ 0 \\ \end{bmatrix*} + t \begin{bmatrix*}[r] 1 \\ 0 \\ 1 \\ \end{bmatrix*} \)

In geometric terms, the solution set is a plane through the origin

\( \text{span} \left \{ \begin{bmatrix*}[r] 4 \\ 1 \\ 0 \\ \end {bmatrix*}, \begin{bmatrix*}[r] 1 \\ 0 \\ 1 \\ \end {bmatrix*} \right\} \)

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Describing solution sets with parametric vectors

Parametric vectors can be used to describe solution sets containing more than a single solution. For example:

\(\mathbf{x} = s\mathbf{u} + t\mathbf{v}\) where \(\mathbf{u, v}\) are two solutions and \(s, t \in \mathbb{R}\) are parameters.

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Theorem

Theorem

1. Every homogeneous linear system \(\mathbf{Ax = b}\) admits the trivial solution \(\mathbf{x = 0}\). (Besides the trivial solution and depending on its coefficient matrix a homogeneous linear system may admit non trivial solutions.)

2. If \(\mathbf{u, v}\) are two solutions of the homogeneous linear system \(\mathbf{Ax = b}\) then \(\mathbf{u + v}\) is also a solution of the system.

3. If \(\mathbf{u}\) is a solution of the homogeneous linear system \(\mathbf{Ax = b}\) and \(c\) is a scalar then \(c\mathbf{u}\) is also a solution of the system.

Proof

1. \(\mathbf{A0 = 0}\)

2. \(\mathbf{A(u + v) = Au + Av = 0 + 0 = 0}\)

3. \(\mathbf{A}(c\mathbf{u}) = c\mathbf{Au} = c\mathbf{0} = \mathbf{0}\)

Corollary

For a homogeneous linear system any linear combination of solutions is itself a solution.

Corollary

The solution set of a homogenous linear system is either

\(\{\mathbf{0}\}\)

if there is no free variable or the linear span

\(\text{span}\{\mathbf{u, v, \dots, w}\}\)

of a number of vectors if there is at least one free variable.

Note

\(\text{span}\{\mathbf{0}\} = \{\mathbf{0}\}\)

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Definition of non homogeneous linear system

Nonhomogeneous Linear System

A linear system is said to be non homogeneous if it can be written in the form \(\mathbf{Ax = b}\) where \(\mathbf{A}\) is an \(m \times n\) matrix, \(\mathbf{x} \in \mathbb{R}^n\) is an unknown vector, and \(\mathbf{b} \in \mathbb{R}^m\) is a non zero vector of \(\mathbb{R}^m\).

Note

The solution set of a non homogeneous linear system cannot contain the zero vector.

To see why, let the zero vector be a solution the non homogenous linear system \(\mathbf{Ax = b}\)

\( \underbrace{ \begin{bmatrix*}[r] a_{11} & a_{12} & \dots & a_{1n} \\ a_{21} & a_{22} & \dots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{m1} & a_{m2} & \dots & a_{mn} \\ \end {bmatrix*} }_{\mathbf{A}} \underbrace{ \begin{bmatrix} 0 \\ 0 \\ \vdots \\ 0 \\ \end{bmatrix} }_{\mathbf{x}} = 0 \begin{bmatrix*}[r] a_{11} \\ a_{12} \\ \vdots \\ a_{1n} \\ \end{bmatrix*} + 0 \begin{bmatrix*}[r] a_{21} \\ a_{22} \\ \vdots \\ a_{2n} \\ \end{bmatrix*} + \dots + 0 \begin{bmatrix*}[r] a_{m1} \\ a_{m2} \\ \vdots \\ a_{mn} \\ \end{bmatrix*} = \begin{bmatrix*}[r] 0 \\ 0 \\ \vdots \\ 0 \\ \end{bmatrix*} + \begin{bmatrix*}[r] 0 \\ 0 \\ \vdots \\ 0 \\ \end{bmatrix*} + \dots + \begin{bmatrix*}[r] 0 \\ 0 \\ \vdots \\ 0 \\ \end{bmatrix*} = \underbrace{ \begin{bmatrix*}[r] 0 \\ 0 \\ \vdots \\ 0 \\ \end{bmatrix*} }_{\mathbf{b}} \)

This doesn’t work because vector \(\mathbf{b}\) by definition is non zero.

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Theorem

Theorem

1. The solution set of a non homogeneous linear system \(\mathbf{Ax = b}\) does not contain the zero vector.

2. If \(\mathbf{p, q}\) are two distinct solutions of a non homogeneous linear system \(\mathbf{Ax = b}\) then their difference \(\mathbf{q - p}\) is a non trivial solution of the homogeneous linear system \(\mathbf{Ax = 0}\) (sharing the same coefficient matrix \(\mathbf{A}\)).

3. If a linear system \(\mathbf{Ax = b}\) admits two distinct solutions \(\mathbf{p, q}\) then the system \(\mathbf{Ax = b}\) actually admits infinitely many solutions.

Proof

1. \(\mathbf{A0 = 0 \ne b}\)

2. \(\mathbf{A(q - p)}\) \(\mathbf{= Aq - Ap = b - b = 0}\)

3. \(\forall t \in \mathbb{R}\) the vector \(\mathbf{p} + t(\mathbf{q - p})\) is a solution of the non homogeneous linear system \(\mathbf{Ax = b}\):

\( \mathbf{A}(\mathbf{p} + t(\mathbf{q - p})) = \mathbf{Ap} + t\mathbf{A(q - p)} = \mathbf{b} + t\mathbf{0} = \mathbf{b} \)

Corollary

A non homogeneous linear system \(\mathbf{Ax = b}\) has

either no solution

or one, unique (necessarily non trivial) solution

or infinitely many solutions which can be written in parametric vector form

\(\mathbf{p} + r\mathbf{u} + s\mathbf{v} + \dots + t\mathbf{w}\)

where \(\mathbf{p}\) is a particular (necessarily non trivial) solution of the non homogeneous linear system \(\mathbf{Ax = b}\) and \(r\mathbf{u} + s\mathbf{v} + \dots + t\mathbf{w}\) is the solution set of the associated homogeneous linear system \(\mathbf{Ax = 0}\) (sharing the same coefficient matrix \(\mathbf{A}\)) in parametric vector form

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Example

PROBLEM STATEMENT

Describe the solution set of the following linear system in parametric vector form.

\( \begin{aligned} 2x_1 && + && 4x_2 && - && 6x_3 && = && 0 \\ 3x_1 && + && 6x_2 && - && 9x_3 && = && -3 \\ \end {aligned} \)

HOMOGENEITY

The linear system is non homogeneous \(\mathbf{Ax = b}\).

ROW REDUCE

\( \left [ \begin{matrix*}[r] \boxed{2} & 4 & -6 \\ 3 & 6 & -9 \\ \end {matrix*} \left | \begin{matrix*}[r] 0 \\ -3 \\ \end {matrix*} \right. \right] \underset{r_2 \leftarrow\,\,\, r_2 + (-\frac{3}{2})r_1}{\rightarrow} \underbrace{ \left [ \begin{matrix*}[r] \boxed{2} & 4 & -6 \\ 0 & 0 & 0 \\ \end {matrix*} \left | \begin{matrix*}[r] 0 \\ \boxed{-3} \\ \end {matrix*} \right. \right] }_{\textbf{row echelon}} \)

CONSISTENCY

The linear system is inconsistent and so has no solution.

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Example

PROBLEM STATEMENT

Describe the solution set of the following linear system in parametric vector form.

\( \begin{aligned} 2x_1 && + && 4x_2 && - && 6x_3 && = && 0 \\ 4x_1 && + && 8x_2 && - && 10x_3 && = && 4 \\ \end {aligned} \)

HOMOGENEITY

The linear system is non homogeneous \(\mathbf{Ax = b}\).

ROW REDUCE

\( \left [ \begin{matrix*}[r] \boxed{2} & 4 & -6 \\ 4 & 8 & -10 \\ \end {matrix*} \left | \begin{matrix*}[r] 0 \\ 4 \\ \end {matrix*} \right. \right] \underset{r_2 \leftarrow\,\,\, r_2 + (-2)r_1}{\rightarrow} \underbrace{ \left [ \begin{matrix*}[r] \boxed{2} & 4 & -6 \\ 0 & 0 & \boxed{2} \\ \end {matrix*} \left | \begin{matrix*}[r] 0 \\ 4 \\ \end {matrix*} \right. \right] }_{\textbf{row echelon}} \underset{r_2 \leftarrow\,\,\, \frac{1}{2} r_2}{\rightarrow} \left [ \begin{matrix*}[r] \boxed{2} & 4 & -6 \\ 0 & 0 & \boxed{1} \\ \end {matrix*} \left | \begin{matrix*}[r] 0 \\ 2 \\ \end {matrix*} \right. \right] \)

\( \left [ \begin{matrix*}[r] \boxed{2} & 4 & -6 \\ 0 & 0 & \boxed{1} \\ \end {matrix*} \left | \begin{matrix*}[r] 0 \\ 2 \\ \end {matrix*} \right. \right] \underset{r_1 \leftarrow\,\,\, r_1 + 6r_2}{\rightarrow} \left [ \begin{matrix*}[r] \boxed{2} & 4 & 0 \\ 0 & 0 & \boxed{1} \\ \end {matrix*} \left | \begin{matrix*}[r] 12 \\ 2 \\ \end {matrix*} \right. \right] \underset{r_1 \leftarrow\,\,\, \frac{1}{2} r_1}{\rightarrow} \underbrace{ \left [ \begin{matrix*}[r] \boxed{1} & 2 & 0 \\ 0 & 0 & \boxed{1} \\ \end {matrix*} \left | \begin{matrix*}[r] 6 \\ 2 \\ \end {matrix*} \right. \right] }_{\textbf{reduced}} \)

CONSISTENCY

The linear system is consistent and has infinitely many solutions since there is at least one free variable.

\( \begin{aligned} x_1 + 2x_2 &= 6 && \iff && x_1 = -2x_2 + 6 \\ & && && x_2 \,\,\,\text{free} \\ & && && x_3 = 2 \\ \end {aligned} \)

\( \mathbf{x} = \begin{bmatrix*}[r] -2t + 6 \\ t \\ 2 \\ \end{bmatrix*} = \begin{bmatrix*}[r] 6 \\ 0 \\ 2 \\ \end{bmatrix*} + \begin{bmatrix*}[r] -2t \\ t \\ 0 \\ \end{bmatrix*} = \begin{bmatrix*}[r] 6 \\ 0 \\ 2 \\ \end{bmatrix*} + t \begin{bmatrix*}[r] -2 \\ 1 \\ 0 \\ \end{bmatrix*} \)

The solution set is a line through the point \((6, 0, 2)\) and parallel to

\(\text{span}\left\{\begin{bmatrix*}[r] -2 \\ 1 \\ 0 \\ \end{bmatrix*}\right\}\)

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Example

PROBLEM STATEMENT

Describe the solution set of the following linear system in parametric vector form.

\( \begin{aligned} x_1 && - && 4x_2 && - && x_3 && = && 5 \\ \end {aligned} \)

ROW REDUCE

\( \underbrace{ \left [ \begin{matrix*}[r] \boxed{1} & -4 & -1 \\ \end {matrix*} \left | \begin{matrix*}[r] 5 \\ \end {matrix*} \right. \right] }_{\textbf{reduced row echelon}} \)

CONSISTENCY

The linear system is consistent and has infinitely many solutions since there is at least one free variable.

\( \begin{aligned} x_1 - 4x_2 - x_3 &= 5 && \iff && x_1 = 4x_2 + x_3 + 5 \\ & && && x_2 \,\,\,\text{free} \\ & && && x_3 \,\,\,\text{free} \\ \end {aligned} \)

\( \mathbf{x} = \begin{bmatrix*}[r] 4s + t + 5 \\ s \\ t \\ \end{bmatrix*} = \begin{bmatrix*}[r] 5 \\ 0 \\ 0 \\ \end{bmatrix*} + \begin{bmatrix*}[r] 4s \\ s \\ 0 \\ \end{bmatrix*} + \begin{bmatrix*}[r] t \\ 0 \\ t \\ \end{bmatrix*} = \begin{bmatrix*}[r] 5 \\ 0 \\ 0 \\ \end{bmatrix*} + s \begin{bmatrix*}[r] 4 \\ 1 \\ 0 \\ \end{bmatrix*} + t \begin{bmatrix*}[r] 1 \\ 0 \\ 1 \\ \end{bmatrix*} \)

The solution set is a plane through the point \((5, 0, 0)\) and parallel to

\(\text{span} \left \{ \begin{bmatrix*}[r] 4 \\ 1 \\ 0 \\ \end{bmatrix*}, \begin{bmatrix*}[r] 1 \\ 0 \\ 1 \\ \end{bmatrix*} \right\} \)

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Geometric interpretations

If the linear system \(\mathbf{Ax = b}\) admits one free variable and its solution set in parametric form is

\(\{\mathbf{p} + s\mathbf{u} \mid s \in \mathbb{R}\}\)

then the solution set can be described geometrically as the line through the tip of the vector \(\mathbf{p}\) parallel to the vector \(\mathbf{u}\), and the line spanned by the vector \(\mathbf{u}\) is the solution set of \(\mathbf{Ax = 0}\).

If the linear system \(\mathbf{Ax = b}\) admits two free variables and its solution set in parametric form is

\(\{\mathbf{p} + s\mathbf{u} + t\mathbf{v} \mid s, t \in \mathbb{R}\}\)

then the solution set can be described geometrically as the plane through the tip of the vector \(\mathbf{p}\) parallel to the vectors \(\mathbf{u, v}\), and the plane spanned by the vectors \(\mathbf{u, v}\) is the solution set of \(\mathbf{Ax = 0}\).

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Example

Find the parametric vector describing the points on the line through the tips of the vectors

\( \mathbf{p} = \begin{bmatrix*}[r] -3 \\ 2 \\ \end{bmatrix*}, \mathbf{q} = \begin{bmatrix*}[r] 0 \\ -3 \\ \end{bmatrix*} \)

Every point on the line through the tips of the vectors \(\mathbf{p, q}\) is the tip of a vector of the form

\( \mathbf{p} + t\mathbf{(q - p)} = \begin{bmatrix*}[r] -3 \\ 2 \\ \end{bmatrix*} + t \left(\begin{bmatrix*}[r] 0 \\ -3 \\ \end{bmatrix*} - \begin{bmatrix*}[r] -3 \\ 2 \\ \end{bmatrix*}\right) = \begin{bmatrix*}[r] -3 \\ 2 \\ \end{bmatrix*} + t \begin{bmatrix*}[r] 3 \\ -5 \\ \end{bmatrix*} \)

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Summary

The non homogeneous linear system \(\mathbf{Ax = b}\)	…has no free variable	…has at least one free variable	.
…is consistent	\(\mathbf{Ax = 0}\) only admits the trivial solution \(\mathbf{Ax = b}\) admits the single non trivial solution \(\mathbf{p}\)	The solution set of \(\mathbf{Ax = 0}\) is of the form \(\text{span}\{\mathbf{v_1, v_2, \dots, v_k}\}\) The solution set of \(\mathbf{Ax = b}\) is of the form \(\mathbf{p} + \text{span}\{\mathbf{v_1, v_2, \dots, v_k}\}\) for some \(\mathbf{p \ne 0}\)	…is necessarily consistent
…is inconsistent	\(\mathbf{Ax = 0}\) only admits the trivial solution \(\mathbf{Ax = b}\) admits no solution	The solution set of \(\mathbf{Ax = 0}\) is of the form \(\text{span}\{\mathbf{v_1, v_2, \dots, v_k}\}\) The solution set of \(\mathbf{Ax = b}\) is empty	…is necessarily consistent
.	…has no free variable	…has at least one free variable	The associated homogeneous system \(\mathbf{Ax = 0}...\)

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