Recall:

Theorem 2 (Existence and Uniqueness Theorem)

Existence: A linear system is consistent if and only if the rightmost column of the augmented matrix is not a pivot column. That is, if and only if an echelon form has no row of the form:

\[ \begin{bmatrix} 0 & \dots & 0 & | & b \end{bmatrix} \]

with \(b \neq 0\).

Uniqueness: If it's consistent, then the solution set contains:

  1. a unique solution when there is no free variable.
  2. infinitely many solutions when there is at least one free variable.

(I) Homogeneous Linear System

Definition

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

It always has at least one solution, namely \(\vec{x}=\vec{0}\). This is called the trivial solution, the zero vector in \(\mathbb{R}^n\).

Uniqueness question: Does \(A\vec{x}=\vec{0}\) have a nontrivial solution?

Existence and uniqueness theorem tells us:

Proposition

\(A\vec{x}=\vec{0}\) has a nontrivial solution if and only if the equation has at least one free variable.

Example 1

Determine if the following homogeneous equation has a nontrivial solution. Then describe the solution set.

\[ \begin{cases} x_1 + 3x_2 + x_3 = 0 \\ -4x_1 - 9x_2 + 2x_3 = 0 \\ -3x_2 - 6x_3 = 0 \end{cases} \]

Solution: We can use either the coefficient matrix or the augmented matrix for homogeneous Linear Systems.

\[ A = \begin{bmatrix} 1 & 3 & 1 \\ -4 & -9 & 2 \\ 0 & -3 & -6 \end{bmatrix} \rightarrow \begin{bmatrix} 1 & 3 & 1 \\ 0 & 3 & 6 \\ 0 & -3 & -6 \end{bmatrix} \rightarrow \begin{bmatrix} 1 & 3 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 0 \end{bmatrix} \rightarrow \begin{bmatrix} 1 & 0 & -5 \\ 0 & 1 & 2 \\ 0 & 0 & 0 \end{bmatrix} \]

Translating this back to equations:

\[ \begin{cases} x_1 - 5x_3 = 0 \implies x_1 = 5x_3 \\ x_2 + 2x_3 = 0 \implies x_2 = -2x_3 \\ x_3 \text{ is } x_3 \text{ (free)} \end{cases} \]

Writing the solution in parametric vector form:

\[ \vec{x} = x_3 \begin{bmatrix} 5 \\ -2 \\ 1 \end{bmatrix} = x_3 \vec{v} \]

Description: The solution set is a line through \(\vec{0}\) and \(\vec{v}\) in \(\mathbb{R}^3\). This represents \(\text{Span}\{\vec{v}\}\).

A 3D Cartesian coordinate plot illustrating the solution set for a homogeneous linear system. A single solid blue line extends infinitely through the x1, x2, and x3 axes. The line passes exactly through the origin (0,0,0) and a highlighted red point at (-5, -2, 1), which represents the direction vector v. This visualizes that a solution space with one free variable forms a one-dimensional line in 3D space.
Figure: Geometric Solution as a Line Spanned by vector \( \vec{v} = \begin{bmatrix} -5 \\ -2 \\ 1 \end{bmatrix} \)
Example 2

A single Linear equation can be treated as a very simple system of equations. Describe the solution of the homogeneous "system":

\[ 10x_1 - 3x_2 - 2x_3 = 0 \]

Solution: There is no need for matrix notation. Solve for the basic variable \(x_1\) in term of the free variables \(x_2\) and \(x_3\).

\[ \begin{cases} x_1 = 0.3x_2 + 0.2x_3 \\ x_2 = x_2 \\ x_3 = x_3 \end{cases} \]

In parametric vector form:

\[ \vec{x} = x_2 \begin{bmatrix} 0.3 \\ 1 \\ 0 \end{bmatrix} + x_3 \begin{bmatrix} 0.2 \\ 0 \\ 1 \end{bmatrix} = x_2\vec{u} + x_3\vec{v} \]

Description: Every solution is a linear combination of \(\vec{u}\) and \(\vec{v}\), i.e., \(\text{Span}\{\vec{u}, \vec{v}\}\). Since neither \(\vec{u}\) nor \(\vec{v}\) is a scalar multiple of the other, the solution set is a plane through \(\vec{u}\), \(\vec{v}\) and the origin in \(\mathbb{R}^3\).

Conclusion

The solution set of a homogeneous equation \(A\vec{x}=\vec{0}\) can always be expressed explicitly as \(\text{Span}\{\vec{v}_1, \dots, \vec{v}_p\}\) for suitable vectors \(\vec{v}_1, \dots, \vec{v}_p\). Specifically:

  • only zero vector: \(\text{Span}\{\vec{0}\}\) (the origin).
  • one free variable: \(\text{Span}\{\vec{v}\}\) (a line through the origin and \(\vec{v}\)).
  • two free variables: \(\text{Span}\{\vec{u}, \vec{v}\}\) (a plane through \(\vec{u}\), \(\vec{v}\) and the origin).

(II) Solutions of Nonhomogeneous Systems

Example 3

Describe the solutions of the following system in parametric vector form. Also, give a geometric description of the solution set and compare it to that in Example 1.

\[ \begin{cases} x_1 + 3x_2 + x_3 = 1 \\ -4x_1 - 9x_2 + 2x_3 = -1 \\ -3x_2 - 6x_3 = -3 \end{cases} \]

(Matrix A is the same as that in Example 1)

Solution augmented matrix:

\[ \begin{bmatrix} A & | & \vec{b} \end{bmatrix} = \begin{bmatrix} 1 & 3 & 1 & 1 \\ -4 & -9 & 2 & -1 \\ 0 & -3 & -6 & -3 \end{bmatrix} \rightarrow \begin{bmatrix} 1 & 3 & 1 & 1 \\ 0 & 3 & 6 & 3 \\ 0 & -3 & -6 & -3 \end{bmatrix} \rightarrow \begin{bmatrix} 1 & 3 & 1 & 1 \\ 0 & 1 & 2 & 1 \\ 0 & 0 & 0 & 0 \end{bmatrix} \rightarrow \begin{bmatrix} 1 & 0 & -5 & -2 \\ 0 & 1 & 2 & 1 \\ 0 & 0 & 0 & 0 \end{bmatrix} \]

Extracting the equations:

\[ \begin{cases} x_1 = 5x_3 - 2 \\ x_2 = -2x_3 + 1 \\ x_3 \text{ is free} \end{cases} \]

Writing as a vector equation:

\[ \vec{x} = x_3 \begin{bmatrix} 5 \\ -2 \\ 1 \end{bmatrix} + \begin{bmatrix} -2 \\ 1 \\ 0 \end{bmatrix} = x_3\vec{v} + \vec{p} \]

Notice that \(\vec{v}\) satisfies \(A\vec{x}=\vec{0}\) in Example 1, and \(\vec{p}\) satisfies \(A\vec{x}=\vec{b}\).

Description: The solution set is a line through \(\vec{p}\), parallel to the line that is the solution set of the associated homogeneous System \(A\vec{x}=\vec{0}\). This represents a translation of the line.

A 3D Cartesian coordinate plot comparing the solution sets of a homogeneous and nonhomogeneous system. A dashed gray line passes through the origin, representing the homogeneous solution space. A solid blue line represents the nonhomogeneous solution space, running perfectly parallel to the gray line. A solid red arrow, labeled as translation vector p = (-1, 2, 0), points from the origin to a specific point on the blue line. This visualizes the theorem that the solution to a nonhomogeneous system Ax=b is simply a geometric translation of the associated homogeneous solution Ax=0.
Figure: Solution of \( A\vec{x} = \vec{b} \) as a Translation of \( A\vec{x} = \vec{0} \)

Theorem

Suppose \(A\vec{x}=\vec{b}\) is consistent for some given \(\vec{b}\), and let \(\vec{p}\) be a solution. Then the solution set of \(A\vec{x}=\vec{b}\) is the set of all vectors of the form:

\[ \vec{x} = \vec{p} + \vec{v}_h \]

Where \(\vec{v}_h\) is any solution of the homogeneous equation \(A\vec{x}=\vec{0}\).