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The Structure of Solutions and Fundamental Systems of Solutions

The Structure of Solutions

The Solution Structure of a Homogeneous Linear System

The Solution Space of a Homogeneous Linear System

The solution space of the homogeneous linear system $A\vec{x} = \vec{0}$ is a vector space.

Proof

Proof:

The zero solution $\vec{0}$ is a solution
If $\vec{x}_1, \vec{x}_2$ are solutions, then $\vec{x}_1 + \vec{x}_2$ is also a solution
If $\vec{x}$ is a solution and $k$ is a constant, then $k\vec{x}$ is also a solution

The Solution Structure of a Non-Homogeneous Linear System

The Solution Structure of a Non-Homogeneous System

Let $\vec{x}_0$ be a particular solution of the non-homogeneous linear system $A\vec{x} = \vec{b}$ , and let $\vec{x}_h$ be any solution of the corresponding homogeneous system $A\vec{x} = \vec{0}$ . Then $\vec{x}_0 + \vec{x}_h$ is a solution of the non-homogeneous system.

Proof

Proof: $A(\vec{x}_0 + \vec{x}_h) = A\vec{x}_0 + A\vec{x}_h = \vec{b} + \vec{0} = \vec{b}$

The Structure Theorem for Solutions

The General Solution of a Non-Homogeneous System

The general solution of the non-homogeneous linear system $A\vec{x} = \vec{b}$ is: $\vec{x} = \vec{x}_0 + \vec{x}_h$

where:

$\vec{x}_0$ is a particular solution of the non-homogeneous system
$\vec{x}_h$ is the general solution of the corresponding homogeneous system

Fundamental Systems of Solutions

The Definition of a Fundamental System of Solutions

The set of all solutions of the homogeneous linear system $A\vec{x} = \vec{0}$ is called a fundamental system of solutions.

Properties:

Every solution can be expressed as a linear combination of the vectors in this system
The fundamental system of solutions is not unique, but any two fundamental systems of solutions are equivalent

The Properties of a Fundamental System of Solutions

Properties:

The vectors in a fundamental system of solutions are linearly independent
Every solution of the homogeneous system can be expressed as a linear combination of a fundamental system of solutions
The number of vectors in a fundamental system of solutions equals the number of free variables

How to Find a Fundamental System of Solutions

Steps:

Use Gaussian elimination to reduce the coefficient matrix to row echelon form
Identify the pivots and the free variables
For each free variable in turn, set it equal to 1 and the other free variables equal to 0, solve for the pivot variables, and thereby obtain the fundamental system of solutions

Example

Example 1: find the fundamental system of solutions of the system $\begin{cases} x + y + z = 0 \\ 2x + 2y + 2z = 0 \end{cases}$ .

Solution:

The augmented matrix: $\begin{pmatrix} 1 & 1 & 1 & | & 0 \\ 2 & 2 & 2 & | & 0 \end{pmatrix}$
Reduce to row echelon form: $\begin{pmatrix} 1 & 1 & 1 & | & 0 \\ 0 & 0 & 0 & | & 0 \end{pmatrix}$
Pivot: $x$ ; free variables: $y, z$
Setting $y = 1, z = 0$ gives $\vec{v}_1 = (-1, 1, 0)$
Setting $y = 0, z = 1$ gives $\vec{v}_2 = (-1, 0, 1)$
The fundamental system of solutions: $\{(-1, 1, 0), (-1, 0, 1)\}$

The General Solution

The Definition of the General Solution

The set of all solutions of the homogeneous linear system $A\vec{x} = \vec{0}$ is called the general solution.

Properties:

The general solution is a vector space
If $\vec{x}_1, \vec{x}_2$ are two distinct solutions, then $k_1\vec{x}_1 + k_2\vec{x}_2$ is also a solution
The general solution can be expressed as a linear combination of a fundamental system of solutions

The General Solution of a Homogeneous System

Form: $\vec{x} = k_1\vec{v}_1 + k_2\vec{v}_2 + \dots + k_r\vec{v}_r$

where:

$\{\vec{v}_1, \vec{v}_2, \dots, \vec{v}_r\}$ is a fundamental system of solutions
$k_1, k_2, \dots, k_r$ are arbitrary constants

The General Solution of a Non-Homogeneous System

Form: $\vec{x} = \vec{x}_0 + k_1\vec{v}_1 + k_2\vec{v}_2 + \dots + k_r\vec{v}_r$

where:

$\vec{x}_0$ is a particular solution
$\{\vec{v}_1, \vec{v}_2, \dots, \vec{v}_r\}$ is a fundamental system of solutions of the corresponding homogeneous system
$k_1, k_2, \dots, k_r$ are arbitrary constants

Analysis of the Structure of Solutions

The Unique Solution

Condition: the system has a unique solution when the rank of the coefficient matrix equals the number of unknowns.

Characteristics:

The homogeneous system has only the zero solution
The non-homogeneous system has a unique solution

Infinitely Many Solutions

Condition: the system has infinitely many solutions when the rank of the coefficient matrix is less than the number of unknowns.

Characteristics:

The homogeneous system has infinitely many solutions
The non-homogeneous system has infinitely many solutions
The number of free variables equals the number of unknowns minus the rank of the coefficient matrix

No Solution

Condition: the system has no solution when the rank of the coefficient matrix is less than the rank of the augmented matrix.

The Dimension of the Solution Space

Computing the Dimension of the Solution Space

The Dimension of the Solution Space

The dimension of the solution space of the homogeneous linear system $A\vec{x} = \vec{0}$ equals $n - r(A)$ , where $n$ is the number of unknowns and $r(A)$ is the rank of the coefficient matrix.

Example

Example 2: analyse the dimension of the solution space of the system $\begin{cases} x + y + z = 0 \\ 2x + 2y + 2z = 0 \end{cases}$ .

Solution:

The coefficient matrix: $A = \begin{pmatrix} 1 & 1 & 1 \\ 2 & 2 & 2 \end{pmatrix}$
$r(A) = 1$ , $n = 3$
The dimension of the solution space: $3 - 1 = 2$

Parametric Solutions

The Definition of a Parametric Solution

A solution written in terms of parameters is called a parametric solution.

How to Find a Parametric Solution

Steps:

Reduce to row echelon form by Gaussian elimination
Identify the pivots and the free variables
Express the remaining variables in terms of the free variables used as parameters

Example

Example 3: find the parametric solution of the system $\begin{cases} x + y + z = 0 \\ 2x + 2y + 2z = 0 \end{cases}$ .

Solution:

Reduce to row echelon form: $\begin{pmatrix} 1 & 1 & 1 & | & 0 \\ 0 & 0 & 0 & | & 0 \end{pmatrix}$
Pivot: $x$ ; free variables: $y, z$
Let $y = t$ and $z = s$ , then $x = -t - s$
The parametric solution: $x = -t - s$ , $y = t$ , $z = s$ (where $t, s$ are arbitrary constants)

Exercises

Exercise 1

Write down the fundamental system of solutions of $\begin{cases} x + y + z = 0 \\ 2x + 2y + 2z = 0 \end{cases}$ .

Reference Answer

structure of solutionsfundamental system of solutionslinear dependenceaugmented matrix

Approach: Follow the steps for finding a fundamental system of solutions.

Steps:

The augmented matrix: $\begin{pmatrix} 1 & 1 & 1 & | & 0 \\ 2 & 2 & 2 & | & 0 \end{pmatrix}$
Reduce to row echelon form: $\begin{pmatrix} 1 & 1 & 1 & | & 0 \\ 0 & 0 & 0 & | & 0 \end{pmatrix}$
Pivot: $x$ ; free variables: $y, z$
Setting $y = 1, z = 0$ gives $\vec{v}_1 = (-1, 1, 0)$
Setting $y = 0, z = 1$ gives $\vec{v}_2 = (-1, 0, 1)$

Answer: the fundamental system of solutions is $\{(-1, 1, 0), (-1, 0, 1)\}$

Exercise 2

Find the general solution of the system $\begin{cases} x + y + z = 6 \\ 2x + 3y + z = 13 \\ x - y + 2z = 4 \end{cases}$ .

Reference Answer

structure of solutionsfundamental system of solutionslinear dependenceGaussian elimination

Approach: First find a particular solution, then find the fundamental system of solutions of the corresponding homogeneous system.

Steps:

Gaussian elimination gives the particular solution: $x = 5$ , $y = 1$ , $z = 0$
The corresponding homogeneous system: $\begin{cases} x + y + z = 0 \\ 2x + 3y + z = 0 \\ x - y + 2z = 0 \end{cases}$
Gaussian elimination gives the fundamental system of solutions: the empty set (only the zero solution)
The general solution: $x = 5$ , $y = 1$ , $z = 0$

Answer: the unique solution $x = 5$ , $y = 1$ , $z = 0$

Exercise 3

Find the parametric solution of the system $\begin{cases} x + y + z = 1 \\ 2x + 2y + 2z = 2 \end{cases}$ .

Reference Answer

structure of solutionsfundamental system of solutionslinear dependenceGaussian elimination

Approach: Reduce to row echelon form by Gaussian elimination, then express the solution in terms of parameters.

Steps:

The augmented matrix: $\begin{pmatrix} 1 & 1 & 1 & | & 1 \\ 2 & 2 & 2 & | & 2 \end{pmatrix}$
Reduce to row echelon form: $\begin{pmatrix} 1 & 1 & 1 & | & 1 \\ 0 & 0 & 0 & | & 0 \end{pmatrix}$
Pivot: $x$ ; free variables: $y, z$
Let $y = t$ and $z = s$ , then $x = 1 - t - s$

Answer: $x = 1 - t - s$ , $y = t$ , $z = s$ (where $t, s$ are arbitrary constants)

Exercise 4

Analyse the structure of the solutions of the system $\begin{cases} x + y = 1 \\ x + y = 2 \end{cases}$ .

Reference Answer

structure of solutionsfundamental system of solutionslinear dependencerank

Approach: Use Gaussian elimination to analyse the structure of the solutions.

Steps:

The augmented matrix: $\begin{pmatrix} 1 & 1 & | & 1 \\ 1 & 1 & | & 2 \end{pmatrix}$
Reduce to row echelon form: $\begin{pmatrix} 1 & 1 & | & 1 \\ 0 & 0 & | & 1 \end{pmatrix}$
The rank of the coefficient matrix is 1, the rank of the augmented matrix is 2
The rank of the coefficient matrix is less than the rank of the augmented matrix

Answer: there is no solution

Exercise 5

Prove that: the general solution of a non-homogeneous linear system equals a particular solution plus the general solution of the corresponding homogeneous system.

Reference Answer

structure of solutionsfundamental system of solutionslinear dependence

Approach: Use the linearity properties to prove it.

Steps:

Let $\vec{x}_0$ be a particular solution of the non-homogeneous system
Let $\vec{x}_h$ be any solution of the corresponding homogeneous system
Verify that $\vec{x}_0 + \vec{x}_h$ is a solution of the non-homogeneous system: $A(\vec{x}_0 + \vec{x}_h) = A\vec{x}_0 + A\vec{x}_h = \vec{b} + \vec{0} = \vec{b}$
Verify that every solution can be written in this form

Answer: the proof is complete

Summary

Symbols Used in This Article

Symbol	Type	Pronunciation / Note	Meaning in This Article
$\vec{x} = \vec{x}_0 + \vec{x}_h$	Mathematical symbol	x equals x0 plus xh	The general solution of the non-homogeneous system = a particular solution + the general solution of the corresponding homogeneous system
$\vec{x}_0$	Mathematical symbol	x nought	A particular solution of the non-homogeneous system
$\vec{x}_h$	Mathematical symbol	x h	A solution of the corresponding homogeneous system $A\vec{x}=\vec{0}$
$\vec{x} = k_1\vec{v}_1 + \dots + k_r\vec{v}_r$	Mathematical symbol	linear combination	The general solution of the homogeneous system expressed as a linear combination of a fundamental system of solutions
$\{\vec{v}_1, \dots, \vec{v}_r\}$	Mathematical symbol	v vectors	A fundamental system of solutions, that is, a basis of the homogeneous solution space
$k_1, \dots, k_r$	Mathematical symbol	k constants	The arbitrary constants that appear in the general solution
$n - r(A)$	Mathematical symbol	n minus rank of A	The dimension of the homogeneous solution space (the number of free variables)

Cramer's Rule

Comprehensive Exercises on Systems of Linear Equations

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The Structure of Solutions and Fundamental Systems of Solutions