Basic Concepts of Systems of Linear Equations

The Definition of a System of Linear Equations

The Definition of a Linear Equation

Linear Equation

A linear equation in nn unknowns x1,x2,…,xnx_1, x_2, \dots, x_n is an equation of the form: a11x1+a12x2+⋯+a1nxn=ba_{11}x_1 + a_{12}x_2 + \dots + a_{1n}x_n = b

where aija_{ij} are real numbers, bib_i is a real number, and a1,a2,…,ana_1, a_2, \dots, a_n are not all zero.

The Definition of a System of Linear Equations

System of Linear Equations

A system of mm linear equations in nn unknowns x1,x2,…,xnx_1, x_2, \dots, x_n:

{a11x1+a12x2+⋯+a1nxn=b1a21x1+a22x2+⋯+a2nxn=b2⋮am1x1+am2x2+⋯+amnxn=bm\begin{cases} a_{11}x_1 + a_{12}x_2 + \dots + a_{1n}x_n = b_1 \\ a_{21}x_1 + a_{22}x_2 + \dots + a_{2n}x_n = b_2 \\ \vdots \\ a_{m1}x_1 + a_{m2}x_2 + \dots + a_{mn}x_n = b_m \end{cases}

where aija_{ij} are real numbers and bib_i are real numbers.

Classification of Systems of Linear Equations

Homogeneous Linear Systems

Homogeneous Linear System

If all the constant terms bi=0b_i = 0 in a system of linear equations, that is:

{a11x1+a12x2+⋯+a1nxn=0a21x1+a22x2+⋯+a2nxn=0⋮am1x1+am2x2+⋯+amnxn=0\begin{cases} a_{11}x_1 + a_{12}x_2 + \dots + a_{1n}x_n = 0 \\ a_{21}x_1 + a_{22}x_2 + \dots + a_{2n}x_n = 0 \\ \vdots \\ a_{m1}x_1 + a_{m2}x_2 + \dots + a_{mn}x_n = 0 \end{cases}

then the system is called a homogeneous linear system.

Non-Homogeneous Linear Systems

Non-Homogeneous Linear System

If at least one constant term bi≠0b_i \neq 0 occurs in a system of linear equations, then the system is called a non-homogeneous linear system.

Matrix Representation

The Coefficient Matrix

Coefficient Matrix

The coefficient matrix of the system of linear equations is:

A=(a11a12…a1na21a22…a2n⋮⋮⋱⋮am1am2…amn)A = \begin{pmatrix} 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{pmatrix}

The Augmented Matrix

Augmented Matrix

The augmented matrix of the system of linear equations is:

(A∣b⃗)=(a11a12…a1n∣b1a21a22…a2n∣b2⋮⋮⋱⋮∣⋮am1am2…amn∣bm)(A|\vec{b}) = \begin{pmatrix} a_{11} & a_{12} & \dots & a_{1n} & | & b_1 \\ a_{21} & a_{22} & \dots & a_{2n} & | & b_2 \\ \vdots & \vdots & \ddots & \vdots & | & \vdots \\ a_{m1} & a_{m2} & \dots & a_{mn} & | & b_m \end{pmatrix}

The Matrix Form

Matrix Form

A system of linear equations can be written in matrix form: Ax⃗=b⃗A\vec{x} = \vec{b}

where:

  • AA is the coefficient matrix
  • x⃗=(x1,x2,…,xn)T\vec{x} = (x_1, x_2, \dots, x_n)^T is the vector of unknowns
  • b⃗=(b1,b2,…,bm)T\vec{b} = (b_1, b_2, \dots, b_m)^T is the vector of constant terms

The Notion of a Solution

The Definition of a Solution

Solution

A set of values of the variables that satisfies every equation of the system is called a solution of the system.

The Solution Set

Solution Set

The set of all solutions of a system of linear equations is called the solution set of that system.

Classification of Solutions

  1. Unique solution: the system has exactly one solution
  2. Infinitely many solutions: the system has infinitely many solutions
  3. No solution: the system has no solution

Equivalence of Systems of Linear Equations

The Definition of Equivalent Systems

Equivalent Systems

If two systems of linear equations Ax⃗=b⃗A\vec{x} = \vec{b} and Cx⃗=d⃗C\vec{x} = \vec{d} have the same solution set, the two systems are called equivalent.

Equivalent Transformations

Equivalent Transformations

Applying any of the following transformations to a system of linear equations yields a new system equivalent to the original one:

  1. Interchanging the positions of two equations
  2. Multiplying an equation by a nonzero constant
  3. Adding a multiple of one equation to another equation

Exercises

Exercise 1

Write the system of linear equations {x+2y=53x−y=4\begin{cases} x + 2y = 5 \\ 3x - y = 4 \end{cases} in matrix form.

system of linear equationsmatrix representationcriterion for solutionscoefficient matrix

Approach: Identify the coefficient matrix, the vector of unknowns and the vector of constant terms.

Steps:

  1. The coefficient matrix: A=(123−1)A = \begin{pmatrix} 1 & 2 \\ 3 & -1 \end{pmatrix}

  2. The vector of unknowns: x⃗=(xy)\vec{x} = \begin{pmatrix} x \\ y \end{pmatrix}

  3. The vector of constant terms: b⃗=(54)\vec{b} = \begin{pmatrix} 5 \\ 4 \end{pmatrix}

  4. The matrix form: (123−1)(xy)=(54)\begin{pmatrix} 1 & 2 \\ 3 & -1 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 5 \\ 4 \end{pmatrix}

Answer: Ax⃗=b⃗A\vec{x} = \vec{b}, where A=(123−1)A = \begin{pmatrix} 1 & 2 \\ 3 & -1 \end{pmatrix}, x⃗=(xy)\vec{x} = \begin{pmatrix} x \\ y \end{pmatrix}, b⃗=(54)\vec{b} = \begin{pmatrix} 5 \\ 4 \end{pmatrix}

Exercise 2

Decide whether the system {x+y=22x+2y=4\begin{cases} x + y = 2 \\ 2x + 2y = 4 \end{cases} is homogeneous or non-homogeneous.

system of linear equationsmatrix representationcriterion for solutionsconstant terms

Approach: Check whether all the constant terms are zero.

Steps:

  1. The first equation: x+y=2x + y = 2, the constant term is 2, which is not zero
  2. The second equation: 2x+2y=42x + 2y = 4, the constant term is 4, which is not zero
  3. Neither equation has a zero constant term

Answer: it is a non-homogeneous linear system

Exercise 3

Write the augmented matrix of the system {x+y+z=02x+3y+z=0\begin{cases} x + y + z = 0 \\ 2x + 3y + z = 0 \end{cases}.

system of linear equationsmatrix representationcriterion for solutionsconstant terms

Approach: Arrange the coefficients and the constant terms into a matrix in order.

Steps:

(111∣0231∣0)\begin{pmatrix} 1 & 1 & 1 & | & 0 \\ 2 & 3 & 1 & | & 0 \end{pmatrix}

Answer: (A∣b⃗)=(111∣0231∣0)(A|\vec{b}) = \begin{pmatrix} 1 & 1 & 1 & | & 0 \\ 2 & 3 & 1 & | & 0 \end{pmatrix}

Exercise 4

Decide whether the system {x+y=1x+y=2\begin{cases} x + y = 1 \\ x + y = 2 \end{cases} has a solution.

system of linear equationsmatrix representationcriterion for solutions

Approach: Analyse the relation between the two equations.

Steps:

  1. The first equation: x+y=1x + y = 1
  2. The second equation: x+y=2x + y = 2
  3. The two equations have the same left-hand side but different right-hand sides
  4. There are no x,yx, y that satisfy both equations at once

Answer: there is no solution

Exercise 5

Prove that: the homogeneous linear system Ax⃗=0⃗A\vec{x} = \vec{0} always has a solution.

system of linear equationsmatrix representationcriterion for solutionszero vector

Approach: Use the properties of the zero vector.

Steps:

  1. Let x⃗=0⃗=(0,0,…,0)T\vec{x} = \vec{0} = (0, 0, \dots, 0)^T
  2. Substitute into the equation: A0⃗=0⃗A\vec{0} = \vec{0}
  3. A matrix times the zero vector equals the zero vector
  4. Therefore x⃗=0⃗\vec{x} = \vec{0} is a solution of the homogeneous system

Answer: the proof is complete; the zero solution always exists

The Definition of Linear Dependence

Linear Dependence

Linear dependence: let v⃗1,v⃗2,…,v⃗m\vec{v}_1, \vec{v}_2, \dots, \vec{v}_m be a system of vectors in a vector space. If there exist numbers k1,k2,…,kmk_1, k_2, \dots, k_m that are not all zero such that: k1v⃗1+k2v⃗2+⋯+kmv⃗m=0⃗k_1\vec{v}_1 + k_2\vec{v}_2 + \dots + k_m\vec{v}_m = \vec{0}

then this system of vectors is called linearly dependent; otherwise they are called linearly independent.

Summary

Symbols Used in This Article

SymbolTypePronunciation / NoteMeaning in This Article
x1,x2,…,xnx_1, x_2, \dots, x_nMathematical symbolx sub 1, x sub 2The nn unknowns of the system
aija_{ij}Mathematical symbola sub i jThe entry in row ii, column jj of the coefficient matrix
AAMathematical symbolmatrix AThe coefficient matrix, formed by all the coefficients aija_{ij}
(A∣b⃗)(A\mid \vec{b})Mathematical symbolaugmented matrixThe augmented matrix: the coefficient matrix with the vector of constant terms appended
Ax⃗=b⃗A\vec{x} = \vec{b}Mathematical symbolA x equals bThe matrix form of the system of linear equations
x⃗=(x1,x2,…,xn)T\vec{x} = (x_1, x_2, \dots, x_n)^TMathematical symbolx vectorThe vector of unknowns
Ax⃗=0⃗A\vec{x} = \vec{0}Mathematical symbolA x equals zeroThe homogeneous system (all constant terms zero)

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