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If a consistent system has no '''free variables''', there will be one unique solution. Conversely, if there is at least one free variable, there will be infinitely many solutions. If a consistent system has no free variables, there will be one unique solution. This can generalized in terms of [[LinearAlgebra/Rank|rank]]: a full column rank matrix has one unique solution.

Conversely, if there is at least one free variable, there will be infinitely many solutions.
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 * [[LinearAlgebra/ParticularSolution/2Equations2Unknowns|system of 2 equations, 2 unknowns, and one solution]]
 * [[LinearAlgebra/ParticularSolution/2Equations2Unknowns|system of 3 equations, 3 unknowns, and one solution]]		  * [[LinearAlgebra/ParticularSolution/2Equations2Unknowns|system of 2 equations, 2 unknowns, and one particular solution]]
 * [[LinearAlgebra/ParticularSolution/2Equations2Unknowns|system of 3 equations, 3 unknowns, and one particular solution]]
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Linear algebra introduces solution strategies such as [[LinearAlgebra/Elimination|elimination]] to find particular solutions in more complex systems. Linear algebra introduces solution strategies such as [[LinearAlgebra/Elimination|elimination]] to find particular solutions in more complex systems. Reproducing the example from that page:

{{{
x + 2y + z = 2
3x + 8y + z = 12
4y + z = 2
}}}

This system is formulated as a matrix and eliminated into:

{{{
┌ ┐ ┌ ┐ ┌ ┐
│[1] 2 1 │ │ x│ │ 2│
│ 0 [2] -2 │ │ y│ = │ 6│
│ 0 0 [5]│ │ z│ │-10│
└ ┘ └ ┘ └ ┘
}}}

This matrix has full column rank, so there is one unique solution: ''[2, 1, -2]''. This can be considered a particular solution.

Considering instead a system with free variables:

{{{
w + 2x + 2y + 2z = 1
2w + 4x + 6y + 8z = 5
3w + 6x + 8y + 10z = 6
}}}

This system is formulated as a matrix and eliminated into:

{{{
┌ ┐ ┌ ┐ ┌ ┐
│[1] 2 2 2│ │ w│ │ 1│
│ 0 0 [2] 4│ │ x│ = │ 3│
│ 0 0 0 0│ │ y│ │ 0│
└ ┘ │ z│ └ ┘
              └ ┘
}}}

There are infinitely many solutions. To find one of them, try setting all free variables to 0 and solving the equations.

{{{
2y + 4z = 3
2y + 4(0) = 3
2y = 3
y = 3/2

w + 2x + 2y + 2z = 1
w + 2(0) + 2y + 2(0) = 1
w + 2y = 1
w + 2(3/2) = 1
w + 3 = 1
w = -2
}}}

This reveals that a particular solution is ''[-2 0 3/2 0]''.

Particular Solution

A particular solution of a linear system is a point that satisfies the system. This is a vector of values notated as xp.

Contents

  1. Particular Solution
    1. Description
      1. Number of solutions
    2. Solutions

Description

A consistent linear system has either one or infinitely many solutions. Any one unique solution is called a particular solution.

If a system has no solutions, it is inconsistent.

Number of solutions

If a consistent system has no free variables, there will be one unique solution. This can generalized in terms of rank: a full column rank matrix has one unique solution.

Conversely, if there is at least one free variable, there will be infinitely many solutions.

One way to identify if a system is inconsistent is to compute the reduced augmented matrix. If it has a pivot in the right-most column, then the system is inconsistent.

Solutions

The idea of particular solution follows from basic algebra.

Linear algebra introduces solution strategies such as elimination to find particular solutions in more complex systems. Reproducing the example from that page: 

x + 2y + z = 2
3x + 8y + z = 12
4y + z = 2

This system is formulated as a matrix and eliminated into: 

┌           ┐ ┌  ┐   ┌   ┐
│[1]  2   1 │ │ x│   │  2│
│ 0  [2] -2 │ │ y│ = │  6│
│ 0   0  [5]│ │ z│   │-10│
└           ┘ └  ┘   └   ┘

This matrix has full column rank, so there is one unique solution: [2, 1, -2]. This can be considered a particular solution.

Considering instead a system with free variables: 

w + 2x + 2y + 2z = 1
2w + 4x + 6y + 8z = 5
3w + 6x + 8y + 10z = 6

This system is formulated as a matrix and eliminated into: 

┌           ┐ ┌  ┐   ┌  ┐
│[1] 2  2  2│ │ w│   │ 1│
│ 0  0 [2] 4│ │ x│ = │ 3│
│ 0  0  0  0│ │ y│   │ 0│
└           ┘ │ z│   └  ┘
              └  ┘

There are infinitely many solutions. To find one of them, try setting all free variables to 0 and solving the equations. 

2y + 4z = 3
2y + 4(0) = 3
2y = 3
y = 3/2

w + 2x + 2y + 2z = 1
w + 2(0) + 2y + 2(0) = 1
w + 2y = 1
w + 2(3/2) = 1
w + 3 = 1
w = -2

This reveals that a particular solution is [-2 0 3/2 0].

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LinearAlgebra/ParticularSolution (last edited 2026-01-07 00:01:45 by DominicRicottone)