Particular Solution

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

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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.

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Solutions

The idea of particular solution follows from basic algebra.

• system of 2 equations, 2 unknowns, and one particular solution

• system of 3 equations, 3 unknowns, and one particular solution

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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