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= Solution = = General Solution =
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[[LinearAlgebra/Elimination|Elimination]] is a fundamental step in solving systems of equations, but the method demonstrated on that page relies upon the right hand side being known values. This page instead supposes that the right hand side is unknown. A '''complete solution''' is a generalization of a [[LinearAlgebra/ParticularSolution|particular solution]]. It is generally notated as ''x,,c,,''.
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== Introduction == == Description ==
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{{{
w + 2x + 2y + 2z = a
2w + 4x + 6y + 8z = b
3w + 6x + 8y + 10z = c
}}}
A consistent linear system has either one or infinitely many solutions. The general solution describes all of them.
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This corresponds to the following [[LinearAlgebra/Elimination|eliminated]] augmented matrix: A complete solution is formalized as ''x,,c,, = x,,p,, + x,,0,,''. That is, the [[LinearAlgebra/NullSpace#Solution|null space]] must be solved and added to the [[LinearAlgebra/ParticularSolution|particular solution]].
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{{{
┌ ┐
│ [1] 2 2 2 a│
│ 0 0 [2] 4 b-2a│
│ 0 0 0 0 c-b-a│
└ ┘
}}}

The immediate question then is: for what values of ''a'', ''b'', and ''c'' is this system solvable?
The key is that any combination of the null space vectors can be added to a particular solution and give a new particular solution, because they have an identity property.
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== Trivial Solvability == == Solution ==
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The above system can be easily shown to be solvable for some values. Consider the system:
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0w + 0x + 0y + 0z = c - b - a
0 = c - b - a
w + 2x + 2y + 2z = 1
2w + 4x + 6y + 8z = 5
3w + 6x + 8y + 10z = 6
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In other words, for any values of ''a'' and ''b'' which sum to ''c'', this system is solvable. It was noted [[LinearAlgebra/ParticularSolution#Solutions|here]] that a particular solution is ''[-2 0 3/2 0]''.
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----



== General Solvability ==

----



== Solutions ==

The first step to computing the '''complete solution''' (''x,,c,,'') is computing a '''particular solution''' (''x,,p,,'').

Find a valid set of values for the right hand side. In this system, for example, [1 5 6].

Set all '''free varaibles''' to zero. Substitute to find all '''pivot variables'''.
Furthermore, it was noted [[LinearAlgebra/NullSpace#Solutions|here]] that the null space vectors are ''[-2 1 0 0]'' and ''[2 0 -2 1]''. The null space solution is ''any'' linear combination of these vectors. Consider:
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2y = 3
y = 3/2

w + 2y = 1
w + 2(3/2) = 1
w + 3 = 1
w = -2
       ┌ ┐ ┌ ┐
       │ -2 │ │ 2 │
x = C │ 1 │ + C │ 0 │
 0 1│ 0 │ 2│ -2 │
       │ 0 │ │ 1 │
       └ ┘ └ ┘
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The particular solution is: Altogether, the complete solution for the second example above is:
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[-2 0 3/2 0] x = x + x
 c p 0

     ┌ ┐ ┌ ┐ ┌ ┐
     │ -2 │ │ -2 │ │ 2 │
x = │ 0 │ + C │ 1 │ + C │ 0 │
 c │ 3/2│ 1│ 0 │ 2│ -2 │
     │ 0 │ │ 0 │ │ 1 │
     └ ┘ └ ┘ └ ┘
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Next, compute the [[LinearAlgebra/NullSpaces|null space]]. If the matrix is '''full rank''', the particular solution is the complete solution.

Finally, the complete solution is the particular solution plus any combination of the null space.

{{{
┌ ┐ ┌ ┐ ┌ ┐
│ -2 │ │ -2 │ │ 2 │
│ 0 │ + C │ 1 │ + C │ 0 │
│ 3/2│ 1│ 0 │ 2│ -1 │
│ 0 │ │ 0 │ │ 1 │
└ ┘ └ ┘ └ ┘
}}}

The values ''C,,1,,'' and ''C,,2,,'' are any number, because any combination of the two vectors is valid for the complete solution.

General Solution

A complete solution is a generalization of a particular solution. It is generally notated as xc.


Description

A consistent linear system has either one or infinitely many solutions. The general solution describes all of them.

A complete solution is formalized as xc = xp + x0. That is, the null space must be solved and added to the particular solution.

The key is that any combination of the null space vectors can be added to a particular solution and give a new particular solution, because they have an identity property.


Solution

Consider the system:

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

It was noted here that a particular solution is [-2 0 3/2 0].

Furthermore, it was noted here that the null space vectors are [-2 1 0 0] and [2 0 -2 1]. The null space solution is any linear combination of these vectors. Consider:

       ┌    ┐     ┌    ┐
       │ -2 │     │  2 │
x  = C │  1 │ + C │  0 │
 0    1│  0 │    2│ -2 │
       │  0 │     │  1 │
       └    ┘     └    ┘

Altogether, the complete solution for the second example above is:

x  = x  + x
 c    p    0

     ┌    ┐     ┌    ┐     ┌    ┐
     │ -2 │     │ -2 │     │  2 │
x  = │  0 │ + C │  1 │ + C │  0 │
 c   │ 3/2│    1│  0 │    2│ -2 │
     │  0 │     │  0 │     │  1 │
     └    ┘     └    ┘     └    ┘


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LinearAlgebra/GeneralSolution (last edited 2026-02-04 04:32:14 by DominicRicottone)