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26 Chapter One. <strong>Linear</strong> SystemsThat single vector is, of course, a particular solution. The associated homogeneoussystem reduces via the same row operationsx + 2y − z = 02x + 4y = 0y − 3z = 0x + 2y − z = 0−2ρ 1 +ρ 2 ρ 2 ↔ρ 3−→ −→ y − 3z = 02z = 0to also give a singleton set.⎛ ⎞0{ ⎝0⎠}0As the theorem states, and as discussed at the start of this subsection, in thissingle-solution case the general solution results from taking the particular solutionand adding to it the unique solution of the associated homogeneous system.3.10 Example Also discussed there is that the case where the general solutionset is empty fits the ‘General = Particular+Homogeneous’ pattern. This systemillustrates. Gauss’ methodx + z + w = −12x − y + w = 3x + y + 3z + 2w = 1x + z + w = −1−2ρ 1 +ρ 2−→ −y − 2z − w = 5−ρ 1 +ρ 3y + 2z + w = 2shows that it has no solutions. The associated homogeneous system, of course,has a solution.x + z + w = 02x − y + w = 0x + y + 3z + 2w = 0−2ρ 1 +ρ 2 ρ 2 +ρ 3−→−ρ 1 +ρ 3−→ −y − 2z − w = 0x + z + w = 00 = 0In fact, the solution set of the homogeneous system is infinite.⎛ ⎞ ⎛ ⎞−1 −1{ ⎜−2⎟⎝ 1 ⎠ z + ⎜−1⎟⎝ 0 ⎠ w ∣ z, w ∈ R}0 1However, because no particular solution of the original system exists, the generalsolution set is empty — there are no vectors of the form ⃗p + ⃗ h because there areno ⃗p ’s.3.11 Corollary Solution sets of linear systems are either empty, have oneelement, or have infinitely many elements.Proof. We’ve seen examples of all three happening so we need only prove thatthose are the only possibilities.First, notice a homogeneous system with at least one non-⃗0 solution ⃗v hasinfinitely many solutions because the set of multiples s⃗v is infinite — if s ≠ 1then s⃗v − ⃗v = (s − 1)⃗v is easily seen to be non-⃗0, and so s⃗v ≠ ⃗v.
Section I. Solving <strong>Linear</strong> Systems 27Now, apply Lemma 3.8 to conclude that a solution set{⃗p + ⃗ h ∣ ∣ ⃗ h solves the associated homogeneous system}is either empty (if there is no particular solution ⃗p), or has one element (if thereis a ⃗p and the homogeneous system has the unique solution ⃗0), or is infinite (ifthere is a ⃗p and the homogeneous system has a non-⃗0 solution, and thus by theprior paragraph has infinitely many solutions).QEDThis table summarizes the factors affecting the size of a general solution.number of solutions of theassociated homogeneous systemparticularsolutionexists?yesnooneuniquesolutionnosolutionsinfinitely manyinfinitely manysolutionsnosolutionsThe factor on the top of the table is the simpler one. When we performGauss’ method on a linear system, ignoring the constants on the right side andso paying attention only to the coefficients on the left-hand side, we either endwith every variable leading some row or else we find that some variable does notlead a row, that is, that some variable is free. (Of course, “ignoring the constantson the right” is formalized by considering the associated homogeneous system.We are simply putting aside for the moment the possibility of a contradictoryequation.)A nice insight into the factor on the top of this table at work comes from consideringthe case of a system having the same number of equations as variables.This system will have a solution, and the solution will be unique, if and only if itreduces to an echelon form system where every variable leads its row, which willhappen if and only if the associated homogeneous system has a unique solution.Thus, the question of uniqueness of solution is especially interesting when thesystem has the same number of equations as variables.3.12 Definition A square matrix is nonsingular if it is the matrix of coefficientsof a homogeneous system with a unique solution. It is singular otherwise,that is, if it is the matrix of coefficients of a homogeneous system with infinitelymany solutions.3.13 Example The systems from Example 3.3, Example 3.5, and Example 3.9each have an associated homogeneous system with a unique solution. Thus thesematrices are nonsingular.( )3 42 −1⎛⎝ 3 2 1⎞6 −4 0⎠0 1 1⎛⎝ 1 2 −1⎞2 4 0 ⎠0 1 −3
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