Linear Algebra - Free Books

26 Chapter One. Linear 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.