Linear Algebra

Section I. Solving Linear Systems 17Another thing shown plainly is that setting both w and u to zero gives thatthis⎛ ⎞ ⎛ ⎞x 0y⎜z⎟⎝w⎠ = 4⎜0⎟⎝0⎠u 0is a particular solution of the linear system.2.14 Example In the same way, this systemreduces⎛⎝ 1 −1 1 1⎞3 0 1 35 −2 3 5⎛⎠ −3ρ 1+ρ 2−→−5ρ 1+ρ 3x − y + z = 13x + z = 35x − 2y + 3z = 5⎝ 1 −1 1 1⎞0 3 −2 00 3 −2 0⎠ −ρ 2+ρ 3−→to a one-parameter solution set.⎛ ⎞ ⎛ ⎞{ ⎝ 1 0⎠ + ⎝ −1/32/3 ⎠ z ∣ z ∈ R}0 1⎛⎝ 1 −1 1 1⎞0 3 −2 0⎠0 0 0 0Before the exercises, we pause to point out some things that we have yet todo.The first two subsections have been on the mechanics of Gauss’ method.Except for one result, Theorem 1.4 — without which developing the methoddoesn’t make sense since it says that the method gives the right answers — wehave not stopped to consider any of the interesting questions that arise.For example, can we always describe solution sets as above, with a particularsolution vector added to an unrestricted linear combination of some other vectors?The solution sets we described with unrestricted parameters were easilyseen to have infinitely many solutions so an answer to this question could tellus something about the size of solution sets. An answer to that question couldalso help us picture the solution sets, in R 2 , or in R 3 , etc.Many questions arise from the observation that Gauss’ method can be donein more than one way (for instance, when swapping rows, we may have a choiceof which row to swap with). Theorem 1.4 says that we must get the samesolution set no matter how we proceed, but if we do Gauss’ method in twodifferent ways must we get the same number of free variables both times, sothat any two solution set descriptions have the same number of parameters?Must those be the same variables (e.g., is it impossible to solve a problem oneway and get y and w free or solve it another way and get y and z free)?