Linear Algebra

12 Chapter One. Linear SystemsCompared with (∗), the advantage of (∗∗) is that z can be any real number.This makes the job of deciding which tuples are in the solution set much easier.For instance, taking z = 2 shows that (1/2, −5/2, 2) is a solution.2.2 Definition In an echelon form linear system the variables that are not leadingare free.2.3 Example Reduction of a linear system can end with more than one variablefree. On this system Gauss’s Methodx + y + z − w = 1y − z + w = −13x + 6z − 6w = 6−y + z − w = 1−→y − z + w = −1−3y + 3z − 3w = 3x + y + z − w = 1−y + z − w = 1−3ρ 1 +ρ 33ρ 2 +ρ 3 y − z + w = −1−→ρ 2 +ρ 4 0 = 0x + y + z − w = 10 = 0leaves x and y leading, and both z and w free. To get the description that weprefer we work from the bottom. We first express the leading variable y in termsof z and w, with y = −1 + z − w. Moving up to the top equation, substitutingfor y gives x + (−1 + z − w) + z − w = 1 and solving for x leaves x = 2 − 2z + 2w.The solution set{(2 − 2z + 2w, −1 + z − w, z, w) ∣ ∣ z, w ∈ R} (∗∗)has the leading variables in terms of the free variables.2.4 Example The list of leading variables may skip over some columns. Afterthis reduction2x − 2y = 0z + 3w = 23x − 3y = 0x − y + 2z + 6w = 4−(3/2)ρ 1 +ρ 3z + 3w = 2−→−(1/2)ρ 1 +ρ 4 0 = 02x − 2y = 02z + 6w = 4−→z + 3w = 20 = 02x − 2y = 00 = 0−2ρ 2 +ρ 4x and z are the leading variables, not x and y. The free variables are y and wand so we can describe the solution set as {(y, y, 2 − 3w, w) ∣ ∣ y, w ∈ R}. Forinstance, (1, 1, 2, 0) satisfies the system — take y = 1 and w = 0. The four-tuple(1, 0, 5, 4) is not a solution since its first coordinate does not equal its second.A variable that we use to describe a family of solutions is a parameter. Wesay that the solution set in the prior example is parametrized with y and w.