Linear Algebra

48 Chapter One. Linear Systemsof the reduced echelon form. Consider, for example, this system that is shownbrought to echelon form and then to reduced echelon form.⎛⎝ 2 6 1 2 5⎞0 3 1 4 10 3 1 2 5⎠ −ρ2+ρ3−→⎛⎝ 2 6 1 2 5⎞0 3 1 4 1⎠0 0 0 −2 4⎛⎞1 0 −1/2 0 −9/2−3ρ 2+ρ 1−→ −→ ⎝0 1 1/3 0 3 ⎠−ρ 3+ρ 10 0 0 1 −2(1/2)ρ 1 (4/3)ρ 3+ρ 2−→(1/3)ρ 2−(1/2)ρ 3Starting with the middle matrix, the echelon form version, back substitutionproduces −2x 4 = 4 so that x 4 = −2, then another back substitution gives3x 2 + x 3 + 4(−2) = 1 implying that x 2 = 3 − (1/3)x 3 , and then the finalback substitution gives 2x 1 + 6(3 − (1/3)x 3 ) + x 3 + 2(−2) = 5 implying thatx 1 = −(9/2) + (1/2)x 3 . Thus the solution set is this.⎛ ⎞ ⎛ ⎞ ⎛ ⎞x 1 −9/2 1/2S = { ⎜x 2⎟⎝x 3⎠ = ⎜ 3⎟⎝ 0 ⎠ + ⎜−1/3⎟⎝ 1 ⎠ x ∣3 x 3 ∈ R}x 4 −2 0Now, considering the final matrix, the reduced echelon form version, note thatadjusting the parametrization by moving the x 3 terms to the other side doesindeed give the description of this infinite solution set.Part of the reason that this works is straightforward. While a set can havemany parametrizations that describe it, e.g., both of these also describe theabove set S (take t to be x 3 /6 and s to be x 3 − 1)⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎛ ⎞−9/2 3{ ⎜ 3⎟⎝ 0 ⎠ + ⎜−2⎟⎝ 6 ⎠ t ∣ −4 1/2t ∈ R} { ⎜8/3⎟⎝ 1 ⎠ + ⎜−1/3⎟⎝ 1 ⎠ s ∣ s ∈ R}−2 0−2 0nonetheless we have in this book stuck to a convention of parametrizing usingthe unmodified free variables (that is, x 3 = x 3 instead of x 3 = 6t). We caneasily see that a reduced echelon form version of a system is equivalent to aparametrization in terms of unmodified free variables. For instance,⎛x 1 = 4 − 2x 3⇐⇒ ⎝ 1 0 2 4 ⎞0 1 1 3⎠x 2 = 3 − x 30 0 0 0(to move from left to right we also need to know how many equations are in thesystem). So, the convention of parametrizing with the free variables by solvingeach equation for its leading variable and then eliminating that leading variablefrom every other equation is exactly equivalent to the reduced echelon formconditions that each leading entry must be a one and must be the only nonzeroentry in its column.