Linear Algebra

Section I. Solving Linear Systems 7Another way that linear systems can differ from the examples shown earlieris that some linear systems do not have a unique solution. This can happen intwo ways.The first is that it can fail to have any solution at all.1.12 Example Contrast the system in the last example with this one.x + 3y = 12x + y = −32x + 2y = 0x + 3y = 1−2ρ 1 +ρ 2−→ −5y = −5−2ρ 1 +ρ 3−4y = −2Here the system is inconsistent: no pair of numbers satisfies all of the equationssimultaneously. Echelon form makes this inconsistency obvious.The solution set is empty.−(4/5)ρ 2+ρ 3−→x + 3y = 1−5y = −50 = 21.13 Example The prior system has more equations than unknowns, but thatis not what causes the inconsistency — Example 1.11 has more equations thanunknowns and yet is consistent. Nor is having more equations than unknownsnecessary for inconsistency, as is illustrated by this inconsistent system with thesame number of equations as unknowns.x + 2y = 82x + 4y = 8−2ρ 1+ρ 2−→x + 2y = 80 = −8The other way that a linear system can fail to have a unique solution is tohave many solutions.1.14 Example In this systemx + y = 42x + 2y = 8any pair of numbers satisfying the first equation automatically satisfies the second.The solution set {(x, y) ∣ x + y = 4} is infinite; some of its membersare (0, 4), (−1, 5), and (2.5, 1.5). The result of applying Gauss’ method herecontrasts with the prior example because we do not get a contradictory equation.−2ρ 1 +ρ 2−→x + y = 40 = 0Don’t be fooled by the ‘0 = 0’ equation in that example. It is not the signalthat a system has many solutions.