Linear Algebra

Section I. Solving Linear Systems 7Here 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−→ −5y = −5x + 3y = 10 = 21.14 Example The prior system has more equations than unknowns, but thatis not what causes the inconsistency — Example 1.12 has more equations thanunknowns and yet is consistent. Nor is having more equations than unknownsnecessary for inconsistency, as we see with this inconsistent system that has 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, besideshaving no solutions, is to have many solutions.1.15 Example In this systemx + y = 42x + 2y = 8any pair of real numbers (s 1 , s 2 ) satisfying the first equation also satisfies thesecond. 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’s Method here contrasts with the prior examplebecause we do not get a contradictory equation.−2ρ 1 +ρ 2−→x + y = 40 = 0Don’t be fooled by the final system in that example. A ‘0 = 0’ equation isnot the signal that a system has many solutions.1.16 Example The absence of a ‘0 = 0’ does not keep a system from havingmany different solutions. This system is in echelon form has no ‘0 = 0’, but hasinfinitely many solutions.x + y + z = 0y + z = 0Some solutions are: (0, 1, −1), (0, 1/2, −1/2), (0, 0, 0), and (0, −π, π). There areinfinitely many solutions because any triple whose first component is 0 andwhose second component is the negative of the third is a solution.Nor does the presence of a ‘0 = 0’ mean that the system must have manysolutions. Example 1.12 shows that. So does this system, which does not have