In conclusion, if αβ = 12 and β ≠ 4 or α ≠ 3 then no solution If αβ = 12 and β = 4 or α = 3 then infinitely many solutions x=λ x − 2 y + 4 z = 3 2 − 3λ 3x + 2 z = 2 3x + 2 z = 2 z = λ arbitrary 5x + 2 y = 1 2 5x + 2 y = 1 1 − 5λ y= 2
15. Decide whether the following system of linear equations is consistent and find the solution in dependence on parameters α and β:
α x + (α + β ) y + β z = 3α + 5β
β x + αβ y + α z = α ( 2 β + 3) + β α x + β y + β z = α + 5β
Solution
α x + (α + β ) y + β z = 3α + 5β
α β x + αβ y + α z = α ( 2 β + 3) + β ⇔ β α α x + β y + β z = α + 5β
α + β β x 3α + 5β αβ α y = α ( 2β + 3) + β β β z α + 5β
α α +β β 0 α 0 β α = α (β 2 −α 2 ) β αβ α = β αβ α = α α β α β β α β β The expression α ( β 2 − α 2 ) will be 0 when or α
= 0 or when β2 = α2. Otherwise,
α (β 2 −α 2 ) ≠ 0 Then If α ≠ 0 and β2 ≠ α2 then rang(A) = rang (A*) = 3 => CONSISTENT with a unique solution If α = 0
β y + β z = 5β y + z = 5 x = 1 βx = β ⇔ x =1 y = λ λ ∈ ℝ z = 5− λ β y + β z = 5β Which has rang(A) = rang(A*) = 2 => CONSISTENT with infinitely many solutions | Ejercicios Propuestos 33