Matrices and Determinants Part 2

Given, system of equation
2x+y−z=3
x−y−z=α
3x+3y+βz=3
has infinitely many solutions,
if Δ=0,Δ1=Δ2=Δ3=0
Now,Δ=|

2	1	−1
1	−1	−1
3	3	β

|=0
⇒2(−β+3)−1(β+3)−1(3+3)=0
⇒ β = −1
Δ1=|

3	1	−1
α	−1	−1
3	3	−1

|=0
⇒3(1+3)−1(−α+3)−1(3α+3)=0
⇒12+α−3−3α−3=0
⇒ α = 3
Also, Δ2=|

2	3	−1
1	α	−1
3	3	−1

|=0
⇒2(−α+3)−3(−1+3)−1(3−3α)=0
⇒ α = 3
and Δ3=|

2	1	3
1	−1	α
3	3	3

|=0
⇒2(−3−3α)−1(3−3α)+3(3+3)=0
⇒−3α+9=0
⇒ α = 3
∴ α = 3, β = −1
So, α+β−αβ=3−1−3(−1)=5