Q.

Consider the following system of equations

αx+2y+z=1

2αx+3y+z=1

3x+αy+2z=β

for some α,β∈ℝ. Then which of the following is NOT correct               [2023]

1 It has no solution if α=-1 and β≠2  
2 It has no solution for α=-1 and for all β∈ℝ  
3 It has a solution for all α≠-1 and β=2  
4 It has no solution for α=3 and for all β≠2  

Ans.

(2)

Given system of equations has no solution if

|α212α313α2|=0

⇒α(6-α)-2(4α-3)+(2α2-9)=0 ⇒α=3,-1

For no solution, any one of Δx,Δy,Δz should be non-zero.

Now, Δx=|211311α2β|≠0⇒β≠2

Now, Δy=|α112α1132β|≠0 ⇒-αβ+2α≠0

For α=-1⇒β≠2

So, option (2) is incorrect as it has no solution for α=-1 and β∈R-{2}.