JEE Main 2023 · 29 January · morning · Single correct · +4 / −1
Consider the following system of equations
α x+2y+z=1
2α x+3y+z=1
3x+α y+2z=β
for some α,β∈ ℝ. Then which of the following is NOT correct.
- (a) It has a solution for all α≠-1 and β=2
- (b) It has no solution if α=-1 and β≠2
- (c) It has no solution for α=-1 and for all β ∈ ℝ
- (d) It has no solution for α=3 and for all β≠2
Answer and solution
Answer (c) It has no solution for α=-1 and for all β ∈ ℝ
Solution
Maths sir · 4 steps
Find where the determinant vanishes, then test those two values of α to see what β must be.
- Coefficient determinant. Expand along the first row: D = α(6-α) - 2(4α-3) + (2α²-9) = α²-2α-3. If D ≠ 0, the system has exactly one solution whatever β is.
D = |α, 2, 1; 2α, 3, 1; 3, α, 2| = (α-3)(α+1)
- Case alpha = −1. The second equation minus the first gives -x+y=0, so y=x. Then the first equation becomes x+z=1, and the third becomes 2x+2z=β, that is 2=β. So β=2 gives infinitely many solutions and β≠ 2 gives none.
- Case alpha = 3. The second equation minus the first gives 3x+y=0, so y=-3x, and the first then gives z=1+3x. Put these in the third: 3x-9x+2+6x = 2, so it reads 2=β. Again β=2 gives infinitely many solutions and β≠ 2 gives none.
Option What actually happens Statement is (a) α≠-1, β=2 unique solution if α≠3, infinitely many if α=3 correct (b) α=-1, β≠2 third equation demands β=2, so no solution correct (c) α=-1, every β at β=2 there are infinitely many solutions not correct (d) α=3, β≠2 third equation demands β=2, so no solution correct - So. With α=-1 the system fails only when β≠2, not for every real β. The statement that is NOT correct is (c).
The idea to keep. A zero determinant does not mean no solution. Eliminate and look at the leftover equation: 0=0 means infinitely many solutions, 0= nonzero means none. That choice usually depends on the constant term.
Answer from the NTA final answer key. Solution by PrepWiser.