JEE Main 2023 · 30 January · evening · Single correct · +4 / −1
For α, β ∈ ℝ, suppose the system of linear equations
x-y+z=5 ; 2 x+2 y+α z=8 ; 3 x-y+4 z=β
has infinitely many solutions. Then α and β are the roots of :
- (a) x²+18 x+56=0
- (b) x²-10 x+16=0
- (c) x²+14 x+24=0
- (d) x²-18 x+56=0
Answer and solution
Answer (d) x²-18 x+56=0
Solution
Maths sir · 4 steps
Make the determinant zero to get α, then write the third equation as a mix of the first two to get β.
- Determinant must vanish. Infinitely many solutions needs the coefficient determinant Δ to be zero; a non-zero Δ would give exactly one solution.
Δ = |1, -1, 1; 2, 2, α; 3, -1, 4| = (8+α) + (8-3α) + (-2-6) = 8 - 2α
- Find alpha. 8 - 2α = 0 gives α = 4. The second equation becomes 2x + 2y + 4z = 8.
- Find beta. Look for a and b with a(1, -1, 1) + b(2, 2, 4) = (3, -1, 4). From a + 2b = 3 and -a + 2b = -1 you get b = 1/2, a = 2, and the z-coefficients check: 2 + 2 = 4. The right sides must follow the same mix, so β = 2 × 5 + 1/2 × 8 = 14.
- So. α + β = 18 and αβ = 56, so α and β are the roots of x² - 18x + 56 = 0, option (d).
The idea to keep. For infinitely many solutions, set Δ = 0 first. Then the right-hand sides must obey the same combination that links the left-hand sides.
Answer from the NTA final answer key. Solution by PrepWiser.