JEE Main 2023 · 31 January · morning · Single correct · +4 / −1
For the system of linear equations
x+y+z=6
α x+β y+7 z=3
x+2 y+3 z=14
which of the following is NOT true ?
- (a) If α=β=7, then the system has no solution
- (b) For every point (α, β) ≠(7,7) on the line x-2 y+7=0, the system has infinitely many solutions
- (c) There is a unique point (α, β) on the line x+2 y+18=0 for which the system has infinitely many solutions
- (d) If α=β and α ≠ 7, then the system has a unique solution
Answer and solution
Answer (b) For every point (α, β) ≠(7,7) on the line x-2 y+7=0, the system has infinitely…
Solution
Maths sir · 3 steps
D = 0 only puts (α, β) on a line. Infinitely many solutions also needs Dₓ, D_y, D_z to vanish.
- Main determinant. Expand the coefficient determinant along the first row. The system has a unique solution exactly when D ≠ 0.
D = |1, 1, 1; α, β, 7; 1, 2, 3| = 2β - α - 7
- Other determinants. Replacing columns with (6, 3, 14) gives Dₓ = 4β + 11, D_y = -4α - 50 and D_z = 8β - 2α - 3. All three are zero only at α = -25/2, β = -11/4. This point lies on α - 2β + 7 = 0, so D = 0 there too.
Option Check Result Statement (a) α = β = 7 D = 0, Dₓ = 39 ≠ 0 no solution true (b) on α - 2β + 7 = 0 D = 0, but Dₓ, D_y, D_z vanish only at (-25/2, -11/4) every other point gives no solution NOT true (c) on α + 2β + 18 = 0 -25/2 - 11/2 + 18 = 0 exactly one such point true (d) α = β ≠ 7 D = β - 7 ≠ 0 unique solution true - So. On the line x - 2y + 7 = 0, only the single point (-25/2, -11/4) gives infinitely many solutions. Every other point on it gives none. So (b) is the statement that is NOT true.
The idea to keep. D = 0 only tells you the solution is not unique. Then check Dₓ, D_y, D_z: if any one is nonzero there is no solution, and if all are zero there are infinitely many.
Answer from the NTA final answer key. Solution by PrepWiser.