JEE Main 2023 · 30 January · morning · Single correct · +4 / −1

Let the system of linear equations
x+y+kz=2
2x+3y-z=1
3x+4y+2z=k
have infinitely many solutions. Then the system
(k+1)x+(2k-1)y=7
(2k+1)x+(k+5)y=10
has :

  1. (a) unique solution satisfying x-y=1
  2. (b) infinitely many solutions
  3. (c) no solution
  4. (d) unique solution satisfying x+y=1
Answer and solution

Answer (d) unique solution satisfying x+y=1

Solution

Maths sir · 5 steps

Use the infinite-solutions condition to pin down k = 3, then the second system is an easy 2 by 2 solve.

  1. Determinant must vanish. For infinitely many solutions, the coefficient determinant must be zero. Expanding along the first row gives 1(6+4) - 1(4+3) + k(8-9) = 3 - k, so k = 3.
  2. Check consistency. With k = 3, adding the first two equations gives 3x + 4y + 2z = 3, which is exactly the third equation. So the third equation adds nothing new, and the system really does have infinitely many solutions.
  3. Second system. Put k = 3: the equations become 4x + 5y = 7 and 7x + 8y = 10. The determinant is 4 × 8 - 5 × 7 = -3 ≠ 0, so the solution is unique.
  4. Solve it. Multiply the first equation by 7 and the second by 4, then subtract: 35y - 32y = 49 - 40, so y = 3. Then 4x = 7 - 15 = -8, so x = -2.
  5. So. The unique solution is (-2, 3), and x + y = -2 + 3 = 1. The answer is (d).

The idea to keep. Infinitely many solutions needs two things: the coefficient determinant is zero, and the equations stay consistent. Find the parameter from Δ = 0, then confirm one equation is a combination of the others.

Answer from the NTA final answer key. Solution by PrepWiser.

Study Determinants

Asked about 190 times in JEE Main since 2020. This question is from 2023.