JEE Main 2025 · 23 January · evening · Single correct · +4 / −1

The system of equations
x+y+z=6, ; x+2 y+5 z=9, ; x+5 y+λ z=μ,
has no solution if

  1. (a) λ=17, μ=18
  2. (b) λ=17, μ ≠ 18
  3. (c) λ=15, μ ≠ 17
  4. (d) λ ≠ 17, μ ≠ 18
Answer and solution

Answer (b) λ=17, μ ≠ 18

Solution

Maths sir · 3 steps

Eliminate x, then y, and the last equation reduces to (λ-17)z = μ-18. Read everything off that.

  1. Remove x. Subtract the first equation from the second: y + 4z = 3. Subtract it from the third: 4y + (λ-1)z = μ - 6.
  2. Remove y. Take 4 times y + 4z = 3 away from 4y + (λ-1)z = μ - 6. The y terms cancel and one equation in z alone is left.

    (λ - 17) z = μ - 18

    CaseThe last equation becomesSolutions
    λ ≠ 17z = (μ-18)/(λ-17)exactly one
    λ = 17, μ = 180 = 0infinitely many
    λ = 17, μ ≠ 180 = μ - 18 ≠ 0none
  3. So. The system has no solution exactly when λ = 17 and μ ≠ 18, which is option (b). As a check, the coefficient determinant works out to λ - 17, which is zero only at λ = 17.

The idea to keep. For no solution, the coefficient determinant must be zero and the right-hand sides must disagree. Reduce the system to one equation like 0 · z = c and ask whether c is zero.

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 2025.