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JEE Mains Maths · Determinants

No Solution and Inconsistent Systems

Systems that have no solution: the coefficient determinant is zero but the right-hand sides do not fit, and the values of a parameter for which that happens.

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Why this matters

Seventeen PYQs, sixteen of them multiple choice, and two from 2026. Nine ask for the value of a constant that leaves the system with no solution; eight count or collect such values, for an angle in an interval or a set of numbers. Two ideas cover the page.

Concept 1 of 2: The value that gives no solution

No solution needs Δ=0\Delta=0, so find the parameter values that make it zero. Then check each: if some Cramer determinant is non-zero, or elimination ends in 0=c0=c with c≠0c\neq0, the system has no solution. A value where everything vanishes gives infinitely many instead.

Definition

  • No solution: Δ=0\Delta=0 and at least one of Δx,Δy,Δz≠0\Delta_x,\Delta_y,\Delta_z\neq0.
  • Equivalently, elimination gives 0=c0=c, c≠0c\neq0.
  • Two equations in two unknowns: parallel lines, a1a2=b1b2≠c1c2\frac{a_1}{a_2}=\frac{b_1}{b_2}\neq\frac{c_1}{c_2}.

Inconsistency

Δ=0,(Δx,Δy,Δz)≠(0,0,0)\Delta=0,\quad(\Delta_x,\Delta_y,\Delta_z)\neq(0,0,0)

Worked example

For which λ\lambda has x+y+z=1x+y+z=1, x+2y+3z=3x+2y+3z=3, x+3y+λz=4x+3y+\lambda z=4 no solution?
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 2 Apr 2026 Shift 1 · Q53Moderate

Example 1 · Determinants · No Solution and Inconsistent Systems

Let α,β∈R\alpha,\beta\in R be such that the system of linear equations
x+2y+z=5x+ 2y+z= 5
2x+y+αz=52x+y+\alpha z= 5
8x+4y+βz=188x+ 4y+\beta z= 18
has no solution. Then βα\frac{\beta}{\alpha} is equal to :

A root of the determinant may give infinitely many

Each value with Δ=0\Delta=0 must be tested. In kx+y=1, x+ky=1kx+y=1,\ x+ky=1, k=1k=1 gives the same line twice, and only k=−1k=-1 gives no solution.
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Concept 2 of 2: Counting the values

When the parameter is an angle or ranges over a set, solve Δ=0\Delta=0 as an equation in it, list every solution in the given range, and keep those where the system is inconsistent. The count is the number of values kept.

Definition

  • Solve Δ(θ)=0\Delta(\theta)=0 in the interval; list every solution.
  • Keep only those that are inconsistent.
  • Check both ends of a closed interval.

Count what survives

#{θ:Δ(θ)=0, inconsistent}\#\{\theta:\Delta(\theta)=0,\ \text{inconsistent}\}

Worked example

For how many θ∈[0,2π]\theta\in[0,2\pi] has x+y=1x+y=1, x+(cos⁡θ)y=2x+(\cos\theta)y=2 no solution?
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2023 · 1 February 2023 · Q65Moderate

Example 2 · Determinants · No Solution and Inconsistent Systems

Let SS denote the set of all real values of λ\lambda such that the system of equations
λx+y+z=1\lambda x + y + z = 1
x+λy+z=1x + \lambda y + z = 1
x+y+λz=1x + y + \lambda z = 1
is inconsistent, then ∑λ∈S(∣λ∣2+∣λ∣)\sum_{\lambda \in S} \left( |\lambda|^{2}+ |\lambda| \right) is equal to

Candidates are not answers

Every root of Δ\Delta is only a candidate. Discard those that give infinitely many solutions before counting.
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Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

  • The value that gives no solution

    Inconsistency

    Δ=0,(Δx,Δy,Δz)≠(0,0,0)\Delta=0,\quad(\Delta_x,\Delta_y,\Delta_z)\neq(0,0,0)
  • Counting the values

    Count what survives

    #{θ:Δ(θ)=0, inconsistent}\#\{\theta:\Delta(\theta)=0,\ \text{inconsistent}\}

Watch out for (2)

Test yourself on Determinants

20 past JEE Mains questions from this chapter, timed at 48 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

Take the Determinants chapter testDrill all 17 questions