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Consistency questions give you a system with a letter in it, λ or k, and ask when it has no solution, one solution or infinitely many. The matrix method answers the first question in one line and the rest in two. Most lost marks come from the case where the determinant is zero. This post explains the test, works through the classic two-parameter problem, and shows the case where the usual shortcut fails.
What does the syllabus include?
The JEE (Main) 2026 syllabus lists "Test of consistency and solution of simultaneous linear equations in two or three variables using matrices". The JEE (Advanced) 2026 brochure lists solutions of simultaneous linear equations in two or three variables. The method below is from Section 4.6 of the current NCERT Class 12 chapter on determinants. Check the documents for your exam year.
What does NCERT's test say?
NCERT defines a system as consistent if it has at least one solution, and inconsistent if it has none. Write the system as AX = B, with A the coefficient matrix. Then:
|A| ≠ 0: a unique solution, X = A⁻¹B.
|A| = 0 and (adj A)B ≠ O: no solution.
|A| = 0 and (adj A)B = O: in NCERT's words, the system "may be either consistent or inconsistent", with infinitely many solutions or none.
The third case is the one to respect. It is a reason to keep going, not an answer.
What does it mean geometrically?
With two variables, each equation is a line. One solution means the lines cross. No solution means they are parallel. Infinitely many means they are the same line.
Take 2x + 3y = 5 and 4x + 6y = k. Here |A| = 12 − 12 = 0, so the lines are parallel or identical. adj A has rows (6, −3) and (−4, 2), so (adj A)B is the column (30 − 3k, 2k − 20). This is zero only when k = 10. For k = 10, the second equation is twice the first: one line, infinitely many solutions. For any other k, the lines are parallel and there is no solution.
With three variables, each equation is a plane, and the same three outcomes appear. Drawing even a rough picture helps: Larkin and Simon (1987) argued that a diagram shows information that is only implicit in equations and would otherwise have to be worked out.
Worked example: find λ and μ
For what values of λ and μ does this system have a unique solution, no solution, or infinitely many?
x + y + z = 6
x + 2y + 3z = 10
x + 2y + λz = μ
Step 1: the determinant. Expand along the first row: |A| = 1(2λ − 6) − 1(λ − 3) + 1(2 − 2) = λ − 3. So the solution is unique whenever λ ≠ 3, whatever μ is. For instance, λ = 4 and μ = 13 give x = 5, y = −2, z = 3; substitute to check all three equations.
Step 2: λ = 3. Now (adj A)B works out to the column (μ − 10, 20 − 2μ, μ − 10). If μ ≠ 10, this is not zero, so there is no solution.
Step 3: λ = 3 and μ = 10. Now (adj A)B = O, which by itself settles nothing. Subtract the second equation from the third: 0 = μ − 10 = 0, so the third equation adds nothing. Subtract the first from the second: y + 2z = 4. Choose z freely, then y = 4 − 2z and x = 6 − y − z = 2 + z. Infinitely many solutions.
Answer: unique for λ ≠ 3; none for λ = 3, μ ≠ 10; infinitely many for λ = 3, μ = 10.
When does (adj A)B = O still mean no solution?
Take x + y + z = 1, x + y + z = 2, x + y + z = 3. The coefficient matrix is all ones. Every 2×2 minor of it is zero, so adj A is the zero matrix and (adj A)B = O for any B. Yet the system clearly has no solution: the same left side cannot equal 1 and 2.
That is why NCERT says "may be". When (adj A)B = O, reduce the equations by elimination, as in step 3, before you decide.
Which mistakes cost the most marks?
Writing "no solution" as soon as |A| = 0.
Writing "infinitely many" as soon as (adj A)B = O, without eliminating.
Forgetting the unique case covers all μ. In the example, λ ≠ 3 gives a unique solution even when μ = 10.
Computing A⁻¹B when |A| = 0.
Sign slips in the cofactors used for adj A. Our guide to adjoint and inverse properties has a check you can apply.
A practice routine
Three-outcome table (twice a week). Take four systems with a parameter. For each, write the values that give one, none and infinitely many solutions.
Both methods. For one system a week, find the answer by (adj A)B and again by elimination. They must agree.
Picture it. For two-variable systems, sketch the lines for each case.
Mix chapters. Put consistency questions into mixed sets. In Rohrer and Taylor's experiment, performance a week later was "vastly superior after mixed practice" compared with practising one type at a time.
If you are not sure why a case analysis went wrong, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning, not only the final values.
Today, redo the λ and μ example with the third equation changed to x + 3y + λz = μ, and find the new critical values.
