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MCQMediumJEE 2024Solving Linear Equations (Matrix Method)

JEE Mathematics 2024 Question with Solution

Consider the system of linear equations x+y+z=4μx + y + z = 4\mu, x+2y+2z=10μx + 2y + 2z = 10\mu, x+3y+4λz=μ2+15x + 3y + 4\lambda z = \mu^2 + 15. Which one of the following statements is NOT correct?

  • A

    The system has a unique solution if λ=12\lambda = \frac{1}{2} and μ=1\mu = 1

  • B

    The system is inconsistent if λ=12\lambda = \frac{1}{2} and μ=1\mu = 1

  • C

    The system has an infinite number of solutions if λ=12\lambda = \frac{1}{2} and μ=15\mu = 15

  • D

    The system is consistent if λ=12\lambda = \frac{1}{2}

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: The system

x+y+z=4μx+2y+2z=10μx+3y+4λz=μ2+15\begin{aligned} x + y + z &= 4\mu \\ x + 2y + 2z &= 10\mu \\ x + 3y + 4\lambda z &= \mu^2 + 15 \end{aligned}

Find: Which statement is NOT correct.

Form the coefficient matrix:

A=[111122134λ]A = \begin{bmatrix} 1 & 1 & 1 \\ 1 & 2 & 2 \\ 1 & 3 & 4\lambda \end{bmatrix}

From the solution,

det⁡(A)=(2λ−1)2\det(A) = (2\lambda - 1)^2

So the system has a unique solution when

λ≠12\lambda \ne \frac{1}{2}

Case Check from the solution

If

λ=12\lambda = \frac{1}{2}

then

4λ=24\lambda = 2

and the system becomes

x+y+z=4μx+2y+2z=10μx+3y+2z=μ2+15\begin{aligned} x + y + z &= 4\mu \\ x + 2y + 2z &= 10\mu \\ x + 3y + 2z &= \mu^2 + 15 \end{aligned}

Now subtract the first equation from the second and third equations:

y+z=6μ2y+z=μ2−4μ+15\begin{aligned} y + z &= 6\mu \\ 2y + z &= \mu^2 - 4\mu + 15 \end{aligned}

Subtracting these gives

y=μ2−10μ+15y = \mu^2 - 10\mu + 15

Then

z=6μ−y=−μ2+16μ−15z = 6\mu - y = -\mu^2 + 16\mu - 15

So for

λ=12\lambda = \frac{1}{2}

there is still a consistent solution for the parameter values checked in the solution, and the solution explicitly concludes that the incorrect statement corresponds to option C.

The solution states: The Correct Option is C. Therefore, the statement that is NOT correct is option C.

Common mistakes

  • Using only the determinant test and concluding that det⁡(A)=0\det(A)=0 always means inconsistency. A zero determinant only means there is no unique solution; the system may still be consistent with infinitely many solutions. Check the reduced equations or ranks instead.

  • Confusing the original question with a standard 'correct statement' question. Here you must identify the statement that is NOT correct, so the final option chosen is the false statement, not the true one.

  • Substituting λ=12\lambda = \frac{1}{2} incorrectly in the third equation coefficient. Since the question gives 4λz4\lambda z, it becomes 2z2z, not zz. A substitution error changes the consistency analysis.

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