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MCQMediumJEE 2023Solving Linear Equations (Matrix Method)

JEE Mathematics 2023 Question with Solution

For the system of linear equations αx+y+z=1\alpha x + y + z = 1, x+αy+z=1x + \alpha y + z = 1, x+y+αz=βx + y + \alpha z = \beta, which one of the following statements is NOT correct?

  • A

    It has infinitely many solutions if α=2\alpha = 2 and β=−1\beta = -1.

  • B

    It has no solution if α=−2\alpha = -2 and β=1\beta = 1.

  • C

    x+y+z=34x + y + z = \frac{3}{4} if α=2\alpha = 2 and β=1\beta = 1.

  • D

    It has infinitely many solutions if α=1\alpha = 1 and β=1\beta = 1.

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: The system is

αx+y+z=1\alpha x + y + z = 1 x+αy+z=1x + \alpha y + z = 1 x+y+αz=βx + y + \alpha z = \beta

Find: Which statement is not correct.

The coefficient matrix is

[α111α111α]\begin{bmatrix} \alpha & 1 & 1 \\ 1 & \alpha & 1 \\ 1 & 1 & \alpha \end{bmatrix}

and its determinant is

Δ=α(α2−1)−1(α−1)+1(1−α)\Delta = \alpha(\alpha^2-1) - 1(\alpha-1) + 1(1-\alpha)

So,

Δ=α3−3α+2\Delta = \alpha^3 - 3\alpha + 2

Factoring,

Δ=(α−1)2(α+2)\Delta = (\alpha - 1)^2(\alpha + 2)

Hence, Δ=0\Delta = 0 when α=1\alpha = 1 or α=−2\alpha = -2.

For α=1,β=1\alpha = 1, \beta = 1, the system becomes

x+y+z=1x + y + z = 1 x+y+z=1x + y + z = 1 x+y+z=1x + y + z = 1

so it has infinitely many solutions. Therefore, statement D is correct.

For α=2,β=1\alpha = 2, \beta = 1, the determinant is

Δ=4\Delta = 4

Also,

Δ1=1,Δ2=1,Δ3=1\Delta_1 = 1, \quad \Delta_2 = 1, \quad \Delta_3 = 1

Thus,

x=14,y=14,z=14x = \frac{1}{4}, \quad y = \frac{1}{4}, \quad z = \frac{1}{4}

and therefore

x+y+z=34x + y + z = \frac{3}{4}

So statement C is correct.

For α=2,β=−1\alpha = 2, \beta = -1, we still have

Δ=4≠0\Delta = 4 \neq 0

so the system has a unique solution, not infinitely many solutions. The solution states that this case is inconsistent, but in either case it does not have infinitely many solutions. Therefore, statement A is not correct.

For α=−2,β=1\alpha = -2, \beta = 1, since Δ=0\Delta = 0, the system is singular, and the provided solution identifies this case as having no solution. Hence statement B is treated as correct according to the solution.

Therefore, the correct option is A.

Option Check by Determinant

A quick way is to first check when infinitely many solutions are even possible. That can happen only when the determinant is zero. Since

Δ=(α−1)2(α+2)\Delta = (\alpha - 1)^2(\alpha + 2)

we must have α=1\alpha = 1 or α=−2\alpha = -2 for non-unique behavior.

Now test the options:

  • Option A uses α=2\alpha = 2, for which Δ=4≠0\Delta = 4 \neq 0. So infinitely many solutions are impossible.
  • Option D uses α=1\alpha = 1 and β=1\beta = 1, giving the same equation three times, hence infinitely many solutions.
  • Option C is verified directly from Cramer's rule in the solution.

Therefore, option A is the statement that is not correct.

Common mistakes

  • Assuming infinitely many solutions can occur even when Δ≠0\Delta \neq 0 is incorrect. If the determinant is non-zero, the system has a unique solution. First check Δ\Delta before classifying the system.

  • Confusing the parameters α\alpha and β\beta can lead to testing the wrong case. α\alpha affects the coefficient matrix, while β\beta appears only in the constant term of the third equation. Separate these roles carefully.

  • Stopping after finding Δ=0\Delta = 0 is incomplete. A zero determinant does not automatically mean infinitely many solutions; the system may also be inconsistent. After Δ=0\Delta = 0, compare the equations or examine consistency.

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