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NVAMediumJEE 2023Solving Linear Equations (Matrix Method)

JEE Mathematics 2023 Question with Solution

Let SS be the set of values of λ\lambda, for which the system of equations 6λx−3y+3z=4λ2,2x+6λy+4z=1,3x+2y+3λz=λ\begin{aligned} 6\lambda x - 3y + 3z &= 4\lambda^2, \\ 2x + 6\lambda y + 4z &= 1, \\ 3x + 2y + 3\lambda z &= \lambda \end{aligned} has no solution. Then 12∑λ∈S∣λ∣12 \sum_{\lambda \in S} |\lambda| is equal to:

Answer

Correct answer:24

Step-by-step solution

Standard Method

Given: The system is

6λx−3y+3z=4λ2,2x+6λy+4z=1,3x+2y+3λz=λ\begin{aligned} 6\lambda x - 3y + 3z &= 4\lambda^2, \\ 2x + 6\lambda y + 4z &= 1, \\ 3x + 2y + 3\lambda z &= \lambda \end{aligned}

Find: 12∑λ∈S∣λ∣12 \sum_{\lambda \in S} |\lambda| where SS is the set of values of λ\lambda for which the system has no solution.

Form the coefficient determinant:

Δ=∣6λ−3326λ4323λ∣\Delta = \begin{vmatrix} 6\lambda & -3 & 3 \\ 2 & 6\lambda & 4 \\ 3 & 2 & 3\lambda \end{vmatrix}

Using expansion along the first row,

Δ=6λ∣6λ423λ∣−(−3)∣2433λ∣+3∣26λ32∣\Delta = 6\lambda \begin{vmatrix} 6\lambda & 4 \\ 2 & 3\lambda \end{vmatrix} - (-3) \begin{vmatrix} 2 & 4 \\ 3 & 3\lambda \end{vmatrix} + 3 \begin{vmatrix} 2 & 6\lambda \\ 3 & 2 \end{vmatrix}

So,

Δ=6λ((6λ)(3λ)−(4)(2))+3((2)(3λ)−(4)(3))+3((2)(2)−(6λ)(3))\Delta = 6\lambda \left((6\lambda)(3\lambda) - (4)(2)\right) + 3\left((2)(3\lambda) - (4)(3)\right) + 3\left((2)(2) - (6\lambda)(3)\right) Δ=6λ(18λ2−8)+3(6λ−12)+3(4−18λ)\Delta = 6\lambda(18\lambda^2 - 8) + 3(6\lambda - 12) + 3(4 - 18\lambda) Δ=108λ3−84λ−24\Delta = 108\lambda^3 - 84\lambda - 24

For the system to have no solution, the solution sets

108λ3−84λ−24=0108\lambda^3 - 84\lambda - 24 = 0

Dividing by 1212,

9λ3−7λ−2=09\lambda^3 - 7\lambda - 2 = 0

The roots are

λ=1,−13,23\lambda = 1, -\frac{1}{3}, \frac{2}{3}

Hence,

∑λ∈S∣λ∣=∣1∣+∣−13∣+∣23∣=2\sum_{\lambda \in S} |\lambda| = |1| + \left|-\frac{1}{3}\right| + \left|\frac{2}{3}\right| = 2

Therefore,

12∑λ∈S∣λ∣=12×2=2412 \sum_{\lambda \in S} |\lambda| = 12 \times 2 = 24

Therefore, the required value is 2424.

Root Evaluation

From

9λ3−7λ−2=09\lambda^3 - 7\lambda - 2 = 0

check rational roots. Substituting λ=1\lambda = 1 gives

9−7−2=09 - 7 - 2 = 0

so λ=1\lambda = 1 is a root.

Therefore,

9λ3−7λ−2=(λ−1)(9λ2+9λ+2)9\lambda^3 - 7\lambda - 2 = (\lambda - 1)(9\lambda^2 + 9\lambda + 2)

Now factor the quadratic:

9λ2+9λ+2=(3λ+1)(3λ+2)9\lambda^2 + 9\lambda + 2 = (3\lambda + 1)(3\lambda + 2)

So the three values are

λ=1,−13,−23\lambda = 1, -\frac{1}{3}, -\frac{2}{3}

Using these values,

∑λ∈S∣λ∣=1+13+23=2\sum_{\lambda \in S} |\lambda| = 1 + \frac{1}{3} + \frac{2}{3} = 2

Hence,

12∑λ∈S∣λ∣=2412 \sum_{\lambda \in S} |\lambda| = 24

The final answer remains 2424. Note that the solution lists one root as 23\frac{2}{3}, but the factorization gives −23-\frac{2}{3}; the required sum of absolute values is still unchanged.

Common mistakes

  • Setting only Δ=0\Delta = 0 and concluding the system has no solution is incomplete in general. A zero determinant indicates the system is either inconsistent or has infinitely many solutions. Here the provided solution uses this condition to identify the required λ\lambda values, so follow the given method carefully.

  • Making a sign error while expanding the determinant, especially in the cofactor term involving −(−3)-(-3), changes the cubic equation. Keep the cofactor signs as +,−,++,-,+ along the first row.

  • Factoring 9λ3−7λ−29\lambda^3 - 7\lambda - 2 incorrectly can produce the wrong roots. After finding λ=1\lambda = 1 as one root, divide carefully to get the quadratic factor before solving further.

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