Logo
Jeeify
MCQMediumJEE 2025Solving Linear Equations (Matrix Method)

JEE Mathematics 2025 Question with Solution

If the system of equations:

3x+y+βz=32x+αy+z=2x+2y+z=4\begin{aligned} 3x + y + \beta z &= 3 \\ 2x + \alpha y + z &= 2 \\ x + 2y + z &= 4 \end{aligned}

has infinitely many solutions, then the value of 22β−9α22\beta - 9\alpha is:

  • A

    4949

  • B

    3131

  • C

    4343

  • D

    3737

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: A system of three linear equations is given and it has infinitely many solutions.

Find: The value of 22β−9α22\beta - 9\alpha.

For infinitely many solutions of a system of three linear equations, we use the determinant conditions stated in the solution:

Δ=0,Δx=0,Δy=0,Δz=0\Delta = 0, \quad \Delta_x = 0, \quad \Delta_y = 0, \quad \Delta_z = 0

Using the equations from the solution:

3x+y+βz=33x + y + \beta z = 3 2x+αy−z=−32x + \alpha y - z = -3 x+2y+z=4x + 2y + z = 4

The coefficient matrix is

A=(31β2α−1121)A = \begin{pmatrix} 3 & 1 & \beta \\ 2 & \alpha & -1 \\ 1 & 2 & 1 \end{pmatrix}

So,

Δ=∣31β2α−1121∣=0\Delta = \begin{vmatrix} 3 & 1 & \beta \\ 2 & \alpha & -1 \\ 1 & 2 & 1 \end{vmatrix} = 0

Expanding along the first row,

3(α⋅1−(−1)⋅2)−1(2⋅1−(−1)⋅1)+β(2⋅2−α⋅1)=03(\alpha \cdot 1 - (-1) \cdot 2) - 1(2 \cdot 1 - (-1) \cdot 1) + \beta(2 \cdot 2 - \alpha \cdot 1) = 0 3(α+2)−1(2+1)+β(4−α)=03(\alpha + 2) - 1(2 + 1) + \beta(4 - \alpha) = 0 3α+6−3+4β−αβ=03\alpha + 6 - 3 + 4\beta - \alpha\beta = 0 3α+4β−αβ+3=0⋯(1)3\alpha + 4\beta - \alpha\beta + 3 = 0 \quad \cdots (1)

Now set Δy=0\Delta_y = 0. Replacing the second column by the constants vector B=(3−34)B = \begin{pmatrix} 3 \\ -3 \\ 4 \end{pmatrix},

Δy=∣33β2−3−1141∣=0\Delta_y = \begin{vmatrix} 3 & 3 & \beta \\ 2 & -3 & -1 \\ 1 & 4 & 1 \end{vmatrix} = 0

Expanding,

3(−3⋅1−(−1)⋅4)−3(2⋅1−(−1)⋅1)+β(2⋅4−(−3)⋅1)=03(-3 \cdot 1 - (-1) \cdot 4) - 3(2 \cdot 1 - (-1) \cdot 1) + \beta(2 \cdot 4 - (-3) \cdot 1) = 0 3(−3+4)−3(2+1)+β(8+3)=03(-3 + 4) - 3(2 + 1) + \beta(8 + 3) = 0 3(1)−3(3)+11β=03(1) - 3(3) + 11\beta = 0 3−9+11β=03 - 9 + 11\beta = 0 −6+11β=0-6 + 11\beta = 0 β=611\beta = \frac{6}{11}

Substitute β=611\beta = \frac{6}{11} into equation (1):

3α+4(611)−α(611)+3=03\alpha + 4\left(\frac{6}{11}\right) - \alpha\left(\frac{6}{11}\right) + 3 = 0 3α+2411−6α11+3=03\alpha + \frac{24}{11} - \frac{6\alpha}{11} + 3 = 0

Multiplying by 1111,

33α+24−6α+33=033\alpha + 24 - 6\alpha + 33 = 0 27α+57=027\alpha + 57 = 0 27α=−5727\alpha = -57 α=−5727=−199\alpha = -\frac{57}{27} = -\frac{19}{9}

Now compute

22β−9α=22(611)−9(−199)22\beta - 9\alpha = 22\left(\frac{6}{11}\right) - 9\left(-\frac{19}{9}\right) =2⋅6−(−19)= 2 \cdot 6 - (-19) =12+19= 12 + 19 =31= 31

Therefore, the correct option is B.

There is a discrepancy between the question and the solution: the second equation in the question is 2x+αy+z=22x + \alpha y + z = 2, while the solution works with 2x+αy−z=−32x + \alpha y - z = -3. The extracted answer follows the solution, which explicitly concludes that the correct option is B.

Determinant Condition Breakdown

Given: The system has infinitely many solutions.

Find: 22β−9α22\beta - 9\alpha.

The key idea used in the solution is that for infinitely many solutions, the relevant determinant conditions must vanish. The solution directly evaluates Δ\Delta and Δy\Delta_y to determine α\alpha and β\beta.

From

Δ=0\Delta = 0

we get

3α+4β−αβ+3=03\alpha + 4\beta - \alpha\beta + 3 = 0

From

Δy=0\Delta_y = 0

we get

−6+11β=0-6 + 11\beta = 0

so

β=611\beta = \frac{6}{11}

Substituting into the first relation,

3α+4(611)−α(611)+3=03\alpha + 4\left(\frac{6}{11}\right) - \alpha\left(\frac{6}{11}\right) + 3 = 0 3α+2411−6α11+3=03\alpha + \frac{24}{11} - \frac{6\alpha}{11} + 3 = 0 33α+24−6α+33=033\alpha + 24 - 6\alpha + 33 = 0 27α+57=027\alpha + 57 = 0 α=−199\alpha = -\frac{19}{9}

Finally,

22β−9α=22(611)−9(−199)=3122\beta - 9\alpha = 22\left(\frac{6}{11}\right) - 9\left(-\frac{19}{9}\right) = 31

Hence, the value is 3131, so the correct option is B.

Common mistakes

  • Using only Δ=0\Delta = 0 and stopping there is incorrect. For infinitely many solutions, the system must also be consistent, so the determinant conditions involving the constants must agree with the coefficient matrix conditions. Follow the full consistency test used in the solution.

  • Copying the equations incorrectly leads to a wrong answer. The solution uses the second equation as 2x+αy−z=−32x + \alpha y - z = -3, not the form shown in the question. Use one consistent system while computing, and note any source discrepancy.

  • Making sign errors while expanding determinants is a common issue. In cofactor expansion, the middle term carries a minus sign. Write each minor carefully before simplifying.

Practice more Solving Linear Equations (Matrix Method) questions

Get unlimited AI-adaptive practice, mastery tracking, and an AI tutor that explains every step — free to start.

Start free trialMore on Solving Linear Equations (Matrix Method)

Related questions