MCQMediumJEE 2023Solving Linear Equations (Matrix Method)

JEE Mathematics 2023 Question with Solution

If the system of equations x+2y+3z=3(i)x + 2y + 3z = 3 \quad \cdots (i) 4x+3y4z=4(ii)4x + 3y - 4z = 4 \quad \cdots (ii) 8x+4yλz=9+μ(iii)8x + 4y - \lambda z = 9 + \mu \quad \cdots (iii) has infinitely many solutions, then the ordered pair (λ,μ)(\lambda, \mu) is equal to:

  • A

    (725,215)\left(\frac{72}{5}, \frac{21}{5}\right)

  • B

    (725,215)\left(\frac{-72}{5}, \frac{-21}{5}\right)

  • C

    (725,215)\left(\frac{72}{5}, \frac{-21}{5}\right)

  • D

    (725,215)\left(\frac{-72}{5}, \frac{21}{5}\right)

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given:

x+2y+3z=3x + 2y + 3z = 3 4x+3y4z=44x + 3y - 4z = 4 8x+4yλz=9+μ8x + 4y - \lambda z = 9 + \mu

Find: The ordered pair (λ,μ)(\lambda, \mu) for which the system has infinitely many solutions.

Eliminate xx from the first two equations:

4(i)(ii)5y+16z=8(iv)4(i) - (ii) \Rightarrow 5y + 16z = 8 \quad \cdots (iv)

Now eliminate xx from the second and third equations:

2(ii)(iii)2y+(λ8)z=1μ(v)2(ii) - (iii) \Rightarrow 2y + (\lambda - 8)z = -1 - \mu \quad \cdots (v)

Next, eliminate yy using equations (iv)(iv) and (v)(v):

2(iv)5(v)2(iv) - 5(v) (10y+32z)(10y+5(λ8)z)=165(1μ)\Rightarrow \left(10y + 32z\right) - \left(10y + 5(\lambda - 8)z\right) = 16 - 5(-1 - \mu) (325(λ8))z=21+5μ\Rightarrow \left(32 - 5(\lambda - 8)\right)z = 21 + 5\mu

For infinitely many solutions, this reduced equation must become an identity. Therefore both coefficient and constant term must be zero:

325(λ8)=032 - 5(\lambda - 8) = 0 725λ=072 - 5\lambda = 0 λ=725\lambda = \frac{72}{5}

Also,

21+5μ=021 + 5\mu = 0 μ=215\mu = -\frac{21}{5}

Hence,

(λ,μ)=(725,215)(\lambda, \mu) = \left(\frac{72}{5}, -\frac{21}{5}\right)

Therefore, the correct option is C. The solution incorrectly labels the option as A, but the derived ordered pair matches option C.

Elimination and consistency condition

Given:

x+2y+3z=3x + 2y + 3z = 3 4x+3y4z=44x + 3y - 4z = 4 8x+4yλz=9+μ8x + 4y - \lambda z = 9 + \mu

Find: Values of λ\lambda and μ\mu such that the system has infinitely many solutions.

From

4(x+2y+3z)(4x+3y4z)=1244(x + 2y + 3z) - (4x + 3y - 4z) = 12 - 4

we get

5y+16z=85y + 16z = 8

From

2(4x+3y4z)(8x+4yλz)=8(9+μ)2(4x + 3y - 4z) - (8x + 4y - \lambda z) = 8 - (9 + \mu)

we get

2y+(λ8)z=1μ2y + (\lambda - 8)z = -1 - \mu

Multiply the second reduced equation by 55 and the first by 22:

10y+5(λ8)z=55μ10y + 5(\lambda - 8)z = -5 - 5\mu 10y+32z=1610y + 32z = 16

Subtracting,

(10y+32z)(10y+5(λ8)z)=16(55μ)\left(10y + 32z\right) - \left(10y + 5(\lambda - 8)z\right) = 16 - (-5 - 5\mu) (325λ+40)z=21+5μ\left(32 - 5\lambda + 40\right)z = 21 + 5\mu (725λ)z=21+5μ(72 - 5\lambda)z = 21 + 5\mu

For infinitely many solutions, this must hold for all permissible values in the dependent system, so

725λ=0,21+5μ=072 - 5\lambda = 0, \qquad 21 + 5\mu = 0

Thus,

λ=725,μ=215\lambda = \frac{72}{5}, \qquad \mu = -\frac{21}{5}

Therefore,

(λ,μ)=(725,215)(\lambda, \mu) = \left(\frac{72}{5}, -\frac{21}{5}\right)

So the correct option is C.

Common mistakes

  • Assuming the option label stated in the solution heading is correct without checking the worked value. Here the heading says A, but the derived ordered pair matches option C. Always compare the final computed expression with the listed options.

  • Eliminating equations incorrectly in 2(ii)(iii)2(ii) - (iii) and writing the coefficient of zz as 77 instead of λ8\lambda - 8. The parameter λ\lambda must be retained until the consistency condition is applied.

  • Using only the determinant-zero condition and forgetting consistency of the augmented system. For infinitely many solutions, the reduced equation must become an identity, so both the coefficient term and constant term must vanish.

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