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MCQMediumJEE 2025Solving Linear Equations (Matrix Method)

JEE Mathematics 2025 Question with Solution

The system of equations x+y+z=6,x + y + z = 6, x+2y+5z=9,x + 2y + 5z = 9, x+5y+λz=μ,x + 5y + \lambda z = \mu, has no solution if:

  • A

    λ=17,μ≠18\lambda = 17, \mu \neq 18

  • B

    λ≠17,μ≠18\lambda \neq 17, \mu \neq 18

  • C

    λ=15,μ≠17\lambda = 15, \mu \neq 17

  • D

    λ=17,μ=18\lambda = 17, \mu = 18

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given:

x+y+z=6x + y + z = 6 x+2y+5z=9x + 2y + 5z = 9 x+5y+λz=μx + 5y + \lambda z = \mu

Find: The condition on λ\lambda and μ\mu for which the system has no solution.

From the first equation,

x=6−y−zx = 6 - y - z

Substitute this in the second and third equations.

Using the second equation,

(6−y−z)+2y+5z=9(6 - y - z) + 2y + 5z = 9 6+y+4z=96 + y + 4z = 9 y+4z=3y + 4z = 3

Using the third equation,

(6−y−z)+5y+λz=μ(6 - y - z) + 5y + \lambda z = \mu 6+4y+(λ−1)z=μ6 + 4y + (\lambda - 1)z = \mu

From y+4z=3y + 4z = 3, we get

y=3−4zy = 3 - 4z

Substitute this into

6+4y+(λ−1)z=μ6 + 4y + (\lambda - 1)z = \mu

Then,

6+4(3−4z)+(λ−1)z=μ6 + 4(3 - 4z) + (\lambda - 1)z = \mu 6+12−16z+(λ−1)z=μ6 + 12 - 16z + (\lambda - 1)z = \mu 18+(λ−17)z=μ18 + (\lambda - 17)z = \mu

For no solution, the coefficient of zz must be zero, but the constant relation must be contradictory. Hence,

λ−17=0\lambda - 17 = 0

so

λ=17\lambda = 17

Then the equation becomes

18=μ18 = \mu

Therefore the system is inconsistent when

λ=17andμ≠18\lambda = 17 \quad \text{and} \quad \mu \ne 18

Elimination-Based Explanation

Given: The same three equations.

Find: When the equations become inconsistent.

A linear system has no solution when reduction leads to a false statement. Here the reduced relation is

18+(λ−17)z=μ18 + (\lambda - 17)z = \mu

If λ≠17\lambda \ne 17, then zz can be determined, so the system remains consistent.

Only when

λ=17\lambda = 17

does the variable term vanish, and the equation reduces to

18=μ18 = \mu

This is impossible if μ≠18\mu \ne 18. Hence the system has no solution exactly in that case.

Therefore, the correct option is A.

Common mistakes

  • Setting only λ=17\lambda = 17 and forgetting the condition on μ\mu. This is wrong because when λ=17\lambda = 17 and μ=18\mu = 18, the system is still consistent. After making the coefficient of zz zero, also check whether the remaining constant equation is contradictory.

  • Assuming determinant zero automatically means no solution. This is wrong because determinant zero can also give infinitely many solutions. After finding the singular case, reduce the equations further to test consistency.

  • Making an algebra error while substituting x=6−y−zx = 6 - y - z into the second or third equation. This changes the condition on λ\lambda incorrectly. Carefully simplify to obtain y+4z=3y + 4z = 3 and then 18+(λ−17)z=μ18 + (\lambda - 17)z = \mu.

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