Verified Solution • 2025–26 Session moderate

QJEEM22Ma4c4d: JEE Main 2022 Mathematics - Class 12 JEEM22Ma4c4d Solution: If the system of linear equations where, R, has no solution, then

If the system of linear equations

2x+3y−z=−22x + 3y - z = - 2

x+y+z=4x + y + z = 4

x−y+∣λ∣z=4λ−4x - y + |\lambda |z = 4\lambda - 4

where, λ\lambda ∈\in R, has no solution, then

Multiple Choice Options

  • A λ\lambda = 7
  • B λ\lambda = −- 7
  • C λ\lambda = 8
  • D λ\lambda 2 = 1

Step-by-Step Verified Solution

Δ=|23−11111−1|λ||=0⇒|λ|=7

But at λ=7, Dx=Dy=Dz=0\lambda = 7,\,{D_x} = {D_y} = {D_z} = 0

P1:2x+3y−z=−2{P_1}:2x + 3y - z = - 2

P2:x+y+z=4{P_2}:x + y + z = 4

P3:x−y+∣λ∣z=4λ−4{P_3}:x - y + |\lambda |z = 4\lambda - 4

So clearly 5P2−2P1=P35{P_2} - 2{P_1} = {P_3} , so at λ=7\lambda = 7 , system of equation is having infinite solutions.

So λ=−7\lambda = - 7 is correct answer.

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