Verified Solution • 2025–26 Session moderate

QJEEM20Ma734e: JEE Main 2020 Mathematics - Class 12 JEEM20Ma734e Solution: If the system of equations x+y+z=2 2x+4y–z=6 3x+2y+ z= has…

If the system of equations
x+y+z=2
2x+4y–z=6
3x+2y+ λ\lambda z= μ\mu
has infinitely many solutions, then

Multiple Choice Options

  • A 2 λ\lambda - μ\mu = 5
  • B λ\lambda - 2 μ\mu = -5
  • C 2 λ\lambda + μ\mu = 14
  • D λ\lambda + 2 μ\mu = 14

Step-by-Step Verified Solution

D=0|11124−132λ|=0

⇒\Rightarrow (4λ+2)−1(2λ+3)+1(4−12)=0(4\lambda + 2) - 1(2\lambda + 3) + 1(4 - 12) = 0

⇒\Rightarrow 4λ+24\lambda + 2 −2λ−3- 2\lambda - 3 −- 8 =0= 0

⇒\Rightarrow 2λ=9⇒λ=922\lambda = 9 \Rightarrow \lambda = {9 \over 2}

Dx=|21164−1μ2−9/2|=0

⇒μ=5\Rightarrow \mu = 5

By checking all the options we find (C) is correct

2λ+μ=142\lambda + \mu = 14

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