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JEE Main 20268 AprilEvening Shift
Exam
JEE Main
Year
2026
Date
8 April
Shift
Evening Shift
Subject
Mathematics
Chapter
Determinants

JEE Main 2026 · 8 April · Evening Shift · Mathematics · Determinants

If the system of linear equations: x+y+z=6 , x+2y+5z=10 , 2x+3y+λ...

If the system of linear equations:
x+y+z=6x+y+z=6,
x+2y+5z=10x+2y+5z=10,
2x+3y+λz=μ2x+3y+\lambda z=\mu
has infinitely many solutions, then the value of λ+μ\lambda+\mu equals:

Options

  1. A1212
  2. B1616
  3. C2222
  4. D2828

Correct answer

C. 2222

Step-by-step solution

For the system of linear equations to have infinitely many solutions, the determinant of the coefficient matrix Δ\Delta must be zero, and Δx=Δy=Δz=0\Delta_x = \Delta_y = \Delta_z = 0.

Δ=∣11112523λ∣=0\Delta = \begin{vmatrix} 1 & 1 & 1 \\ 1 & 2 & 5 \\ 2 & 3 & \lambda \end{vmatrix} = 0

Expanding the determinant:

1(2λ−15)−1(λ−10)+1(3−4)=01(2\lambda - 15) - 1(\lambda - 10) + 1(3 - 4) = 0

λ−6=0⇒λ=6\lambda - 6 = 0 \Rightarrow \lambda = 6

Now, setting Δz=0\Delta_z = 0:

Δz=∣116121023μ∣=0\Delta_z = \begin{vmatrix} 1 & 1 & 6 \\ 1 & 2 & 10 \\ 2 & 3 & \mu \end{vmatrix} = 0

Expanding the determinant:

1(2μ−30)−1(μ−20)+6(3−4)=01(2\mu - 30) - 1(\mu - 20) + 6(3 - 4) = 0

μ−16=0⇒μ=16\mu - 16 = 0 \Rightarrow \mu = 16

Therefore, λ+μ=6+16=22\lambda + \mu = 6 + 16 = 22.

Answer: 2222
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