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JEE Main 20264 AprilEvening Shift
Exam
JEE Main
Year
2026
Date
4 April
Shift
Evening Shift
Subject
Mathematics
Chapter
Determinants

JEE Main 2026 · 4 April · Evening Shift · Mathematics · Determinants

If the system of equations: x+y+z=5 x+2y+3z=9 x+3y+λ z=μ has...

If the system of equations:
x+y+z=5x+y+z=5
x+2y+3z=9x+2y+3z=9
x+3y+λz=μx+3y+\lambda z=\mu
has infinitely many solutions, then the value of λ+μ\lambda+\mu is:

Options

  1. A1616
  2. B1818
  3. C1919
  4. D2121

Correct answer

B. 1818

Step-by-step solution

For the system to have infinitely many solutions, Δ=0\Delta = 0 and Δx=Δy=Δz=0\Delta_x = \Delta_y = \Delta_z = 0.

Δ=∣11112313λ∣=1(2λ−9)−1(λ−3)+1(3−2)\Delta = \begin{vmatrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 3 & \lambda \end{vmatrix} = 1(2\lambda - 9) - 1(\lambda - 3) + 1(3 - 2)

Δ=2λ−9−λ+3+1=λ−5\Delta = 2\lambda - 9 - \lambda + 3 + 1 = \lambda - 5

Setting Δ=0\Delta = 0 gives λ=5\lambda = 5.

Now, calculating Δz\Delta_z:

Δz=∣11512913μ∣=1(2μ−27)−1(μ−9)+5(3−2)\Delta_z = \begin{vmatrix} 1 & 1 & 5 \\ 1 & 2 & 9 \\ 1 & 3 & \mu \end{vmatrix} = 1(2\mu - 27) - 1(\mu - 9) + 5(3 - 2)

Δz=2μ−27−μ+9+5=μ−13\Delta_z = 2\mu - 27 - \mu + 9 + 5 = \mu - 13

Setting Δz=0\Delta_z = 0 gives μ=13\mu = 13.

For λ=5\lambda = 5 and μ=13\mu = 13, we can verify that Δx=0\Delta_x = 0 and Δy=0\Delta_y = 0 as well.

Therefore, λ+μ=5+13=18\lambda + \mu = 5 + 13 = 18.

Answer: 1818
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