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JEE MAIN 2026 · Matrices and Determinants

JEE Main 2026 Session 2 Question Paper: 8 Apr, Shift 2 (B.E./B.Tech), question 3

Updated
Single correct+4 / -1Answer matches NTA’s final key

If the system of linear equations:

x+y+z=6,x+2y+5z=10,2x+3y+λz=μx + y + z = 6,\qquad x + 2y + 5z = 10,\qquad 2x + 3y + \lambda z = \mu

has infinitely many solutions, then the value of λ+μ\lambda + \mu equals:

Choose one

Solution

Solution

For infinitely many solutions the coefficient determinant must vanish:

∣11112523λ∣=(2λ−15)−(λ−10)+(3−4)=λ−6=0⇒λ=6.\begin{vmatrix} 1 & 1 & 1 \\ 1 & 2 & 5 \\ 2 & 3 & \lambda \end{vmatrix} = (2\lambda - 15) - (\lambda - 10) + (3 - 4) = \lambda - 6 = 0 \Rightarrow \lambda = 6.

With λ=6\lambda = 6 the left side of the third equation is exactly the sum of the first two: (x+y+z)+(x+2y+5z)=2x+3y+6z(x + y + z) + (x + 2y + 5z) = 2x + 3y + 6z. The system stays consistent only if the right sides add the same way: μ=6+10=16\mu = 6 + 10 = 16.

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

Answer: 22.

Answer checked against the official answer key. The solution is our own working.

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