JEE Main 2026 — Mathematics Algebra

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

$$\Delta = \begin{vmatrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 3 & \lambda \end{vmatrix} = 1(2\lambda - 9) - 1(\lambda - 3) + 1(3 - 2)$$Δ=​111​123​13λ​​=1(2λ−9)−1(λ−3)+1(3−2)

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

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

Now, calculating $\Delta_z$Δz​:

$$\Delta_z = \begin{vmatrix} 1 & 1 & 5 \\ 1 & 2 & 9 \\ 1 & 3 & \mu \end{vmatrix} = 1(2\mu - 27) - 1(\mu - 9) + 5(3 - 2)$$Δz​=​111​123​59μ​​=1(2μ−27)−1(μ−9)+5(3−2)

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

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

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

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

Answer: $18$18