JEE Main 2026 — Mathematics Algebra

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

$$\Delta = \begin{vmatrix} 1 & 1 & 1 \\ 1 & 2 & 5 \\ 2 & 3 & \lambda \end{vmatrix} = 0$$Δ=​112​123​15λ​​=0

Expanding the determinant:

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

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

Now, setting $\Delta_z = 0$Δz​=0:

$$\Delta_z = \begin{vmatrix} 1 & 1 & 6 \\ 1 & 2 & 10 \\ 2 & 3 & \mu \end{vmatrix} = 0$$Δz​=​112​123​610μ​​=0

Expanding the determinant:

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

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

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

Answer: $22$22