JEE Main 2024 — Mathematics Algebra

Given: $x+2y+3z=5$x+2y+3z=5, $2x+3y+z=9$2x+3y+z=9 and $4x+3y+\lambda z=\mu$4x+3y+λz=μ.

For infinite solutions, $\Delta ={\Delta }_{1}={\Delta }_{2}={\Delta }_{3}=0$Δ=Δ1​=Δ2​=Δ3​=0.

$$\Rightarrow \Delta =|\begin{matrix}1 & 2 & 3 \\ 2 & 3 & 1 \\ 4 & 3 & \lambda \end{matrix}|=0$$⇒Δ=∣124​233​31λ​∣=0

$$\Rightarrow 3\lambda -3-4\lambda +8+18-36=0$$⇒3λ−3−4λ+8+18−36=0

$$\Rightarrow \lambda =-13$$⇒λ=−13

Now, finding

$$\Rightarrow {\Delta }_{1}=|\begin{matrix}5 & 2 & 3 \\ 9 & 3 & 1 \\ \mu & 3 & -13\end{matrix}|=0$$⇒Δ1​=∣59μ​233​31−13​∣=0

$$\Rightarrow \mu =15$$⇒μ=15

Now, finding

$$\Rightarrow {\Delta }_{2}=|\begin{matrix}1 & 5 & 3 \\ 2 & 9 & 1 \\ 4 & \mu & \lambda \end{matrix}|$$⇒Δ2​=∣124​59μ​31λ​∣

$$\Rightarrow {\Delta }_{2}=|\begin{matrix}1 & 5 & 3 \\ 2 & 9 & 1 \\ 4 & 15 & -13\end{matrix}|$$⇒Δ2​=∣124​5915​31−13​∣

$$\Rightarrow {\Delta }_{2}=0$$⇒Δ2​=0

And

$$\Rightarrow {\Delta }_{3}=|\begin{matrix}1 & 2 & 5 \\ 2 & 3 & 9 \\ 4 & 3 & \mu \end{matrix}|$$⇒Δ3​=∣124​233​59μ​∣

$$\Rightarrow {\Delta }_{3}=|\begin{matrix}1 & 2 & 5 \\ 2 & 3 & 9 \\ 4 & 3 & 15\end{matrix}|$$⇒Δ3​=∣124​233​5915​∣

$$\Rightarrow {\Delta }_{3}=0$$⇒Δ3​=0

For $\lambda =-13,\mu =15$λ=−13,μ=15 system of equation has infinite solutions.

$\Rightarrow \lambda +2\mu =17$⇒λ+2μ=17.