If the system of equations x+4 y-z=λ , 7 x+9 y+μ z=-3,5 x+y+2 z=-1 has infinitely many solutions, then (2 μ .+3 λ) is equal to :

Mathematics · Determinants

JEE Main 2024 — 8 April, Shift 2 — Question 7

If the system of equations $x+4 y-z=\lambda$ , $7 x+9 y+\mu z=-3,5 x+y+2 z=-1$ has infinitely many solutions,

then $(2 \mu .+3 \lambda)$ is equal to :

\text{ }\Delta \text{ }=\left| \begin{matrix}1 & 4 & -1 \\7 & 9 & \mu \\5 & 1 & 2 \\\end{matrix} \right|=0

$\Rightarrow(18-\mu)-4(14-5 \mu)-(7 45)=0 \Rightarrow \mu=0$

$\Delta=\Delta_{\mathrm{x}}=\Delta_{\mathrm{y}}=\Delta_{\mathrm{z}}=0$ (For infinite solution)

${{\text{}\!\!\Delta\!\!\text{}}_{\text{x}}}=\left|\begin{matrix}\lambda & 4 & -1 \\-3 & 9 & \mu \\-1 & 1 & 2 \\\end{matrix} \right|=0$

$\lambda(18-\mu)-4(-6+\mu)-1(-3+9)=0$

$18 \lambda+24-6=0 \Rightarrow \lambda=-1$

$(2 \mu .+3 \lambda)=-3$

Answer key and solution verified before publishing.

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