JEE MAIN - Mathematics (2025 - 24th January Evening Shift - No. 8)
If the system of equations
$$ \begin{aligned} & x+2 y-3 z=2 \\ & 2 x+\lambda y+5 z=5 \\ & 14 x+3 y+\mu z=33 \end{aligned} $$
has infinitely many solutions, then $\lambda+\mu$ is equal to :
$$ \begin{aligned} & x+2 y-3 z=2 \\ & 2 x+\lambda y+5 z=5 \\ & 14 x+3 y+\mu z=33 \end{aligned} $$
has infinitely many solutions, then $\lambda+\mu$ is equal to :
13
10
12
11
Explanation
$$\begin{aligned}
&\begin{aligned}
& \mathrm{D}=\left|\begin{array}{rrr}
1 & 2 & -3 \\
2 & \lambda & 5 \\
14 & 3 & \mu
\end{array}\right|=0, \lambda \mu+42 \lambda-4 \mu+107=0 \\
& \mathrm{D}_1=2 \lambda \mu+99 \lambda-10 \mu+255 \\
& \mathrm{D}_2=13-\mu \\
& D_3=5 \lambda+5 \\
& D_2=0 \Rightarrow \mu=13 \& D_3=0 \Rightarrow \lambda=-1
\end{aligned}\\
&\text { check & verify for these values } \mathrm{D} \& \mathrm{D}_2=0
\end{aligned}$$
Comments (0)
