JEE MAIN - Mathematics (2025 - 24th January Morning Shift - No. 19)
If the system of equations
$$\begin{aligned} & 2 x-y+z=4 \\ & 5 x+\lambda y+3 z=12 \\ & 100 x-47 y+\mu z=212 \end{aligned}$$
has infinitely many solutions, then $\mu-2 \lambda$ is equal to
56
59
57
55
Explanation
$$\begin{aligned}
& \Delta=0 \Rightarrow\left|\begin{array}{ccc}
2 & -1 & 1 \\
5 & \lambda & 3 \\
100 & -47 & \mu
\end{array}\right|=0 \\
& 2(\lambda \mu+141)+(5 \mu-300)-235-100 \lambda=0 \ldots (1)\\
& \Delta_3=0 \Rightarrow\left|\begin{array}{ccc}
2 & -1 & 4 \\
5 & \lambda & 12 \\
100 & -47 & 212
\end{array}\right|=0 \\
& 6 \lambda=-12 \Rightarrow \lambda=-2 \\
& \text { Put } \lambda=2 \text { in }(1) \\
& 2(-2 \mu+141)+5 \mu-300-235+200=0 \\
& \mu=53 \\
& \therefore 57
\end{aligned}$$
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