Mathematics · 2020
JEE Main · 5 September 2020, Shift 2 · Q54
If the system of linear equations x+y+3 z=0; x+3 y+ k^2 z=0; 3 x+y+3 z=0 has a non-zero solution (x, y, z) for some k ∈ R, then x+(y/z) is equal to:
If the system of linear equations
$$\begin{aligned}
& x+y+3 z=0 \\
& x+3 y+\mathrm{k}^2 z=0 \\
& 3 x+y+3 z=0
\end{aligned}
$$
has a non-zero solution $\displaystyle (x, y, z)$ for some $\displaystyle \mathrm{k} \in \mathbf{R}$, then $\displaystyle x+\left(\frac{y}{z}\right)$ is equal to:
Official answer
(2)
$\displaystyle -3$
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JEE Main 2020 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.