JEE Main · 2020 · 05 Sep 2020 · Shift 2 · Matrices

A non-trivial solution exists only when the coefficient determinant vanishes, $\left|\begin{array}{ccc}1& 1& 3\\ 1& 3& k^2\\ 3& 1& 3\end{array}\right|=0$, which on expansion forces $k^2=9$. Subtracting the first equation from the third, $x+y+3z=0$ and $3x+y+3z=0$ give $2x=0$, so $x=0$. The first equation then reduces to $y+3z=0$, so $\frac{y}{z}=-3$. Hence $x+\frac{y}{z}=0+(-3)=-3$.