Question

For what values of $k$, the following system of equations possess a nontrivial solution over the set of rationals: $x+ky+3z=0,3x+ky-2z=0,2x+3y-4z=0$. Find $\frac{2k}{11}$.

Answer 1

$x+ky+3z=0$

$3x+ky-2z=0$

$2x+3y-4z=0$

$\left|\begin{array}{ccc}1& k& 3\\ 3& k& -2\\ 2& 3& -4\end{array}\right|=0$

$\Rightarrow 1(-4k+6)-k(-12+4)+3(9-2k)=0$

$\Rightarrow -4k+6+12k-4k+27-6k=0$

$\Rightarrow -2k+33=0$

$\Rightarrow k=\frac{33}{2}$

$\Rightarrow \frac{2k}{11}=\frac{2\frac{\left(33\right)}{2}}{11}=3$

$\therefore$The correct answer is $3$