Question

For what value of k do the following system of equations possess a non-trivial solution over the set of rationals.
$x+ky+3z=0,3x+ky-2z=0,2x+3y-4z=0$
Also, find all the solutions of the system.

Answer 1

For non-trivial solution $D=0$
or $\left|\begin{array}{ccc}1& k& 3\\ 3& k& -2\\ 2& 3& -4\end{array}\right|=0$
Apply $R_2-3R_1,R_3-2R_1$
$\therefore △=\left|\begin{array}{ccc}1& k& 3\\ 0& -2k& -11\\ 0& 3-2k& -10\end{array}\right|=0$
or $20k+11(3-2k)=0$
or $33-2k=0$
$\therefore k=33/2$
Putting the value of k, the equations are
$x+\frac{33}{2}y+3z=0$ .......(1)
$3x+\frac{33}{2}y-2z=0$ .......(2)
$2x+3y-4z=0$ ........(3)
Mutliply (1) by (3) and subtract from (2) and similarly multiply (1) by 2 and subtract from (3). Thus we get the equivalent system of equations as
$x+\frac{33}{2}y+3z=0$
$-33y-11z=0$
$-30y-10z=0$
From any of the last two, we get $3y=-z$
or $\frac{y}{1}=\frac{z}{-3}=\lambda$, say.
$\therefore y=\lambda ,z=-3\lambda .$
From $\left(1\right),$ we get

$x+\frac{33}{2}y+3z=0$
$x+\frac{33}{2}\lambda -9\lambda =0\therefore x=-\frac{15}{2}\lambda$
$\therefore x:y:z=-\frac{15}{2}:1:-3$