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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.


Solution

For non-trivial solution $$D = 0$$
or $$\begin{vmatrix} 1 & k & 3 \\ 3 & k & -2 \\ 2 & 3 & -4\end{vmatrix} = 0$$
Apply $$R_2 - 3R_1, R_3 - 2R_1$$
$$\therefore \triangle = \begin{vmatrix} 1 & k & 3 \\
0 & -2k & -11 \\ 0 & 3-2k & -10\end{vmatrix} = 0$$
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 $$(1),$$ we get 
$$x + \frac{33}{2}y + 3z = 0$$
$$x + \frac{33}{2}\lambda - 9 \lambda = 0 
\therefore x = - \frac{15}{2} \lambda$$
$$\therefore x : y : z = -\frac{15}{2} : 1 : -3$$

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