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

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