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 $$\dfrac{2k}{11}$$.

Solution

## $$x+ky+3z=0$$$$3x+ky-2z=0$$$$2x+3y-4z=0$$$$\begin{vmatrix} 1 & k & 3 \\ 3 & k & -2 \\ 2 & 3 & -4 \end{vmatrix}=0$$$$\Rightarrow 1\left( -4k+6 \right) -k\left( -12+4 \right) +3\left( 9-2k \right) =0$$$$\Rightarrow -4k+6+12k-4k+27-6k=0$$$$\Rightarrow -2k+33=0$$$$\Rightarrow k=\cfrac { 33 }{ 2 }$$$$\Rightarrow \cfrac { 2k }{ 11 } =\cfrac { 2\cfrac { (33) }{ 2 } }{ 11 } =3$$$$\therefore$$The correct answer is $$3$$Mathematics

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