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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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