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If the system of equations 2x+3ky+(3k+4)z=0, x+(k+4)y+(4k+2)z = 0, x+2(k+4)y+(3k+4)z=0 has non trivial solution, then K =

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8 or $- \frac{1}{2}$

–4 or $\frac{1}{2}$

4 or $- \frac{1}{2}$

–8 or $\frac{1}{2}$

answer is A.

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Detailed Solution

$| \begin{matrix} 2 & 3 k & 3 k + 4 \\ 1 & k + 4 & 4 k + 2 \\ 1 & 2 (k + 4) & 3 k + 4 \end{matrix} | = 0$

$\Rightarrow k = - 8 o r \frac{1}{2}$

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