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Q.

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 trival solution then k=

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a

-4 or 1/2

b

8 or -1/2

c

4 or -1/2

d

-8 or 1/2

answer is A.

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Detailed Solution

$\begin{array}{l} |\begin{array}{ccc} 2 & 3 k & 3 k + 4 \\ 1 & k + 4 & 4 k + 2 \\ 1 & 2 (k + 4) & 3 k + 4 \end{array}| = 0 \\ 2 [(k + 4) (3 k + 4) - (2 k + 8) (4 k + 2)] - 3 k [3 k + 4 - 4 k - 2] + (3 k + 4) [2 k + 8 - k - 4] = 0, \\ k = \frac{1}{2} or - 8 \end{array}$

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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 trival solution then k=