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Question

The number of values of k for which the linear equations
4x+ky+2z=0
kx+4y+z=0
2x+2y+z=0
possess a non-
zero solution is : ?

A
3
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B
2
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C
1
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D
0
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Solution

The correct option is B 2
Here D=∣ ∣4k2k41221∣ ∣
For non-zero solution
D=0
On expanding the above determinant we get,
k26k+8=0
k=2,4
Therefore, there are 2 values of k for non-zero solutions.
Hence, option 'B' is correct.

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