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Question

Mathematics Question on Determinants

The number of values of k for which the linear equations 4x+ky+2z=04x + ky + 2z = 0 , kx+4y+z=0kx + 4y + z = 0 and 2x+2y+z=02x + 2y + z = 0 possess a non-zero solution is

A

2

B

1

C

5

D

3

Answer

2

Explanation

Solution

Δ=0\Delta = 0 4k2 k41 221=0\Rightarrow\begin{vmatrix}4&k&2\\\ k&4&1\\\ 2&2&1\end{vmatrix} = 0  4(42)k(k2)+2(2k8)=0\Rightarrow \ 4(4 - 2) - k(k - 2) + 2(2k - 8) = 0  8k2+2k+4k16=0\Rightarrow \ 8 - k^2 + 2k + 4k - 16 = 0 k26k+8=0k^2 - 6k + 8 = 0  (k4)(k2)=0  k=4,2\Rightarrow \ (k - 4 )(k-2) = 0 \ \Rightarrow \ k = 4 , 2