Question

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

• A. 3
• B. 2
• C. 1
• D. zero

Answer 1

Answer: (B)

2

For the system to have non-zero solution

$\left|\begin{array}{ccc}4& k& 2\\ k& 4& 1\\ 2& 2& 1\end{array}\right|=0$

Expanding the determinant, we get

$4(4-2)-k(k-2)+2(2k-8)$

$\Rightarrow k^2-6k+8=0$

$\Rightarrow k=2$ or $4$

Therefore, two values of $k$ exist.

Hence, option 'B' is correct.