Question bank/JEE Main/Mathematics

If the trivial solution is the only solution of the system of
equations -
$$\begin{matrix} x - ky + z = 0 \\ kx + 3y - kz = 0 \\ 3x + y - z = 0 \end{matrix}$$
then set of all values of k is

A$R - \{ 2\}$B$R - \{ - 3\}$C$R - \{ + 2, - 3\}$D$\{ - 2, - 3\}$

Show answer and solution▾

$\Delta = k - 3 + 2k^{2} + k - 9$
$\Delta = 2k^{2} + 2k - 12$
$\Delta = 2((k + 3)(k - 2))$
for trivial solution $\Delta \neq 0$ $\therefore(k + 3)(k - 2) \neq 0$
$k \neq - 3$ and $k \neq + 2$ $\therefore k \in R - \{ + 2, - 3\}$

Source: Grass-G / M-17 Mathematics Paper 26 Nov Second.docx