Determinants Question - None

Question

If the trivial solution is the only solution of the system of equations $x - ky + z = 0$, $kx + 3y - kz = 0$, $3x + y - z = 0$, then the set of all values of $k$ is :

Accepted Answer

$x - ky + z = 0$
$kx + 3y - kz = 0$
$3x + y - z = 0$
this system will have non trivial solution if (non-trivial solution)
$\Rightarrow \begin{vmatrix} 1 & -k & 1 \ k & 3 & -k \ 3 & 1 & -1 \end{vmatrix} = 0$
$\Rightarrow 1(-3 + k) + k(-k + 3k) + 1(k - 9) = 0$
$\Rightarrow k - 3 + 2k^2 + k - 9 = 0$
$\Rightarrow 2k^2 + 2k - 12 = 0$
$\Rightarrow k^2 + k - 6 = 0$
$\Rightarrow k = -3, k = 2$
So the system of equations will have only trivial solution when $k \in R - \{2, -3\}$.

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