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Q. 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 :

Determinants

Solution:

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