value of k so that equation have non zero soution

Question

\begin{equation} x+ky-z=0 \end{equation} \begin{equation} 3x-ky+z=0 \end{equation} \begin{equation} x-3y+z=0 \end{equation}

the problem that I am having is that it's in the form homogeneous equation so $x=y=z=0$. Thus, my answer is that it can take value but my books mention the answer as 1.

Answer 1

Your system depends on $k$. If for some value of $k$ the determinant of your system is zero, then you may find more than one solution.

Try $k=3$ and check $ (x,y,z)=(0,1,3)$ as an answer different from $(0,0,0)$

As a result if your determinant is zero, then the solutions are not necessarily unique.

Answer 2

Adding the first two equations, $4x = 0$, so $x = 0$.

After substituting $x = 0$, the first two equations become:

$$ky-z=0$$ $$-ky+z=0$$

which is the same line.