Given the following linear system in $3$ parameters, above $\mathbb{R}:$ $$\begin{array}{l}x+ky+z=0\\kx+y+kz=0\\(k+1)x-y+z=0\end{array}$$ For which $k$ values for this system:
Is there a single solution?
There are infinite number of solutions?
I eventually, I came across this matrix: $$\left(\begin{array}{ccc}1&k&1\\0&(1-k^2)&0\\0&(-k^2-k-1)&-k\end{array}\right)$$ but stopped, because I can no longer divide by numbers which include $k$ (because I don't want to assume $k$ is not equal to a certain number).
I'd be glad if someone could point me out on this.
from the first equation we get $$x=-ky-z$$ plugging this in the second equation we obtain $$k(-ky-z)+y+kz=0$$ mupltiplying this out $$y(-k^2+1)=0$$ can you finish now?