How to tell if a linear system is consistent

Question

So I have a list of equations and have made it into REF which gives me

$$\left[\begin{matrix}1 & -1 & 1 \\ 0 & 1 & -1 \\ 0 & 0 & 0\end{matrix}\middle|\begin{matrix}1 \\ 2k-k^{2} \\ -2k^{2}+k-3\end{matrix}\right]$$

How do I tell what values this is consistent for? I think it's $k=1$, $k=3/2$ and $k=2$ however, when I put $k=1$ into the system I get the following:

$x-y+z=1$, $y-z=1$ and $0=-4$ the last part of which obviously doesn't make much sense? Some help please!

EDIT: the original matrix was:

$$\left[\begin{matrix}1 & -1 & 1 \\ 1 & -2 & 2 \\ 2 & -1 & 1\end{matrix}\middle|\begin{matrix}1 \\ (k-1)^{2} \\ -k^{2}+3k-1\end{matrix}\right]$$

Answer 1

The system is consistent if $-2k^2 + k - 3 = 0 \iff 2k^2 - k + 3 = 0$, but this equation has no real solution since $\triangle = (-1)^2 - 4\cdot 2\cdot 3 = -23 < 0$, which means it is inconsistent.

Answer 2

When is $-2k^2 + k - 3 = 0$?

Answer that, and you'll have the value that makes the system consistent.

If there is no solution (no value of $k$ which makes the entry zero), then the system of equations is never consistent (hence, is inconsistent), whatever $k$ may happen to be.

Answer 3

The system is inconsistent if your matrix contains any of this: $$\begin{bmatrix} 0 & 0 & 0 &| &\text{non-zero number} \end{bmatrix}$$ Thus, we need the right side to be $0$ in order to make the system consistent. Hence, we need $$-2k^2+k+3=0.$$ Solving for $k$, I used the quadratic formula (or you can factor too), $$k = \frac{-1 \pm \sqrt{(1)^2-4(-2)(3)}}{2(-2)}=\frac{-1 \pm 5}{-4}=-\frac{3}{2},-1$$ We need $k=\boxed{-\frac{3}{2}}$ or $k=\boxed{-1}$ for this system to be consistent.