For what values of k is the vector u in the span of the columns of A? Let $$u = \begin{pmatrix} -1 \\1 \\-1 \end{pmatrix}$$ 
and let$$A = \begin{pmatrix} 1&0&1 \\1&1&0 \\0&1&k \end{pmatrix}$$ 
For what values of k is the vector u in the span of the columns of A?
Here is my work so far:
$$Au = \begin{pmatrix} 1&0&1&-1 \\1&1&0&1 \\0&1&k&-1 \end{pmatrix}$$$$Au = \begin{pmatrix} 1&0&1&-1 \\0&1&-1&2 \\0&1&k&-1 \end{pmatrix}$$$$Au = \begin{pmatrix} 1&0&1&-1 \\0&1&-1&2 \\0&0&k+1&-3 \end{pmatrix}$$
I've reduced it as much as I can, but I'm stuck at this point. Does this mean that k can be equal to -4?

$(-1, 1, -1) = x (1, 1, 0) + y (0, 1, 1) + z (1, 0, k) $
\begin{equation*}
 \left\{\begin{aligned}
-1 = x 1 + y 0 + z 1 \\
1 = x 1 + y 1 + z 0 \\
-1 = x 0 + y 1 + z k
 \end{aligned}
 \right.
\end{equation*}
\begin{equation*}
 \left\{\begin{aligned}
-1 - x &= z(x) \\
1 - x &= y(x) \\
-1 &= y + z k = 1-x + (-1-x) k
 \end{aligned}
 \right.
\end{equation*}
$- \frac{x-2}{x + 1} = k(x)$
