For what values of k is the vector u in the span of the columns of A?

Question

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?

Answer 1

Recall what we're trying to do here: the vector $u$ is in the span of the columns of $A$ if there is an $x$ such that $Ax = u$. Under other circumstances you might already know all the entries of $A$ (i.e. there is no $k$ to investigate), and you would write down the augmented matrix $(A\vert u)$ (as you have done here), and then row-reduce.

What is the significance of that process? Why do you want to get $1$s on the diagonal? (what would that tell you about $x$?) What is the significance of a $0$ on the diagonal?

Answer 2

$(-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*}

$- \frac{x-2}{x + 1} = k(x)$