Linear combination of others Write one of the vectors as a linear combination of the others.
$$ \left[
    \begin{array}{cc|c}
      0\\
      1\\
      1
    \end{array}
\right],\left[
    \begin{array}{cc|c}
      1\\
      0\\
      -1
    \end{array}
\right],\left[
    \begin{array}{cc|c}
      4\\
      5\\
      1
    \end{array}
\right] $$ 
So far, this is what I have done:
I took the 3 vectors and made them into a matrix.
$$ \left[
    \begin{array}{ccc}
      0&1&4\\
      1&0&5\\
      1&-1&1
    \end{array}
\right] $$
Then I put it into RREF and got the matrix below.
$$ \left[
    \begin{array}{ccc}
      1&0&5\\
      0&1&4\\
      0&0&0
    \end{array}
\right] $$
I don't know how to continue this problem.
 A: The last column is the coefficients in your linear combination. So 
$$5(0,1,1)+4(1,0,-1)=(4,5,1)$$
A: If one can be expressed as a linear combination of the others, they are linearly dependent. This means that we can find constants $x,y,z $ s.t.
$$ \left[
    \begin{array}{cc|c}
      0\\
      1\\
      1
    \end{array}
\right]x+
 \left[
    \begin{array}{cc|c}
      1\\
      0\\
      -1
    \end{array}
\right]y + \left[
    \begin{array}{cc|c}
      4\\
      5\\
      1
    \end{array}
\right]z = 
\left[
    \begin{array}{cc|c}
      0\\
      0\\
      0
    \end{array}
\right]$$
$$ \left[
    \begin{array}{ccc|c}
      0&1&4\\
      1&0&5\\
      1&-1&1\\
    \end{array}
\right] \left[
    \begin{array}{cc|c}
      x\\
      y\\
      z\\
    \end{array}
\right] = \left[
    \begin{array}{cc|c}
      0\\
      0\\
      0\\
    \end{array}
\right]$$
As you found, rref tells us:
$$ \left[
    \begin{array}{ccc|c}
      1&0&5\\
      0&1&4\\
      0&0&0\\
    \end{array}
\right] \left[
    \begin{array}{cc|c}
      x\\
      y\\
      z\\
    \end{array}
\right] = \left[
    \begin{array}{cc|c}
      0\\
      0\\
      0\\
    \end{array}
\right]$$
This leads to $2$ equation:
$$x + 5z = 0$$
$$y + 4z = 0$$
If we let $z = -1$
$$x = 5 \space\text { and } \space y = 4$$
So:
$$ 5\left[
    \begin{array}{cc|c}
      0\\
      1\\
      1
    \end{array}
\right]+
 4\left[
    \begin{array}{cc|c}
      1\\
      0\\
      -1
    \end{array}
\right] - \left[
    \begin{array}{cc|c}
      4\\
      5\\
      1
    \end{array}
\right] = 
\left[
    \begin{array}{cc|c}
      0\\
      0\\
      0
    \end{array}
\right]$$
Or, equivalently,
$$ 5\left[
    \begin{array}{cc|c}
      0\\
      1\\
      1
    \end{array}
\right]+
 4\left[
    \begin{array}{cc|c}
      1\\
      0\\
      -1
    \end{array}
\right] = \left[
    \begin{array}{cc|c}
      4\\
      5\\
      1
    \end{array}
\right] $$
