Identifying when vectors are linearly dependent Let u = (1, 2, 1), v = (0, 1, s) and w = (2, 0, t). Find the condition ons and t which makes the set {u, v, w} linearly dependent.
I set up the following matrix, and started to row reduce said matrix:
$ \left[
    \begin{array}{ccc|c}
      1&0&2&0\\
      2&1&0&0\\
      1&s&t&0
    \end{array}
\right] $
$\rightarrow$$ \left[
    \begin{array}{ccc|c}
      1&0&2&0\\
      0&1&-4&0\\
      0&s&t-2&0
    \end{array}
\right] $
$\rightarrow$ $ \left[
    \begin{array}{ccc|c}
      1&0&2&0\\
      0&1&-4&0\\
      0&s&t-2&0
    \end{array}
\right] $
$\rightarrow$ $ \left[
    \begin{array}{ccc|c}
      1&0&2&0\\
      0&1&-4&0\\
      0&0&4s+(t-2)&0
    \end{array}
\right] $
I then got confused as to where to go. I divided the last row by a factor of (4s+(t-2)) so that the matrix became
$ \left[
    \begin{array}{ccc|c}
      1&0&2&0\\
      0&1&-4&0\\
      0&0&1&0
    \end{array}
\right] $
This of course would reduce to the identity, giving only the trivial solutions. That would suggest that irrespective of what s and t are the vectors could not be linearly dependent. Have I gone wrong some where, as the questions seems to suggest that there probably should be some case where the vectors are linear dependent
Thanks in advance!
 A: You need that the original matrix does not have full rank. You reduce it to a matrix where it is easy to determine the rank. This is good. 
In your second to last matrix it is true that if you can divide by $4s+(t-2)$ then you get that your matrix has full rank. 
Therefore, for you matrix not to have full rank you need that you cannot divide by  $4s+(t-2)$, that is $4s+(t-2)=0$, which is basically the condition you are looking for. 
A: The set $\{u,v,w\}$ is linearly independent if and only if there exist three not all zero real coefficients  $\alpha,\beta,\gamma$ such that
$$\alpha u+\beta v= w$$
Since the first component of $v$ is zero, we can easily conclude that $\alpha=2$. Knowing this and observing that the second coefficient of $w$ is zero, we have $\beta=-4$. Now,
$$\alpha \begin{pmatrix} 1 \\ 2 \\1 \end{pmatrix}+\beta \begin{pmatrix} 0 \\ 1 \\s \end{pmatrix} =\begin{pmatrix} 2 \\ 4 \\2 \end{pmatrix}+\begin{pmatrix} 0 \\ -4 \\-4s \end{pmatrix}= \begin{pmatrix} 2 \\ 0 \\t \end{pmatrix}$$
$$\implies \begin{pmatrix} 2 \\ 0 \\2-4s \end{pmatrix}= \begin{pmatrix} 2 \\ 0 \\t \end{pmatrix}$$ 
For the vectors to be linearly dependent, we must have $t=2-4s$.
