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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!

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    $\begingroup$ Hint: why could you divide by $4s + (t-2)$? $\endgroup$ – lulu Sep 24 '15 at 11:27
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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.

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    $\begingroup$ Awesome thanks for your help! $\endgroup$ – Cameron Sep 24 '15 at 12:01
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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 $t=2-4s$.

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