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Determine the base space S of all solutions to the equation system =

\begin{cases} 1q +2w +7e-9r+31t =0 \\ 2q +4w+ 7e-11r+34t =0\\ 3q+6w+5e-11r+29t=0 \\ \end{cases}

$$ \left[ \begin{array}{ccccc|c} 1&2&7&-9&31&0\\ 2&4&7&-11&34&0\\ 3&6&5&-11&29&0 \end{array} \right] $$

I'm not really sure what is the question about... I've already made: $$ \left[ \begin{array}{ccccc|c} 1&2&0&-2&3&0\\ 0&0&1&-1&4&0\\ 0&0&0&-0&0&0 \end{array} \right] $$

so:

\begin{cases} q=-2w+2r-3t \\ e=r-4t\\ \end{cases}

So what should I give as answer? w,r and t?

1 Answers 1

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So the set S = { $ \lambda * (-2, 1, 0, 0, 0) + \mu * (2, 0, 1, 1, 0) + \nu * (-3, 0, 0, 0, 1) $ } is the space of the solutions.

Now go from here and try determining a basis for S.

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    More directly, to get the 3 basis vectors,take in turn each of the independent variables *w*, *r*, *t* to be 1 and the other two to be 0. – 2017-01-22
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    Isn't that exactly what I did ? But you still have to consider the dependent variables q and e. Sorry if my notation is a bit unclear. But it's supposed to be in that order (q,w,e,r,t) – 2017-01-22
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    I was just indicating one does not have to explicitly form the nullspace, as you did, in order to list a basis for it. Yes, put the variables in the order $(q, w, e, r, t)$ -- or any permutation of them that's convenient. Then assign 1 to one of the (in this case, 3) independent variables (here $w$, $r$, $t$) and 0 to the other independent variables, calculating the resulting values of the dependent variables ($q$ and $e$), and form the corresponding vector of these values. It's just a shortcut that works in general. – 2017-01-23
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    Ok year, I can see what you mean. Thing is I wanted to give more of a hint and not the solution itself, your way is much easier though if one already knows how to work with vectorspaces/bases/etc – 2017-01-23