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So, I have the following System of linear equations and I have to determine the set of solutions $\mathcal{L} = (Au)$:

$$\left\{ \begin{array}{ll} a + 2b + c - 3d = 3 \\ -2a + b - 2c + d = -1 \\ 2a + 2c -2d = 2 \\ a + 3b + c -4d = 4 \\ \end{array} \right.$$

If I try to reach row echelon form with the gauß-algorithm I get $c = b - a$ and $d = b - 1$ but what is the solution set now?

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    $\begingroup$ That's it, because $a$ and $b$ are free parameters. $\endgroup$ – user326159 Jan 18 '17 at 21:42
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Write the solutions in vector form: $u+av+bw$. Explicitly: $$\begin{bmatrix}0\\0\\0\\-1\end{bmatrix}+a\begin{bmatrix}1\\0\\-1\\0\end{bmatrix}+b\begin{bmatrix}0\\1\\1\\1\end{bmatrix}.$$

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