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Linear programming: basic solution?

If the matrix consists of $$\begin{bmatrix}1&-2&0&0&0\\-3&6&1&3&0\\0&0&2&6&-1\end{bmatrix},$$ how is it that there are 4 basic solutions? It says that the first and second columns, as well as the third and fourth columns are scalar multiples, but what does that have to do with basic solutions?

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    A basic solution is the intersection of two or more constraints. If these are the only constraints (no sign restrictions), then there is a mistake somewhere. The second comment is about basic variables, not basic solutions (slightly different).2012-10-09
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    Er...what are you talking about here, to begin with? Is the given matrix the extended one of a linear system in 4 unknowns and thus the last *column* is what we have at the right of the equality sign, or it is a coefficients matrix of a homogeneous system in 5 unknowns??2012-10-09

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