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(Found in "Linear Algebra and Its Applications" by Gilbert Strang)

"Find the value of c which makes it possible to solve:

$$ \left\{ \begin{array}{c} u+v+2w=2\\ 2u+3v-w=5\\ 3u+4v+w=c \end{array}\right.$$"

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Thank you very much – Beatriz Vieira Mar 12 '13 at 10:31
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Use the appropriate extended matrix of coefficients: $$\left(\begin{array}{ccc|c}1&1&2&2\\2&3&-1&5\\3&4&1&c \end{array}\right)$$ Row-reduce it, and find the value of $c$ for which you have no rows of the form $\begin{pmatrix}0&0&0&|&\alpha\end{pmatrix}$ for $\alpha\neq0$. For that value there exists a solution.
If there are no zero rows - then the solution is unique. If there is at least one zero-row then there are infinitely many solutions.

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