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I'm looking for the name of this kind of $n$-by-$n$ matrix:

$$\left(\begin{array}{cccc} -s_1 & b_{12} & b_{13} & b_{14} \\ b_{21} & -s_2 & b_{23} & b_{24} \\ b_{31} & b_{32} & -s_3 & b_{34} \\ b_{41} & b_{42} & b_{43} & -s_4 \end{array}\right)$$

where $s_i = \sum_{j\neq i} b_{ij}$. This is basically so I can google it, so by name I really mean an appropriate "search string".

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I think it's the Q-matrix of a continuous time Markov chain, but I'm not sure... – user71815 Jun 12 '13 at 11:29
Sounds about right – Lucas Jun 12 '13 at 15:50

We have diagonally dominant matrices if absolute values of diagonal elements are larger than sum of absolute values of the others in that row.

Such matrices are invertible, ie. nonzero determinant.

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