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Let $A$ be an $n$ by $n$ real matrix such that all entries not on the diagonal are positive, and the sum of the entries in each row is negative. How to show that the determinant of $A$ is non-zero?

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@AndreaOrta: I tried to solve using the determinant function. What's your suggestion? – Sugata Adhya Jan 24 '13 at 17:14
up vote 2 down vote accepted

Since the off-diagonal elements in any row are all positive, their sum is positive. This implies that the diagonal element is of each row is strictly greater than the sum of the off-diagonal elements in magnitude.Thus we have that $A$ is a strictly diagonally dominant matrix. A strictly diagonally dominant matrix is non-singular by the Levy–Desplanques theorem. Therefore the determinant of $A$ is nonzero.

See: link

For a proof of the Levy-Desplanques theorem, see this link:

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