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According to the answer of znt to the previous version, I revise the question as follows:

Is there a real $(n-1)\times n$ matrix $A$ such that $A$ is not a full rank matrix and satisfy $a_{ii}<0$ and $a_{ij}>0$ for $i \neq j$ and $\sum_{i} a_{ij}<0$ for every $j\leq n-1$.

A stronger version:

Is there a singular $n\times n$ matrix $A$ such that diagonal entries are negative, off diagonal entries are positive and each column sum up to a negative number?

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    $\begingroup$ Sure there is. For example set $n=5$ and just make the sum of the first two rows equal the sum of the last two. $\endgroup$
    – znt
    Commented Jul 10, 2016 at 17:00
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    $\begingroup$ Is your matrix supposed to be $(n - 1) \times n$ (as in the original question) or $n \times n$ (as in the equivalent formulation)? $\endgroup$
    – LSpice
    Commented Jul 14, 2016 at 20:32
  • $\begingroup$ @LSpice a sufficient condition for full rankness of $n-1 \times n $ matrix is that every( or at least one) $n-1 \times n-1$ submatrix would be invertible. So putting $ n:=n-1$ leads to the above equivalent formulation. $\endgroup$
    – Ali Taghavi
    Commented Jul 14, 2016 at 20:35
  • $\begingroup$ @DuchampGérardH.E. Thank you. I revise the title. $\endgroup$
    – Ali Taghavi
    Commented Jul 14, 2016 at 20:39

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No, to your stronger question. If an $n \times n$ matrix satisfies the condition you specified, it would be a strictly diagonally dominant matrix which is non-singular.

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  • $\begingroup$ I realy thank you very much for your answer. $\endgroup$
    – Ali Taghavi
    Commented Jul 14, 2016 at 20:54

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