Is a square matrix with positive determinant, positive diagonal entries and negative off-diagonal entries an M-matrix? I'm trying to determine if a certain class of matrices are M-matrices in general. I'm considering square matrices $A$ with the following properties: 


*

*$\det(A) > 0$ (strictly),

*all the diagonal entries are positive,

*all the off diagonal entries are negative.


An M-matrix can be characterized in many ways. I've tried proving this (or finding a counterexample) by looking at the principal minors and have found that $A$ is an M-matrix if it has dimension 2 or 3, but it's hard to make any sort of induction with that. Right now I'm trying two other definitions (they're equivalent) of M-matrices


*

*There is a positive vector such that $Ax > 0$ (component-wise).

*$A$ is monotone (i.e. $Ax \geq 0$ implies $x \geq 0$).


Again, 1 isn't hard to show if the matrix is small, but this is hard to generalize, so I thought an easier approach might be using 2 and try to proceed by contradiction. Does anyone here have some suggestions? This is an outside project for a class I'm working on so I don't know if these matrices are or are not M-matrices in general - mostly just looking for tips here.  
 A: What you are suggesting is that the (necessary and sufficient) conditions


*

*$A$ is a Z-matrix (matrix with non-positive off-diagonal entries),

*all the principal minors of $A$ are positive,


for an $n\times n$ matrix $A$ to be an M-matrix are equivalent to (or implied by) the conditions


*

*$A$ is a Z-matrix (strict negativity of the off-diagonal entries is unnecessary due to the continuity of the determinant),

*all $1\times 1$ principal minors and $\det(A)$ are positive.


Note that the positivity of $1\times 1$ minors means that $A$ has positive diagonal. 
This is true trivially if $n=1$ or $n=2$. It is also true if $n=3$ (positivity of diagonal and non-negativity of off-diagonal entries implies positivity of $2\times 2$ minors). However, this fail to be the case when $n\geq 4$.
For example,
$$
A=\left(\begin{array}{rrrr} 1 & -2 & -1 & -3\\ -1 & 2 & -5 & -2\\ -1 & -5 & 1 & -1\\ -5 & -1 & -1 & 3 \end{array}\right).
$$
Note that the determinants of both the leading and trailing $3\times 3$ submatrices are negative but $\det(A)=30>0$.
