Show that a matrix satisfying certain conditions is non-singular

Question

I have a square matrix $A$ satisfying the following conditions:

1. The elements on the diagonal are negative;
2. All other elements are non-negative;
3. All row sums are less than or equal to $0$;
4. There is at least one row sum that is less than $0$.

I think that $A$ is non-singular, but I cannot prove it. I was thinking to try to show that $\det{A} \neq 0$ by determining that there is no eigenvalue that is zero. Can anyone point me to a book or sketch a proof (not necessarily the one I have in mind)?

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Edit 1 It seems that I am missing a condition. I will try to sketch the situation more closely. There are two more matrices $B$ and $C$ with non-negative entries and at least one element positive such that the matrix defined by

\begin{equation} D := A + B + C \end{equation}

is irreducible and all rows sums of $D$ are $0$. This should immediately exclude the matrix proposed by @MooS.

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Edit 2 Does the first edit imply that $A$ is actually irreducible so that $A$ is irreducibly diagonally dominant and non-singular as shown in the first sentence of this Wikipedia page?

Answer 1

You need all row sums to be $<0$, else you can do the following:

$$\begin{pmatrix}-1&1&0\\1&-1&0\\0&0&-1\end{pmatrix}$$

If you assume all row sums to be $<0$, the statement is true. After multiplying by $-1$, the matrix will be a Diagonally dominant matrix.