Let
$$
A = \begin{bmatrix}
-a_{12}-a_{13}-a_{14} & a_{12} & a_{13} & 1\\
a_{21} & -a_{21}-a_{23}-a_{24} & a_{23} & 1\\
a_{31} & a_{32} & -a_{31} – a_{32} – a_{34} & 1\\
a_{41} & a_{42} & a_{43} & 1
\end{bmatrix},
$$
where all $a_{ij}$'s are positive reals. If we explicitly calculate the determinant of $A$ and factor whole expression, then we can easily see that $\det(A) < 0$.
Is it possible to prove that $\det(A) < 0$ (or that $\det(A) \neq 0$ if it is easier) using some matrix manipulations without calculating it directly?
Best Answer
In general, suppose $A$ is a real $n\times n$ matrix in the form of $\pmatrix{-M&b\\ c^T&a}$ where $a$ is a scalar, $b,c$ are positive vectors and $M$ is a strictly diagonally dominant matrix with a positive diagonal and non-positive off-diagonal entries.
By Geršgorin disc theorem, all eigenvalues of $M$ have positive real parts. Thus $\det(M)>0$ and $M$ is a Z-matrix whose eigenvalues have positive real parts. It follows that $M$ is an M-matrix and $M^{-1}$ is entrywise positive. So, by using Schur complement, we get $$ \det(A)=\det(-M)(a+c^TM^{-1}b)=(-1)^{n-1}\det(M)(a+c^TM^{-1}b) $$ and hence $\operatorname{sgn}(\det(A))=(-1)^{n-1}$. In your case, $n=4$ and hence $\det(A)<0$.