Let $M=\begin{pmatrix} A & 0\\ 0 &B \end{pmatrix}$, where $A=[a_{i,j}]_{n\times n}$ and $B=[b_{i,j}]_{n \times n}$ are two square matrix of size $n$ with all non-negative entries. Assume that $\displaystyle\sum_{j=1}^n a_{i,j} <1$ and $\displaystyle\sum_{j=1}^n b_{i,j} <1$ for all $i=1,\ldots,n$.
Consider two vectors $u=(u_1,\ldots, u_{2n}) \in \mathbb{R}^{2n}_{++}$ and $v=(v_1,\ldots, v_{2n}) \in \mathbb{R}^{2n}_{++}$ such that new matrix $N=M+v^T\cdot u=[n_{i,j}]_{2n \times 2n}$ satisfies that $\displaystyle\sum_{j=1}^{2n} n_{i,j} <1$. This matrix $N$ is also called the adjusted matrix.
Define $N_{i,j}$ as the matrix by eliminating the $i$-th row and $j$-th column in the matrix $I-N$. The conjecture is that for all $1<i<n$, \begin{equation}\label{keyequa} |\det(N_{1,i})| >|\det(N_{1,n+i})| \end{equation}
It's easy to see that with matrix $M$, $|\det(M_{1,i})| >|\det(M_{1,n+i})|$. And then I tried to apply the lemma about the determinant of the adjusted matrix in here, but the second part, which is $u^T \mathrm{adj}(N_{1,i}) v$ and $u^T\mathrm{adj}(N_{1,n+i}) v$ is hard to compare.
Another idea I have thought relates to the network. Consider $N$ as an adjacent matrix of a weighted network. Then we know that $\dfrac{|\det(N_{i,j})|}{\det(I-N)}$ is sum of all possible paths from $j \rightarrow i$, then comparing $|\det(N_{1,i})|$ and $|\det(N_{1,n+i})|$ is equivalent to comparing the path $i \rightarrow 1$ and the path $n+i \rightarrow 1$. Once again, I'm stuck here.
The following information maybe helpful, which is that $n_{1,i}>n_{1,n+1}$ and $n_{i,1}>n_{n+1,1}$ for all $1<i<n$.
Any ideas/discussions/proofs are very welcome. Thank you very much.