On the determinant of a certain matrix with non negative integer entries with fixed row sum Let $d$ be a positive integer. Let $n \ge 3$ be an integer. Let $A$ be a $n \times n$  matrix with non-negative integer entries such that each row sum is $d$ and the determinant is also $d$ . Let $B$ be a $n \times n$ matrix whose first column has all the same positive integer entries say $m$  and all the other entries of the matrix are $0$ . Then is it true that $\det (A+B)=d+m$ ? 
 A: Yes, it's true. By assumption, $A$ has nonzero determinant so has an inverse $A^{-1}$. Then
\begin{equation}
\det(A+B)=\det(A)\det(I+A^{-1}B)=d\det(I+A^{-1}B).
\end{equation}
To prove the claim, it suffices to show that this latter determinant is $1+m/d$. 
Think about what $A^{-1}B$ looks like; it only has nonzero entries on the first column. So $I+A^{-1}B$ is lower triangular, and the diagonal is all $1$s except for the $(1,1)$ entry, where it is $1$ plus the first entry in $m\cdot A^{-1}e$, where $e$ is the all-ones vector (as this is the first column of $B$ by assumption). Because the determinant of a lower triangular matrix is the product of the diagonal entries, it suffices to show that $A^{-1}e=(1/d)e$. But this is true as $e$ is an eigenvector of eigenvalue $d$ for $A$, as $A$ has constant row sum $d$. Thus, $A^{-1}e=(1/d)e$, which completes the proof.
(By the way, this evidently also means that the $n\geq 3$, non-negativity and integrality of $A$, and positivity and integrality of $B$ are not necessary. Actually, even better, if you allow $A$ to have arbitrary nonzero determinant $c$, not necessarily equal to the common row sum $d$, you will get $\det(A+B)=c(1+m/d)$.)
A: A way to see that the last column of $A$ for example is $\begin{pmatrix}s-\sum_{i=1}^{n-1}a_{1,i}\\s-\sum_{i=1}^{n-1}a_{2,i}\\\vdots\\s-\sum_{i=1}^{n-1}a_{n,i}\end{pmatrix}.$ $s$ is the row sum. Your statement is equivalent after spliting $\det(A+B)=\det(A)+\det(C)$, now the first column of $C$ is all $m$ and the rest entries as   $A$ Replacing $m$ by $s$ i.e. $\dfrac{s}{m}\det(C)$ equals $\det(A)$ as one sees the first column all $s$ is the sum (combination) of all columns of $A$.
