# 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 $$\det (A+B)=d+m$$ ?

Yes, it's true. By assumption, $$A$$ has nonzero determinant so has an inverse $$A^{-1}$$. Then $$$$\det(A+B)=\det(A)\det(I+A^{-1}B)=d\det(I+A^{-1}B).$$$$ To prove the claim, it suffices to show that this latter determinant is $$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 $$\det(A+B)=c(1+m/d)$$.)

• This proof is beautiful!XD Nov 10, 2019 at 4:00
• @pooja Thanks much!
– J.G
Nov 10, 2019 at 4:05
• Extraordinary proof(+1). Nov 10, 2019 at 4:08

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 $$A$$.