Sum of each row is $m$, prove that $m$ divides the determinant This is a question from our reviewer for our exam for linear algebra. I just want to have some ideas how to tackle the problem.
If $A$ is an $n\times n$ matrix with integer coefficients, such that the sum of each row's elements is equal to $m$, show that $m$ divides the determinant.
 A: Here's a guide: Why is the "all 1's" vector an eigenvector of $A$? What is its eigenvalue? How does $\det(A)$ relate to the eigenvalues of $A$?
A: The other answer is probably better, but I offer this as an alternative.
Let B be an upper triangular matrix of 1s and 0s.  For example, 4x4:
$B = \begin{array} {c c c c }
1 & 1 & 1 & 1\\
0 & 1 & 1 & 1\\
0 & 0 & 1 & 1\\
0 & 0 & 0 & 1\\
\end{array}$
Let $C = A \times B$.  Since $det(B) = 1$ then $det(C) = det(A)$.  By the definition of matrix multiplication, the last column of C is m.  For example:
$C = \begin{array} {c c c c }
... & ... & ... & m\\
... & ... & ... & m\\
... & ... & ... & m\\
... & ... & ... & m\\
\end{array}$
$det(C) = m \cdot det(some sub matrix) - m \cdot det(someothersubmatrix) + \dots$
Since C is an integer matrix, the determinant of each submatrix is an integer, so m divides det(C), so m divides det(A).
Not exactly a very conceptually useful proof, but not a difficult one either.
A: Given any $n \times n$ matrix $A = (a_{ij})$ whose row sums all equal to $m$, i.e.
$$\sum_{j=1}^n a_{ij} = m, \quad\text{ for } 1 \le i \le n \tag{*1}$$
Let $\text{adj}(A) = (b_{ij})$ be its adjugate matrix.
We know
$$\text{adj}(A) A = \det(A) I_n
\quad\iff\quad
\sum_{j=1}^n b_{ij} a_{jk} = \det(A) \delta_{ik}\quad\text{ for } 1 \le i, k \le n
\tag{*2}$$
where $\delta_{ik}$ is the Kronecker's delta. In RHS of (*2), if we fix $i$ to $1$ and then sum over $k$, $(*1)$ will imply:
$$m \sum_{j=1}^n b_{1j} = \sum_{j=1}^n \sum_{k=1}^n b_{1j} a_{jk} = \det(A) \sum_{k=1}^n \delta_{1k} = \det(A)$$
When $a_{ij}$ are all integers, so do $b_{ij}$. As a result $\displaystyle \sum_{k=1}^n b_{1k}$ is an integer and hence $m$ divides $\det(A)$.
