How to show the determinant of a matrix is zero for which the column elements sum to zero

Question

It is given that the sum of the elements of each column of a $N\times N$ square matrix ${\rm A}$ is zero. How can I show that the determinant is zero without specializing to $N=2$ or $N=3$?

Answer 1

$det(A) = = det \left( \begin{matrix}a_1 \\ \vdots \\ a_{n-1}\\a_n \end{matrix} \right) = det \left( \begin{matrix}a_1 \\ \vdots \\ a_{n-1} \\a_1 + \cdots +a_n \end{matrix} \right) = det \left( \begin{matrix}a_1 \\ \vdots \\ a_{n-1}\\0 \end{matrix} \right) = 0$

Where $a_i$ are the row vectors of $A$.

Of course, within the matrix $0$ stands for the row vector $0 = (0 \ldots 0)$.

Answer 2

Suppose the matrix is $(a_{ij}$, you have $a_{ni}=\sum_{j=1}^{j=n-1}-a_{ji}$, if you sum the first $n-1$ line you have the last line with the opposite sign.