TITLE: prove that $|AA^t|=nk^2$ ; $k\in \mathbb Z$
QUESTION [5 upvotes]: Let $A_{n-1 \times n}=(a_{ij})$ be with entries in $\mathbb Z$ such that $\sum_{j=1}^{n}a_{ij}=0 , \forall i\in \{1,2,\cdots,n-1\}$. 
Show that

$|AA^t|=nk^2$ ; $k\in \mathbb Z$.

REPLY [4 votes]: Another solution is to consider the matrix $B_{n\times n}$ whose first $n-1$ rows equal to the rows of $A$ and the last row equal to $(1,\ldots,1)$. 
Consider also the matrix $C_{n\times n}$  whose first column entries are equal to $1$, the diagonal entries are equal to $1$ and the other entries are equal to $0$. So $\det(BC)=\det(B)\det(C)=\det(B)$. 
Notice that the entries of the first column of $BC$ are $BC_{j1}=0$, for $1\leq j<n$, and $BC_{n1}=n$. Thus, $\det(BC)$ is a multiple of $n$, i.e., $\det(BC)=nk$. Thus, $det(B)=nk$ and $det(BB^t)=n^2k^2$.
Finally,
$BB^t=
\left(\begin{array}{ c c }
     AA^t & 0_{n-1\times 1} \\
     0_{1\times n-1} & n
  \end{array} \right)$. Thus, $n^2k^2=\det(BB^t)=\det(AA^t)n$ and $\det(AA^t)=nk^2$.