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

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

 A: 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 $\det(AA^t)=nk^2$.
A: By the Binet-Cauchy formula, $|AA^t| = \sum_{k = 1}^n |A_k|^2$, where $A_k$ is the matrix obtained from $A$ by deleting the $k$-th column of $A$. For $k = 1,2,\ldots, n-1$, perform the the column operation $C_1 + C_2 + \cdots + \hat{C_k} + \cdots + C_{n-1} \to C_n$ on $A_k$. That is, take the sum of the $j$th columns of $A_k$ for $1 \le j \le n-1$ and $j\neq k$, and add it to the $n$th column. Since $\sum_{j = 1}^n a_{ij} = 0$ for all $i$, the resulting matrix is
$$A_k^{'} = \begin{pmatrix}a_{11} & a_{12} & \cdots & \hat{a_{1k}}  & \cdots & -a_{1k}\\a_{21} & a_{22} & \cdots & \hat{a_{2k}} & \cdots & -a_{2k}\\\vdots & \vdots &  & \vdots &  & \vdots\\a_{(n-1)1} & a_{(n-1)2} & \cdots & \hat{a_{(n-1)k}} & \cdots & -a_{(n-1)k}\end{pmatrix}$$
A cyclic permutation of second, third,. . ., $(n-1)$st columns of $A_k'$ yields the matrix
$$A_k^{''} = \begin{pmatrix}a_{11} & a_{12} & \cdots & -a_{1k} & \cdots & a_{1(n-1)}\\a_{21} & a_{22} & \cdots & -a_{2k} & \cdots & a_{2(n-1)}\\\vdots & \vdots &  & \vdots & &\vdots\\a_{(n-1)1} & a_{(n-1)2} & \cdots & -a_{(n-1)k} & \cdots &a_{(n-1)(n-1)}\end{pmatrix}$$
which has determinant $-|A_n|$. None of the column operations involved affect $|A_k|^2$, so $$|A_k|^2 = |A_k^{''}|^2 = |A_n|^2 \quad (k = 1,2,\ldots,n-1)$$
and hence
$$|AA^t| = \sum_{k = 1}^n |A_n|^2 = n|A_n|^2.$$
Of course, since each entry of $A$ is an integer, $|A_n|$ is an integer.
