$adj(A)$ and odd numbers Let $A\in M_{n}(\mathbb{Z}),n\geq 2.$
Prove that, if $A$ has exactly one odd number on each line and on each column, then $adj(A)$ has the same property.
For example, $A$ could be $ \begin{pmatrix}
2a & 2b & ... &2c-1 \\ 
2d-1 &2e  &...  &2f \\ 
\vdots  & \vdots  & \ddots  &\vdots  \\ 
 2g& 2h-1 &...  &2i 
\end{pmatrix}$, but I haven't found a meaningful idea by writing it in this form so far. 
 A: Note: I'll use the notation $A \equiv B \mod 2$ for matrices to mean that 
all entries of $A$ are equivalent mod $2$ to the corresponding entries of $B$.
The "mod $2$" homomorphism  $\mathbb Z \to \mathbb Z/2\mathbb Z$ induces a homomorphism of rings $M_n(\mathbb Z) \to M_n(\mathbb Z/2\mathbb Z)$.
That $A$ has one odd number in each row and column says $A \equiv P \mod 2$ where $P$ is a permutation matrix. Then $P^{-1}$ is also a permutation matrix, and $A P^{-1} \equiv I \mod 2$.  In particular, $\det(A)$ is odd and since
$A\; \text{adj}(A) = \det(A) I$, 
$$ \text{adj}(A) - P^{-1} = P^{-1} P (\text{adj}(A) - P^{-1}) \equiv P^{-1} A (\text{adj}(A) - P^{-1}) \equiv 0 \mod 2$$
A: We can rearrange the rows and columns so that it suffices to show that
$$\det\left(\begin{matrix}
a_{11} & \cdots & a_{1n} \\ 
\vdots & & \vdots \\ 
a_{n1} & \cdots & a_{nn} \end{matrix}\right)$$
is odd when the diagonal entries $a_{ii}$ are the only odd entries. This is easy to show by induction:
$$\det\left(\begin{matrix}
a_{11} & a_{12} \\ 
a_{21} & a_{22} \end{matrix}\right) = a_{11}a_{22} - a_{12}a_{21}$$ is clearly odd because the first term is odd and the second term is even. Now we assume the result is true for $n-1$. Now 
$$\det\left(\begin{matrix}
a_{11} & \cdots & a_{1n} \\ 
\vdots & & \vdots \\ 
a_{n1} & \cdots & a_{nn} \end{matrix}\right) = \sum_{j = 1}^n a_{1j} \det{A_{1j}}$$
where $A_{1j}$ is a $(n-1)\times (n-1)$ submatrix. This sum is odd because $\det A_{11}$ is odd by induction, $a_{11}$ is odd and $a_{1j}$ is even for $j > 1$.
