How many entries in $3\times 3$ matrix with integer entries and determinant equal to $1$ can be even?

Question

Let $A$ be a $3\times 3$ matrix with integer entries such that $\det(A)=1$. At most how many entries of $A$ can be even?

I get a possible solution as $6$ by considering the $3 \times 3$ identity matrix. But I am not sure about that is it possible to have more than $6$ even entries. Please help me enumerate this problem to prove my answer.

Answer 1

If there are 7 or more even entries, you can show that the determinant is even, since the expansion is a sum of the products of three numbers, one of which is always even.

Answer 2

Using Laplace expansion or Sarrus's rule, we have $$ \begin{vmatrix}a&b&c\\d&e&f\\g&h&i\end{vmatrix}=aei-afh-bdi+bfg+cdh-ceg$$

In order for this expression to be equal to $1$, it must be odd, meaning that at least one of the $6$ products must be odd. And if one of the products is odd, then all three of the terms in the product must be odd.

Therefore there can be at most $6$ even entries, and the identity matrix shows that there can be exactly six.