Even Entries

Obviously, the identity matrix has determinant $1$ and has $6$ even entries.

If a matrix has at least $7$ even entries, then (by the Pigeonhole Principle) there must be at least one row all of whose entries are even. But that would force the determinant to be even. Thus the maximum number of even entries is $\boxed{6}$ .

By Pigeonhole Principle, I get that the matrix would have maximum 6 even number inside that, but I try to figure out an example to prove that it is "exactly" have a solution of that, so I consider a Matric

$M = \begin{bmatrix} 2 & 1 & 2 \\ 2 & 2 & 1\\ a & b & c \end{bmatrix}$

Such that $a$ is an odd number and $b$ , $c$ are even number.

So after some try, I get

$M = \begin{bmatrix} 2 & 1 & 2 \\ 2 & 2 & 1\\ 5 & 4 & 4 \end{bmatrix}$

Which is a solution, So Answer Proved.

We can give an example of such a matrix with 6 even entries. If there are 7 or more even entries, then necessarily there is a row with only even entries, if we calculate the determinant by expanding this line, it is a sum of even terms, thus the determinant is even and can't be 1. . Thus 6 is the maximum possible number of entries of A that are even and such as det(A)=1 .