$A \in M_3(\mathbb Z)$ be such that $\det(A)=1$ ; then what is the maximum possible number of entries of $A$ that are even ?
Let $A \in M_3(\mathbb Z)$ be such that $\det(A)=1$ ; then what is the maximum possible number of entries of $A$ that are even ?
Solution 1:
Clearly $I_3$ is an example where we can have $6$ even entries with $\det(A)=1$, if there are $7$ or more even entries then there must be at least one row having all entries as even integers, expand the $\det$ along that row, you'll get an even $\det(A)$, which is a contradiction. Thus maximum even entries possible are $6.$