What is the largest determinant you can get by filling in 0,1 or 2 into a 4-by-4 matrix? For example
$$\left| \begin{array}{ccc}
2 & 0 & 0 & 2 \\
2 & 0 & 2 & 0 \\
0 & 2 & 1 & 2 \\
2 & 2 & 0 & 0 \end{array} \right|=40$$
Can it get bigger than that? And what's your approach?
 A: More generally, consider an $n \times n$ matrix whose entries are in $[0,k]$ for some $k > 0$.
Since the determinant is an affine function of each matrix element, we may as well assume that each element is $0$ or $k$.  Moreover, by scaling, the answer will be $k^n$ times what we would get by replacing $k$ by $1$.  Now look up "Hadamard maximal determinant problem".
See e.g. https://oeis.org/A003432 and references there.
A: First of all, the option of filling in a "1" can always be avoided because the determinant is always a linear function of any entry. So substituting either 2 or 0 can ensure an equal or bigger determinant.
And if the question is to fill in either 2 or 0, then as Mark Bennet pointed out, it can be simplify into a binary matrix. However, this "simplified" problem is actually a particular case of the very complicated "Hadamard's maximal determinant problem" (MathWorld | Wikipedia). The general solution still has not been discovered yet.
The answer for this particular case is $48=3\times2^4$. And there're 60 different possible matrices that have this determinant.
A: $$\left| \begin{array}{ccc}
2&0  & 2 & 2 \\
2&2 & 0 & 2  \\
2&2 & 2 & 0  \\
0&2 & 2 & 2  \end{array} \right|=48$$
