Maximum determinant of $3 \times 3$ matrix

Question

Good one guys! I'm studying to the maths olympiads in my college and I ran to the following problem:

What is the possible matrix $3 \times 3$, that you can write using digits from $0 $ to $9$, (you can repeat them), that gives you the maximum determinant?

I got by brute force the matrix: \begin{pmatrix} 0 && 9 && 9\\ 9 && 0 && 9 \\ 9 && 9 && 0\\ \end{pmatrix}

Are there any other ways to do it besides brute force?

I looked about Hadamard maximum determinant theorem but I did not get how to apply it.

Thanks in advance =)

Answer 1

The determinant of a $3\times 3$ matrix is just the area of the parallelepiped spanned by it's column vectors. If you think about the problem geometrically, I think it's a bit easier to see why your answer should be correct.

Answer 2

More generally, consider the $n \times n$ case. Since the determinant is a linear function of the entries in any given row or column, it's clear that there is an optimal solution with all entries $0$ or $9$. Dividing by $9$, you have an $n \times n$ $0-1$ matrix with maximum determinant, and that corresponds to a normalized solution of the $(n+1) \times (n+1)$ Hadamard maximal determinant problem.

Answer 3

For a $3\times 3$ matrix you can also easily see it by using The rule of Sarrus.