Find the maximum determinant of the $3 \times 3$ matrix containing only $-1$ or $1$ in it.

Question

This question was found on the math contest provided by Thailand Mathematics Club.

According to the question, there are $2^9 = 512$ ways to write $3 \times 3$ matrices containing only $-1$ or $1$ in it.

The solution (maximum determinant) was revealed to be $0$. Is it true or false?

Answer 1

If you have a condition on real $n\times n$-matrices such that if $M$ fulfills that condition, then $-M$ does too (that is, the condition is symmetric under negation), then the only way that the maximal determinant of matrices with that condition is $0$ is if all matrices fulfilling that condition have determinant $0$. That's because $\det(-M)=-\det M$.

Now it is easy to check that $3$ is odd, and that the condition of all entries being either $1$ or $-1$ is symmetric under negation.

Therefore all you need it to find a matrix that has non-zero determinant. But that one is easy: $$\det\begin{pmatrix} -1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{pmatrix} = 4 \ne 0$$ Indeed, this is already a counterexample to the claim by itself. However I didn't construct it as such; all I did was to ensure that all the rows are obviously linearly independent, by just modifying the diagonal.

Answer 2

There is probably a more elegant solution than this, but since the $3\times 3$ case only has a handful of cases, we can check them by brute force (albeit in a slightly smart way). Obviously $0$ is a possible determinant, so we consider now nonsingular matrices. Note that swapping rows as well as multiplying all entries by $-1$ do not change the absolute value of the determinant. Now we try and break into cases regarding the possible entries in the first row considerng only the magnitude of the determinant, then we only need to dicuss two cases: $1,1,1$ and $1,1,-1$. All other cases can be obtained by swapping columns or multiplying by $-1$.

In the first case, if the determinant is to be nonzero then we only need consider the case where the second row is $1,1,-1$, or $1,-1,-1$ (again, other cases are obtainable by swapping columns). Now the last row has $8$ possible choices each, and by testing all $16$ cases manually we find that the determinant can have absolute values $0,4$ only. We can proceed identically for the case where the first row is $1,1,-1$, noting that we need not dicsuss the case where the second row is $1,1,1$ or $-1,-1,-1$ since that is already covered in the previous case. Eventually we find that the only possible values for the absolute value of the determinant are $0$ and $4$.

It is possible to find by explicit example that $0,4,-4$ are all obtainable values of the determinant, which then proves the result.