Given matrix 3x3 with pre-set conditions and values of {0,-1,1), calculate the maximum value of its determinant

Question

I know a similar question has been posted a number of times BUT this is not the same problem.

Given matrix 3x3 with entries {0,-1,1}, what is the largest value of its determinant when you know that there are exactly three entries of zeroes, at least two entries of 1 and at least one entry of -1.

Since there are exactly three elements equal to zero, that means that at least 3 addendum will be qual to zero so the max value we can think of is l3.

$$ \begin{pmatrix} &a &b &c\\ &d &e &f \\ &g &h &i \\ \end{pmatrix} = x - y $$

where $x = (aei + dhc + gbf)$ and $y = (gec + ahf + dbi)$.

Determinant of value 3 ( $x-y=3$) would require one of these scenarios:

$$ 3 - 0 = 3$$(impossible) $$ 2 - (-1) = 3$$ $$ 1 - (-2) = 3$$

The last two options are also impossible, considering the distribution of zeroes, we reach the conclusion that there are (at most) 2 non-zero addendums, hence we can conclude that the maximum value of the discriminant is 2.

I am a bit worried about the info we were given at the beginning, regarding the minimum number of 1 and -1 = I know that the sign can affect the determinant value, but I think that going with 2 as the max value is reasonable.

Thoughts?

Answer 1

That's not very convincing. Why don't you consider three cases:

1. All three zeros are in the same row or column;
2. Two zeros are in the same row or column;
3. The zeros are in different rows and columns.

Further everything is obvious. In the first case it is always $0$, in the second and third cases it is not more than $2$.

It remains to construct a determinant equal to $2$. Here and only here it is important to observe the conditions for the number of entries equal to $1$ and $-1$.