Maximal and Minimal value of $\bf{3^{rd}}$ order determinant whose elements are from the set $\bf{\{0,1\}}$.
Maximal and Minimal value of $\bf{3^{rd}}$ order determinant whose elements are from the set $\bf{\{-1,0,1\}}$.
My try for first::
Let $A = \begin{pmatrix} a & b & c\\ d & e & f\\ g & h & i \end{pmatrix}$. here $a,b,c,d,e,f,g,h,i\in \{0,1\}$
Now after that how can I calculate Max. and Min. value of Given Determinants.
Thanks
For both parts, if $M$ is the maximum value and $m$ is the minimum value, then it is clear that $m=-M$ (via row swapping). Hence, it suffices to determine $M$ only.
For (1), $\det(A)$ is a sum of $3$ integers in $\{0,1\}$ subtracted by $3$ integers in $\{0,1\}$. Hence, $M\leq 3$. The value $3$ is not achievable, as it would mean every entry of $A$ has to be $1$, but this leads to $\det(A)=0$. Ergo, $M\leq 2$. It is easy to see that $M=2$.
For (2), $\det(A)$ is a sum of $6$ integers belonging in $\{-1,0,1\}$. Hence, $M\leq 6$. Now, the value $6$ is not achievable as it would mean that every entry must be $-1$ or $+1$. Thus, $aei=bfg=cdh=+1$ and $gec=hfa=idb=-1$, so $$+1=(abcdefghi)^2=(aei)(bfg)(cdh)(gec)(hfa)(idb)=(+1)^3(-1)^3=-1\,,$$ which is absurd. The value $5$ is not achievable, as it implies that exactly one entry of $A$ is $0$, whilst the rest are $-1$ or $+1$, but then $\det(A)$ is a sum of at most $4$ integers belonging in $\{-1,+1\}$. Hence, $M\leq 4$. It is easy to see that $M=4$.
I wonder what the answer would be for $n$-by-$n$ matrices, with the same questions and constraints.
(1) My guess is Max 2 and Min -2. A value of 3 cannot be achieved. Think it as $aei+bfg+cdh-gec-hfb-idb$. For 2 take $a=e=i=b=f=g=1$
(2) I am not sure of this, but I got 4 as max value. So -4 should be min. For 4 take $c=e=f=1$ rest are -1.