"If each entry of a 3×3 determinant is either 1 or -1 then find the possible values of such a determinant".
I have tried a comparitively easier variation of the above problem that is instead of +1 and -1 the entries can be +1 and 0, and it was easier because if i take a triplet(triplet here implies the product of 3 elements formed when expanding determinant) and it is 0 then we can say that all of them arent non zero and since each element of determinant occurs in 2 triplets, we can easily analyze to get values as -2,+2,-1,+1,0.
But when dealing with the above shown version of this problem, it is taking a lot of cases to be dealt with, at a point i am becoming unclear, i tried the same approach as the below problem in the above problem like:
|a b c |
|d e f | = det.
|g h i |
aei+bfg+cdh-(gec+hfa+idb)
Each triplet = +1 or -1 If triplet= 1 then either all 3 are 1 or 2 are -1 and one is 1. But when i try to analyze it with the other 3 triplets in which the 3 elements repeat for ex: aei let a=e=-1 and i=1 then analyzing with gec,idb,hfa is becoming very complicated.
I tried searching for similar alliterations to this question on this site for answer but did not found what i exactly needed. Can anyone share an algorithm, process or if possible just the thinking process to solve above questions. (I found a nice way to find number of equivalence relations on a set which thus lead me here again)
Let $M=\begin{vmatrix} a_1&a_2&a_3\\a_4&a_5&a_6\\a_7&a_8&a_9\end{vmatrix}=(a_1a_5a_9+a_2a_6a_7+a_3a_4a_8)-(a_1a_6a_8+a_2a_4a_9+a_3a_5a_7)$ so in this problem we have $$M=(\pm1\pm1\pm1)-(\pm1\pm1\pm1)$$ each of the two numbers in brackets can be equal to $(\pm1\pm3)$ so $$M=(\pm1,\pm3)-(\pm1,\pm3)\in\{0,\pm2,\pm4,\pm6\}$$ $M=0$ when two lines are equal.
$M=4$ when $a_1=a_2=a_3=a_4=a_7=a_8=1$ and $a_5=a_6=a_9=-1$ (interchanging two lines we get $M=-4$).
To have $M=2\text { or } 6$ we need $$159+267+348=b \text { and } 168+249+357=-b \text { with b=1,3, respectively. }$$ We can put $a_1=1$ from which we get in both cases $$9=-37,\space5=-24,\space8=-27,\space6=-34$$ Then $$M=\begin{vmatrix} 1&a_2&a_3\\a_4&-a_2a_4&-a_3a_4\\a_7&-a_2a_7&-a_3a_7\end{vmatrix}=a_4a_7\begin{vmatrix} 1&a_2&a_3\\1&-a_2&-a_3\\1&-a_2&-a_3\end{vmatrix}=0,\text { contradiction. }$$ So $2$ and $6$ never appear and the only possible values of determinants are $0$ and $\pm4.$