Let P be a matrix of order 3 x 3 such that all the entries in P are from the set {-1,0,1} . Then, the maximum possible value of the determinant of P is _____

Answer

4

Explanation

det (p) = `|{:(a_(1),,a_(2),,a_(3)),(b_(1),,b_(2),,b_(3)),(c_(1),,c_(2),,c_(3)):}|`
`=(a_(1)b_(2)c_(3)+a_(2)b_(3)c_(1)+a_(3)b_(1)c_(2)) -(a_(3)b_(2)c_(1) +a_(1)b_(3)c_(2)+a_(2)b_(1)c_(3))`
`=x-y`
where `x=a_(1)b_(2)c_(3)+a_(2)b_(3)c_(1)+a_(3)b_(1)c_(2)` and
`y=a_(3)b_(2)c_(1)+a_(1)b_(3)c_(2)+a_(2)b_(1)c_(3)`
Now `x le 3 " and " y ge -3`
If we consider det (P) =6 then
`a_(1)b_(2)c_(3)=a_(2)b_(3)c_(1)=a_(3)b_(1)c_(2) =1`
n and `a_(3)b_(2)c_(1) =a_(1)b_(3)c_(2)=a_(2)b_(1)c_(3)=-` which is contradiction .
So next possibility is 5 for which one of the terms must be zero but that will make two terms zero.
Hence greatest value of determinant is 4.