Let `P` be a matrix of order `3xx3` 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 ______.

Explanation

Let Det (P) `=|{:(a_(1), b_(1), c_(1)), (a_(2), b_(2), c_(2)), (a_(3), b_(3), c_(3)):}| `
`= a_(1) (b_(2)c_(3) - b_(3)c_(2))-a_(2)(b_(1)c_(3)-b_(3)c_(1)) + a_(3)(b_(1)c_(2) -b_(2)c_(1))`
Now, maximum value of Det (P) = 6
If `a_(1) = 1, a_(2) = -1, a_(3) =1, b_(2)c_(3) = b_(1)c_(3) = b_(1)c_(2) = 1 " and "b_(2)c_(2) = b_(3)c_(1) = b_(2)c_(1) = -1`
But it is not possible as
`(b_(2)c_(3))(b_(3)c_(2)) = -1 " and " (b_(1)c_(3)) (b_(3)c_(2))(b_(2)c_(1))=1`
i.e., `b_(1)b_(2)b_(3)c_(1)c_(2)c_(3) = 1 " and " -1`
Similar contradiction occurs when
`a_(1) = 1, a_(2) = 1, a_(3) = 1, b_(2)c_(1) = b_(3)c_(1) = b_(1)c_(2) = 1 " and " b_(3)c_(2) = b_(1)c_(3)= b_(1)c_(2) =-1`
Now, for value to be 5 one of the terms must be zero but that will make 2 terms zero which means answer cannot be 5
Now, `|{:(1, " "1, 1), (-1, " "1, 1), (1, -1, 1):}| =4`
Hence, maximum value is 4.