Question

Let $P$ be a matrix of order $3 \times 3$ 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 1

$\Delta=\begin{vmatrix}a_{1} & a_{2} & a_{3} \\ b_{1} & b_{2} & b_{3} \\ c_{1} & c_{2} & c_{3}\end{vmatrix}$
$=\underbrace{\left(a_{1} b_{2} c_{3}+a_{2} b_{3} c_{1}+a_{3} b_{1} c_{2}\right)}_{x}-\underbrace{\left(a_{3} b_{2} c_{1}+a_{2} b_{1} c_{3}+a_{1} b_{3} c_{2}\right)}_{y}$
Now if $x \leq 3$ and $y \geq-3$ the $\Delta$ can be maximum $6$
But it is not possible as $x=3$ each term of $x=1$ and $y=3$
each term of $y=-1$
$\Rightarrow \displaystyle\prod_{i=1}^{3} a_{i} b_{i} c_{i}=1$
and $\displaystyle\prod_{i=1}^{3} a_{i} b_{i} c_{i}=1$
Which is contradiction. So now next possibility is $4$