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

Answer: 4

$\Delta =\left|\begin{array}{ccc}{a}_1& a_2& a_3\\ b_1& b_2& b_3\\ c_1& c_2& c_3\end{array}\right|$
$=\underset{x}{\underbrace{\left(a_1{b}_2{c}_3+a_2{b}_3{c}_1+a_3{b}_1{c}_2\right)}}-\underset{y}{\underbrace{\left(a_3{b}_2{c}_1+a_2{b}_1{c}_3+a_1{b}_3{c}_2\right)}}$
Now if $x\le 3$ and $y\ge -3$ 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 $y=-1$
$\Rightarrow \prod _{i=1}^3{a}_i{b}_i{c}_i=1$
and $\prod _{i=1}^3{a}_i{b}_i{c}_i=1$
Which is contradiction. So now next possibility is $4$