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

$△=\left|\begin{array}{ccc}{a}_1& a_2& a_3\\ b_1& b_2& b_3\\ c_1& c_2& c_3\end{array}\right|=\underset{x}{\underset{}{(a_1{b}_2{c}_3+a_2{b}_3{c}_1+a_3{b}_1{c}_2)}}-\underset{y}{\underset{}{(a_3{b}_2{c}_1+a_2{b}_1{c}_3+a_1{b}_3{c}_2)}}$
Now if $x\le 3$ and $y\ge -3$
then $△$ can be maximum $6$
But it is not possible
as $x=3\Rightarrow$ each term of $x=1$
and $y=3\Rightarrow$ each term of $y=-1$
$\Rightarrow {\Pi }_{i=1}^3{a}_i{b}_i{c}_i=1$ and ${\Pi }_{i=1}^3{a}_i{b}_i{c}_i=-1$
which is contradiction
So now next possibility is $4$
which is obtained as $\left|\begin{array}{ccc}1& 1& 1\\ -1& 1& 1\\ 1& -1& 1\end{array}\right|=1(1+1-1(-1-1)+1(1-1)=4$.