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

Let $\Delta =\left|\begin{array}{ccc}{a}_1& a_2& a_3\\ b_1& b_2& b_3\\ c_1& c_2& c_3\end{array}\right|$
Now, let $\overrightarrow{a}=(a_1,a_2,a_3),\overrightarrow{b}=(b_1,b_2,b_3),\overrightarrow{c}=(c_1,c_2,c_3),$
then the volume of parallelopiped formed by vectors $\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$ as their adjecent sides is given by $|\Delta |.$
$\Delta =\overrightarrow{a}.(\overrightarrow{b}\times \overrightarrow{c})=\left[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}\right]$
$\Delta \le |\overrightarrow{a}||\overrightarrow{b}||\overrightarrow{c}|\le 3\sqrt{3}<6$
$\Delta =(a_1{b}_2{c}_3+a_2{b}_3{c}_1+a_3{b}_1{c}_2-a_1{b}_3{c}_2-a_2{b}_1{c}_3-a_3{b}_2{c}_1)$
Now, $\Delta$ can take maximum integeral value $5,$ as all the entries in $\Delta$ are integers
There are six terms in the $\Delta$. To get value $5$, we need one term as $0$ and all other term as $+1$.
But, each element $(a_i,b_i,c_i)$ comes exactly in two term in $\Delta$ so, if we put any one of the element as $0$, the two term becomes $0,$therefore $\Delta$ can not be equal to $5.$
Let's try for $\Delta =4$
$\Delta =(a_1{b}_2{c}_3+a_2{b}_3{c}_1+a_3{b}_1{c}_2-a_1{b}_3{c}_2-a_2{b}_1{c}_3-a_3{b}_2{c}_1)$
Start by making any two term $0.$
Let's make first and last terms $0$, for that take $b_2=0.$
$\Rightarrow \Delta =a_2{b}_3{c}_1+a_3{b}_1{c}_2-a_1{b}_3{c}_2-a_2{b}_1{c}_3{\downarrow }_{\text{positive}}{\downarrow }_{\text{positive}}{\downarrow }_{\text{negative}}{\downarrow }_{\text{negative}}$
In third term $a_1{c}_2{b}_3,$ we have $a_1$ as independent variable so, take $a_2=-1,b_3=-1,c_1=1a_3=1,b_1=1,c_2=1c_3=1,a_1=1,b_2=0$
$\Delta =\left|\begin{array}{ccc}1& -1& 1\\ 1& 0& -1\\ 1& 1& 1\end{array}\right|=4.$