Question

Let $P$ be a matrix of order $3\times 3$ such that all the entries in $P$ are from the set $\left\{ -1,0,1 \right\}$. Then the maximum possible value of the determinant of $P$ is ______.

Answer 1

Hint: Expand the determinant by taking symbolical entries. Separate the positive and negative entries. Proceed with trial and error method for different maximum values of the determinant. \[\]

Complete step-by-step answer:
Let us assume the matrix $P$ has entries as follows \[P=\left[ \begin{matrix}
   {{a}_{1}} & {{a}_{2}} & {{a}_{3}} \\
   {{b}_{1}} & {{b}_{2}} & {{b}_{3}} \\
   {{c}_{1}} & {{c}_{2}} & {{c}_{3}} \\
\end{matrix} \right]\]. As given in the body of solution all the entries in $P$ are taken from the set $\left\{ -1,0,1 \right\}$. Now we shall denote the determinant value of $P$ as $\Delta \left( P \right)$ and calculate $\Delta \left( P \right)$ by expansion from the first row,
\[\Delta \left( P \right)=\left| \begin{matrix}
   {{a}_{1}} & {{a}_{2}} & {{a}_{3}} \\
   {{b}_{1}} & {{b}_{2}} & {{b}_{3}} \\
   {{c}_{1}} & {{c}_{2}} & {{c}_{3}} \\
\end{matrix} \right|={{a}_{1}}\left( {{b}_{2}}{{c}_{3}}-{{b}_{3}}{{c}_{2}} \right)-{{a}_{2}}\left( {{b}_{1}}{{c}_{3}}-{{b}_{3}}{{c}_{1}} \right)+{{a}_{3}}\left( {{b}_{1}}{{c}_{2}}-{{b}_{2}}{{c}_{1}} \right)\]

We shall separate the positive and negative terms in the expression,
\[\Delta \left( P \right)=\left( {{a}_{1}}{{b}_{2}}{{c}_{3}}+{{a}_{2}}{{b}_{3}}{{c}_{1}}+{{a}_{3}}{{b}_{1}}{{c}_{2}} \right)-\left( {{a}_{1}}{{b}_{3}}{{c}_{2}}+{{a}_{2}}{{b}_{1}}{{c}_{3}}+{{a}_{3}}{{b}_{2}}{{c}_{1}} \right)={{D}_{1}}-{{D}_{2}}\text{ }\left( \text{say} \right)\]
We observe that if all the entries are taken from the set $\left\{ -1,0,1 \right\}$then the maximum value of any term inside ${{D}_{1}}$ or ${{D}_{2}}$ is 1.
Case-1: The value of the determinant will be maximum if ${{D}_{1}}$ is maximum at the value 3 and ${{D}_{2}}$ is minimum at the value -3 , so that $\Delta \left( P \right)={{D}_{1}}-{{D}_{2}}=6$. In symbols,
\[\begin{align}
  & {{D}_{1}}=3 \\
 & \Rightarrow {{a}_{1}}{{b}_{2}}{{c}_{3}}+{{a}_{2}}{{b}_{3}}{{c}_{1}}+{{a}_{3}}{{b}_{1}}{{c}_{2}}=3 \\
 & \Rightarrow {{a}_{1}}{{b}_{2}}{{c}_{3}}=1,{{a}_{2}}{{b}_{3}}{{c}_{1}}=1,{{a}_{3}}{{b}_{1}}{{c}_{2}}=1 \\
\end{align}\]
Similarly,
\[\begin{align}
  & {{D}_{2}}=-3 \\
 & \Rightarrow {{a}_{1}}{{b}_{3}}{{c}_{2}}+{{a}_{2}}{{b}_{1}}{{c}_{3}}+{{a}_{3}}{{b}_{2}}{{c}_{1}}=-3 \\
 & \Rightarrow {{a}_{1}}{{b}_{3}}{{c}_{2}}=-1,{{a}_{2}}{{b}_{1}}{{c}_{3}}=-1,{{a}_{3}}{{b}_{2}}{{c}_{1}}=-1 \\
\end{align}\]
We can observe that the values of the terms inside ${{D}_{1}}$ and ${{D}_{2}}$ cannot occur at the same time. In other words to assign ${{D}_{1}}=3$ if we take ${{a}_{1}}={{b}_{2}}={{c}_{3}}={{a}_{2}}={{b}_{3}}={{c}_{1}}={{a}_{3}}={{b}_{1}}={{c}_{2}}=1 $ then putting the same in ${{D}_{2}}$ we cannot get to $-3$. So we reject the possibility of $\Delta \left( P \right)$ being 6.\[\]
Case-2: The next possible maximum value of $\Delta \left( P \right)$ is 4 where at least two terms from ${{D}_{1}}$ and ${{D}_{2}}$ vanishes. We take an example to demonstrate one such possibility we take ${{a}_{1}}={{b}_{2}}={{c}_{3}}={{a}_{2}}={{b}_{3}}={{c}_{1}}={{a}_{3}}=1,{{b}_{1}}={{c}_{2}}=-1$ where two terms from ${{D}_{2}}$ vanishes. Here,
\[\Delta \left( P \right)=\left| \begin{matrix}
   1 & 1 & 1 \\
   -1 & 1 & 1 \\
   1 & -1 & 1 \\
\end{matrix} \right|=4\]
We can also find out also such other possibilities $ \Delta \left( P \right) $ being 4. So the maximum possible value of the determinant of $P$ is 4.\[\]

Note: We note that we can find the determinant value of a matrix when the matrix is a square matrix. We also note that the determinant of the inverse of matrix $A$ is reciprocal of determinant of $A$ which means $\Delta \left( {{A}^{-1}} \right)={{\Delta }^{-1}}\left( A \right)$.