Question

If the entries in a $3\times 3$ determinant are either 0 or 1, then the greatest value of their determinants is:
(a) 1
(b) 2
(c) 3
(d) 9

Answer 1

Hint: For solving this problem firstly, we will assume a $3\times 3$ determinant $\left| A \right|$. After that, we will write the determinant value in terms of variables and then we use our intelligence to solve this problem correctly.

Complete step-by-step answer:
There is a $3\times 3$ determinant such that the value of each element is either 0 or 1. And we have to find the maximum value of the determinant.
Now, let $\left| A \right|$ is a $3\times 3$ determinant such that each element is either 0 or 1. We will use the following formula of determinant to find the determinant value:
$\begin{aligned} & \left| A \right|=\left| \begin{matrix} {{a}_{11}} & {{a}_{12}} & {{a}_{13}} \\\ {{a}_{21}} & {{a}_{22}} & {{a}_{23}} \\\ {{a}_{31}} & {{a}_{32}} & {{a}_{33}} \\\ \end{matrix} \right| \\\ & \Rightarrow \left| A \right|={{a}_{11}}\left( {{a}_{22}}{{a}_{33}}-{{a}_{23}}{{a}_{32}} \right)+{{a}_{12}}\left( {{a}_{23}}{{a}_{31}}-{{a}_{21}}{{a}_{33}} \right)+{{a}_{13}}\left( {{a}_{21}}{{a}_{32}}-{{a}_{22}}{{a}_{31}} \right)....................\left( 1 \right) \\\ \end{aligned}$
Now, we have to find the maximum value of the expression on the right-hand side in the equation (1). And if the value of each element is either 0 or 1 so, the product of any two elements will be either 0 or 1. Moreover, We can conclude that if the product is 0 then at least one of them should be 0 and if the product is 1 then both of them must be 1.
Now, consider the term ${{a}_{11}}\left( {{a}_{22}}{{a}_{33}}-{{a}_{23}}{{a}_{32}} \right)$ . For the maximum value of ${{a}_{11}}\left( {{a}_{22}}{{a}_{33}}-{{a}_{23}}{{a}_{32}} \right)$ the value of ${{a}_{11}}$ , ${{a}_{22}}{{a}_{33}}$ and ${{a}_{23}}{{a}_{32}}$ should be as per the data given below:
$\begin{aligned} & {{a}_{11}}=1 \\\ & {{a}_{22}}{{a}_{33}}=1 \\\ & \Rightarrow {{a}_{22}}={{a}_{33}}=1.....................\left( 2 \right) \\\ & {{a}_{23}}{{a}_{32}}=0..............................\left( 3 \right) \\\ \end{aligned}$
Now, the maximum value of ${{a}_{11}}\left( {{a}_{22}}{{a}_{33}}-{{a}_{23}}{{a}_{32}} \right)$ will be equal to 1. But from equation (3) we conclude that at least one of the term from ${{a}_{23}}$ and ${{a}_{32}}$ will be equal to 0.
Now, consider the term ${{a}_{12}}\left( {{a}_{23}}{{a}_{31}}-{{a}_{21}}{{a}_{33}} \right)$ . For the maximum value of ${{a}_{12}}\left( {{a}_{23}}{{a}_{31}}-{{a}_{21}}{{a}_{33}} \right)$ the value of ${{a}_{12}}$ , ${{a}_{23}}{{a}_{31}}$ and ${{a}_{21}}{{a}_{33}}$ should be as per the data given below:
$\begin{aligned} & {{a}_{12}}=1 \\\ & {{a}_{23}}{{a}_{31}}=1 \\\ & \Rightarrow {{a}_{23}}={{a}_{31}}=1.....................\left( 4 \right) \\\ & {{a}_{21}}{{a}_{33}}=0..............................\left( 5 \right) \\\ \end{aligned}$
Now, the maximum value of ${{a}_{12}}\left( {{a}_{23}}{{a}_{31}}-{{a}_{21}}{{a}_{33}} \right)$ will be equal to 1. But from equation (5) and equation (2) we conclude that the value of ${{a}_{21}}$ will be equal to 0.
Now, consider the term ${{a}_{13}}\left( {{a}_{21}}{{a}_{32}}-{{a}_{22}}{{a}_{31}} \right)$ . From the above discussion, we can write that as ${{a}_{21}}=0$ so, the value of ${{a}_{21}}{{a}_{32}}=0$ . Then, for the maximum value of ${{a}_{13}}\left( {{a}_{21}}{{a}_{32}}-{{a}_{22}}{{a}_{31}} \right)$ the value of ${{a}_{13}}$ should be equal to 0 if ${{a}_{22}}{{a}_{31}}=1$ otherwise, the value of ${{a}_{13}}\left( {{a}_{21}}{{a}_{32}}-{{a}_{22}}{{a}_{31}} \right)$ will be negative if the value of ${{a}_{22}}{{a}_{31}}=1$ and ${{a}_{13}}=1$ . There can be one more possibility that value of ${{a}_{13}}=1$ and ${{a}_{22}}{{a}_{31}}=0$ . And in all scenario, the maximum value of ${{a}_{13}}\left( {{a}_{21}}{{a}_{32}}-{{a}_{22}}{{a}_{31}} \right)$ will be equal to 0.
Now, from the above discussion we conclude the following data:
$\begin{aligned} & {{a}_{11}}\left( {{a}_{22}}{{a}_{33}}-{{a}_{23}}{{a}_{32}} \right)={{a}_{12}}\left( {{a}_{23}}{{a}_{31}}-{{a}_{21}}{{a}_{33}} \right)=1 \\\ & {{a}_{13}}\left( {{a}_{21}}{{a}_{32}}-{{a}_{22}}{{a}_{31}} \right)=0 \\\ \end{aligned}$
Now, as the value of $\left| A \right|={{a}_{11}}\left( {{a}_{22}}{{a}_{33}}-{{a}_{23}}{{a}_{32}} \right)+{{a}_{12}}\left( {{a}_{23}}{{a}_{31}}-{{a}_{21}}{{a}_{33}} \right)+{{a}_{13}}\left( {{a}_{21}}{{a}_{32}}-{{a}_{22}}{{a}_{31}} \right)$ . Then, the maximum value of determinant will be equal to $\left| A \right|=1+1+0=2$.
Thus, the maximum value of determinant will be 2.
Hence, (b) is the correct option.

Note: Here, the student should first understand what is asked in the problem before solving. After that, we should proceed as per the data given in the question and use our intelligence to solve such problems. Moreover, such problems don’t have any set pattern or any direct formula that will give us the answer. But we should solve these types of problems with our intelligence and try to solve them correctly.