Question

Let $\left\{ {{D}_{1}},{{D}_{2}},{{D}_{3}},....,{{D}_{n}} \right\}$ be the set of third-order determinant that can be made with the distinct non-zero real numbers ${{a}_{1}},{{a}_{2}},{{a}_{3}},.....,{{a}_{9}}$ . Then
a) $\sum\limits_{i=1}^{n}{{{D}_{i}}=1}$
b)\[\sum\limits_{i=1}^{n}{{{D}_{i}}=0}\]
c)${{D}_{i}}={{D}_{j}}$ for all i, j
d) none of these

Answer 1

Hint: Since it is given in the question that we have a set of third-order determinants, so we have in total 9 elements of a determinant. Find the total number of ways of arranging all the 9 elements in various configurations. Then add all the configurations to get the summation of determinants formed by using determinant properties.

Complete step by step answer:
As we know that, for a third-order determinant, we have:
$A=\left[ \begin{matrix}
   {{a}_{1}} & {{a}_{2}} & {{a}_{3}} \\
   {{a}_{4}} & {{a}_{5}} & {{a}_{6}} \\
   {{a}_{7}} & {{a}_{8}} & {{a}_{9}} \\
\end{matrix} \right]$
Where, ${{a}_{1}},{{a}_{2}},{{a}_{3}},.....,{{a}_{9}}$ are non-zero real numbers.
Let us assume that, one of the configurations of determinants formed by ${{a}_{1}},{{a}_{2}},{{a}_{3}},.....,{{a}_{9}}$ is:
${{D}_{1}}=\left[ \begin{matrix}
   {{a}_{1}} & {{a}_{2}} & {{a}_{3}} \\
   {{a}_{4}} & {{a}_{5}} & {{a}_{6}} \\
   {{a}_{7}} & {{a}_{8}} & {{a}_{9}} \\
\end{matrix} \right]$
Now, we need to find number of configurations of determinants formed bye ${{a}_{1}},{{a}_{2}},{{a}_{3}},.....,{{a}_{9}}$
Let us exchange row 1 with row 2. We have:
${{D}_{2}}=\left[ \begin{matrix}
   {{a}_{4}} & {{a}_{5}} & {{a}_{6}} \\
   {{a}_{1}} & {{a}_{2}} & {{a}_{3}} \\
   {{a}_{7}} & {{a}_{8}} & {{a}_{9}} \\
\end{matrix} \right]$
Since, Determinant of a matrix changes its sign if we interchange any two rows or columns present in a matrix.
So, we can say that:
${{D}_{2}}=-{{D}_{1}}$
From the result, we can say that whatever configuration of determinant is formed by the non-zero elements ${{a}_{1}},{{a}_{2}},{{a}_{3}},.....,{{a}_{9}}$ , each of them will have its negative counterpart.
Now, we need to find the summation of all the determinants formed.
i.e. $\sum\limits_{i=1}^{n}{{{D}_{i}}}$
We can say that:
$\sum\limits_{i=1}^{n}{{{D}_{i}}}={{D}_{1}}+{{D}_{2}}+{{D}_{3}}+......+{{D}_{n}}$ , where n is the total number of configurations of determinants formed.
As we have known that, ${{D}_{2}}=-{{D}_{1}}$
Therefore, ${{D}_{2}}+{{D}_{1}}=0$
Hence, adding any configuration of determinant formed withs its counterpart we get zero as the answer.
Therefore, \[\sum\limits_{i=1}^{n}{{{D}_{i}}=0}\]

So, the correct answer is “Option D”.

Note: There is another way to solve this question.
Since we know that the total number of elements on a third-order determinant is 9. So, the number of ways of placing 9 elements in 9 places is: $9!$
Since $9!=362880$ is an even number., we can say that the counterparts of each determinant formed are: $\dfrac{9!}{2}$
Now, we have determinants as well as their counterparts. When we add them, we get zero as an answer. No determinant is left behind. So, the summation of determinants formed is zero.