With the numbers 2, 4, 6, 8, all the possible determinants with these four different elements are constructed. What is the sum of the values of all such determinants?

Concept:

Number of ways of arranging n object = n!

Calculation:

Given numbers are 2, 4, 6 & 8.

We can form determinant of order 2.

We know that the number of ways of arranging n object = n!

Number of determinants = 4! = 4 × 3 × 2 × 1 = 24

Let's form some determinants from the given numbers.

\(\begin{vmatrix} 2 & \rm 6 \\ 8 & 4 \end{vmatrix}\) = 2 × 4 - 6 × 8 = 8 - 48 = - 40

\(\begin{vmatrix} 6 & \rm 2 \\ 4 & 8 \end{vmatrix}\) = 48 - 8 = 40

\(\begin{vmatrix} 2 & \rm 8 \\ 6 & 4 \end{vmatrix}\) = 8 - 48 = - 40

\(\begin{vmatrix} 6 & \rm 4 \\ 2 & 8 \end{vmatrix}\) = 48 - 8 = 40

\(\begin{vmatrix} 4 & \rm 8 \\ 6 & 2 \end{vmatrix}\) = 8 - 48 = - 40

\(\begin{vmatrix} 8 & \rm 4 \\ 2 & 6 \end{vmatrix}\) = 48 - 8 = 40

\(\begin{vmatrix} 4 & \rm 6 \\ 8 & 2 \end{vmatrix}\) = 8 - 48 = - 40

\(\begin{vmatrix} 8 & \rm 2 \\ 4 & 6 \end{vmatrix}\) = 48 - 8 = - 40

Hence, we can see that we are getting the pattern where the other value will neutralize each determinant value.

Accordingly, the sum of the values of all determinants = 0

∴ The required sum is zero.