Question

Let {D1, D2, D3, .........Dn} be the
set of third order determinant that can be
made with the distinct non-zero real
numbers a1, a2 ...... a9 (using all), then​

Answer 1

Given:  {D1, D2, D3, .........Dn} is the  set of third order determinant.

To find: Sum of determinant?

Solution:

• Now we have given that  {D1, D2, D3, .........Dn} be the  set of third order determinant that can be  made with the distinct non-zero real  numbers a1, a2 ...... a9.

• So the total number of determinant formed will be: 9!

• Now the number of determinants which are even: 9! / 2

• Now,  there is a determinant formed by interchanging two  rows/columns consecutively which corresponds to the each determinant.

• So we can say that sum of these two pairs will be 0.

• Taking two determinants at a time, sum of them will always be 0.

• So the sum of all the determinant will be:

                  0 + 0 + 0 +..................9! / 2 = 0

• So the summation will be zero.

Answer:

                                 n

So the answer is:    ∑  = D(i) = 0

                                 i=1