Let {D1, D2, D3 ,Dn} be the set of third...

Let `{D_1, D_2, D_3 ,D_n}` be the set of third order determinant that can be made with the distinct non-zero real numbers `a_1, a_2, a_qdot` Then `sum_(i=1)^n D_i=1` b. `sum_(i=1)^n D_i=0` c. `D_i-D_j ,AAi ,j` d. none of these

`overset(n)underset(i=1)(Sigma) D_(i)=1`

`overset(n)underset(i=1)(Sigma)D_(i)=0`

`D_(i)D_(j), AA I, j`

None of these

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The correct Answer is:

the total number of third -order determinants is 9! The number of determinants s even and of these there are `9!//2` pairs of determinants which are obtained by changing two consecutive
`" rows so " overset(n)underset(i=1)(Sigma) D_(i) =0`

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