Let {D1,D2,......,Dk} be the set of third order determinants that can be made with the distinct nonzero real numbers a1,a2,a3.....,a9,then
(a)k=factorial 9 (b) D1+D2+........+Dk=0 (c)at least one Di=0 (d) none of these.
I think ans is (d)
as a1,a2,a3...... are distinct non zero real nos.
so, k< 9!
and we donot know abt values of a1,a2,a3..... so we can't say
D1 + D2 + D3........ = 0
or Di= 0
so, I'll go with (d)
but I m not very much sure..... :(
Lokesh,the answer that I have is option (a) and (b).As far as number of determinants is concerned,it is 9C3*6C3*3C3*(factorial 3 )3 and hence,no. of determinants is factorial 9.
well no doubt ur method is correct.....
but u have found no. of matrices (which is 9!)
but no. of determinants are less..
because here some matrices have same determinant....
for example
these 2 determinants r same but u have counted them as 2 different....
this is what I think.....
may be I am wrong.....
DETERMINANTS.JPG
but I think option b may be correct...
because after removing same determinants....
corrs. to each determinant left , there will be a determinant which is obt. by single transformation(either row or column) of it.
so, both r of opposite sign, hence in addition cancel each other .
in this way D1 +D2 + D3 +.....................Dk = 0
Lokesh,I agree with your reasoning for option (b).For option (a), question has itself said about determinants and not matrices at all.So,whatever calculation I have done is for determinants and not for matrices.
Well, now I m confused.........
total no. of arrangements of a1,a2,a3,.......a9 to form a matrix is 9!
but no. of determinants should be less .......
because in these arrangements many determinants have same values..
so, no. of determinants < 9!
note: determinant is not any arrangement like a matrix but its a value..
so, in 9! values many values r repeating...so <9!
please correct me if I m wrong anywhere.........