Question about determinants.

Suppose you have n numbers of the form (a1+n*a2+n^2*a3+...n^(n-1)*an) where a1, a2, a3, ..., an are all integers in the interval [0,n-1]. Suppose that d divides all of those numbers. Suppose you arrange those a1, a2, a3, ..., an in each row of a matrix. Does d necessarily divide the determinant of that matrix?

consecutive integers?

>>2
No

Not a pro at linear mathematics, but I don't believe that' (a1, a2, a3, ...)/d = (detA)/d.  You could test this rather simply by simply limiting your set to a small matrix greater than 1x1, dividing all members by d and summing them, then finding the determinate and dividing by d.  If they're not equal, it's not a rule. 

>>4
I never said that they were equal. I only meant that if those numbers formed by their rows were all divisible by a number, then does it imply that the determinant is divisible by the same number?

You mean evenly divisible?  In that case yes, it would imply that exactly, because the determinant is the sum of the products, all of which are divisible (evenly?) by d.  This comes down to a problem of:

[(a1*b1) + (a2*b2)]/d = (a1*b1)/d + (a2*b2)/d.

If the original numbers are factors of d, then their products are factors of d, and the sum of those products must also be a factor of d. 

Proof is more than likely in any linear mathematics text.  Should be fairly straight forward.

>>6
That's not what I meant. I mean you find n n-digit numbers, which are divisible by a number, and you arrange them in a matrix. And then determine whether the determinant is divisible by the same number.




For example, take the 2 2-digit numbers, 21 and 12. They are both divisible by three. Then the determinant

| 2 1 |
| 1 2 |

= 2*2-1*1 = 3 which is divisible by 3. The question is whether or not it always holds.

Take the definition of the deteriminant.


can't write it here, don't know VB code or whatever
But it's the sum over the permutation group Sn the sign of e multiplied by the product of A subscript i,e(i)

Now all Aij are divisible by d, therefore the sum of numbers of this form is divisible by d.

>>8
21 (= 2*10 + 1) is divisible by 3, but 2 and 1 are not. Reread OP.