Question about determinants (restated)

Look at this:
1.
The two numbers 21 and 12 are divisible by 3. The determinant

| 2 1 |
| 1 2 |

is equal to 2*2-1*1=3, which is divisible by 3.

Or for another example, the three numbers 112, 763, and 959 are divislbe by 7.

The determinant

|1 1 2|
|7 6 3|
|9 5 9|

= 6*9 + 3*9 + 2*7*5 - 9*6*2 - 5*3 - 7*9 = 54 + 27 + 70 - 108 - 15 - 63 = -35 which is also divisible by 7.

The question is this: does this always work?

No.

>>2
Counterexample plz

>>2 here.

I was just messing with you. The proof TO THE AFFIRMATIVE is easy.

|3 3|
|0 0|

Kaboom...

Having said that, I tried a bunch of 2x2 matrices with no zeroes, and making sure that det > 0, and it worked (for divisible by 3 case). You might be onto something.

It doesn't really have to be determinants though, as I don't think any special property of matrices is causing it. Reformulate it as an equation (writing, for example, 12 as 1*10 + 2*1, and so on) and mess about with it. It'll just be a slightly rearranged version of result that if the digit sum of a number is divisible by three, then the number is divisible by three.

>>5

I was 5, just thought that maybe you will consider 3 to be a multiple of zero, because trivially 0 = 3*0. So yeah, it even works when det = 0 (and probably < 0 if you extend the notion of divisibility to the integers)

|1 2 1|
|1 1 0|
|1 3 2|

>>7

That's zero again... I guess it'd be better to say that determinant is 0 mod 3, because of slightly differing views of divisibility.

>>7
This is equal to 0. 0 is divisible by any number.

You can at least say that the converse isn't true... If I wasn't so tired I would write out a proof (at least for the 2x2, mod 3 case)

Go to definition of determinant as alternating multilinear form and try to get some kind of proof.

PROOF (for 2x2 case):

Suppose a, b, c, d are integers between 0 and 9 inclusive, and that there exist nonnegative integers x, y, and n such that

10a + b = xn

and

10c + d = yn.

I.e., the numbers with digits "ab" and "cd" are multiples of n.

Then

|a b|
|c d|

= ad-bc = a(yn - 10c) - c(xn - 10a) = ayn - cxn - 10ac + 10ac

= (ay-cx)n

an integer multiple of n.

This can be generalized a lot, of course- for example, the integers need not be nonnegative, or in base 10. I'm just validating the OP's intuition.

Someone else can do the 3x3, I'm going to bed.

Thanks OP, I've had a lot of fun messing about with this today, I showed a bunch of people at the maths department and they were impressed. It's like a fucking magic trick

>>1
Yes, it always works.

>>13
Wow really?