Determinant of a Finite Field 2x2 Matrix

rehcarlos
Messages
8
Reaction score
0

Homework Statement



Find the determinant of:
|1 1|
|2 1|

The field is Z3.

Homework Equations



The field is Z3, that is, to multiply two numbers, you first multiply then take the remainder of the division by 3.

The Attempt at a Solution


I tried:
( 1 x 1 ) - ( 1 x 2 )
1 x 1 will be: 1x1 = 1 => then get the remainder of the division by 3 that is 1
1 x 2 will be: 1 x 2 = 2 => then get the remainder of the division by 3 that is 2

So, the determinant would be 1 - 2 = -1 => then get the remainder of the division by 3 that is -1

But the answer is 2
 
Physics news on Phys.org
rehcarlos said:

Homework Statement



Find the determinant of:
|1 1|
|2 1|

The field is Z3.

Homework Equations



The field is Z3, that is, to multiply two numbers, you first multiply then take the remainder of the division by 3.

The Attempt at a Solution


I tried:
( 1 x 1 ) - ( 1 x 2 )
1 x 1 will be: 1x1 = 1 => then get the remainder of the division by 3 that is 1
1 x 2 will be: 1 x 2 = 2 => then get the remainder of the division by 3 that is 2

So, the determinant would be 1 - 2 = -1 => then get the remainder of the division by 3 that is -1

But the answer is 2

In Z3, -1=2. Any two numbers that differ by a multiple of 3 have the same remainder when divided by 3.
 
I know it may be simple, but I'm still confuse,

-1 divided by 3 has a remainder of -1
2 divided by 3 has a remainder of 2

So I don't get how they have the same remainder
 
rehcarlos said:
I know it may be simple, but I'm still confuse,

-1 divided by 3 has a remainder of -1
2 divided by 3 has a remainder of 2

So I don't get how they have the same remainder

To say Z3 is about 'remainders' isn't quite accurate. Two number are equal in Z3 if they differ by a multiple of 3. That's the sense in which they have the same remainder. 2-(-1)=3*1.
 
Ok then,

Thanks!
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top