Prime Field Homework: Find A-1 Over Z5

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Homework Help Overview

The problem involves finding the inverse of a matrix A over the prime field Z5, given the matrix A and its determinant. The original poster presents a matrix and requests assistance specifically with the final part of the problem, which requires working within modular arithmetic.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the implications of working in the prime field Z5, questioning how to handle fractions and negative numbers in modular arithmetic. Some participants explore the reduction of matrix entries modulo 5 and the implications for solving equations.

Discussion Status

Participants are actively engaging with the problem, exploring various interpretations of modular arithmetic and how it applies to the entries of the matrix. There is a focus on checking values and understanding the relationships between the elements in the context of mod 5.

Contextual Notes

There are indications of confusion regarding the treatment of negative numbers and fractions in the context of Z5, as well as the need to verify potential solutions within the constraints of the prime field.

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Homework Statement



Let A = [3 2 1; 0 1 4; 4 2 1]

(i) Find the cofactors C11, C12, C13, C21, C22, C23, C31, C32, C33 of A.
(ii) Given that det(A) = 7, use the adjoint method to find A-1.
(iii) Use the answer to part (ii) to find A-1 over the prime field Z5.

The Attempt at a Solution



For (i) and (ii) I got inverse of A = 1/7[-7 0 7; 16 -1 -12; -4 2 3]
no problems here,

Just wondering how I do part (iii)
I've had a peek at the solutions and relevant textbooks, but can't seem to find anything.

Thanks
 
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A "prime field" contains only positive integers. In particular, Z5 contains only {0, 1, 2, 3, 4}. That is, it does not explicitely contain "16/7" or -1/7. For that matter, it does not contain "-7/7= -1". But since 1+ 6= 7= 0 (mod 5), "-1" is really just "6". Knowing that 16/7= x (mod 5) is the same as 16= 7x what is x? (Remember that 16= 3(5)+ 1 so 16= 1 (mod 5) and 7= 1(5)+ 2 so 7= 2 (mod 5). 16= 7x (mod 5) is the same as 1= 2x (mod 5). The only possible values of x are 0, 1, 2, 3, 4 so just try each.

Reduce all of the values of A to (mod 5) form.
 
I understand the concept of mod 5, but I cannot comprehend all of it.

I can only make sense up to this part:

For that matter, it does not contain "-7/7= -1". But since 1+ 6= 7= 0 (mod 5)

How does 7=0?
isn't 7mod5=0?

"-1" is really just "6"
:confused:

1= 2x (mod 5). The only possible values of x are 0, 1, 2, 3, 4 so just try each

Am I trying to satify the equation using only x values ranging from 1 to 4? Whole numbers only?
 
ok consider the first entry of column 1

1/7 (-7)
implies
-7 = 7a then mod 5 this

so
3 = 2a
note "a" can equal 0 1 2 3 4
so
3 = 2 * 0 = 0 mod 5 = 0 not true
3 = 2 * 1 = 2 mod 5 = 2 not true
3 = 2 * 2 = 4 mod 5 = 4 not true
3 = 2 * 3 = 6 mod 5 = 1 not true
3 = 2 * 4 = 8 mod 5 = 3 true
so first entry is 4

consider the second entry of column 1

16 / 7
implies
-7 = 7a then mod 5 this

so
1 = 2a
note again "a" can equal 0 1 2 3 4
so
1 = 2 * 3 = 6 mod 5 = 1 true

the resultant matrix should be

[401,321,314]
 

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