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(basic) prime field

  1. Jun 11, 2008 #1
    1. The problem statement, all variables and given/known data

    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.

    3. 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
     
  2. jcsd
  3. Jun 11, 2008 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    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.
     
  4. Jun 11, 2008 #3
    I understand the concept of mod 5, but I cannot comprehend all of it.

    I can only make sense up to this part:

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

    :confused:

    Am I trying to satify the equation using only x values ranging from 1 to 4? Whole numbers only?
     
  5. Jun 2, 2009 #4
    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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