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Matrix groups

  1. Jul 31, 2010 #1
    1. The problem statement, all variables and given/known data

    [PLAIN]http://img844.imageshack.us/img844/2975/31668142.gif [Broken]

    3. The attempt at a solution

    I know the inverse would be of the form:

    [tex]M^{-1}_{a,p}=\frac{1}{1-(p-1)(10 \bmod\ p)} \begin{bmatrix} 1&-(a \bmod\ p)\\ -(p-1)& 1 \end{bmatrix}[/tex]

    Where [tex]1-(p-1)(10 \bmod\ p)[/tex] is the determinant. Whatever this equals to, I must find its inverse.

    But how does one evaluate [tex]10 \bmod\ p[/tex] and [tex](p-1)[/tex] when [tex]p \in \{ 3,7 \}[/tex]? Do we take '10 mod 7' or '10 mod 3'? :confused: I appreciate if anyone could show me that.
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Aug 1, 2010 #2
    I believe you are supposed to compute two inverses, for p = 3 and for p = 7 (with a = 10 in both cases).
     
  4. Aug 2, 2010 #3
    Are you sure? Because the question apparently asks for one inverse...

    But in the case we need 2 inverses

    10 mod 3 = 1 & (p-1)=2
    10 mod 7 = 3 & (p-1)=6

    The inverses will be

    [tex]M^{-1}_{10,3}=1^{-1} \begin{bmatrix} 1&-1\\ -2& 1 \end{bmatrix}[/tex]

    [tex]M^{-1}_{10,7}=3^{-1} \begin{bmatrix} 1&-3\\ -6& 1 \end{bmatrix}[/tex]

    Inverse of 1 is 1, and the inverse of 3 is 5 [since (3.5) mod 7 = 1]. When the arithmetic is done modulo 3 and modulo 7 respectively the inverse matrices are:

    [tex]M^{-1}_{10,3}=\begin{bmatrix} 1&2\\ 1& 1 \end{bmatrix}[/tex]

    [tex]M^{-1}_{10,3}=\begin{bmatrix} 5.1&5.4\\ 5.1& 5.1 \end{bmatrix}[/tex][tex]=\begin{bmatrix} 5&6\\ 5& 5 \end{bmatrix}[/tex]

    Is this correct then?
     
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