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Homework Help: Matrix Order

  1. Sep 11, 2010 #1
    1. The problem statement, all variables and given/known data

    The following matrix is an elements of the group GL2(2), that is, the general linear group of 2x2 matrices in [tex]\mathbb{Z}_2[/tex]:

    [tex]A = \begin{pmatrix}1 & 1 \\1 & 0\end{pmatrix}[/tex]

    Find the order of the element A.

    3. The attempt at a solution

    I know that the order of A is 3. Because A3=I, where "I" is the identity. I found this by trial and error:

    [tex]A^1 = \begin{pmatrix}1 & 1 \\1 & 0\end{pmatrix}^1 \neq \begin{pmatrix}1 & 0 \\0 & 1\end{pmatrix}[/tex]

    [tex]A^2 = \begin{pmatrix}1 & 1 \\1 & 0\end{pmatrix} \begin{pmatrix}1 & 1 \\1 & 0\end{pmatrix} \neq \begin{pmatrix}1 & 0 \\0 & 1\end{pmatrix}[/tex]

    [tex]A^3 = \begin{pmatrix}1 & 1 \\1 & 0\end{pmatrix} \begin{pmatrix}1 & 1 \\1 & 0\end{pmatrix} \begin{pmatrix}1 & 1 \\1 & 0\end{pmatrix} = \begin{pmatrix}1 & 0 \\0 & 1\end{pmatrix}[/tex].

    Here is my question, is there a shorthand method for finding "n" in:

    [tex]\begin{pmatrix}1 & 1 \\1 & 0\end{pmatrix}^n = \begin{pmatrix}1 & 0 \\0 & 1\end{pmatrix}[/tex]

    Is there any way of solving for n without going through all the suffering matrix multipications above?
     
  2. jcsd
  3. Sep 11, 2010 #2

    Dick

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    Well, your calculation isn't even right. A^3 isn't equal to I. In fact, A^n is never equal to I.
     
  4. Sep 12, 2010 #3
    We are working in [tex]\mathbb{Z}_2[/tex]!

    [tex]A^3= \begin{pmatrix}3 & 2 \\2 & 1\end{pmatrix}[/tex]

    Which is equal to

    [tex]I=\begin{pmatrix}1 & 0 \\0 & 1\end{pmatrix}[/tex]

    in [tex]\mathbb{Z}_2[/tex]. So, the only way to solve this kind of problem is by trial and error?
     
  5. Sep 12, 2010 #4

    Dick

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    Ooops. Missed the Z_2. No, I don't know of any systematic way. Some times it can help to cleverly write the matrix as a sum of others or look for a pattern is small powers. But A^3 is already a small power.
     
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