Finding Order of A in GL2(2) of \mathbb{Z}_2

[roam]

Homework Statement

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

A = \begin{pmatrix}1 & 1 \\1 & 0\end{pmatrix}

Find the order of the element A.

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:

A^1 = \begin{pmatrix}1 & 1 \\1 & 0\end{pmatrix}^1 \neq \begin{pmatrix}1 & 0 \\0 & 1\end{pmatrix}

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}

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}.

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

\begin{pmatrix}1 & 1 \\1 & 0\end{pmatrix}^n = \begin{pmatrix}1 & 0 \\0 & 1\end{pmatrix}

Is there any way of solving for n without going through all the suffering matrix multipications above?

 

[Dick]

Well, your calculation isn't even right. A^3 isn't equal to I. In fact, A^n is never equal to I.

 

[roam]

Well, your calculation isn't even right. A^3 isn't equal to I. In fact, A^n is never equal to I.

We are working in \mathbb{Z}_2!

A^3= \begin{pmatrix}3 & 2 \\2 & 1\end{pmatrix}

Which is equal to

I=\begin{pmatrix}1 & 0 \\0 & 1\end{pmatrix}

in \mathbb{Z}_2. So, the only way to solve this kind of problem is by trial and error?

 

[Dick]

We are working in \mathbb{Z}_2!

A^3= \begin{pmatrix}3 & 2 \\2 & 1\end{pmatrix}

Which is equal to

I=\begin{pmatrix}1 & 0 \\0 & 1\end{pmatrix}

in \mathbb{Z}_2. So, the only way to solve this kind of problem is by trial and error?

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.