# Homework Help: Matrix Order

1. Sep 11, 2010

### roam

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 $$\mathbb{Z}_2$$:

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

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:

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

$$A^2 = \begin{pmatrix}1 & 1 \\1 & 0\end{pmatrix} \begin{pmatrix}1 & 1 \\1 & 0\end{pmatrix} \neq \begin{pmatrix}1 & 0 \\0 & 1\end{pmatrix}$$

$$A^3 = \begin{pmatrix}1 & 1 \\1 & 0\end{pmatrix} \begin{pmatrix}1 & 1 \\1 & 0\end{pmatrix} \begin{pmatrix}1 & 1 \\1 & 0\end{pmatrix} = \begin{pmatrix}1 & 0 \\0 & 1\end{pmatrix}$$.

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?

2. Sep 11, 2010

### Dick

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

3. Sep 12, 2010

### roam

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?

4. Sep 12, 2010

### Dick

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.