Finding the order of an element in $GL(2,\mathbb{R})$ I am working on a problem involving basic abstract algebra/group theory and am getting confused. I am following an online course by Dr. Bob found here, and am currently on assignment two. 
My difficulty lies with problem 1b where I am given a matrix $A=$
$\left( \begin{array}{ccc}
0 & -1 \\
1 & 0 \\ \end{array} \right)$ and asked to find its order.
Now I am fairly sure that matrix multiplication is not commutative so this makes me suspect that there are either multiple answers or a convention we must adopt (which I dont think he mentioned). If I multiply on the right I get $A\cdot A = -I, A^3 = A\cdot A^2 =
\left( \begin{array}{ccc}
0 & 1 \\
-1 & 0 \\ \end{array} \right)$, and $A^4 = A\cdot A^3 = I$ so $|A| =4$.
Now when I do this on the by multiplying on the left by $A$ I get the same answer, but my intuition says this is a coincidence because of the trivial chosen matrix. 
Is it true in general that the order of elements in $GL(2,\mathbb{R})$ is the same regardless of which side you multiply on, or are there criterion when this property holds? Finally, since I'm guessing that this is just a special case situation, which side do I multiply on when asked to find the order of an element?
Thanks for the help!
 A: A square matrix commutes with itself, with respect to both matrix addition and multiplication, and we have associativity to justify this:  
By associativity, we have, e.g. $$A^4 = A\cdot A^3 = A\cdot (A \cdot A^2) = A\cdot(A\cdot(A\cdot A)) = A\cdot ((A\cdot A)\cdot A) $$ $$= (A \cdot (A \cdot A))\cdot A = ((A\cdot A)\cdot A)\cdot A = (A^2\cdot A) \cdot A = A^3\cdot A = A^4$$
In general:  $A^n = A\cdot A^{n-1} = A^2 \cdot A^{n-2} = \cdots = A^{n-2}\cdot A^2 = A^{n-1} \cdot A$...  
Indeed, note that in your present exercise, after having computed $A^2 = -I$, we have that $$A^4 = A^2\cdot A^2 = -I \cdot -I = (-1)^2 I^2 = I$$ which could have saved you a lot of work! 
A: Matrix multiplication is not commutative, but it is associative, so taking powers make sense: $A^3 = A \cdot (A \cdot A) = (A \cdot A) \cdot A$, and so forth. More generally, $A^n$ is well-defined. 
A: I want to give a different solution to this. Note that there is an isomorphism between $\mathbb C$ and the group of all matrices of the form $\begin{pmatrix} a & -b \\b & a\\\end{pmatrix}$, where $a^2+b^2\neq 0$. The bijection is $a+ib\mapsto \begin{pmatrix} a & -b \\b & a\\\end{pmatrix}$. Now the given matrix corresponds to $i\in \mathbb C$ whose order is $4$. Hence we are done.
