Why when transforming matrices is the transformation matrix first? Is there a reason that the transformation matrix is on the left?
For example:
$$
\begin{bmatrix}2&1\\-1&1\end{bmatrix}\begin{bmatrix}x&z\\y&v\end{bmatrix}
$$
I understand that when multiplying matrices order is important, but is there any other reason other than that that the transformation matrix has to come first?
Is it because there is no other way to 'write' the transformation matrix that allows it to be on the right?
 A: Read up on matrix multiplication.
Yes you can have it on the right, if the vector you are transforming is a row vector:
$$\begin{bmatrix}x&y\end{bmatrix}\begin{bmatrix}2&-1\\1&1\end{bmatrix}$$
But as you can see here $-1$ and $1$ switch places as we need to do this transpose operator we describe below and which is on wikipedia.
If it is a more complicated transformation, say given by the matrix $\bf M$, then a column vector being transformed (multiplied by $\bf M$) from the left is the same as a row vector being transformed (multiplied by ${\bf M}^T$ from the right:
$$({\bf Mv})^T = {\bf v}^T {\bf M}^T$$
This is due to a famous algebra rule of transpose (the $(\cdot)^T$ operator).
A: We have 
$$\begin{bmatrix}-1&0\\0&-1\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}-x\\-y\end{bmatrix}$$
whereas
$$\begin{bmatrix}x\\y\end{bmatrix}\begin{bmatrix}-1&0\\0&-1\end{bmatrix}$$
is undefined because it is a $2\times 1$ matrix multiplying a $2\times 2$ matrix. Matrix multiplication requires the number of columns of the left matrix to equal the number of rows of the right matrix. If the transformation matrix were taking other square matrices then this could be acceptable. e.g. if the transformation were something like
$$\begin{bmatrix}w&x\\y&z\end{bmatrix}\begin{bmatrix}-1&0\\0&-1\end{bmatrix}$$
then that would be allowed. In general though we have the transformation matrix on the left of the multiplication.
