change of coordinate matrix with non-standard basis I want to find the change of coordinate matrix  that changes ${\beta^{'}}$ coordinates into ${\beta}$ coordinates.  
$${\beta^{'}}=\{(0,10),(5,0)\}$$
$${\beta}=\{(-1,3),(2,-1)\}$$
I received a hint that said to just compute $[I]_{\beta^{'}}^\beta$
I don' have any problems writing a linear transformation as a matrix when the basis are the standard basis but when I try this case I don't get the same answer that's in the solution.
I computed $[I]_{\beta^{'}}^\beta$ by first computing the identity function with respect to $\beta^{'}$ and got 
$$I(0,10)=(0,10)$$
$$I(5,0)=(5,0)$$
Then when I try to write these vectors in with respect to $\beta$, similar to how I  would with respect to the standard basis I get
$$\begin{bmatrix}
    30       & -5 \\
    -10      & 10\\
\end{bmatrix}$$
Thanks in advance!
 A: To avoid slightly confusing notations, let's denote $x'=(0,10)$ and $y'=(5,0)$ and $x=(-1,3)$,$y=(2,-1)$. 
Your aim is, given any element $v=ax'+by'=cx+dy$ (i.e. its coordinate is $(a,b)$ with respect to $\beta'$ and $(c,d)$ with respect to $\beta$) you want some matrix $A$ such that $A(a,b)^{t}=(c,d)^{t}$ ($t$ is transpose here. So these are column vectors)
At this point, you might notice that in fact the whole job is dependent on knowing how to represent $x'$ in terms of $x,y$ and the same for $y'$. Because, then, you can write $ax'+by'$ in terms of $x$ and $y$.
But it's easy to see that $x'=2(2x+y)=4x+2y$ and $y'=x+3y$. This means, in fact, 
$$cx+dy=(4x+2y)a+(x+3y)b=(4a+b)x+(2a+3b)y$$
Since $x,y$ are members of a basis, it follows that $4a+b=c$ and $2a+3b=d$. So
$$A=\begin{pmatrix} 4 & 1 \\ 2 & 3\end{pmatrix}$$.
Some simple checks can assure you that this works;
This method also works in more general case other than $\mathbb{R}^2$.
Summary: Write the member of the current basis as a linear combination of members of a basis you want to change to. The coefficients of these linear combinations give you the matrix you want.
