Q:

Find a matrix representation of T with respect to the basis vectors. 

I am not quite sure how to entirely solve this problem. 
Find the matrix representation of the linear map: $T: \mathbb{R}^2 \to \mathbb{R}^2, (x, y) \to (2y, 3x-y)$ be relative to the following bases: $\alpha = \{(1, 0), (0,1)\}$ and $\beta = \{(1,3), (2,5)\}$.
This is how I approached the problem:
$T(\alpha_1) = (0, 3)$ and $T(\alpha_2) = (2, -1)$.
Now I need to solve $T(\alpha_1) = k_1\beta_1 + k_2\beta_2$, $T(\alpha_2) = m_1\beta_1 + m_2\beta_2$ and  but I am not sure if this is the right procedure.  In fact, I am not sure I completely understand why I am doing this.  Any help appreciated.  The scalars are then placed as column vectors in the matrix $[T]_\alpha^\beta$.

A:

In fact, your values are:  $m_1=-12$, $m_2=7$, $k_1=6$ and $k_2=-3$. Replacing your vectors for the second basis $\beta_1=(1,3)$ and $\beta_2=(2,5)$ in 
$$(0,3)=T(1,0)=T(\alpha_1) = k_1\beta_1 + k_2\beta_2$$
$$(2,-1)=T(0,1)=T(\alpha_2) = m_1\beta_1 + m_2\beta_2$$
this will give you:
$$(0,3)=T(\alpha_1) = k_1(1,3) + k_2(2,5)=(k_1+2k_2,3k_1+5k_2)$$
$$(2,-1)=T(\alpha_2) = m_1(1,3) + m_2(2,5)=(m_1+2m_2,3m_1+5m_2)$$
developing the linear combinations you get two linear systems for the $k$ and $m$ unknowns:
$$ k_1+2k_2=0 $$ 
$$ 3k_1+5k_2=3 $$
and
$$ m_1+2m_2=2 $$
$$ 3m_1+5m_2=-1 $$
you must get the values I wrote above. Finally, your matrix shoud be the following:
$$\left(\begin{array}{cc} \ 6 & -12\\ -3&\ 7\end{array}\right)$$
In general, if you want to solve for $k_1$, $k_2$, $m_1$ and $m_2$ you can do the following steps:
Consider an augmented matrix with the columns you mentioned:
$$\left(\begin{array}{cc|c}  
 1 & 2 & \ z\\  
 3 & 5 & \ w  
\end{array}\right)$$
and then apply Guass-Jordan method.  First, multiply by $-3$ the first row and add this result to the second row to get:
$$\left(\begin{array}{cc|c}  
 1 & 2 & \ z\\  
 0 & -1 & \ w-3z  
\end{array}\right)$$
Multiply by $-1$ the second row:
$$\left(\begin{array}{cc|c}  
 1 & 2 & \ z\\  
 0 & 1 & \ 3z-w  
\end{array}\right)$$
Multiply the second row by $-2$ and add the result to the first row to get:
$$\left(\begin{array}{cc|c}  
 1 & 0 & \ z-2(3z-w)\\  
 0 & 1 & \ 3z-w  
\end{array}\right)$$
This give you
$$\left(\begin{array}{cc|c}  
 1 & 0 & \ -5z+2w\\  
 0 & 1 & \ 3z-w  
\end{array}\right)$$
So, when you replace $(z,w)=(0,3)$ in the last matrix
$$\left(\begin{array}{cc|c}  
 1 & 0 & \ \ 6\\  
 0 & 1 & \ -3  
\end{array}\right)$$
you get the values of $k_1=6$ and $k_2=-3$.
When you replace $(z,w)=(2,-1)$ in the same matrix we replaced above you will get:
$$\left(\begin{array}{cc|c}  
 1 & 0 & \ -12 \\  
 0 & 1 & \ 7 
\end{array}\right)$$
so $m_1=-12$ and $m_2=7$.

