Linear transformation with respect to basis problem Let $T: \mathbb{R}^2 \to \mathbb{R}^2$ such that $T(\begin{bmatrix} 3 \\ 1 \end{bmatrix}) = \begin{bmatrix} 1 \\2 \end{bmatrix}$ and $T(\begin{bmatrix} -1 \\ 0 \end{bmatrix}) = \begin{bmatrix} 1 \\1 \end{bmatrix}$. Find the matrix $A$ representing $T$. 
I understand that to approach this problem, I have to view $\begin{bmatrix} 3 \\1 \end{bmatrix}$ and $\begin{bmatrix} -1 \\ 0 \end{bmatrix}$ as a basis, $B = \{ v_1, v_2 \}$, where $[v1 v2]$ is the transition matrix from $[x]B$ to $x$. How do I use $\begin{bmatrix} 1\\2 \end{bmatrix}$ and $\begin{bmatrix} 1 \\1 \end{bmatrix}$? I'm unclear on their connection to the basis vectors $v_1$ and $v_2$.

$\begin {bmatrix} 1\\0 \end{bmatrix} =  - \begin {bmatrix} -1\\0 \end{bmatrix}$
$T(\begin {bmatrix} 1\\0 \end{bmatrix} ) = -T(\begin {bmatrix} -1\\0 \end{bmatrix}) = -\begin {bmatrix} 1\\1 \end{bmatrix} = \begin {bmatrix} -1\\-1 \end{bmatrix}$
$\begin {bmatrix} 0\\1 \end{bmatrix} = \begin {bmatrix} 3\\1 \end{bmatrix} + 3\begin {bmatrix} -1\\0 \end{bmatrix}$
$T(\begin {bmatrix} 1\\0 \end{bmatrix} ) = T(\begin {bmatrix} 3\\1 \end{bmatrix}) + 3T(\begin {bmatrix} -1\\0 \end{bmatrix}) = \begin {bmatrix} 1\\2 \end{bmatrix} + 3\begin {bmatrix} 1\\1 \end{bmatrix} = \begin {bmatrix} 4\\5 \end{bmatrix}$
$T = \begin {bmatrix} -1&4\\-1&5 \end{bmatrix}$
