Finding the formula for a linear transformation given the transformation of the basis vectors. 
Consider the basis $\{\vec{p},\vec{q}\}$ where $\vec{p}=(1,1)$ and
  $\vec{q}=(-1,0)$. Let $T:\mathbb{R}^2\to\mathbb{R}^2$ be the linear
  operator such that $T(\vec{p})=(1,-2)$ and $T(\vec{q})=(4,1)$. Then
  find a formula for $T(x_1,x_2)$ and use this formula to find
  $T(5,-7)$.

I figured this might make the transformation matrix $A$ be $ A =
\left[
\begin{array}{rr}
 1 &  4 \\
-2 &  1 \\
\end{array}
\right]$ such that $T_A(\vec{v}) = A\vec{v}$, but this doesn't seem to pan out.
 A: If we interpret the question as if all vectors are expressed in the standard basis, than we have to find a matrix :
$$
A_s=\begin{bmatrix}
a&b\\
c&d
\end{bmatrix}
$$ 
such that
$$
\begin{bmatrix}
a&b\\
c&d
\end{bmatrix}
\begin{bmatrix}
1\\
1
\end{bmatrix}=
\begin{bmatrix}
1\\
-2
\end{bmatrix}
\quad \land \quad
\begin{bmatrix}
a&b\\
c&d
\end{bmatrix}
\begin{bmatrix}
-1\\
0
\end{bmatrix}=
\begin{bmatrix}
4\\
1
\end{bmatrix}
$$ 
this gives the system:
$$
\begin{cases}
a+b=1\\c+d=-2\\-a=4\\-c=1
\end{cases}
$$
with solutions $a=-4,b=5,c=-1,d=-1$ and the matrix $A$ is:
$$
A_s=\begin{bmatrix}
-4&5\\
-1&-1
\end{bmatrix}
$$
IN this interpretation the fact that $\{\vec p,\vec q\}$ is a basis is true but not relevant.
If the components of $T(\vec p)$ and $T(\vec q)$ in OP are expressed in the basis $t=\{\vec p,\vec q\}$, than ,since in this basis we have $\vec p=[1,0]^T$ and $\vec q=[0,1]^T$, than it is represented , in this basis, by the matrix:
$$
A_t=\begin{bmatrix}
1&4\\
-2&1
\end{bmatrix}
$$
but note that this transformation is different from the other one.
