General Linear Transformation 
Let $v_1 = \begin{bmatrix} 2 \\ -1  \end{bmatrix}$ and $v_2=\begin{bmatrix} 1 \\  -1 \end{bmatrix}$ and let $$A= \begin{bmatrix} 3 & 2 \\ -2 & 1 \end{bmatrix}$$ be a matrix for $T\colon \Bbb R^2\to \Bbb R^2$ relative to the basis $B = \{v_1, v_2\}$. 

From this I found that:

$T(v_1) = \begin{bmatrix} 4 \\ -1 \end{bmatrix}$ and $T(v_2) = \begin{bmatrix} 5 \\ -3 \end{bmatrix}$

How would I find a formula for $T\begin{pmatrix} \begin{bmatrix}x_1 \\ x_2\end{bmatrix} \end{pmatrix}$
The answer in my book is $\begin{bmatrix} -x_1-6x_2 \\ 2x_1+5x_2  \end{bmatrix}$
 A: The real thing to clarify here is how the transformation is being expressed. The answer given is correct if the goal is to rewrite the transformation in terms of the standard basis $w_1=[1,0]^T,w_2=[0,1]^T$.
The change of basis matrix (the one that converts $w$-basis coordinates to $v$-basis coordinates is $C=\begin{bmatrix}1&1\\-1&-2\end{bmatrix}$, which has inverse $C^{-1}=\begin{bmatrix}2&1\\-1&-1\end{bmatrix}$.
Then the new matrix of $T$ with respect to the $w$-basis is 
$$
A'=C^{-1}AC=\begin{bmatrix}-1&-6\\2&5\end{bmatrix}
$$
Thus when the expression $T(x_1w_1+x_2w_2)$ is written in terms of the $w$-basis coordinates, it becomes $A'\begin{bmatrix}x_1\\x_2\end{bmatrix}=\begin{bmatrix}-x_1-6x_2\\2x_1+5x_2\end{bmatrix}$
This quantity on the left hand side must be what is meant by "$T(\begin{bmatrix}x_1\\x_2\end{bmatrix})$". Since people often get lost when they mix the notation of $T$ with coordinate vectors, I prefer to avoid writing them together as this question statement seems to have done.
A: So, we should probably take the inverse of $\begin{pmatrix} 2&1 \\ -1&-1 \end{pmatrix}^{-1} = \begin{pmatrix} 1&1 \\ -1&-2 \end{pmatrix}$. This is called the change of basis. So, $\begin{pmatrix} 4&5 \\ -1&-3 \end{pmatrix}\cdot \begin{pmatrix} x_{1}\\-x_{1} \end{pmatrix} + \begin{pmatrix} 4&5 \\ -1&-3 \end{pmatrix}\cdot \begin{pmatrix} x_{2}\\-2x_{2} \end{pmatrix} = \begin{pmatrix} -x_{1}-6x_{2}\\2x_{1}+5x_{2} \end{pmatrix}$.
