Confused with changes of bases. I have a problem to solve, and it makes me realise that I have a bad understanding of basis change and the linear applications that go along. I have a linear application who's matrix is expressed like this:
$A=\begin{bmatrix}
 \frac{4a-b}{3}&\frac{2a-2b}{3} \\ 
 \frac{-2a+2b}{3}& \frac{-a+4b}{3}
\end{bmatrix}
$
I suppose in order to get the basis of this linear appplication I should compute the values of $Av_{1}$ and $Av_{2}$ where $v_{1}=\binom{1}{0}$ and $v_{2}=\binom{0}{1}$. Am I correct so far?
Now, where my lack of understand lacks is here: how to I determine the matrix $A^{'}$ that is the associated matrix in the base $x_{1} \:x_{2}$? Should I just compute $Ax_{1}$ and $Ax_{2}$?
EDIT 1:
I made a small mistake in the matrix:
it's not$$A=\begin{bmatrix}
 \frac{4a-b}{3}&\frac{2a-2b}{3} \\ 
 \frac{-2a+2b}{3}& \frac{-a+4b}{3}
\end{bmatrix}
$$ but
$$A=\begin{bmatrix}
 \frac{4a-b}{3}&\frac{-2a+2b}{3} \\ 
 \frac{2a-2b}{3}& \frac{-a+4b}{3}
\end{bmatrix}
$$
I calculated the determinant of this matrix, and found that it is equal to:
$3ab$.
I know that to find the inverse of the matrix A, I can use the adjudant matrix. So normally I should be able to compute this:
$$A^{-1}=\frac{1}{3ab}*adj(A)$$
For $adj(A)$ I did this:


*

*Find the cofactor matrix:


$Com(A)$:
$$Com(A)=\begin{bmatrix}
 \frac{-a+4b}{3}&\frac{2a-2b}{3} \\ 
 \frac{2b-2a}{3}& \frac{4a-b}{3}
\end{bmatrix}
$$
The adjugate matrix will be equal to the transpose of the $Com(A)$. So we have:
$$adj(A)=\begin{bmatrix}
 \frac{-a+4b}{3}&\frac{2b-2a}{3} \\ 
 \frac{2a-2b}{3}& \frac{4a-b}{3}
\end{bmatrix}
$$


*

*Find the expression of the inverse:


I have now:
$$A^{-1}=\frac{1}{3ab}*\begin{bmatrix}
 \frac{-a+4b}{3}&\frac{2b-2a}{3} \\ 
 \frac{2a-2b}{3}& \frac{4a-b}{3}
\end{bmatrix}$$
Or in a simplified way:
$$A^{-1}=\frac{1}{9ab}*\begin{bmatrix}
 -a+4b&2b-2a\\ 
 2a-2b& 4a-b
\end{bmatrix}$$
So now all I have to do is to chose the two vectors I want $v_{1},v_{2}$ and calculate $A^{-1}v_{1}$ and $A^{-1}v_{2}$? And this will give me the basis of the linear transformation in the basis ($v_{1},v_{2}$), right?
EDIT 2:
I made a mistake. I should not calculate $A^{-1}v_{1}$ and $A^{-1}v_{2}$ but solve $v_{1}=A^{-1}\begin{bmatrix}
\frac{4a-b}{3}\\ 
\frac{2a-2b}{3}
\end{bmatrix}$ and $v_{2}=A^{-1}\begin{bmatrix}
\frac{-2a-2b}{3}\\ 
\frac{-a+4b}{3}
\end{bmatrix}$

HINT.-The determinant of $A$ is equal to $ab$ so if $ab\ne 0$ then the matrix is invertible. If you want to know the base $\{b_1,b_2\}$ where $b_1=\begin{pmatrix} x_1\\y_1\end{pmatrix}$ and $b_2=\begin{pmatrix} x_2\\y_2\end{pmatrix}$
you have to solve
$$A\begin{pmatrix} x_1\\y_1\end{pmatrix}=\begin{pmatrix}  \frac{4a-b}{3}\\\frac{-2a+2b}{3}\end{pmatrix}$$ 
$$A\begin{pmatrix} x_2\\y_2\end{pmatrix}=\begin{pmatrix}  \frac{2a-2b}{3}\\\frac{-a+4b}{3}\end{pmatrix}$$ 
For this you multiply both equations for the inverse $A^{-1}$ of $A$ (you know that if $ab\ne 0$ then $A$ is invertible). Finally you have $$\begin{pmatrix} x_1\\y_1\end{pmatrix}=A^{-1}\begin{pmatrix}  \frac{2a-2b}{3}\\\frac{-a+4b}{3}\end{pmatrix}$$
$$\begin{pmatrix} x_2\\y_2\end{pmatrix}=A^{-1}\begin{pmatrix}  \frac{2a-2b}{3}\\\frac{-a+4b}{3}\end{pmatrix}$$ 
Do you know calculate the inverse matrix of A?
