Let $f:\mathbb{R}^2 \to \mathbb{R}^2$ be a linear map and consider the matrix representation $$M_{B}^{B} =\begin{pmatrix} 4 & 6\\ 6 & 3 \end{pmatrix}$$ with respect to the basis $B=\bigg\{ \begin{pmatrix} 2 \\ 2 \end{pmatrix} , \begin{pmatrix} 3\\ 2 \end{pmatrix}\bigg\}$. Find the matrix representation of $f$ with respect to the canonical basis.
MY ATTEMPT: Let $b_1=\begin{pmatrix} 2 \\ 2 \end{pmatrix}$, $b_2=\begin{pmatrix} 3\\ 2 \end{pmatrix}$ and $e_j$ the j-th vector of the canonical basis. Then I have $f(b_1)=\begin{pmatrix} 4\\ 6 \end{pmatrix} \Longrightarrow f(2e_1 +2e_2)=2 f(e_1) +2f(e_2) =4e_1 + 6e_2$
$f(b_2)=\begin{pmatrix} 6\\ 3 \end{pmatrix} \Longrightarrow f(3e_1 +2e_2)=3 f(e_1) +2f(e_2) =6e_1 + 3e_2$
Solving this system I obtain that $f(e_1)=2e_1 -3e_2=\begin{pmatrix} 2\\ -3 \end{pmatrix}$ and $f(e_2)= 6e_2=\begin{pmatrix} 0\\ 6 \end{pmatrix}$. So the matrix rappresentation is $M_{E}^{E}=\begin{pmatrix} 2 & 0\\ -3 & 6 \end{pmatrix}$.
BUT my theacher says that the solution is $M_{E}^{E}=\begin{pmatrix} 13 & 10\\ 0 & 0 \end{pmatrix}$. Where I am wrong?
$$f(2e_1 +2e_2)=2 f(e_1) +2f(e_2) \color{red}{=4e_1 + 6e_2}$$
This is where you are wrong, $4$ and $6$ are coordinates in the basis $B$, so there is one more step to get $e_1$ and $e_2$ into the game. Otherwise your approach is nice.