I've stuck in this problem for a couple of hours. I've found examples on the Internet, but it seems to me that the problem I'm dealing with doesn't provide all the information I need. Am I right or there does exist a way to solve it?
Please help me, thanks!
Let $M$ be the desired matrix such that $$M = \begin{pmatrix}M_{1,1}&M_{1,2}\\M_{2,1}&M_{2,2}\end{pmatrix}$$ where $M_{1,1},M_{1,2},M_{2,1}$ and $M_{2,2}$ are constants to be determined. From the problem statement that you provided we are aware what the images of the basis vectors of $T$ are under this Linear Tranformation or (Change of Basis) consequently $$\begin{pmatrix}M_{1,1}&M_{1,2}\\M_{2,1}&M_{2,2}\end{pmatrix} \begin{pmatrix}1\\1\end{pmatrix} = \begin{pmatrix}1\\2\end{pmatrix}\tag{1}$$ and $$\begin{pmatrix}M_{1,1}&M_{1,2}\\M_{2,1}&M_{2,2}\end{pmatrix} \begin{pmatrix}2\\3\end{pmatrix} = \begin{pmatrix}0\\1\end{pmatrix}\tag{2}$$
$(1)$ implies that $$\begin{pmatrix}M_{1,1}+M_{1,2}\\M_{2,1}+M_{2,2}\end{pmatrix} = \begin{pmatrix}1\\2\end{pmatrix}\tag{3}$$
similary $(2)$ implies that $$\begin{pmatrix}2M_{1,1}+3M_{1,2}\\2M_{2,1}+3M_{2,2}\end{pmatrix} = \begin{pmatrix}0\\1\end{pmatrix}\tag{4}$$
Systems $(3)$ and $(4)$ yield the following equations $$M_{1,1}+M_{1,2} = 1\\ 2M_{1,1}+3M_{1,2} = 0$$
and $$M_{2,1}+M_{2,2} = 2\\ 2M_{2,1}+3M_{2,2} = 1$$
solving the above system of equations will yield $$M = \begin{pmatrix}3&-2\\5&-3\end{pmatrix}$$
I suggest you compute the images of the basis vectors of $T$ using $M$ just to be sure