Find the matrix of the transformation conjugate to $f(x, y) = (x - y, 2x)$ in the basis $(\mathbb R^2)^*$ conjugate to the basis $\{(1, 1), (1, 2) )\}$ of the space $\mathbb R^2$.
Unfortunately, I don't quite understand the conjugate transformation. From what I understand, the conjugate transformation matrix is the transposition of the transformation matrix $f$. Then that would mean that it is the matrix tracing in standard bases, so I compute it: $$\begin{bmatrix} a & b \\ c & d \end{bmatrix} \cdot \begin{bmatrix} x \\ y \end{bmatrix}=\begin{bmatrix} x-y \\ 2x \end{bmatrix} \Rightarrow M(f)_{st}^{st}=\begin{bmatrix} a & b \\ c & d \end{bmatrix}=\begin{bmatrix} 1 & 1 \\ 1 & 2 \end{bmatrix}$$ However, I am given a non-standard basis to find the conjugate basis to, so I think I should find the matrix $M(f)_{st}^B$ and its transpose will be the conjugate transformation matrix I am looking for.
But I don't know how to calculate $M(f)_{st}^B$ and if my reasoning is at least correct.
Edit using Anne Bauval hint:
For $B=\{(1, 1), (1, 2) )\}$: $$\begin{cases}f(1,1)=(0,2)=a(1,1)+c(1,2) \\ f(1,2)=(-1,4)=b(1,1)+d(1,2) \end{cases} \Rightarrow \begin{cases}a=-2\\ c=2\\ b=-4\\d=3 \end{cases} \Rightarrow M(f)_B^B=\begin{bmatrix} -2 & -4 \\ 2 & 3\end{bmatrix}$$ So the matrix of the transformation conjugate is: $$A=\left( M(f)_B^B \right)^T= \begin{bmatrix} -2 & 2 \\ -4 & 3\end{bmatrix}$$ Is it correct?