I solved this one example like this :
$$T(a,b,c)=(a+2b,3c-a)$$
Here I written each vetor as basis of $(a,b,c)$ and got: $(a[1,-1],b[2,0],c[3,0])$
and the answer is the matrix : $\bigl(\begin{smallmatrix} 1 &2 &0 \\ -1 &0 &3 \end{smallmatrix}\bigr)$
I can't understand how to apply this principle however for this exercise: $T(a+bi,x+yi)=(a-b-y,x-2y+b,y-b)$
The answer: $\bigl(\begin{smallmatrix} 1 &0 \\ 0 &1 \\ 0 &0 \end{smallmatrix}\bigr)$
how can I reach this solution? I think I know how to solve when there is a linear transformation with simple variables like a,b,c but here I have $a+bi,x+yi$