I am given a linear map with: $$ \begin{align} f(1,1,0) &= (3,7,1) \\ f(1,0,1) &= (3,4,2) \\ f(0,2,1) &= (-1,2,1) \end{align} $$
and I have to find a $3 \times 3$ matrix that belongs to this linear map.
The one I found is: $$ \begin{pmatrix} 4 & -1 & -1 \\ 6 & -1 & -12 \\ 1 & 0 & 0 \end{pmatrix} $$
and I got this by subtracting and adding the other $f$'s from one another and then applying the same formula to the other vectors, to basically get: $f(0,0,1)$, $f(0,1,0)$, $f(1,0,0)$
..but I'm not sure if what I got is right or if I even used the correct method. Every help is greatly appreciated, I'm really having some trouble here.
Edit:
Okay I have found out a method to verify my result and I found out that what I calculated is completely wrong. Right now I'm trying to figure out how to actually do it.
Let's denote the implicitly given basis vectors by $a=(1,1,0)^T,\ ,b=(1,0,1)^T,\ c=(0,2,1)^T$.
Then we have $a-b=(0,1,-1)^T$, $c-2(a-b)=(0,0,3)^T$, thus we can express $e_3=(0,0,1)$ and therefore also $e_1=b-e_3$ and $e_2=a-e_1$.
Note that arranging the resulting coefficients of $a,b,c$ in columns we obtain the inverse of the matrix $(a,b,c)$.