Am I correctly finding the standard matrix? 
Question:
Let $F: \Bbb R^3 \to \Bbb R^3$ be the linear transformation satisfying
  $F(1,0,1)=(-3,-3,1)$,$F(0,1,0)=(0,1,1)$, and $F(0,1,1)=(2,-2,1)$. Find
  the standard matrix $A$ of $F$.

My Approach:
I used a method that I haven't been taught, so I'm not sure if I am correct and allowed to do this. Can you guys see if this is acceptable.
Since:
$$F\begin{pmatrix}1\\0\\1\end{pmatrix}=\begin{pmatrix}3\\-3\\1\end{pmatrix}$$
$$F\begin{pmatrix}0\\1\\0\end{pmatrix}=\begin{pmatrix}0\\1\\1\end{pmatrix}$$
$$F\begin{pmatrix}0\\1\\1\end{pmatrix}=\begin{pmatrix}2\\-2\\1\end{pmatrix}$$
We want to find:
$$\text{standard matrix:}F\begin{pmatrix}1&0&0\\0&1&0\\0&0&1\end{pmatrix}=?$$ 
So what I did was:

$$\therefore \text{standard matrix:}F\begin{pmatrix}1&0&0\\0&1&0\\0&0&1\end{pmatrix}=\begin{pmatrix}3&0&2\\-1&0&-1\\-2&1&-1\end{pmatrix}$$
Is this method correct?

Let's check your solution,
So you got $T(1,0,0)=(3,-1,-2)$, $~T(0,1,0)=(0,0,1)$ and $~T(0,0,1)=(2,-1,-1)$
Now $T(1,0,1)=(3,-3,1)$ 
also $~T(1,0,1)=T(e_1+0.e_2+e_3)=Te_1+Te_3=(3,-1,-2)+(2,-1,-1)=(5,-2,-3) \not=(3,-3,1).$
