Find a $3 \times 3$ matrix $A$ such that $g(x) = AxT$

Question

I am new to linear transormations and I can do the $\Bbb R^m \to \Bbb R^n$ tranformations, however I came up to this question and im confused. I'd be grateful if any of you could guide me how to start it please?

Let $g : \Bbb R^3 \to \Bbb R^3$ be the linear transformation satisfying $g \left( \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix} \right) = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}, \ g \left( \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix} \right) = \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}$ and $g \left( \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix} \right) = \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}$.

Find a $3 \times 3$ matrix $A$ such that $g(x) = AxT$ for all $x \in \Bbb R^3$

Answer 1

$g(e_1)=g(1,0,0)=(g(1,0,1)+g(1,1,0)-g(0,1,1))/2=([1,0,0]+[0,1,0]-[1,1,0])/2=[0,0,0]$
$g(e_2)=g(0,1,0)=g(1,1,0)-g(1,0,0)=[0,1,0]-[0,0,0]=[0,1,0]$
$g(e_3)=g(0,0,1)=g(0,1,1)-g(0,1,0)=[1,1,0]-[0,1,0]=[1,0,0]$
The matrix $A=\begin{bmatrix}0&0&1\\0&1&0\\0&0&0\end{bmatrix}$