Q:

Linear map problem

What are the steps to solve the problem like this:
Let $\mathcal{X} = (\vec{x}_{1}, \vec{x}_{2}, \vec{x}_{3})$ is base $\mathbb{C}^{3}$ and $\mathcal{A} \in \mathcal{L} (\mathbb{C}^{3})$. Find the vector $\mathcal{A}\vec{x}$, provided $\vec{x}= \begin{pmatrix} -2\\ 0 \\ 2 \end{pmatrix}$, $\vec{x}_1= \begin{pmatrix} 1\\ 0 \\ 1 \end{pmatrix}$, $\vec{x}_2= \begin{pmatrix} 0\\ 1 \\ 2 \end{pmatrix}$, $\vec{x}_3= \begin{pmatrix} 3\\ 1 \\1 \end{pmatrix}$ and $\mathcal{A}\vec{x}_1= \begin{pmatrix} 1\\ 1 \\1 \end{pmatrix}$, $\mathcal{A}\vec{x}_2= \begin{pmatrix} 2\\ 0 \\0 \end{pmatrix}$, $\mathcal{A}\vec{x}_3= \begin{pmatrix} 0\\ 1 \\3 \end{pmatrix}$.

A:

Hints:
Write $\;x\;$ as a linear combination of $\;x_1,x_2,x_3\;$ :
$$x=ax_1+bx_2+cx_3\;,\;\;a,b,c\in\Bbb C$$
and now use linearity of the map:
$$Ax=aAx_1+bAx_2+cAx_3$$

