Change of basis of linear transformation $ℝ^2 \to ℝ^3$ Linear transformation $T:ℝ^2\to ℝ^3$ in bases
$\left\{ \begin{bmatrix} 1 \\ 1 \end{bmatrix}, \begin{bmatrix} 1 \\ 3 \end{bmatrix}\right\}$ and $\left\{ \begin{bmatrix} 2 \\ 1 \\ 1 \end{bmatrix},\begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix},\begin{bmatrix} 1 \\ -1 \\ 1 \end{bmatrix}\right\}$ has matrix:
$\begin{bmatrix} 
2 & 5 \\
1 & 1 \\
8 & 1 \\
\end{bmatrix}$. What is the standard matrix? 
I thought this problem will be easy, but it got me horribly confused. I don't know where to start.

let $x_1, x_2$ be the canonical basis in \mathbb R^2.
and
$y_1, y_2, y_3$ be the canonical basis in \mathbb R^3.
$T(x_1 + x_2) = T(x_1) + T(x_2) = $$(2\cdot 2 + 1\cdot 1 + 8\cdot 1) y_1 + (2\cdot 1 + 1\cdot 0 + 8\cdot -1) y_2 + (2+1+8) y_3\\
13 y_1 - 6y_2 + 11y_3$
$T(x_1) + 3T(x_2) = 12y_1 + 4y_2 + 7y_3\\
2T(x_2) = -1y_1 +10y_2-4 y_3\\
T(x_2) = \begin{bmatrix}{- \frac 12\\5\\-2}\end{bmatrix}\\
T(x_1) = \begin{bmatrix}{13\\-6\\11}\end{bmatrix}-\begin{bmatrix}{- \frac 12\\5\\-2}\end{bmatrix} = \begin{bmatrix}{13.5\\-11\\13}\end{bmatrix}$
Alternatively
$\begin{bmatrix} 2&1&1\\1&0&-1\\1&1&1\end{bmatrix}\begin{bmatrix}2&5\\1&1\\8&1\end{bmatrix}\begin{bmatrix}1&1\\1&3\end{bmatrix}^{-1} = \begin{bmatrix} 13.5 &-0.5\\-11&5\\13&-2\end{bmatrix}$
