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.

We have the matrices
$$
B^{e\to b} = 
B^{-1} =
\begin{bmatrix} 
1 & 1\\ 
1 & 3 
\end{bmatrix}^{-1}
\quad
T^{b\to b'}
=
\begin{bmatrix} 
2 & 5 \\
1 & 1 \\
8 & 1 \\
\end{bmatrix}
\quad
B^{b'\to e'} = 
B' =
\begin{bmatrix} 
2 & 1 & 1\\
1 & 0 & -1  \\
1 & 1 & 1 
\end{bmatrix}
$$
where $e$ means the canonical basis in $\mathbb{R}^2$, $e'$ the canonical basis in $\mathbb{R}^3$, $b$ and $b'$ the other two given basis sets, so we get
\begin{align}
T^{e\to e'} 
&= B^{b'\to e'} \, T^{b\to b'} B^{e\to b} \\
&=
\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} 
2 & 1 & 1\\
1 & 0 & -1  \\
1 & 1 & 1 
\end{bmatrix}
\begin{bmatrix} 
2 & 5 \\
1 & 1 \\
8 & 1 \\
\end{bmatrix}
\begin{bmatrix} 
3/2 & -1/2\\ 
-1/2 & 1/2 
\end{bmatrix}
\\
&=
\begin{bmatrix} 
13 & 12 \\
-6 & 4 \\
11 & 7 \\
\end{bmatrix}
\begin{bmatrix} 
3/2 & -1/2\\ 
-1/2 & 1/2 
\end{bmatrix}
\\
&=
\begin{bmatrix} 
27/2 & -1/2\\ 
-11 & 5 \\
13 & -2 
\end{bmatrix}
\end{align}
