What is the transformation matrix of the linear mapping? $ f:M_{2.2}(\mathbb{R}) \rightarrow \mathbb{R^3}, $$
\begin{pmatrix}
a & b \\
c & d 
\end{pmatrix} 
\mapsto
\begin{pmatrix}
2a - 4b \\
-6d \\
8a-16b+2d
\end{pmatrix}\beta =\big
(\begin{pmatrix}
1 & 0 \\
0 & 0 
\end{pmatrix}
,\begin{pmatrix}
0 & 1 \\
0 & 0 
\end{pmatrix}
\big),
\begin{pmatrix}
0 & 0 \\
1 & 0 
\end{pmatrix}
\big),
\begin{pmatrix}
0 & 0 \\
0 & 1 
\end{pmatrix}
\big)$
$\beta'= ( e_{2}, 2e_{1}, -e_{3})$ of $\mathbb{R^{3}}$ (
these are the standard unit vectors of $\mathbb{R^{3}}$)
What is the transformation matrix $M_{\beta'}^{\beta}(f)= ?$
Well my thought is that
$f(\left(
\begin{array}{c}
0\\
1\\
0
\end{array} 
\right)) =\left(
\begin{array}{c}
-4\\
0\\
-16
\end{array} 
\right) = \square \left(
\begin{array}{c}
1\\
0\\
8
\end{array} 
\right) +$$\square \left(
\begin{array}{c}
-4\\
0\\
2
\end{array} 
\right) +$
$\square \left(
\begin{array}{c}
0\\
0\\
0
\end{array} 
\right) +$
$\square \left(
\begin{array}{c}
0\\
-6\\
2
\end{array} 
\right)$
$f(\left(
\begin{array}{c}
2\\
0\\
0
\end{array} 
\right)) =\left(
\begin{array}{c}
4\\
0\\
16
\end{array} 
\right) = \square \left(
\begin{array}{c}
1\\
0\\
8
\end{array} 
\right) +$$\square \left(
\begin{array}{c}
-4\\
0\\
2
\end{array} 
\right) +$
$\square \left(
\begin{array}{c}
0\\
0\\
0
\end{array} 
\right) +$
$\square \left(
\begin{array}{c}
0\\
-6\\
2
\end{array} 
\right) $
$f(\left(
\begin{array}{c}
0\\
0\\
-1
\end{array} 
\right)) =\left(
\begin{array}{c}
0\\
6\\
-2
\end{array} 
\right) = \square \left(
\begin{array}{c}
1\\
0\\
8
\end{array} 
\right) +$$\square \left(
\begin{array}{c}
-4\\
0\\
2
\end{array} 
\right) +$
$\square \left(
\begin{array}{c}
0\\
0\\
0
\end{array} 
\right) +$
$\square \left(
\begin{array}{c}
0\\
-6\\
2
\end{array} 
\right) $
And from the linear combinations of the $squares$ I can fill the transformation matrix in the columns and in this way I will get the transformation matrix. Is my idea correct? And how does the transformation matrix look like?
Thank you in advance

Your idea is not correct, since the domain of $f$ is $\Bbb R^{2\times2}$; so it makes no sense to talk about, say$$f\left(\begin{bmatrix}0\\1\\0\end{bmatrix}\right).$$
Note that\begin{align}f\left(\begin{bmatrix}1&0\\0&0\end{bmatrix}\right)&=\begin{bmatrix}2\\0\\8\end{bmatrix}\\&=0\times e_2+1\times(2e_1)-8(-e_3).\end{align}Therefore, the entries of the first column of the matrix that you're after are $0$, $1$, and $-8$.
Can you take it from here?
