That's the way I would do it. Someone please correct me if I do something incorrect. I'm still studying Linear Algebra.
$[T1]_{b}^{\gamma}$ It's a linear transformation that get some vector in the standard basis ($\gamma$) for $R^2$, apply the transformation, and then return it in the basis $b$.
Starting with that, we can affirm that:
$$
[T1]_{b}^{\gamma} = [I]_{b}^{\gamma}\cdot [T1]_{\gamma}^{\gamma}
$$
$[I]_{b}^{\gamma}$ is a change of basis from $\gamma$ to $b$.
Now, all we need to do is get the matrix that represents $[I]_{b}^{\gamma}$ and multiply to the matrix of the transformation $[T1]_{\gamma}^{\gamma}$.
The column vectors for the matrix of transformation $[T1]_{\gamma}^{\gamma}$ is going to be the transformed vectors that compose the basis $\gamma$:
$$
(1,0) \rightarrow (1+0,2\cdot 1 + 4\cdot 0) = (1,2)\\
(0,1) \rightarrow (0+1,2\cdot 0 + 4\cdot 1) = (0,4)\\
\left[\begin{matrix}
1 & 0 \\
2 & 4
\end{matrix}\right]
$$
Now for the matrix of change of basis $\gamma$ to $b$, we need to decompose $\gamma$ vectors in terms of $b$ vectors:
$$
(1,0) = a(1,2)+b(0,1) \rightarrow a = 1 , b = -2\\
(0,1) = c(1,2)+d(0,1) \rightarrow c = 0 , d = 1\\
\left[\begin{matrix}
1 & 0 \\
-2 & 1
\end{matrix}\right]
$$
Now multiplying the matrices, we get:
$$
\left[\begin{matrix}
1 & 0 \\
-2 & 1
\end{matrix}\right] \cdot
\left[\begin{matrix}
1 & 0 \\
2 & 4
\end{matrix}\right] =
\left[\begin{matrix}
1 & 0 \\
0 & 4
\end{matrix}\right]
$$
Now, to check if everything is correct, let's pick a random vector from $R^2$ describe it in terms of $\gamma$:
$$
V_{\gamma} = (a,b)
$$
Then we apply the transformation $[T1]_{b}^{\gamma}$:
$$
\left[\begin{matrix}
1 & 0 \\
0 & 4
\end{matrix}\right]\cdot
\left[\begin{matrix}
a \\
b
\end{matrix}\right] =
\left[\begin{matrix}
a \\
4b
\end{matrix}\right]
$$
Now we just need to check if we apply $[T1]_{\gamma}^{\gamma}$ and then apply $[I]_{b}^{\gamma}$ is going to give us the same result:
$$
\left[\begin{matrix}
1 & 2 \\
0 & 4
\end{matrix}\right] \cdot
\left[\begin{matrix}
a \\
b
\end{matrix}\right] =
\left[\begin{matrix}
a \\
2a + 4b
\end{matrix}\right]\\
\left[\begin{matrix}
1 & 0 \\
-2 & 1
\end{matrix}\right] \cdot
\left[\begin{matrix}
a \\
2a + 4b
\end{matrix}\right] =
\left[\begin{matrix}
a \\
4b
\end{matrix}\right]
$$
So it's indeed correct.
I really don't know if I was clear enough, of if it's 100% correct. I hope to have contributed a little with your learning proccess. Feel free to leave a comment if you have had any doubts on what I've done...
Let's share some knowledge :) Thanks!
