Find the result of T given matrices against ordered basises I am trying to solve this problem:

Consider in $\mathbb{R}^2$ the ordered basis $v_1 = (1, 3)$, $v_2 = (2, 5)$.
Let $T : \mathbb{R}^2 \to \mathbb{R}^2$ be a linear image with as the matrix with respect to $v_1, v_2$:
$A = \begin{bmatrix}2 & 3 \\ 0 & 1 \end{bmatrix}$
After calculation we find the following matrix of T with respect to the standard basis:
$B = \begin{bmatrix}1 & 1 \\ 4 & 2 \end{bmatrix}$
Find $T((3, 8))$ using the matrix $B$
Then find $T((3, 8))$ using $A$

What I can't figure out is how you can correctly determine $T((3, 8))$ using $A$.
.
(Apologies for any errors, this is translated from a non-english textbook without the solution to this problem)
It is my understanding that any matrix $C$ of $T$ with respect to standard basises (bases?) is simply equal to the matrix you would use to represent the linear image $T$ to begin with, $T_C = T$.
So therefore, since $B$ is the matrix of $T$ with respect to the standard basis, $T_B = T$ and we get $T_B((3,8)) = T((3, 8)) = (11, 28)$
I thought I could use this:
If $ \mathcal{B} = v_1, v_2 $ and $ \mathcal{B}' = e_1, e_2 $ the standard basis, then the relationship between the matrices and the basises is given by the schema:
$
\begin{matrix}
 &  & A = m(T)_{\mathcal{B}}^{\mathcal{B}} &  &  \\
 & \mathcal{B} & \longrightarrow  & \mathcal{B} &  \\
m(I)_{\mathcal{B}'}^{\mathcal{B}} & \uparrow  &  & \downarrow  & m(I)_{\mathcal{B}}^{\mathcal{B}'} \\
 & \mathcal{B}' & \longrightarrow & \mathcal{B}' &  \\
 &  & B = m(T)_{\mathcal{B}'}^{\mathcal{B}'} &  & 
\end{matrix}
$
So $
B = m(T)_{\mathcal{B}'}^{\mathcal{B}'} = m(I)_{\mathcal{B}}^{\mathcal{B}'} m(T)_{\mathcal{B}}^{\mathcal{B}} m(I)_{\mathcal{B}'}^{\mathcal{B}} = m(I)_{\mathcal{B}}^{\mathcal{B}'} A m(I)_{\mathcal{B}'}^{\mathcal{B}}
$
So I could figure out the solution to $T((3,1))$ "using $A$" by going through essentially translating the problem into $A$, and then back to $B$. I think that this way I would never be using the solution they've given for $B$ directly in my solution "using $A$".
I then determine $m(I)_{\mathcal{B}}^{\mathcal{B}'}$ first, since it's easy to determine the coordinates with respect to the standard basis, they're just equal to $v_1, v_2$. I get
$ 
m(I)_{\mathcal{B}}^{\mathcal{B}'} = 
\begin{bmatrix}
1 & 2 \\
3 & 5
\end{bmatrix}
$
And then I determine $ m(I)_{\mathcal{B}'}^{\mathcal{B}} = (m(I)_{\mathcal{B}}^{\mathcal{B}'})^{-1}  
= \begin{bmatrix}
-5 & 2 \\
3 & -1
\end{bmatrix}
$
And now it would have to be true that
$
B = m(I)_{\mathcal{B}}^{\mathcal{B}'} A m(I)_{\mathcal{B}'}^{\mathcal{B}}
$
I get:
$B = \begin{bmatrix} 1 & 1 \\ 4 & 2 \end{bmatrix} \neq \begin{bmatrix} 1 & 2 \\ 3 & 5 \end{bmatrix} \begin{bmatrix} 2 & 3 \\ 0 & 1\end{bmatrix} \begin{bmatrix}-5 & 2 \\3 & -1\end{bmatrix} = \begin{bmatrix}5 & -1 \\12 & -2\end{bmatrix}$
But it isn't, so I must've done something wrong.
But I cannot figure out what this would be.
 A: It seems like there is an error in the question. If $B$ is the matrix of the transformation $T$ with respect to the standard basis, then the matrix with respect to the ordered basis $(v_{1},v_{2})$ should be $A=\begin{bmatrix}0&1\\2&3\end{bmatrix}$, not $\begin{bmatrix}2&3\\0&1\end{bmatrix}$.
Indeed, $T(v_{1})=Bv_{1}=\begin{bmatrix}4\\10\end{bmatrix}=0\begin{bmatrix}1\\3\end{bmatrix}+2\begin{bmatrix}2\\5\end{bmatrix}$, so $A\begin{bmatrix}1\\0\end{bmatrix}$ must be $\begin{bmatrix}0\\2\end{bmatrix}$.
A: There are two ways that I can think of that you can solve the question.
$1$. Simply reconstruct $T$ using the columns of the matrix. Consider the following example on how to do that.
Example: Consider the matrix $A=\begin{bmatrix} 1 & 4\\ 3 & 6\end{bmatrix}$. Let's assume that this matrix is the matrix of a linear map $T$ with respect to the standard basis of $R^2$. Then by definition of the matrix of a linear map, we know that $T(1, 0)=(1, 3)$ and $T(0, 1)=(4, 6)$. Thus, $T(x, y)=T(x(1, 0)+y(0, 1))=(x,3x)+(4y,6y)=(x+4y,3x+6y)$. Thus, the linear map this matrix represents is defined by $T(x, y)=(x+4y, 3x+6y)$. You can simply evaluate any vector using this. Recall that a linear map is completely determined on it's basis.
$2$. Let $M(T)$ denote the matrix of $T$ with respect to some basis. Then, the matrix of a vector $v\in V$ is defined to be the $n\times 1$ matrix whose columns contain the scalars needed to write $v$ as a linear combination of the basis of your choice. For example, consider the vector $(2, 8)\in R^2$. let's pick the basis to be $(1, 0), (0, 1)$. Then the matrix of the vector $(2, 8)$ denoted $M(2, 8)=\begin{bmatrix} 2 \\ 8\end{bmatrix}$ as the scalars needed to write $(2, 8)$ with respect to the standard basis are $2$ and $8$ as $(2, 8)=2(1, 0)+8(0, 1)$.
Then find the matrix of $(3, 8)$ with respect to the basis for $A$ and the basis of $B$. Then use the following result to get an answer.
Theorem: Let $v_1,\dots,v_n$ be a basis of $V$. Let $T\in L(V, W)$ and $v\in V$. Then we have that $M(Tv)=M(T)M(V)$. Where $M(T)$ denotes the matrix of $T$ with respect to the basis $v_1,\dots,v_n$ and $M(v)$ denotes the matrix of $v$ with respect to the basis $v_1,\dots,v_n$. Where of course $M(Tv)$ is computed with respect to the basis of $W$.
Example: Define a linear map $T:R^2\to R^2$ by $T(x, y)=(2x, 2y)$. Then the matrix of this linear map with respect to the standard basis is $M(T)=\begin{bmatrix} 2 & 0 \\ 0 & 2\end{bmatrix}$. The above result tells me that for the vector $(3, 4)\in R^2$, I should have that $M(Tv)=M(T)M(v)$. In other word, $M(6, 8)=\begin{bmatrix} 2 & 0 \\ 0 & 2\end{bmatrix} \begin{bmatrix} 3 \\ 4\end{bmatrix}=\begin{bmatrix} 6 \\ 8\end{bmatrix}$.
This was all done with respect to the standard basis. For a different basis, you should be able to change the results.
