Finding the matrix of a linear transformation from an upper triangular matrix to an upper triangular matrix. The question that I am trying to solve is as follows:

Find the matrix $A$ of the linear transformation
  $T(M)=
\begin{bmatrix} 
7 & 3 \\ 
0 & 1 
\end{bmatrix}
M$
  from $U^{2×2}$ to $U^{2×2}$ (upper triangular matrices) with respect to the basis
  $\left\{ \begin{bmatrix} 
1 & 0 \\ 
0 & 0 
\end{bmatrix},\begin{bmatrix}1&1\\ 0&0\end{bmatrix},\begin{bmatrix}0&0\\0&1\end{bmatrix}\right\}$
$A=\begin{bmatrix} \cdot&\cdot&\cdot\\\cdot&\cdot&\cdot\\ \cdot&\cdot&\cdot\end{bmatrix}$

I don't know how to interpret this question. I would have thought that the matrix $A$ is $\begin{bmatrix} 
7 & 3 \\ 
0 & 1 
\end{bmatrix}$ because that is the matrix which defines the linear transformation $T$. I am surprised that the required answer is $A\in M_{3\times 3}$. How can a basis with elements $\in M_{2\times 2}$ form a matrix $\in M_{3\times 3}$?
 A: Upper triangular matrices
$
\begin{bmatrix}
a&b\\
0&c
\end{bmatrix}
$
form a vector space with canonical basis:
$
e_1=\begin{bmatrix}
1&0\\
0&0
\end{bmatrix}
\quad
e_2=\begin{bmatrix}
0&1\\
0&0
\end{bmatrix}
\quad
e_3=\begin{bmatrix}
0&0\\
0&1
\end{bmatrix}
$
that is isomorphic to $\mathbb{R}^3$ by:
$
e_1\rightarrow\vec e_1=\begin{bmatrix}
1\\0\\0
\end{bmatrix}
\quad
e_2\rightarrow\vec e_2=\begin{bmatrix}
0\\1\\0
\end{bmatrix}
\quad
e_3\rightarrow \vec e_3=\begin{bmatrix}
0\\0\\1
\end{bmatrix}
$
Your linear transformation $T$ is defined as a matrix multiplication:
$
T(M)=\begin{bmatrix}
7&3\\
0&1
\end{bmatrix}
\begin{bmatrix}
a&b\\
0&c
\end{bmatrix}
=
\begin{bmatrix}
7a&7b+3c\\
0&c
\end{bmatrix}
$
so, in this canonical representation, it is given by the matrix $T_e$ such that:
$
T_e\vec M=\begin{bmatrix}
7&0&0\\
0&7&3\\
0&0&1
\end{bmatrix}
\begin{bmatrix}
a\\
b\\
c
\end{bmatrix}=
\begin{bmatrix}
7a\\
7b+3c\\
c
\end{bmatrix}
$
Now you want a representation of the same transformation in  a new basis:
$
e'_1=
\begin{bmatrix}
1&0\\
0&0
\end{bmatrix}
\rightarrow\vec e'_1=\begin{bmatrix}
1\\0\\0
\end{bmatrix}
\quad
e'_2=\begin{bmatrix}
1&1\\
0&0
\end{bmatrix}\rightarrow\vec e'_2=\begin{bmatrix}
1\\1\\0
\end{bmatrix}
\quad
e'_3=\begin{bmatrix}
0&0\\
0&1
\end{bmatrix}\rightarrow \vec e'_3=\begin{bmatrix}
0\\0\\1
\end{bmatrix}
$
This transformation of basis is represented by the matrices
$
S=
\begin{bmatrix}
1&1&0\\
0&1&0\\
0&0&1
\end{bmatrix}
\qquad
S^{-1}=
\begin{bmatrix}
1&-1&0\\
0&1&0\\
0&0&1
\end{bmatrix}
$
So the matrix that represents the transformation $T$ in the new basis is:
$
T_{e'}=S^{-1}T_eS=
\begin{bmatrix}
7&0&-3\\
0&7&3\\
0&0&1
\end{bmatrix}
$
A: Let $(X,Y,Z)$ be the basis of $U$, compute $T(X)$, $T(Y)$, $T(Z)$ and express them as a linear combination of $X$, $Y$ and $Z$.
