I was just writing up notes for this in connection with a section on Smith Normal Form in my abstract algebra course. I'll summarize the results here.
Row operations on the matrix change the input basis and column operations on the matrix change the output basis.
First, let $\{u_1, u_2, \ldots, u_n\}$ denote the input basis.
Row operation: $r_i \to r_i + k r_j$. Effect: Replace $u_j$ with $u_j - k u_i$. (Note that the $i$ and $j$ switch, and there is a sign change.)
Row operation: $r_i \leftrightarrow r_j$. Effect: Swap $u_i$ and $u_j$.
Row operation: $r_i \to k r_i$, where $k$ is a unit. Effect: Replace $u_i$ with $k^{-1} u_i$.
Next, let $\{v_1, v_2, \ldots, v_m\}$ be the output basis.
Column operation: $c_i \to c_i + k c_j$. Effect: Replace $v_i$ with $v_i - k v_j$. (Here there's just a sign change.)
Column operation: $c_i \leftrightarrow c_j$. Effect: Swap $v_i$ and $v_j$.
Column operation: $c_i \to k c_i$, where $k$ is a unit. Effect: Replace $v_i$ with $k^{-1} v_i$.
Actually, for Smith Form if you know the Smith form and either the final input basis or the final output basis, you can find the other basis easily. All of this assumes you're doing the ``easy'' case of a Euclidean domain; you need a fourth type of operation over a PID.
Short answer: you are changing basis ... twice!
The transformation $T$ is described geometrically in terms of the standard basis $\mathcal{E} = \{e_1, e_2, e_3\}$ for $\mathbb{R}^3$, so the matrix is
$$ \big[\,T\,\big]_{\mathcal{E}} = \left[ \begin{array}{*{3}{c}} 1&0&0 \\ 0&1&0 \\ 0&0&-1 \end{array} \right].$$
To obtain the matrix in $\mathcal{B}$-coordinates, you must first change basis to the standard coordinates, then apply the tranformation there, and finally change basis back to the new basis. This is called conjugation (or a similarity transformation, as you likely know):
$$ \big[\,T\,\big]_{\mathcal{B}} = \underset{\mathcal{B} \leftarrow \mathcal{E}}{\mathcal{P}} \; \big[\,T\,\big]_{\mathcal{E}} \; \underset{\mathcal{E} \leftarrow \mathcal{B}}{\mathcal{P}}$$
The trick is that when you write a vector $b_1$, you are implicitly expressing it in terms of the standard basis, so $b_1 = [\,b_1\,]_{\mathcal{E}}$, which is why the change-of-basis matrix from $\mathcal{B}$ to $\mathcal{E}$ is
$$\underset{\mathcal{E} \leftarrow \mathcal{B}}{\mathcal{P}} = \big[ \big[\,b_1\,\big]_{\mathcal{E}} \;\; \big[\,b_2\,\big]_{\mathcal{E}} \;\; \big[\,b_3\,\big]_{\mathcal{E}} \big] = \big[ b_1 \;\; b_2 \;\; b_3 \big].$$
Now, apply the matrix $T$ and give the result back in $\mathcal{B}$-coordinates, and you obtain
$$
\big[\,T\,\big]_{\mathcal{B}} = \big[ \big[\,T(b_1)\,\big]_{\mathcal{B}} \;\; \big[\,T(b_2)\,\big]_{\mathcal{B}} \;\; \big[\,T(b_3)\,\big]_{\mathcal{B}} \big]
.$$
Best Answer
The first column of $M_\mathcal{B}(L) = (m_{ij})_{1\le i,j\le 4}$ is defined as the coordinates (with respect to the basis $\mathcal{B}=(\mathcal{B}_1,\mathcal{B}_2,\mathcal{B}_3,\mathcal{B}_4)$) of $A\mathcal{B}_1$.
Since $$\begin{array}{rcl} A\mathcal{B}_1 &=& \begin{bmatrix}1&2\\0&3\end{bmatrix} \begin{bmatrix}1&0\\0&0\end{bmatrix} \\ &=& \begin{bmatrix}1&0\\0&0\end{bmatrix} \\ &=& 1\times \mathcal{B}_1 + 0\times \mathcal{B}_2+ 0\times \mathcal{B}_3+ 0\times \mathcal{B}_4 \\ A\mathcal{B}_1&=& m_{11}\times \mathcal{B}_1 + m_{21}\times \mathcal{B}_2+ m_{31}\times \mathcal{B}_3+ m_{41}\times \mathcal{B}_4. \end{array}$$ This gives us the first column. I let you do the same for the rest of the matrix.