Linear transformation matrix derivation

$A=\begin{bmatrix}1&2\\0&3\end{bmatrix} \in \Bbb R^{2 \times 2}$

$L: \space \Bbb R^{2 \times 2} \longrightarrow \Bbb R^{2 \times 2}; \space X \mapsto AX$

I want to find the transformation matrix with respect to the basis

$\mathcal B_1=\begin{bmatrix}1&0\\0&0\end{bmatrix}, \space \mathcal B_2=\begin{bmatrix}0&0\\1&0\end{bmatrix}, \space \mathcal B_3=\begin{bmatrix}0&1\\0&0\end{bmatrix}, \space \mathcal B_4=\begin{bmatrix}0&0\\0&1\end{bmatrix}$

I know the answer is: $M_{\mathcal B}(L)=\begin{bmatrix}1&2&0&0\\0&3&0&0\\0&0&1&2\\ 0&0&0&3\end{bmatrix}$

I don't know how to get to that matrix.

$$\begin{array}{rcl} A\mathcal{B}_1 &=& \begin{bmatrix}1&2\\0&3\end{bmatrix} \begin{bmatrix}1&0\\0&0\end{bmatrix} \\ &=& \begin{bmatrix}1&0\\0&0\end{bmatrix} \\ &=& 1\times \mathcal{B}_1 + 0\times \mathcal{B}_2+ 0\times \mathcal{B}_3+ 0\times \mathcal{B}_4 \\ A\mathcal{B}_1&=& m_{11}\times \mathcal{B}_1 + m_{21}\times \mathcal{B}_2+ m_{31}\times \mathcal{B}_3+ m_{41}\times \mathcal{B}_4. \end{array}$$