Let $q=1-p$ and $r = \frac pq$. Then $ A_{ij} = \begin{cases}{j\choose i} r^i q^j, & i\le j,\\0 & i>j.\end{cases} $ Therefore $A = \mathrm{diag}(r,r^2,\ldots,r^n)\ B\,\ \mathrm{diag}(q,q^2,\ldots,q^n)$ where $ B_{ij} = \begin{cases}{j\choose i}, & i\le j,\\0 & i>j.\end{cases} $ The matrix $B$ is intimately related to the definition of Pascal matrix. The entries of $C=B^{-1}$ are known to take the following form: $ C_{ij} = \begin{cases}{j\choose i} (-1)^{i+j}, & i\le j,\\0 & i>j.\end{cases} $ Hence the entries of the $M=A^{-1}$ are given by $ M_{ij} = \begin{cases}{j\choose i} (-1)^{i+j} q^{-i}r^{-j}, & i\le j,\\0 & i>j.\end{cases} $