I have a matrix $A$ ($n \times n$) defined as follows:
$$A = \{ 0 \text{ if } i<j,\ \mathrm{Binom}(x=i, \mathrm{size}=j, \mathrm{prob})\text{ if } j \ge i\}$$
This is an upper triangular matrix, and I want to solve a system $Ax =b$ -- thus in a sense invert $A$.
I was wondering if a general inverse exists for this problem.
Any help appreciated, thanks in advance..
On
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 $$ M_{ij} = \begin{cases}{j\choose i} (-1)^{i+j} q^{-i}r^{-j}, & i\le j,\\0 & i>j.\end{cases} $$
I suspect it is a lower triangular matrix. Anyway, append it with an identity matrix and start eliminating from bottom to top (top to bottom). See Gauss-Jordan elimination.