TITLE: Determinant of a generalized Pascal matrix
QUESTION [22 upvotes]: Let $M$ denote the infinite matrix defined recursively by 
$$
M_{ij} = 
\begin{cases}
  1, &  \text{if } i=1 \text{ and } j=1; \\
  aM_{i-1,j}+bM_{i,j-1}+cM_{i-1,j-1}, & \mbox{otherwise}.\\ \end{cases}
$$
($M_{i,0}$ and $M_{0,j}$ are both defined to be $0$.) 
(Added: I just discovered that the numbers in the $M$ matrix are called weighted Delannoy numbers.)
Let $M_n$ denote the $n \times n$ upper-left submatrix of $M$. 
For example, with $a = b = c = 1$, 
$M_1 = \begin{bmatrix} 1 \end{bmatrix}$, $M_2= \begin{bmatrix} 1 & 1 \\ 1 & 3 \end{bmatrix}$, and $M_3 = \begin{bmatrix} 1 & 1 & 1 \\ 1 & 3 & 5 \\ 1 & 5 & 13 \end{bmatrix}$, and $M_4 = \begin{bmatrix}
    1 & 1 & 1 & 1 \\
    1 & 3 & 5 & 7 \\
    1 & 5 & 13 & 25 \\
    1 & 7 & 25 & 63
   \end{bmatrix}.$  
A few years ago one of my students proved, by induction, that $$\det M_n = (ab+c)^{n(n-1)/2}.$$
My question is

Is there a noninductive proof that $\det M_n = (ab+c)^{n(n-1)/2}$ that gives more insight into why the determinant works out so nicely?

For example, when $a = b = 1$, $c = 0$, $M$ is the symmetric Pascal matrix.  I've seen more than one way to prove that $\det M_n = 1$ in this case.  For example, Edelman and Strang give four proofs of an LU-decomposition that does it.  I also once saw, at a conference, a combinatorial proof using the interpretation of the determinant in terms of nonintersecting paths in a directed graph.  (I think the talk was given by Art Benjamin, but it was several years ago, and I may be misremembering.)  So I know that there are some nice proofs in the special case of the Pascal matrix.  But what about the general case?

REPLY [6 votes]: Inspired by the solution of Sivaram Ambikasaran, I would just like to furnish some concision. Given
$$
M_{i,j} = 
\begin{cases}
  a^{i-1}b^{j-1} &  \text{if } i=1 \text{ or } j=1, \\
  aM_{i-1,j}+bM_{i,j-1}+cM_{i-1,j-1} & \mbox{if }i>1 \text{ and } j>1,\\ \end{cases}
$$
one may (inspired by what is said in the Edelman-Strang paper) start subtracting from each row (except the first) $a$ times the previous row. This is to be done simultaneously, or working from the bottom up, so that it is the original value of the row that is subtracted. This amounts to left-multiplying by the elimination matrix $E_{-a}$ with diagonal entries $1$ and subdiagonal entries $-a$. One obtains a matrix $M'$ given by 
$$
M'_{i,j} = 
\begin{cases}
  b^{j-1} & \text{if } i=1, \\
  M_{i,j}-aM_{i-1,j} & \mbox{if }i>1.\\ \end{cases}
$$
The latter expression is $0$ if $j=1$, and otherwise
$$
  M'_{i,j} = bM_{i,j-1}+cM_{i-1,j-1}\qquad\text{if }i,j>1
$$
using the recursion defining $M$. Now proceed similarly by columns, right-multiplying by the transpose $E_{-b}^\top$ of $E_{-b}$, giving a new matrix $M''$ given by
$$
M''_{i,j} = 
\begin{cases}
  \delta_{i,j} & \text{if } j=1, \\
  M'_{i,j}-bM'_{i,j-1} & \mbox{if }j>1.\\ \end{cases}
$$
The latter expression is $0$ if $i=1$, and otherwise
$$\begin{align}
  M''_{i,j} &= bM_{i,j-1}+cM_{i-1,j-1}-b(M_{i,j-1}-aM_{i-1,j-1})\\
           & =(ab+c)M_{i-1,j-1}
\end{align}\qquad\text{if }i,j>1,
$$
using the two expressions for $M'_{i,j}$ given above. In other words one has in block form
$$
  M''=\begin{pmatrix}1&0\\ 0&(ab+c)M_{(n-1)}\end{pmatrix}
$$
where $M_{(n-1)}$ is the $(n-1)\times(n-1)$ top-left submatrix of $M$, which is its counterpart of size one less. Therefore
$$
  \det M_{(n)} = \det M = \det M''= (ab+c)^{n-1} \det M_{(n-1)}
$$
from which it follows by induction that $\det M=(ab+c)^{\tbinom{n}2}$.
The scalar multplication of $M_{(n-1)}$ by $ab+c$ can of course be realised as the multiplication by a multiple of the identity matrix, which commutes with with the matrices obtained in a recursive decomposition of $M_{(n-1)}$. It is then easy to conclude that one has a decomposition
$$
  M_{(n)} = L_{(n)} D_{(n)} U_{(n)}
$$
where $D_{(n)}$ is diagonal with entries $(D_{(n)})_{i,i}=(ab+c)^{i-1}$ and $L_{(n)}$ and $U_{(n)}$ are unitriangular and therefore have determinant $1$. Their explicit form can be obtained from the recurrences
$$
  L_{(n)} = (E_{-a})^{-1}\cdot \begin{pmatrix}1&0\\ 0&L_{(n-1)}\end{pmatrix},
  \qquad
  U_{(n)} = \begin{pmatrix}1&0\\ 0&U_{(n-1)}\end{pmatrix}\cdot(E_{-b}^\top)^{-1},
$$
which can easily found to be solved by $(L_{(n)})_{i,j}=\binom{i-1}{j-1}a^{i-j}$ and $(U_{(n)})_{i,j}=\binom{j-1}{i-1}b^{j-i}$, either using $((E_{-a})^{-1})_{i,j}=a^{i-j}$ or by solving first $((L_{(n)})^{-1})_{i,j}=\binom{i-1}{j-1}(-a)^{i-j}$.