Determinant of a generalized Pascal matrix

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?

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In addition to the recent article by Ik-Pyo Kim cited by Sivaram Ambikasaran, I just came across an 2002 paper by Christian Krattenthaler that solves this problem (its Theorem 1 on page 2, which is for $a=b=1$ but that can easily be seen to imply the general formula; on the other hand the initial row/column are taken of a more general form there). It appears that Roland Bacher also investigated this problem around the same time. –  Marc van Leeuwen Dec 4 '11 at 16:38
@Marc: See also this paper by Zakrajšek and Petkovšek, which appears to build on the work by Krattenthaler and Bacher. –  Mike Spivey Dec 4 '11 at 21:25

EDIT:

So after all the effort I put in couple of days back to prove this, I found out today that this result was already published earlier this year. The article is "LDU decomposition of an extension matrix of the Pascal matrix," by Ik-Pyo Kim (Linear Algebra and Its Applications 434(10) 2187-2196, 2011) and can be found here.

The $LU$ factorization of $M_{n+1}$ is given below. Let $x$ be such that $x^2 = 1 + \frac{c}{ab}$. Then we have that $$M_{n+1} = L_{n+1} U_{n+1}$$ where $$\small L_{n+1} = \begin{bmatrix}1 & 0 & 0 & 0 & 0 & \cdots & 0 & 0\\a & ax & 0 & 0 & 0 & \cdots & 0 & 0\\ a^2 & 2a^2x & a^2x^2 & 0 & 0 & \cdots & 0 & 0\\ a^3 & 3a^3x & 3a^3x^2 & a^3x^3 & 0 & \cdots & 0 & 0\\ a^4 & 4a^4x & 6a^4x^2 & 4a^4x^3 & a^4x^4 & \cdots & 0 & 0\\ \vdots & \vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\ \vdots & \vdots & \vdots & \vdots & \vdots & \cdots & \ddots & \vdots \\ a^n & \binom{n}{1}a^nx^1 & \binom{n}{2}a^nx^2 & \binom{n}{3}a^nx^3 & \binom{n}{4}a^nx^4 & \cdots & \binom{n}{n-1}a^nx^{n-1} & a^nx^n \end{bmatrix}$$

$$\small U_{n+1}^T = \begin{bmatrix}1 & 0 & 0 & 0 & 0 & \cdots & 0 & 0\\b & bx & 0 & 0 & 0 & \cdots & 0 & 0\\ b^2 & 2b^2x & b^2x^2 & 0 & 0 & \cdots & 0 & 0\\ b^3 & 3b^3x & 3b^3x^2 & b^3x^3 & 0 & \cdots & 0 & 0\\ b^4 & 4b^4x & 6b^4x^2 & 4b^4x^3 & b^4x^4 & \cdots & 0 & 0\\ \vdots & \vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\ \vdots & \vdots & \vdots & \vdots & \vdots & \cdots & \ddots & \vdots \\ b^n & \binom{n}{1}b^nx^1 & \binom{n}{2}b^nx^2 & \binom{n}{3}b^nx^3 & \binom{n}{4}b^nx^4 & \cdots & \binom{n}{n-1}b^nx^{n-1} & b^nx^n \end{bmatrix}$$

This gives us $\det(L_n) = (ax)^{\frac{n(n-1)}{2}}$ and $\det(U_n) = (bx)^{\frac{n(n-1)}{2}}$.

Hence, $\det(M_n) = (ax)^{\frac{n(n-1)}{2}}(bx)^{\frac{n(n-1)}{2}} = (abx^2)^{\frac{n(n-1)}{2}} = (ab+c)^{\frac{n(n-1)}{2}}$.

Also note that $$L_n = A_n X_n$$ and $$U_n^T = B_n X_n$$ where $$A_{n+1} = \begin{bmatrix} 1 & 0 & 0 & 0 & 0 & \cdots & 0 & 0\\ 0 & a & 0 & 0 & 0 & \cdots & 0 & 0\\ 0 & 0 & a^2 & 0 & 0 & \cdots & 0 & 0\\ 0 & 0 & 0 & a^3 & 0 & \cdots & 0 & 0\\ 0 & 0 & 0 & 0 & a^4 & \cdots & 0 & 0\\ \vdots & \vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\ \vdots & \vdots & \vdots & \vdots & \vdots & \cdots & \ddots & \vdots\\ 0 & 0 & 0 & 0 & 0& \cdots & 0 & a^n\\ \end{bmatrix}$$ $$B_{n+1} = \begin{bmatrix} 1 & 0 & 0 & 0 & 0 & \cdots & 0 & 0\\ 0 & b & 0 & 0 & 0 & \cdots & 0 & 0\\ 0 & 0 & b^2 & 0 & 0 & \cdots & 0 & 0\\ 0 & 0 & 0 & b^3 & 0 & \cdots & 0 & 0\\ 0 & 0 & 0 & 0 & b^4 & \cdots & 0 & 0\\ \vdots & \vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\ \vdots & \vdots & \vdots & \vdots & \vdots & \cdots & \ddots & \vdots\\ 0 & 0 & 0 & 0 & 0& \cdots & 0 & b^n\\ \end{bmatrix}$$ $$X_{n+1} = \begin{bmatrix}1 & 0 & 0 & 0 & 0 & \cdots & 0 & 0\\1 & x & 0 & 0 & 0 & \cdots & 0 & 0\\ 1 & 2x & x^2 & 0 & 0 & \cdots & 0 & 0\\ 1 & 3x & 3x^2 & x^3 & 0 & \cdots & 0 & 0\\ 1 & 4x & 6x^2 & 4x^3 & x^4 & \cdots & 0 & 0\\ \vdots & \vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\ \vdots & \vdots & \vdots & \vdots & \vdots & \cdots & \ddots & \vdots \\ 1 & \binom{n}{1}x^1 & \binom{n}{2}x^2 & \binom{n}{3}x^3 & \binom{n}{4}x^4 & \cdots & \binom{n}{n-1}x^{n-1} & x^n \end{bmatrix}$$ Further $X_n$ can be written as $C_n \tilde{X}_n$ where $$C_{n+1} = \begin{bmatrix}1 & 0 & 0 & 0 & 0 & \cdots & 0 & 0\\1 & 1 & 0 & 0 & 0 & \cdots & 0 & 0\\ 1 & 2 & 1 & 0 & 0 & \cdots & 0 & 0\\ 1 & 3 & 3 & 1 & 0 & \cdots & 0 & 0\\ 1 & 4 & 6 & 4 & 1 & \cdots & 0 & 0\\ \vdots & \vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\ \vdots & \vdots & \vdots & \vdots & \vdots & \cdots & \ddots & \vdots \\ 1 & \binom{n}{1} & \binom{n}{2} & \binom{n}{3} & \binom{n}{4} & \cdots & \binom{n}{n-1} & 1 \end{bmatrix}$$ and $$\tilde{X}_{n+1} = \begin{bmatrix} 1 & 0 & 0 & 0 & 0 & \cdots & 0 & 0\\ 0 & x & 0 & 0 & 0 & \cdots & 0 & 0\\ 0 & 0 & x^2 & 0 & 0 & \cdots & 0 & 0\\ 0 & 0 & 0 & x^3 & 0 & \cdots & 0 & 0\\ 0 & 0 & 0 & 0 & x^4 & \cdots & 0 & 0\\ \vdots & \vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\ \vdots & \vdots & \vdots & \vdots & \vdots & \cdots & \ddots & \vdots\\ 0 & 0 & 0 & 0 & 0& \cdots & 0 & x^n\\ \end{bmatrix}$$ $$M_n = A_n C_n \tilde{X}_n^2 C_n^T B_n^T$$ Setting $D_n = \tilde{X}_n^2$, we get a nice decomposition for $M_n$ as $$M_n = A_n C_n D_n C_n^T B_n.$$ Note that $A_n,B_n,D_n$ are diagonal matrices where $A_n(i,i) = a^i$,$B_n(i,i) = b^i$ and $D_n(i,i) = \left( 1+\frac{c}{ab} \right)^i$.

$C_n$ is a triangular matrix containing the binomial coefficients where $C_n(i,j) = \left \{ \begin{array}{lr} \binom{i-1}{j-1} & j \leq i\\ 0 & j > i \end{array}\right.$.

EDIT

I computed this decomposition by the usual $LU$ algorithm. But given that now we have this, it should be possible to reverse engineer other proofs, similar to the proofs for the Pascal's matrix.

Proof

Let me first write out the proof based on the algorithm I used to construct these $L$ and $U$ factors.

Consider the elimination matrix $$E_{n+1}^y = \begin{pmatrix} 1 & 0 & 0 & 0 & \cdots\\ -y & 1 & 0 & 0 & \cdots\\ 0 & -y & 1 & 0 & \cdots\\ \vdots & \vdots & \vdots & \ddots & \cdots\\ \vdots & \vdots & \vdots & -y & 1 \end{pmatrix}$$

Note that $E_{n+1}^a L_{n+1} = \begin{pmatrix}1 & 0\\ 0 & ax L_n \end{pmatrix}$ and $E_{n+1}^b U_{n+1}^T = \begin{pmatrix}1 & 0\\ 0 & bx U_n^T \end{pmatrix}$

Hence, $E_{n+1}^a L_{n+1} U_{n+1} (E_{n+1}^b )^T = \begin{pmatrix}1 & 0\\ 0 & abx^2 M_n \end{pmatrix}$.

Our hope is that the above matrix should match with $E_{n+1}^a M_{n+1} (E_{n+1}^b)^T$.

Consider the $(i,j)^{th}$ entry of $E_{n+1}^a M_{n+1}$. This is $M_{n+1}(i,j)-a M_{n+1}(i-1,j)$.

Now the $(i,j)^{th}$ entry of $E_{n+1}^a M_{n+1} (E_{n+1}^b)^T$ is \begin{align} M_{n+1}(i,j)-a M_{n+1}(i-1,j) -b \left( M_{n+1}(i,j-1)-a M_{n+1}(i-1,j-1) \right)\\ = M_{n+1}(i,j) - a M_{n+1}(i-1,j) -b M_{n+1}(i,j-1) + ab M_{n+1}(i-1,j-1) \\ = \left( M_{n+1}(i,j) - a M_{n+1}(i-1,j) -b M_{n+1}(i,j-1) \right) + ab M_{n+1}(i-1,j-1) \\ = c M_{n+1}(i-1,j-1) + ab M_{n+1}(i-1,j-1) = abx^2 M_{n+1}(i-1,j-1) \end{align}

But $M_{n+1}(i-1,j-1) = M_n(i-1,j-1)$. Hence, we get that $$M_{n+1}(i,j)-a M_{n+1}(i-1,j) -b \left( M_{n+1}(i,j-1)-a M_{n+1}(i-1,j-1) \right) = abx^2 M_{n}(i-1,j-1).$$

Hence, coupling with induction we get the complete proof. (Note that the matrix $E_n^y$ is always invertible enabling us to conclude what we want.)

Proof of the decomposition $M_n = A_n C_n D_n C_n^T B_n$ based on a combinatorial argument.

Let us first fix what we are counting.

Consider a positive integer lattice. Starting at a point $(m,n)$ the allowed options to move are either horizontal towards right to $(m+1,n)$ (or) vertical towards top to $(m,n+1)$ (or) diagonal towards right top $(m+1,n+1)$.

The cost for a horizontal movement is $a$, the cost for a vertical movement is $b$ and the cost for a diagonal movement is $c$. Further, the total cost of a path is the product of the costs of individual movements i.e. if a path takes $x$ horizontal right movement, $y$ vertical top movements and $z$ diagonal right top movements, the total cost of this path is $a^xb^yc^z$. Note that by this the cost to stay where you are is $1$.

Claim: $M(i,j)$ gives the sum of the costs over all paths to move from $(1,1)$ to $(i,j)$.

Proof: This immediately follows from the recursive definition of $M(i,j)$. The last step before hitting $(i,j)$, we will be either at $(i-1,j)$ (or) $(i-1,j-1)$ (or) $(i,j-1)$. Hence, the total cost over all paths from $(1,1)$ to $(i,j)$ is $aM(i-1,j) + bM(i,j-1) + cM(i-1,j-1)$.

So all we need to prove is $(A_n C_n D_n C_n^T B_n) (i,j)$ also gives the cost of moving from $(1,1)$ to $(i,j)$. Note that $A_n,B_n,D_n$ are diagonal matrices which essentially means though the product has $4$ summations, they can be collapsed to just one i.e. \begin{align} (A_n C_n D_n C_n^T B_n) (i,j) & = \sum_{k1} \sum_{k2} \sum_{k3} \sum_{k4} A_n(i,k1) C_n(k1,k2) D_n(k2,k3) C_n(k4,k3) B_n(k4,j)\\ & = \sum_{k1} \sum_{k} \sum_{k4} A_n(i,k1) C_n(k1,k) D_n(k,k) C_n(k4,k) B_n(k4,j)\\ & = \sum_{k} A_n(i,i) C_n(i,k) D_n(k,k) C_n(j,k) B_n(j,j)\\ & = \sum_{k=1}^{\min(i,j)} a^{i-1} \binom{i-1}{k-1} (x^2)^{k-1} \binom{j-1}{k-1} b^{j-1}\\ & = \sum_{k=1}^{\min(i,j)} a^{i-1} \binom{i-1}{k-1} \left( 1 + \frac{c}{ab} \right)^{k-1} \binom{j-1}{k-1} b^{j-1}\\ & = \sum_{k=1}^{\min(i,j)} a^{i-k} \binom{i-1}{k-1} \left( ab + c \right)^{k-1} \binom{j-1}{k-1} b^{j-k}\\ & = \sum_{k=1}^{\min(i,j)} \sum_{r=0}^{k-1} a^{i-k} \binom{i-1}{k-1} \binom{k-1}{r} (ab)^{k-1-r} c^r \binom{j-1}{k-1} b^{j-k}\\ & = \sum_{r=0}^{\min(i,j)-1} a^{i-1-r} b^{j-1-r} c^r \left( \sum_{k=r+1}^{\min(i,j)} \binom{i-1}{k-1} \binom{k-1}{r} \binom{j-1}{k-1} \right) \end{align} All we need to prove now is that $\displaystyle \left( \sum_{k=r+1}^{\min(i,j)} \binom{i-1}{k-1} \binom{k-1}{r} \binom{j-1}{k-1} \right)$ counts the number of paths with $r$ top right diagonal movements.

Equivalently, we need to prove that $\displaystyle \left( \sum_{t=0}^{\min(i,j)-r-1} \binom{i-1}{t+r} \binom{t+r}{r} \binom{j-1}{t+r} \right)$ counts the number of paths with $r$ top right diagonal movements.

Counting the number of paths from $(1,1)$ to $(i,j)$ with $r$ top right diagonal movements. If we denote the right horizontal movement by $h$, top vertical movement by $v$ and top right diagonal movement by $d$, then the path we are interested in contains $(i-1-r)$ $h$'s, $(j-1-r)$ $v$'s and $(r)$ $d$'s. Any arrangement of this gives us the desired path. Hence, the total number of paths from $(1,1)$ to $(i,j)$ with $r$ top right diagonal movements is $$\frac{(i+j-r-2)!}{(i-1-r)!(j-1-r)!r!}.$$

Hence all we need to prove is that $$\displaystyle \sum_{t=0}^{\min(i,j)-r-1} \binom{i-1}{t+r} \binom{t+r}{r} \binom{j-1}{t+r} = \frac{(i+j-r-2)!}{(i-1-r)!(j-1-r)!r!}$$

Claim:$\displaystyle \sum_{t=0}^{\min(i,j)-r-1} \binom{i-1}{t+r} \binom{t+r}{r} \binom{j-1}{t+r} = \frac{(i+j-r-2)!}{(i-1-r)!(j-1-r)!r!}$

First note that $\displaystyle \binom{i-1}{t+r} \binom{t+r}{r} = \binom{i-1}{r} \binom{i-1-r}{t}$.

(Can be proved algebraically or through a combinatorial argument of choosing $3$ sets of things in different orders.)

Hence, we get that $$\displaystyle \sum_{t=0}^{\min(i,j)-r-1} \binom{i-1}{t+r} \binom{t+r}{r} \binom{j-1}{t+r} = \binom{i-1}{r} \sum_{t=0}^{\min(i,j)-r-1} \binom{i-1-r}{t} \binom{j-1}{t+r}$$

Now $\displaystyle \sum_{t=0}^{\min(i,j)-r-1} \binom{i-1-r}{t} \binom{j-1}{t+r} = \sum_{t=0}^{\min(i,j)-r-1} \binom{i-1-r}{t} \binom{j-1}{j-1-t+r} = \binom{i+j-r-2}{j-r-1}$

In the first step above, the first step is nothing but $\binom{n}{r} = \binom{n}{n-r}$ while the second step splits things into two different sets and does the same counting.

Hence, we get that $$\displaystyle \sum_{t=0}^{\min(i,j)-r-1} \binom{i-1}{t+r} \binom{t+r}{r} \binom{j-1}{t+r} = \binom{i-1}{r} \binom{i+j-r-2}{j-r-1}$$

But $\displaystyle \binom{i-1}{r} \binom{i+j-r-2}{j-r-1} = \frac{(i+j-r-2)!}{(i-1-r)!(j-1-r)!r!}$ which can be seen through direct algebra or again a combinatorial argument which relies on picking things in different order.

Hence, we get that $\displaystyle \sum_{t=0}^{\min(i,j)-r-1} \binom{i-1}{t+r} \binom{t+r}{r} \binom{j-1}{t+r} = \frac{(i+j-r-2)!}{(i-1-r)!(j-1-r)!r!}$.

This completes the combinatorial proof.

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@MikeSpivey: Correct. I used the usual LU decomposition to identify the underlying pattern and have not proved it rigorously yet. I think I have a combinatorial argument, which is the same as Strang's to get to decomposition in terms of $A_n,B_n,C_n$ and $D_n$, with diagonals included. If we take $x$ horizontal, $y$ vertical and $z$ diagonal paths in a square, the cost of the path is given by $a^xb^yc^z$. $M_n(i,j)$ counts the total cost across all paths. This $(i,j)$ term can also be expressed in terms of coefficients of the matrices $A_n,B_n,C_n$ and $D_n$. –  user17762 Nov 28 '11 at 17:08
@MikeSpivey: Getting the LU directly by a combinatorial argument seems to be messy since the coefficients involve $\sqrt{}$'s which is hard to interpret for me in a combinatorial sense. Also, the term $\frac{c}{ab}$ appears a path horizontal and vertical can be replaced by a diagonal path and hence the cost $ab$ gets replaced by $c$. Anyway I will write out all these in the answers and hope that I will be able to get a combinatorial interpretation for the product in terms of $A_n,B_n,C_n$ and $D_n$. But first I need to go to my lab. Once I am there, I will spend the rest of the day on this :). –  user17762 Nov 28 '11 at 17:12
@MikeSpivey: I think the proof is now done. The combinatorial argument is also almost done though I am not clear how to prove the last statement I have mentioned. I shall finish it later tonight or tomorrow. In the meantime, your comments are welcome. –  user17762 Nov 28 '11 at 21:19
@MikeSpivey: Hope the proof is complete now. Just as an additional generalization, I realize the problem can be further generalized by assuming the transition at every step is not constant i.e. replace $a$ by $a_i$, $b$ by $b_j$ and $c$ by $c_{ij}$. I don't see where the arguments will break down. For instance, terms like $a^k$ will be replaced by terms like $\prod_{i=1}^{k} a_i$ and similar terms. –  user17762 Nov 29 '11 at 0:27
–  user17762 Nov 29 '11 at 0:31

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 $((L_{(n)})^{-1})_{i,j}=\binom{i-1}{j-1}(-a)^{i-j}$.

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+1. And thanks for taking the time to write this up, Marc. –  Mike Spivey Dec 4 '11 at 21:18

This is not an answer but rather a comment of Sivaram's answer.
[update] I've to correct the numerical examples because I used a slightly distorted version of M. But this does not affect the basic idea of using the LDU-decompostion instead of the LU-decomposition. Finally, this is now not much more than a slight optimization of Sivaram's ansatz

If I decompose into $\small L \cdot D \cdot U$ instead, where D is diagonal, then the pattern of the construction of the factors L and U is more "simple"(contains only a resp. b) but their diagonals are 1 and so can be neglected for the determinant. So we need only the determinant of D. Then we find, that D can be described by
$\qquad \small D=[1,c+ab, (c+ab)^2, (c+ab)^3,\ldots ,(c+ab)^{n-1} ] = dV(c+ab,n)$

and the determinant for matrix-size n x n is then the product of these terms up to n-1 by an obvious simple expression. Again the found pattern in the LDU-decomposition must be proven, but the approach might possibly allow a shorter path...

[added] : The actual LDU-decomposition can be written as

$\qquad \small M = P^{\quad a} \cdot \quad dV(c+ab) \cdot (P^{\quad b} \sim )$

where P is the lower triangular Pascalmatrix, "~" means the transpose and dV(x) is the diagonalmatrix containing the consecutive powers of its argument x beginning at $\small x^0$

(remark: I also prefer this LDU often because it avoids introducing sqaures and/or squareroots, like for instance does the cholesky-decomposition for the symmetric case)

in case your software has a LDU-decomposition not available, here is some code usable for Pari/GP

 LDU(Y) = local(dim=#Y, D, MR, ML, dx); \\ Y must be square,
\\ no errorchecks in     demo-documentation
D=matrix(dim,dim); MR=matid(dim); ML=matid(dim);
for(p=1,dim,
D[p,p]=(dx=Y[p,p]);
for(c=p+1,dim,
MR[p,c]=if(dx==0,0,Y[p,c]/dx)
);
for(r=p+1,dim,
ML[r,p]=if(dx==0,0,Y[r,p]/dx)
);
for(r=p+1,dim,
for(c=p+1,dim,
Y[r,c]-=ML[r,p]*dx*MR[p,c]
));
);
return([ML,D,MR]);


The matrix L has the form [now corrected!] $\small L = P^a$

$\small \begin{array} {lllll} 1 & . & . & . & . & . \\ a & 1 & . & . & . & . \\ a^2 & 2a & 1 & . & . & . \\ a^3 & 3a^2 & 3a & 1 & . & . \\ a^4 & 4a^3 & 6a^2 & 4a & 1 & . \\ a^5 & 5a^4 & 10a^3 & 10a^2 & 5a & 1 \end{array}$

while U is simply the transposed and a replaced by b.

[update] Here is the Pari/GP code to generate the matrix M (corrected) and the call for the LDU-decomposition:

M = matrix(6,6,r,c); for(r=1,6,M[r,1]='a^(r-1)); for(c=1,6,M[1,c]='b^(r-1));
for(r=2,rows(M),for(c=2,cols(M),M[r,c]='a*M[r-1,c]+'b*M[r,c-1]+'c*M[r-1,c-1]))
M_ldu = LDU(M); L=M_ldu[1];D=diag(M_ldu[2]);U=M_ldu[3];
print (D)  \\ check the diagonal-component
%1164 = [1,
c + a*b ,
c^2 + 2*b*a*c + b^2*a^2,
... (snipped) ] ~

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Are you sure about the $a+b+c-1$ in your formula for $D[1]$? The only kind of factor that should end up in the determinant is $ab+c$, and this excludes that any diagonal factor of $M$ should have a coefficient of the form you give. –  Marc van Leeuwen Dec 4 '11 at 7:46
Marc, I've checked that formula a couple of times; I don't see any error. If I recombine L.D.U I get the original matrix. Just look at the top-left 2x2-submatrix of M, then we have the diagonal-element $\small m_{1,1}=a+b+c$ and so it is not so surprising to get a+b+c-1 in the diagonal of the LDU. I've just checked the determinant of the 2x2-matrix M using Pari/GP: $\small \begin{array} {lll} matdet([1,1;1,a+b+c]) \\ %1169 = a+b+c-1\end{array}$ –  Gottfried Helms Dec 4 '11 at 8:20
Ah, I see the source of the discrepancy now. The left column and top row are completely 1 in my interpretation of the M matrix, while in the first answers they begin with the vectors of powers of a resp b - so I've just the unit-vectors prefixed. I've read the condition in the OP "if j=1 AND i=1" as "if j=1 and if i = 1" and then as "or". With that correction I get now the expected diagonal. The principal thought of my post concerning the LDU instead of LU-decomposition still stands; however. Thanks for the criticism! –  Gottfried Helms Dec 4 '11 at 8:50

The identity follows almost immediately from the determinant formula for counting non-intersecting paths:

• on one hand, the determinant counts the number of non-intersecting paths from points $a_i=(-i,0)$ to points $b_i=(0,i)$ on the lattice with $a$ edges going right, $b$ up, and $c$ up-right (for each vertex);

• on the other hand, to choose non-intersecting paths means to choose path from each of $n(n-1)/2$ "internal" vertices to it's upper-right neighbor — which can be done in exactly $ab+c$ ways: one can go either up-then-right or diagonally (for example, when $c=0$ there is, obviously, just one "geometric" path with some weight). Upd: that's not quite true; I'll try to fix it.

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I don't see how this leads to an easy proof, but it gives a very intriguing combinatorial question. Counting non-crossing $n$-paths from the bottom of an $(n-1)\times(n-1)$ grid to its right side (the bottom-rightmost path having length $0$) with up/diagonal/right steps, by the number of diagonal steps, explain that there are $\binom{\tbinom{n}2}k$ solutions with $k$ diagonal steps. Even for $k=2$ and $k=\tbinom{n}2-2$ the solutions don't appear to correspond to $k$-subsets of some $\tbinom{n}2$-set in any obvious way. –  Marc van Leeuwen Nov 28 '11 at 13:28
For the record, this failed attempt at a proof inspired a discussion that I had with a collegue (Frédéric Bosio), and eventually he found a beautiful procedure to estabish a bijection, curently pubished as a preprint on arXiv. –  Marc van Leeuwen Feb 10 '13 at 12:51