How to understand an example in Krattenthaler's “Advanced Determinant Calculus”? In C. Krattenthaler's 1999 paper entitled “Advanced Determinant Calculus”, an example is given on p. 8 to illustrate Lemma 3 (p. 7). I don't quite understand this example, and I'm trying to wrap my head around it.
The author illustrates the aforementioned lemma by means of the computation of $$\det_{1 \leq i,j \leq n} \bigg{(} {a+b \choose a-i+j} \bigg{)} .$$
To make everything a bit more palpable, let's take $a=2, b=3,$ and $n=3$.
The first step of the 'recipe' described right above the example is to “take as many factors out of rows and/or columns of your determinant, so that all denominators are cleared”. In this case, we do so by multiplying the matrix with the product \begin{align*} \prod_{i=1}^{n} \frac{(a+b)!}{(a-i+n)!(b+i-1)!} &= \prod_{i=1}^{n} \frac{(2+3)!}{(2-i+3)(3+i-1)!} \\
&= \prod_{i=1}^{3} \frac{5!}{(5-i)!(2+i)!} \\
&= \frac{5!}{4!3!}\cdot\frac{5!}{3!4!}\cdot\frac{5!}{2!5!}. \end{align*}
So far, so good. The trouble starts with what happens inside the new matrix that is multiplied by this product. I don't understand what happens inside this matrix, of which the determinant can be calculated. Upon trying to understand it, I assumed that every term from the product corresponds to the respective rows of the matrix. So $\frac{5!}{4!3!}$ belongs to the first row, $ \frac{5!}{3!4!}$ belongs to the second one, and $ \frac{5!}{2!5!}$ corresponds to the third and last one. In order for the equality to hold, we must divide the rows inside the new matrix by their respective terms in the product. According to my calculations (which may or may not be right, please correct me if the latter is the case), the matrix becomes: $$ A = 
\begin{pmatrix} 4\cdot 3 & 3\cdot4 & 3\cdot 2 \\
3\cdot2 & 3\cdot 4 & 4\cdot3 \\
2\cdot1 & 2\cdot5 & 5\cdot4 \end{pmatrix} .$$
Now, I don't understand how this relates to the author's description of the new matrix $$B = \big{(} (a-i+n)(a-i+n-1) \dots (a-i+j+1) \cdot (b+i-j+1)(b+i-j+2) \dots (b+i-1) \big{)} ,$$ and how it arises after the product is introduced like we did above. In other words, my overall question would be: why is matrix $A$ equal to matrix $B$? Or perhaps: why are their determinants equal?
I'm surprised this should be the case, because I computed the entries that correspond to the different terms in the product inside matrix $B$ and the term-wise multiplications of the resulting matrices don't seem to be equal to matrix $A$.
To illustrate, let's consider some terms.

*

*For term $(a-i+n)$, we have the matrix $b_{a_{n}} := \begin{pmatrix} 4 & 4 & 4 \\
3 & 3 & 3 \\
2 & 2 & 2 \end{pmatrix}  $.

*For term $(a - i + n -1)$, there is the matrix $b_{a_{n-1}} := \begin{pmatrix} 3 & 3 & 3 \\
2 & 2 & 2 \\
1 & 1 & 1  \end{pmatrix} $.

*For term $(a-i+j+1)$, the corresponding matrix is $b_{a_{j+1}} := \begin{pmatrix} 3 & 4 & 5 \\
2 & 3 & 4 \\
1 & 2 & 3 \end{pmatrix} $

*For term $(i+b-j+1)$, we find the matrix $b_{b_{j-1}} :=\begin{pmatrix} 4 & 3 & 2 \\
5 & 4 & 3 \\
6 & 5 & 4 \end{pmatrix} $

*Etc.

It seems to me that the element-wise multiplication of these matrices does not equal matrix $B$.
Questions

*

*Have I made a mistake in my calculations somewhere? Or am I making other errors? If so, please let me know.

*Is the author indeed performing row-wise multiplications of the matrix in question?

*If so, are they multiplied by the terms that I indicated? Or do different rows correspond to other term-wise multiplications?

*Why does the element-wise multiplication of the matrices $b_{a_{n}} \dots b_{b_{1}} $ not equal matrix $A$ according to my computations? How -- if at all -- should I interpret the products inside matrix $B$ differently?

*Could you please explain why matrix $\det(A)$ is equal to matrix $\det(B)$ ? Or why are their determinants equal? Both in the general case, and in the specific case of my toy example?

 A: In summary: you have the correct matrix $A$, and the correct idea for how to get it, and it is indeed equal in every term to the matrix $B$, but you are being misled by confusing notation in the author's description of $B$, and everything you wrote down for the factors whose element-wise product is supposed to be $B$ is wrong.
When you take your starting matrix
$$
\begin{bmatrix}
  \frac{5!}{2!3!} & \frac{5!}{3!2!} & \frac{5!}{4!1!} \\
  \frac{5!}{1!4!} & \frac{5!}{2!3!} & \frac{5!}{3!2!} \\
  \frac{5!}{0!5!} & \frac{5!}{1!4!} & \frac{5!}{2!3!}
\end{bmatrix}
$$
and factor out $\frac{5!}{4!3!}$, $\frac{5!}{3!4!}$, and $\frac{5!}{2!5!}$ from the three rows respectively, we are indeed left with the matrix you gave, but it's more helpful to write it down as:
$$
\begin{bmatrix}
  \frac{4!3!}{2!3!} & \frac{4!3!}{3!2!} & \frac{4!3!}{4!1!} \\
  \frac{3!4!}{1!4!} & \frac{3!4!}{2!3!} & \frac{3!4!}{3!2!} \\
  \frac{2!5!}{0!5!} & \frac{2!5!}{1!4!} & \frac{2!5!}{2!3!}
\end{bmatrix} = 
\begin{bmatrix}
  \frac{4!}{2!}\cdot\frac{3!}{3!} & \frac{4!}{3!}\cdot\frac{3!}{2!} & \frac{4!}{4!} \cdot \frac{3!}{1!} \\
  \frac{3!}{1!}\cdot\frac{4!}{4!} & \frac{3!}{2!}\cdot \frac{4!}{3!} & \frac{3!}{3!} \cdot \frac{4!}{2!} \\
  \frac{2!}{0!}\cdot\frac{5!}{5!} & \frac{2!}{1!}\cdot\frac{5!}{4!} & \frac{2!}{2!} \cdot \frac{5!}{3!}
\end{bmatrix}
$$
In general, the $(i,j)$ entry was $\frac{(a+b)!}{(a-i+j)! (b-j+i)!}$ from which we factored out $\frac{(a+b)!}{(a-i+n)! (b-1+i)!}$ leaving $\frac{(a-i+n)! (b-1+i)!}{(a-i+j)! (b-j+i)!} = \frac{(a-i+n)!}{(a-i+j)!} \cdot \frac{(b-1+i)!}{(b-j+i)!}$.
We lose some information when we write each entry of this matrix down as a product of two (in general, $n-1$) numbers. In the first column, each entry is a product of two "left factors"; in the second column, it's a product of a "left factor" and a "right factor", and in the third column, it's a product of two "right factors". By "left factors", I mean factors from $\frac{(a-i+n)!}{(a-i+j)!}$ and by "right factors" I mean factors from $\frac{(b-1+i)!}{(b-j+i)!}$.
The author's description of the new matrix $B$ comes from writing
$$\frac{(a-i+n)!}{(a-i+j)!} = (a-i+n)(a-i+n-1) \dotsb (a-i+j+1)$$
and
$$\frac{(b-1+i)!}{(b-j+i)!} = (b-1+i)(b-2+i) \dotsb (b-(j-1)+i) \\= (b+i-(j-1))(b+i-(j-2))\dotsb (b+i-1).$$
This is misleading because these products vary in length depending on the column, and the $\dotsb$ style of writing them is confusing. It's especially silly to write out the products as "first factor, second factor, ..., last factor" because there's many cases when there's just zero, or one, or two factors in the product! That's what's throwing you off when you write down the matrices representing the individual factors: not all of those factors appear for every entry.
So, for example, let's look at entry $(1,2)$. It consists of

*

*The factor $4$, corresponding to the falling product $(a-i+n)(a-i+n-1)\dotsb(a-i+j+1)$ which is in fact the product starting and ending at $4$.

*The factor $3$, corresponding to the rising product $(b+i-j+1)(b+i-j+2)\dotsb (b+i-1)$ which is in fact the product starting and ending at $3$.

For entry $(1,1)$, which is also a product $4 \cdot 3$, the breakdown is

*

*The product $4 \cdot 3$, corresponding to the falling product $(a-i+n)(a-i+n-1)\dotsb(a-i+j+1)$ which multiplies the numbers from $4$ all the way down to $3$.

*The empty product, corresponding to the rising product $(b+i-j+1)(b+i-j+2)\dotsb (b+i-1)$, which multiplies the numbers from $4$ "all the way up to $3$".

It would be more accurate to write the $(i,j)$ entry of the new matrix as
$$
    \left(\prod_{k=j+1}^n (a-i+k)\right) \left(\prod_{\ell=1}^{j-1}(b+i-\ell) \right)
$$
with the understanding that the first product is $1$ (an empty product) when $j=n$, and the second product is $1$ (an empty product) when $j=1$.
