Determinants question with factorials

Question

$$A= \begin{pmatrix} n! & (n+1)! & (n+2)!\\ (n+1)! & (n+2)! & (n+3)! \\ (n+2)! & (n+3)! & (n+4)! \end{pmatrix}$$

And $D=\det(A)$, We have to prove that $D/(n!)^3 - 4$ is divisible by $n$.

I did it by simplifying the determinant using some row and column operations and finally ended up in getting a cubic polynomial in $n$. Then I subtracted 4 from the polynomial and got the result.

Is there any other way I could prove the result quicker and in a better manner?

Answer 1

Factorizing the matrix might help a bit.

$$A= \begin{pmatrix} n! & (n+1)! & (n+2)!\\ (n+1)! & (n+2)! & (n+3)! \\ (n+2)! & (n+3)! & (n+4)! \end{pmatrix}=n!\underbrace{\begin{pmatrix} 1 & (n+1) &(n+1)^{\overline{2}}\\ (n+1) & (n+1)^{\overline{2}} & (n+1)^{\overline{3}} \\ \;\;(n+1)^{\overline{2}} & (n+1)^{\overline{3}} & (n+1)^{\overline{4}} \end{pmatrix}}_B$$ using the Pochhammer notation for a rising factorial.

$D/(n!)^3$ is then just the determinant of $B$.

Answer 2

$$ \begin{align} D&=\det\begin{bmatrix} n!&(n+1)!&(n+2)!\\ (n+1)!&(n+2)!&(n+3)!\\ (n+2)!&(n+3)!&(n+4)! \end{bmatrix}\\ &=n!(n+1)!(n+2)! \det\small\begin{bmatrix} 1&1&1\\ n+1&n+2&n+3\\ (n+1)(n+2)&(n+2)(n+3)&(n+3)(n+4)\\ \end{bmatrix}\tag{1}\\ &=n!(n+1)!(n+2)! \det\begin{bmatrix} 1&1&1\\ 0&1&2\\ 0&2n+4&4n+10\\ \end{bmatrix}\tag{2}\\ &=n!(n+1)!(n+2)! \det\begin{bmatrix} 1&1&1\\ 0&1&2\\ 0&0&2\\ \end{bmatrix}\tag{3}\\[12pt] &=2n!(n+1)!(n+2)!\tag{4} \end{align} $$ Explanation:
$(1)$: divide columns by $n!$, $(n+1)!$, $(n+2)!$ respectively
$(2)$: subtract $n+1$ times row $1$ from row $2$
$\phantom{\text{(2):}}$ subtract $(n+1)(n+2)$ times row $1$ from row $3$
$(3)$: subtract $2n+4$ times row $2$ from row $3$
$(4)$: evaluate the determinant

Therefore, $$ \begin{align} \frac{D}{n!^3}-4 &=2(n+1)^2(n+2)-4\\[6pt] &=n\left(2n^2+8n+10\right)\tag{5} \end{align} $$

Answer 3

Following hypergeometric's answer, we have an exact expression for $D/(n!)^3$. All that remains is to show $D/(n!)^3 \equiv 4 \pmod{n}$. Since we just want the determinant of $B$ mod $n$ we can reduce the matrix entries mod $n$:

$$\begin{bmatrix} 1&1&2\\ 1&2&6\\ 2&6&24 \end{bmatrix},$$

and a routine calculation gives a determinant $4$.