Calculating the determinant of a matrix

Question

During the past week, I have been trying to calculate the determinant of the following matrix:

Here is what I have tried so far. I replaced each row starting from the thrid with the difference of the one before and it. In this way i transformed it into an almost diagonal matrix but still if I try to expand it through the elements of the first column I always get almost diagonal minors and it doesnt work. :(

Please, help me.

Answer 1

Elaboration on the attempt so-far:

By your row operations, we end up with the matrix $$ \pmatrix{ -8 & 6 & 6 & 6 & 6 & \cdots & 6\\ -9 & 2 + n & 2 & 2 & 2& \cdots & 2\\ \hline 0 & -n & 1 \cdot 2 & 0 & 0 & \cdots & 0\\ 0 & 0 & -1 \cdot 2 & 2 \cdot 3 & 0 & \cdots & 0\\ \vdots &&&\ddots & \ddots \qquad\ddots&&\vdots\\ \\ \\ 0&&&&&&0\\ 0&\cdots&0&0&&-(n-2)(n-1) & (n-1)n } $$ which must have the same determinant.

Answer 2

how about $$ 38\ n!(n+1)! $$ ?

Note that I take $n=0$ for a $2\times 2$ matrix. For example, $n=0$ $$ \begin{pmatrix} -8&6\\-9&2\end{pmatrix} $$ with $\text{det}=38$

$n=1$ $$ \begin{pmatrix} -8&6&6\\-9&2&2\\-9&2&4\end{pmatrix} $$ with $\text{det}=76=38.2$

$n=2$ $$ \begin{pmatrix}-8&6&6&6\\-9&2&2&2\\-9&2&4&2\\-9&2&2&8\end{pmatrix} $$ with $\text{det}=456=28.2.6$

$n=3$: $$\begin{pmatrix}-8&6&6&6&6\\-9&2&2&2&2\\-9&2&4&2&2\\-9&2&2&8&2\\-9&2&2&2&14\end{pmatrix}$$, with $\text{det}=5472=38.6.24$.

etc.