Finding the pattern to a general matrix Consider the following $(n+1) \times (n+1)$ matrix:
$$\begin{bmatrix} n & 1 & 0 & 0&\cdots &0& 0 & 0 & 0 \\ -n & n-2 & 2 &0& \cdots &0& 0 & 0 & 0 \\ 0 & 1-n & n-4 &3& \cdots &0& 0 & 0 & 0 \\ 0 & 0 & 2-n &n-6& \cdots &0& 0 & 0 & 0 \\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \vdots\\ 0 & 0 & 0 & 0 & \cdots & 6-n & n-2 & 0 & 0 \\ 0 & 0 & 0 & 0 & \cdots &-3& 4-n & n-1 & 0 \\ 0 & 0 & 0 & 0 & \cdots & 0 & -2 & 2-n & n \\ 0 & 0 & 0 & 0 & \cdots & 0 & 0 & -1 & -n \\\end{bmatrix}$$
Show the eigenvalue is $0$ with multiplicity $n+1$.
I have tried row reducing to no avail, it is doable for any particular instance but surely there is a simpler way to find the determinant with a more connected pattern across all integers $n$.
For clarity, the problem is equivalent to showing:
$\begin{bmatrix}x+ n & 1 & 0 & 0&\cdots &0& 0 & 0 & 0 \\ -n &x+ n-2 & 2 &0& \cdots &0& 0 & 0 & 0 \\ 0 & 1-n &x+ n-4 &3& \cdots &0& 0 & 0 & 0 \\ 0 & 0 & 2-n &x+n-6& \cdots &0& 0 & 0 & 0 \\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \vdots\\ 0 & 0 & 0 & 0 & \cdots &x+ 6-n & n-2 & 0 & 0 \\ 0 & 0 & 0 & 0 & \cdots &-3&x+ 4-n & n-1 & 0 \\ 0 & 0 & 0 & 0 & \cdots & 0 & -2 &x+ 2-n & n \\ 0 & 0 & 0 & 0 & \cdots & 0 & 0 & -1 &x -n \\\end{bmatrix} = x^{n+1}$

Add row 1 to row 2
and row 2 is
[0, n-1, 2, 0, ...].
Then add row 2 to row 3
and row 3 is
[0, 0, n-2, 3, 0, ...].
Then add row 3 to row 4
and row 4 is
[0, 0, 0, n-3, 4, 0, ...].
I think that,
with a proper specification
of each row,
this can reduce the matrix
to upper triangular
with the main diagonal being
[n, n-1, n-2, ...]
and the diagonal above it
being
[1, 2, 3, ...],
and the rest of the matrix being zero.
This might be enough
to show what is wanted.
