Finding the pattern to a general matrix Consider the following $(n+1) \times (n+1)$ matrix:

$$\begin{bmatrix} n &amp; 1 &amp; 0 &amp; 0&amp;\cdots &amp;0&amp; 0 &amp; 0 &amp; 0 \\ -n &amp; n-2 &amp; 2 &amp;0&amp; \cdots &amp;0&amp; 0 &amp; 0 &amp; 0 \\ 0 &amp; 1-n &amp; n-4 &amp;3&amp; \cdots &amp;0&amp; 0 &amp; 0 &amp; 0 \\ 0 &amp; 0 &amp; 2-n &amp;n-6&amp; \cdots &amp;0&amp; 0 &amp; 0 &amp; 0 \\ \vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \ddots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \vdots\\ 0 &amp; 0 &amp; 0 &amp; 0 &amp; \cdots &amp; 6-n &amp; n-2 &amp; 0 &amp; 0 \\ 0 &amp; 0 &amp; 0 &amp; 0 &amp; \cdots &amp;-3&amp; 4-n &amp; n-1 &amp; 0 \\ 0 &amp; 0 &amp; 0 &amp; 0 &amp; \cdots &amp; 0 &amp; -2 &amp; 2-n &amp; n \\ 0 &amp; 0 &amp; 0 &amp; 0 &amp; \cdots &amp; 0 &amp; 0 &amp; -1 &amp; -n \\\end{bmatrix}$$

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:

$\begin{bmatrix}x+ n &amp; 1 &amp; 0 &amp; 0&amp;\cdots &amp;0&amp; 0 &amp; 0 &amp; 0 \\ -n &amp;x+ n-2 &amp; 2 &amp;0&amp; \cdots &amp;0&amp; 0 &amp; 0 &amp; 0 \\ 0 &amp; 1-n &amp;x+ n-4 &amp;3&amp; \cdots &amp;0&amp; 0 &amp; 0 &amp; 0 \\ 0 &amp; 0 &amp; 2-n &amp;x+n-6&amp; \cdots &amp;0&amp; 0 &amp; 0 &amp; 0 \\ \vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \ddots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \vdots\\ 0 &amp; 0 &amp; 0 &amp; 0 &amp; \cdots &amp;x+ 6-n &amp; n-2 &amp; 0 &amp; 0 \\ 0 &amp; 0 &amp; 0 &amp; 0 &amp; \cdots &amp;-3&amp;x+ 4-n &amp; n-1 &amp; 0 \\ 0 &amp; 0 &amp; 0 &amp; 0 &amp; \cdots &amp; 0 &amp; -2 &amp;x+ 2-n &amp; n \\ 0 &amp; 0 &amp; 0 &amp; 0 &amp; \cdots &amp; 0 &amp; 0 &amp; -1 &amp;x -n \\\end{bmatrix} = x^{n+1}$
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.
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.