I have to task to determine whether the following matrix $A_r\in\mathbb{R}^{39\times 39}$ defined as
$$ A_r := \begin{pmatrix} 2+4!+r^2 & 2r & -1 & 0 & \ldots & \ldots & 0 \\ 2^5 & 2+5!+r^2 & -2r & 1 &\ddots &\ddots & \vdots\\ 0 & 2^6 & 2+6!+r^2 & 2r & -1 &\ddots & \vdots \\ \vdots &\ddots & \ddots & \ddots & \ddots &\ddots & 0\\ \vdots & \ddots & \ddots &2^{40} & 2+40!+r^2 & 2r & -1 \\ \vdots &\ddots &\ddots & \ddots &2^{41} & 2+41!+r^2 & -2r\\ 0 &\ldots &\ldots &\ldots & 0 &2^{42} & 2+42!+r^2 \end{pmatrix} $$ Does there exist some $r\in\mathbb{R}$ for that $A_r$ has a full rank? If yes, define these $r\in\mathbb{R}$.
My ansatz was to find two vectors which are linearly dependent, but it's difficult to choose two vectors and check whether they are linearly independent or not. To determine $\det(A_r)$ is not a good choice or to find $(A_r)^{-1}$. Is it possible to find some $r\in\mathbb{R}$ for that the matrix $A_r$ has a full rank? What should I do to answer the above questions?
The matrix has full rank since it is strictly diagonally dominant. This result is known as the Levy–Desplanques theorem. https://en.wikipedia.org/wiki/Diagonally_dominant_matrix