How does evaluate the determinant of the following $(2n+1)\times (2n+1)$ matrix? \begin{equation} \det A = \begin{array}{|cccccccccc|cc} 1 & -1 & 0 & \dots & 0 & 0 & 0 & \dots & 0 & 0 & & {\color{blue}{\text{row }1}}\\ -1 & 2 & -1 & \dots & 0 & 0 & 0 & \dots & 0 & 0& & {\color{blue}{\text{row }2}} \\ 0 & -1 & 2 & \dots & 0 & 0 & 0 & \dots & 0 & 0& & {\color{blue}{\text{row }3}} \\ \vdots & \vdots& \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & \vdots& \vdots & &{\color{blue}{\vdots}}\\ 0 & 0 & 0 & \dots & 2 & -1 & 0 & \dots & 0 & 0 && {\color{blue}{\text{row }j-1}}\\ 0 & 0 & 0 & \dots & -1 & 3 & -1 & \dots & 0 & 0 &{\color{blue}{\rightarrow}}& {\color{blue}{\text{row }j}}\\ 0 & 0 & 0 & \dots & 0 & -1 & 2 & \dots & 0 & 0 && {\color{blue}{\text{row }j+1}} \\ \vdots & \vdots& \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & \vdots& \vdots & &{\color{blue}{\vdots}} \\ 0 & 0 & 0 & \dots & 0 & 0 & 0 & \dots & 2 & -1 && {\color{blue}{\text{row }2n}} \\ 0 & 0 & 0 & \dots & 0 & 0 & 0 & \dots & -1 & 1 && {\color{blue}{\text{row }2n+1}}\\ \end{array} \end{equation} The coefficient $3$ is at the $j$-th row for some $2\leq j\leq 2n$. Therefore, $$ \det (L_{A_{11}}) = (-1)^n 2^{n-1}. $$ If $n+1\leq i\leq 3n+1$ we denote the resulting matrix by $A_{12}$ after deleting the $i$-th row and column of $L_A$. Then, by a similar computation, we have $$ \det (L_{A_{12}}) = (-1)^n 2^{n}. $$
Should we thus conclude that $$ \det A = (-1)^n 2^{n-1}n + (-1)^n 2^{n}(2n+1)=(-1)^n 2^{n-1}(5n+2) $$ is equal to $(-1)^n 2^{n-1}(5n+2)$ ?