How to determine direct solution of determinant? How to show that the determinant of the following $(2n+1)×(2n+1)$ matrix $A$? 
  \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}
 By a direct calculation of  determinant. here $3$ 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 $\det (L_{A_{12}}) = (-1)^n 2^{n}.$
$ \det A = (-1)^n 2^{n-1}n + (-1)^n 2^{n}(2n+1)=(-1)^n 2^{n-1}(5n+2)$
equal  to $(-1)^n 2^{n-1}(5n+2)$ ?

For each $X\in\text{Mat}_{m\times m}(\mathbb{R})$, $a,b\in\mathbb{R}$, and $i,j=1,2,\ldots,m$, the notation $$Y:=(R_i\leftleftarrows  a\, R_i+b\, R_j)(X)$$
means $Y$ is obtained from $X$ by changing the $i$-th row of $X$ to $a$ times the $i$-th row of $X$ plus $b$ times the $j$-th row of $X$.  Let $m:=2n+1$.  Consider
$$A[1]:=(R_2\leftleftarrows R_2+R_1)\big(A\big)\,,$$
$$A[2]:=(R_3\leftleftarrows R_3+R_2)\big(A[1]\big)\,,$$
$$A[3]:=(R_4\leftleftarrows R_4+R_3)\big(A[2]\big)\,,$$
$$\vdots$$
$$A[j-1]:=(R_j\leftleftarrows R_j+R_{j-1})\big(A[j-2]\big)\,.$$
Then, 
$$A[j]:=\left(R_{j+1}\leftleftarrows R_{j+1}+\frac{1}{2}\,R_j\right)\big(A[j-1]\big)\,,$$
$$A[j+1]:=\left(R_{j+2}\leftleftarrows R_{j+2}+\frac{2}{3}\,R_{j+1}\right)\big(A[j]\big)\,,$$
$$A[j+2]:=\left(R_{j+3}\leftleftarrows R_{j+3}+\frac{3}{4}\,R_{j+2}\right)\big(A[j+1]\big)\,,$$
$$\vdots$$
$$A[m-1]:=\left(R_m\leftleftarrows R_m+\frac{m-j+1}{m-j+2}\,R_{m-1}\right)\big(A[m-2]\big)\,.$$
The final matrix $A[m-1]$ is upper-triangular with diagonal entries
$$\underbrace{1\,,\,\,1\,,\,\,1\,,\,\,\ldots\,,\,\,1}_{j-1\text{ ones }}\,\,,\,\,2\,,\,\,\frac{3}{2}\,,\,\,\frac{4}{3}\,,\,\,\ldots\,,\,\,\frac{m-j+1}{m-j}\,,\,\,\frac{1}{m-j+1}\,.$$
Therefore,
$$\begin{align}\det\big(A\big)&=\det\big(A[1]\big)=\det\big(A[2]\big)=\ldots=\det\big(A[m-1]\big)\\&=1^{j-1}\cdot 2\cdot \frac{3}{2}\cdot\ldots \cdot\frac{m-j+1}{m-j+2}\cdot\frac{1}{m-j+1}=1\,.\end{align}$$

Let $x\in\mathbb{C}$. If we want to evaluate
\begin{equation}
    \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 }m-1}} \\ 
       0       & 0 & 0 & \dots & 0 & 0 & 0 & \dots & -1 & {\color{red}x}&& {\color{blue}{\text{row }m}}\\
    \end{array}
  \end{equation}
instead, the answer turns out to be
$$1^{j-1}\cdot 2\cdot \frac{3}{2}\cdot\ldots \cdot\frac{m-j+1}{m-j+2}\cdot\left((x-1)+\frac{1}{m-j+1}\right)=(m-j+1)\,(x-1)+1\,.$$
