How to determine direct solution of determinant?

Question

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 $(-1)^n 2^{n-1}(5n+2)$ ?

Answer 1

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, $$\begin{align}\det\big(A\big)&=\det\big(A[1]\big)=\det\big(A[2]\big)=\ldots=\det\big(A[m-1]\big)\\&=1^{j-1}\cdot 2\cdot \frac{3}{2}\cdot\ldots \cdot\frac{m-j+1}{m-j+2}\cdot\frac{1}{m-j+1}=1\,.\end{align}$$

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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 $$1^{j-1}\cdot 2\cdot \frac{3}{2}\cdot\ldots \cdot\frac{m-j+1}{m-j+2}\cdot\left((x-1)+\frac{1}{m-j+1}\right)=(m-j+1)\,(x-1)+1\,.$$

Answer 2

As first pointed out by Jean Marie in comment, for current version of matrix $A$ (in revision 4 of question), the determinant is always $1$.

For any list of $m$ numbers $\alpha = (\alpha_1,\ldots,\alpha_m)$, let $M(\alpha)$ be the $m \times m$ matrix with diagonal elements ($\alpha_1,\ldots,\alpha_m)$, having $-1$ at sub/super diagonals and $0$ elsewhere. Let $\Delta(\alpha)$ be corresponding determinant.

For any $\beta \in \mathbb{R}$, $\ell \in \mathbb{N}$, let $\beta^{\times \ell}$ be a short hand of the list of $\beta$ repeated $\ell$ times.

Matrix $A$ differs form $M(1,2^{\times(2n-1)},1)$ by a single $\lambda = 1$ at the $(j,j)$ entry for some $2 \le j \le 2n$. Treat $\lambda$ as parameter and expand $A$ against $\lambda$, we obtain

$$\det A = \Delta(1,2^{\times(2n-1)},1) + \Delta(1,2^{\times(j-2)})\Delta(2^{\times(2n-j)},1)$$

Expand $\Delta(1,2^{\times(2n-1)},1)$ against first row/column, we find

$$\Delta(1,2^{\times(2n-1)},1) = \Delta(2^{\times(2n-1)},1) - \Delta(2^{\times(2n-2)},1)$$ Let $f_k = \Delta(1,2^{\times k}) = \Delta(2^{\times k},1)$, we have

$$\det A = f_{2n-1} - f_{2n-2} + f_{j-2}f_{2n-j}$$

It is easy to see $f_0 = f_1 = 1$. For $k > 1$, if one expand $\Delta(2^{\times k},1)$ along the first row/column, we obtain the recurrence relation $$f_k = 2f_{k-1} - f_{k-2}$$ Solving this give us $f_k = 1$ for all $k \in \mathbb{N}$. As a result, $$\det A = 1 - 1 + 1\cdot 1 = 1\quad\text{ for } n \in \mathbb{Z}_{+}$$