修改了文件：

编写小组/试卷/1231线性代数考试卷（解析版）.md

@@ -280,8 +280,109 @@ D_n = (-1)^{n-1} \cdot 2^{n-2} \cdot (n+1)
 
 $$
 ---
+（2）加边
 
 
+$$
+\tilde{D} = 
+\begin{vmatrix}
+1 & 0 & 0 & \cdots & 0 \\
+1 & 1 + x_1 & 1 + x_1^2 & \cdots & 1 + x_1^n \\
+1 & 1 + x_2 & 1 + x_2^2 & \cdots & 1 + x_2^n \\
+\vdots & \vdots & \vdots & \ddots & \vdots \\
+1 & 1 + x_n & 1 + x_n^2 & \cdots & 1 + x_n^n
+\end{vmatrix}
+
+$$
+
+$$
+\tilde{D} = 
+\begin{vmatrix}
+1 & -1 & -1 & \cdots & -1 \\
+1 & x_1 & x_1^2 & \cdots & x_1^n \\
+1 & x_2 & x_2^2 & \cdots & x_2^n \\
+\vdots & \vdots & \vdots & \ddots & \vdots \\
+1 & x_n & x_n^2 & \cdots & x_n^n
+\end{vmatrix}
+$$
+
+将第一行拆为 $(2,0,0,\dots,0)$ 与 $(-1,-1,\dots,-1)$ 之和：
+
+$$
+\tilde{D} = 
+\begin{vmatrix}
+2 & 0 & 0 & \cdots & 0 \\
+1 & x_1 & x_1^2 & \cdots & x_1^n \\
+1 & x_2 & x_2^2 & \cdots & x_2^n \\
+\vdots & \vdots & \vdots & \ddots & \vdots \\
+1 & x_n & x_n^2 & \cdots & x_n^n
+\end{vmatrix}
++
+\begin{vmatrix}
+-1 & -1 & -1 & \cdots & -1 \\
+1 & x_1 & x_1^2 & \cdots & x_1^n \\
+1 & x_2 & x_2^2 & \cdots & x_2^n \\
+\vdots & \vdots & \vdots & \ddots & \vdots \\
+1 & x_n & x_n^2 & \cdots & x_n^n
+\end{vmatrix}
+$$
+
+令左边为 $A$，右边为 $B$。
+
+计算 $A$，按第一行展开：
+
+$$
+A = 2 \cdot 
+\begin{vmatrix}
+x_1 & x_1^2 & \cdots & x_1^n \\
+x_2 & x_2^2 & \cdots & x_2^n \\
+\vdots & \vdots & \ddots & \vdots \\
+x_n & x_n^2 & \cdots & x_n^n
+\end{vmatrix}
+= 2 \cdot \left( \prod_{i=1}^{n} x_i \right) \cdot 
+\begin{vmatrix}
+1 & x_1 & \cdots & x_1^{n-1} \\
+1 & x_2 & \cdots & x_2^{n-1} \\
+\vdots & \vdots & \ddots & \vdots \\
+1 & x_n & \cdots & x_n^{n-1}
+\end{vmatrix}
+$$
+
+右边为范德蒙德行列式：
+
+$$
+A = 2 \prod_{i=1}^{n} x_i \cdot \prod_{1 \leq i < j \leq n} (x_j - x_i)
+$$
+
+计算 $B$，提出第一行的因子 $-1$：
+
+$$
+B = (-1) \cdot 
+\begin{vmatrix}
+1 & 1 & 1 & \cdots & 1 \\
+1 & x_1 & x_1^2 & \cdots & x_1^n \\
+1 & x_2 & x_2^2 & \cdots & x_2^n \\
+\vdots & \vdots & \vdots & \ddots & \vdots \\
+1 & x_n & x_n^2 & \cdots & x_n^n
+\end{vmatrix}
+$$
+
+该行列式为 $n+1$ 阶范德蒙德行列式，变量为 $1, x_1, x_2, \dots, x_n$：
+
+$$
+B = (-1) \cdot \prod_{i=1}^{n} (x_i - 1) \cdot \prod_{1 \leq i < j \leq n} (x_j - x_i)
+$$
+
+因此：
+
+$$
+\tilde{D} = A + B = \left( 2 \prod_{i=1}^{n} x_i - \prod_{i=1}^{n} (x_i - 1) \right) \cdot \prod_{1 \leq i < j \leq n} (x_j - x_i)
+$$
+
+$$
+\boxed{\tilde{D} = \left(2\prod\limits_{i=1}^{n}x_i - \prod\limits_{i=1}^{n}(x_i-1)\right) \prod\limits_{1\leq i<j\leq n}(x_j-x_i)}
+$$
+
 
 ---
 14. 设$A=\begin{bmatrix}1 & -1 & 0 & -1 \\ 1 & 1 & 0 & 3 \\ 2 & 1 & 2 & 6\end{bmatrix},B=\begin{bmatrix}1 & 0 & 1 & 2 \\ 1 & -1 & a & a-1 \\ 2 & -3 & 2 & -2\end{bmatrix}$,向量$\alpha=\begin{bmatrix}0\\2\\3\end{bmatrix},\beta=\begin{bmatrix}1\\0\\-1\end{bmatrix}$.