From fb32fd9e2f07fe98420c4e61e28267aa620c4ab7 Mon Sep 17 00:00:00 2001
From: Elwood <3286545699@qq.com>
Date: Mon, 29 Dec 2025 21:43:28 +0800
Subject: [PATCH] vault backup: 2025-12-29 21:43:28

---
 ...231线性代数考试卷（解析版）.md | 84 +------------------
 1 file changed, 1 insertion(+), 83 deletions(-)

diff --git a/编写小组/试卷/1231线性代数考试卷（解析版）.md b/编写小组/试卷/1231线性代数考试卷（解析版）.md
index 9f5cde8..4cc5a95 100644
--- a/编写小组/试卷/1231线性代数考试卷（解析版）.md
+++ b/编写小组/试卷/1231线性代数考试卷（解析版）.md
@@ -202,7 +202,7 @@ k=     \underline{\qquad\qquad\qquad\qquad}.
 
 ---
 
-12. （20 分）计算 下面的两个$n$阶行列式
+13. （20 分）计算 下面的两个$n$阶行列式
 
     $$
     K_n = \begin{vmatrix}
@@ -281,89 +281,7 @@ D_n = (-1)^{n-1} \cdot 2^{n-2} \cdot (n+1)
 $$
 ---
 
-$$
-A = (\alpha_1, \alpha_2, \alpha_3) = 
-\begin{bmatrix} 
-1 & 2 & 1 \\ 
-0 & 1 & 1 \\ 
--1 & 1 & 1 
-\end{bmatrix},
-$$
-
-$$
-B = (\beta_1, \beta_2, \beta_3) = 
-\begin{bmatrix} 
-0 & -1 & 0 \\ 
-1 &  1 & 2 \\ 
-1 &  0 & 1 
-\end{bmatrix}.
-$$
-
-设 $u$ 在基 $\alpha_1, \alpha_2, \alpha_3$ 下的坐标为 $x = (1, 2, -3)^T$，在基 $\beta_1, \beta_2, \beta_3$ 下的坐标为 $y$，则
-
-$$
-u = (\alpha_1, \alpha_2, \alpha_3) x = (\beta_1, \beta_2, \beta_3) y,
-$$
-
-即
-
-$$
-Ax = By.
-$$
-
-因为 $B$ 可逆，所以
-
-$$
-y = B^{-1} A x.
-$$
-
-用增广矩阵求解 $y$：
-
-$$
-(B, Ax) = 
-\begin{bmatrix}
-0 & -1 & 0 & \vert & 2 \\
-1 &  1 & 2 & \vert & -1 \\
-1 &  0 & 1 & \vert & -2
-\end{bmatrix}
-$$
 
-作行初等变换：
-
-$$
-\begin{aligned}
-&\rightarrow
-\begin{bmatrix}
-1 & 0 & 1 & \vert & -2 \\
-0 & 1 & 1 & \vert & 1 \\
-0 & -1 & 0 & \vert & 2
-\end{bmatrix} \\
-&\rightarrow
-\begin{bmatrix}
-1 & 0 & 1 & \vert & -2 \\
-0 & 1 & 1 & \vert & 1 \\
-0 & 0 & 1 & \vert & 3
-\end{bmatrix} \\
-&\rightarrow
-\begin{bmatrix}
-1 & 0 & 0 & \vert & -5 \\
-0 & 1 & 0 & \vert & -2 \\
-0 & 0 & 1 & \vert & 3
-\end{bmatrix}.
-\end{aligned}
-$$
-
-因此向量
-
-$$
-u = \alpha_1 + 2\alpha_2 - 3\alpha_3
-$$
-
-在基 $\beta_1, \beta_2, \beta_3$ 下的坐标为
-
-$$
-y = (-5, -2, 3)^T.
-$$
 
 ---
 13. 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$\alpha=\begin{bmatrix}0\\2\\3\end{bmatrix},\beta=\begin{bmatrix}1\\0\\-1\end{bmatrix}$.