From d1e0d0e235b63490232f89c63c5ff2aef70d2981 Mon Sep 17 00:00:00 2001
From: idealist999 <2974730459@qq.com>
Date: Wed, 31 Dec 2025 20:26:37 +0800
Subject: [PATCH] vault backup: 2025-12-31 20:26:37

---
 .../试卷/1231线性代数考试卷（解析版）.md         | 3 +--
 1 file changed, 1 insertion(+), 2 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 ee3669b..19664aa 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
@@ -220,8 +220,7 @@ $$
 
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-解析：由方程 $XA = B$ 有解可知 $\text{rank} \begin{bmatrix} A \\ B \end{bmatrix} = \text{rank}B=\text{rank}A=k$，由初等变换不改变秩得
-$\text{rank} \begin{bmatrix} B & O \\ A & E \end{bmatrix} =\text{rank} \begin{bmatrix} B & O \\ O & E \end{bmatrix}=n+k$
+解析：类似于方程 $AX = B$ 有解的充要条件是$\text{rank} \begin{bmatrix} A & B \end{bmatrix} = \text{rank}A$，由方程 $XA = B$ 有解可知 $\text{rank} \begin{bmatrix} A \\ B \end{bmatrix} = \text{rank}B=\text{rank}A=k$，由初等变换不改变秩得$$\text{rank} \begin{bmatrix} B & O \\ A & E \end{bmatrix} =\text{rank} \begin{bmatrix} B & O \\ O & E \end{bmatrix}=n+k$$
 
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 ## 三、解答题，共五道，共64分