From 7f28d161bfc2a47a1036746e6fc5e974f06fa019 Mon Sep 17 00:00:00 2001 From: idealist999 <2974730459@qq.com> Date: Wed, 31 Dec 2025 19:27:06 +0800 Subject: [PATCH] vault backup: 2025-12-31 19:27:06 --- .../试卷/1231线性代数考试卷（解析版）.md | 7 +++++++ 1 file changed, 7 insertions(+) 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 26a0a51..4a15329 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 @@ -203,12 +203,19 @@ $$A^n = 6^{n-1}\begin{bmatrix}3&-1\\-9&3\end{bmatrix} $$ \end{bmatrix}_{(mn) \times (mn)}$$其中第一行有$m$个$0$.若$A^k=O$，则 $k$ 的最小值为$\underline{\qquad\qquad\qquad\qquad}.$ --- + 解析：观察得每乘一次第一行少 $m$ 个0，故最少进行 $n$ 次即可 +--- 12. 设 $A, B$ 均为 $n$ 阶方阵，满足$\text{rank} \begin{bmatrix} A \\ B \end{bmatrix} = \text{rank}B,$ 且方程 $XA = B$ 有解。若 $\operatorname{rank} A = k$，则$\text{rank} \begin{bmatrix} B & O \\ A & E \end{bmatrix} =\underline{\hspace{3cm}}.$ +--- +又方程 $XA = B$ 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 +$\text{rank} \begin{bmatrix} B & O \\ A & E \end{bmatrix} =\text{rank} \begin{bmatrix} B & O \\ O & E \end{bmatrix}=n+k$ + +--- ## 三、解答题，共五道，共64分 ---