From c9e1f14b6a1994967fa289eb6ca8b3af124edf0a Mon Sep 17 00:00:00 2001 From: idealist999 <2974730459@qq.com> Date: Wed, 31 Dec 2025 19:26:18 +0800 Subject: [PATCH 1/7] vault backup: 2025-12-31 19:26:18 --- 编写小组/试卷/1231线性代数考试卷（解析版）.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) 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 645fe8b..26a0a51 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 @@ -203,7 +203,7 @@ $$A^n = 6^{n-1}\begin{bmatrix}3&-1\\-9&3\end{bmatrix} $$ \end{bmatrix}_{(mn) \times (mn)}$$其中第一行有$m$个$0$.若$A^k=O$，则 $k$ 的最小值为$\underline{\qquad\qquad\qquad\qquad}.$ --- -解析： +解析：观察得每乘一次第一行少 $m$ 个0，故最少进行 $n$ 次即可 12. 设 $A, B$ 均为 $n$ 阶方阵，满足$\text{rank} \begin{bmatrix} A \\ B \end{bmatrix} = \text{rank}B,$ 且方程 $XA = B$ 有解。若 $\operatorname{rank} A = k$，则$\text{rank} \begin{bmatrix} B & O \\ A & E \end{bmatrix} =\underline{\hspace{3cm}}.$ From 7f28d161bfc2a47a1036746e6fc5e974f06fa019 Mon Sep 17 00:00:00 2001 From: idealist999 <2974730459@qq.com> Date: Wed, 31 Dec 2025 19:27:06 +0800 Subject: [PATCH 2/7] vault backup: 2025-12-31 19:27:06 --- .../试卷/1231线性代数考试卷（解析版）.md | 7 +++++++ 1 file changed, 7 insertions(+) diff --git a/编写小组/试卷/1231线性代数考试卷（解析版）.md b/编写小组/试卷/1231线性代数考试卷（解析版）.md index 26a0a51..4a15329 100644 --- a/编写小组/试卷/1231线性代数考试卷（解析版）.md +++ b/编写小组/试卷/1231线性代数考试卷（解析版）.md @@ -203,12 +203,19 @@ $$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 --- From 0c1ffd9a966f08adc669818d79ab10cd595eb1f9 Mon Sep 17 00:00:00 2001 From: idealist999 <2974730459@qq.com> Date: Wed, 31 Dec 2025 19:27:17 +0800 Subject: [PATCH 3/7] vault backup: 2025-12-31 19:27:17 --- 编写小组/试卷/1231线性代数考试卷（解析版）.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) 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 4a15329..f08d752 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 @@ -212,7 +212,7 @@ $$A^n = 6^{n-1}\begin{bmatrix}3&-1\\-9&3\end{bmatrix} $$ --- -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 +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 $\text{rank} \begin{bmatrix} B & O \\ A & E \end{bmatrix} =\text{rank} \begin{bmatrix} B & O \\ O & E \end{bmatrix}=n+k$ --- From 60a8ba6da0b91b27128986234deff64acccd7238 Mon Sep 17 00:00:00 2001 From: idealist999 <2974730459@qq.com> Date: Wed, 31 Dec 2025 19:28:05 +0800 Subject: [PATCH 4/7] vault backup: 2025-12-31 19:28:05 --- 编写小组/试卷/1231线性代数考试卷（解析版）.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) 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 f08d752..34a0edf 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 @@ -212,7 +212,7 @@ $$A^n = 6^{n-1}\begin{bmatrix}3&-1\\-9&3\end{bmatrix} $$ --- -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 +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 $\text{rank} \begin{bmatrix} B & O \\ A & E \end{bmatrix} =\text{rank} \begin{bmatrix} B & O \\ O & E \end{bmatrix}=n+k$ --- From 10b6879ece9ff901d88e116c3544b155c027c50a Mon Sep 17 00:00:00 2001 From: idealist999 <2974730459@qq.com> Date: Wed, 31 Dec 2025 19:39:44 +0800 Subject: [PATCH 5/7] vault backup: 2025-12-31 19:39:44 --- .../1231线性代数考试卷（解析版）.md | 15 ++++++++++++++- 1 file changed, 14 insertions(+), 1 deletion(-) 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 34a0edf..e875ebf 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 @@ -162,6 +162,19 @@ $$A^n = 6^{n-1}\begin{bmatrix}3&-1\\-9&3\end{bmatrix} $$ $$ a = \underline{\qquad\qquad}. $$ +--- + +【答】5. + +【解析】（方法一）依题意知 $\beta_1, \beta_2, \beta_3$ 线性相关，否则，若 $\beta_1, \beta_2, \beta_3$ 线性无关，则 $\beta_1, \beta_2, \beta_3$ 为向量空间 $R^3$ 的一组基，$\alpha_1, \alpha_2, \alpha_3$ 能由 $\beta_1, \beta_2, \beta_3$ 线性表示，矛盾。记 $B = [\beta_1, \beta_2, \beta_3]$，则 $|B| = 0$，解得 $a = 5$。 + +（方法二）令 +$$A = [\beta_1, \beta_2, \beta_3, \alpha_1, \alpha_2, \alpha_3] = \begin{bmatrix} 1 & 1 & 3 & 1 & 0 & 1 \\ 1 & 2 & 4 & 0 & 1 & 3 \\ 1 & 3 & a & 1 & 1 & 5 \end{bmatrix}$$ +对其进行初等行变换 + +$$A \to \begin{bmatrix} 1 & 1 & 3 & 1 & 0 & 1 \\ 0 & 1 & 1 & -1 & 1 & 2 \\ 0 & 2 & a-3 & 0 & 1 & 4 \end{bmatrix} \to \begin{bmatrix} 1 & 1 & 3 & 1 & 0 & 1 \\ 0 & 1 & 1 & -1 & 1 & 2 \\ 0 & 0 & a-5 & 2 & -1 & 0 \end{bmatrix} = \begin{bmatrix} \gamma_1 & \gamma_2 & \gamma_3 & \gamma_4 & \gamma_5 & \gamma_6 \end{bmatrix}$$ + +由 $\alpha_1, \alpha_2, \alpha_3$ 不能由 $\beta_1, \beta_2, \beta_3$ 线性表示可知 $\gamma_4, \gamma_5, \gamma_6$ 不能由 $\gamma_1, \gamma_2, \gamma_3$ 线性表示，从而 $a = 5$。 --- @@ -212,7 +225,7 @@ $$A^n = 6^{n-1}\begin{bmatrix}3&-1\\-9&3\end{bmatrix} $$ --- -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 +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 $\text{rank} \begin{bmatrix} B & O \\ A & E \end{bmatrix} =\text{rank} \begin{bmatrix} B & O \\ O & E \end{bmatrix}=n+k$ --- From 4f70171fc76b0b1792a0433f1d8250e157a24dcc Mon Sep 17 00:00:00 2001 From: idealist999 <2974730459@qq.com> Date: Wed, 31 Dec 2025 19:49:32 +0800 Subject: [PATCH 6/7] vault backup: 2025-12-31 19:49:32 --- ...231线性代数考试卷（解析版）.md | 38 +++++++++++++++++-- 1 file changed, 35 insertions(+), 3 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 e875ebf..2c4fd4a 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 @@ -194,10 +194,42 @@ $$A \to \begin{bmatrix} 1 & 1 & 3 & 1 & 0 & 1 \\ 0 & 1 & 1 & -1 & 1 & 2 \\ 0 & 2 \underline{\qquad\qquad\qquad\qquad}. $$ --- -解析： -一眼顶针，鉴定为： $$ - x = (1,0,0,0)^T$$ +**答**：$(1,0,0,0)^T$。 + +**解析**：由范德蒙行列式的性质可知 $|A| \neq 0$，从而线性方程组 $Ax = b$ 有唯一解。 + +又由 +$$ +\begin{bmatrix} +1 & a_1 & a_1^2 & a_1^3 \\ +1 & a_2 & a_2^2 & a_2^3 \\ +1 & a_3 & a_3^2 & a_3^3 \\ +1 & a_4 & a_4^2 & a_4^3 +\end{bmatrix} +\begin{bmatrix} +1 \\ +0 \\ +0 \\ +0 +\end{bmatrix} += +\begin{bmatrix} +1 \\ +1 \\ +1 \\ +1 +\end{bmatrix} +$$ +可知 $Ax = b$ 的解为 +$$ +\begin{bmatrix} +1 \\ +0 \\ +0 \\ +0 +\end{bmatrix} +$$ --- From eae2299e451deb33224d6fc1594302829c94d7d6 Mon Sep 17 00:00:00 2001 From: idealist999 <2974730459@qq.com> Date: Wed, 31 Dec 2025 20:02:39 +0800 Subject: [PATCH 7/7] vault backup: 2025-12-31 20:02:39 --- .../试卷/1231线性代数考试卷（解析版）.md | 8 ++++++++ 1 file changed, 8 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 2c4fd4a..bac74eb 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 @@ -125,12 +125,20 @@ $\quad T^{-1} = \begin{bmatrix} 1 & 0 \\ -5 & 1 \end{bmatrix}$是坐标变换矩 $$ \langle \alpha_1 + 2\alpha_2,\, 2\alpha_1 + \alpha_3 \rangle = \underline{\qquad\qquad}. $$ +--- + +【答】4 + +【解析】由内积的性质可知 $$ \begin{aligned} \langle \alpha_1 + 2\alpha_2, 2\alpha_1 + \alpha_3 \rangle &= \langle \alpha_1, 2\alpha_1 + \alpha_3 \rangle + \langle 2\alpha_2, 2\alpha_1 + \alpha_3 \rangle \\[1em] &= \langle \alpha_1, 2\alpha_1 \rangle + \langle \alpha_1, \alpha_3 \rangle + \langle 2\alpha_2, 2\alpha_1 \rangle + \langle 2\alpha_2, \alpha_3 \rangle \\[1em] &= 2\langle \alpha_1, \alpha_1 \rangle + \langle \alpha_1, \alpha_3 \rangle + 4\langle \alpha_2, \alpha_1 \rangle + 2\langle \alpha_2, \alpha_3 \rangle, \end{aligned} $$ 再由题意，可知 $$ \langle \alpha_1, \alpha_1 \rangle = 2,\quad \langle \alpha_1, \alpha_3 \rangle = -2,\quad \langle \alpha_2, \alpha_1 \rangle = 2,\quad \langle \alpha_2, \alpha_3 \rangle = -3. $$ 从而 $$ \langle \alpha_1 + 2\alpha_2, 2\alpha_1 + \alpha_3 \rangle = 4. $$ + +--- 8. 设2阶矩阵A=$\begin{bmatrix}3&-1\\-9&3\end{bmatrix}$，n为正整数，则$A^n=\underline{\quad\quad}$。 --- 解析： + 先计算$A^2$： $$A^2