## 张祖锦2023年数学专业真题分类70天之第44天
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990、 3、 (10 分) 给定数域 $\displaystyle \mathbb\{P\}$ 内 $\displaystyle n$ 个数 $\displaystyle a\_1,\cdots,a\_n$, 令 $\displaystyle s\_k=a\_1^k+\cdots+a\_n^k\ (k=0,1,2,\cdots)$, 其中约定 $\displaystyle a\_i^0=1$, 计算下面行列式
\begin\{aligned\} D=\left|\begin\{array\}\{cccccccccc\}s\_0&s\_1&\cdots&s\_\{n-1\}\\\\ s\_1&s\_2&\cdots&s\_n\\\\ \vdots&\vdots&&\vdots\\\\ s\_\{n-1\}&s\_n&\cdots&s\_\{2n-2\}\end\{array\}\right|. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(长安大学2023年高等代数考研试题) [行列式 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) /
\begin\{aligned\} \mbox\{原式\}=&\left| \left(\begin\{array\}\{cccccccccccccccccccc\} 1&1&\cdots&1\\\\ x\_1&x\_2&\cdots&x\_n\\\\ \vdots&\vdots&\ddots&\vdots\\\\ x\_1^\{n-1\}&x\_2^\{n-1\}&\cdots&x\_n^\{n-1\} \end\{array\}\right)\left(\begin\{array\}\{cccccccccccccccccccc\} 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\{array\}\right)\right| =\prod\_\{1\leq i < j\leq n\}(x\_j-x\_i)^2. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
跟锦数学微信公众号. [在线资料](
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---
991、 3、 计算 $\displaystyle n$ 阶行列式 $\displaystyle \left|\begin\{array\}\{cccccccccc\}a&4&4&4&\cdots&4\\\\ 1&a&2&2&\cdots&2\\\\ 1&2&a&2&\cdots&2\\\\ 1&2&2&a&\cdots&2\\\\ \vdots&\vdots&\vdots&\vdots&&\vdots\\\\ 1&2&2&2&\cdots&a\end\{array\}\right|$. (中国矿业大学(北京)2023年高等代数考研试题) [行列式 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) /
\begin\{aligned\} \mbox\{原式\}=&2\left|\begin\{array\}\{cccccccccc\}\frac\{a\}\{2\}&2&2&2&\cdots&2\\\\ 1&a&2&2&\cdots&2\\\\ 1&2&a&2&\cdots&2\\\\ 1&2&2&a&\cdots&2\\\\ \vdots&\vdots&\vdots&\vdots&&\vdots\\\\ 1&2&2&2&\cdots&a\end\{array\}\right| =2\left|\begin\{array\}\{cccccccccc\}\frac\{a\}\{2\}&2&2&2&\cdots&2\\\\ 1-\frac\{a\}\{2\}&a-2&&&&\\\\ 1-\frac\{a\}\{2\}&&a-2&&&\\\\ 1-\frac\{a\}\{2\}&&&a-2&&\\\\ \vdots&&&&&\\\\ 1-\frac\{a\}\{2\}&&&&&a-2\end\{array\}\right|. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
第 $\displaystyle i$ 列 $\displaystyle \cdot \frac\{1\}\{2\}$ 加到第 $\displaystyle 1$ 列, $\displaystyle 2\leq i\leq n$, 得
\begin\{aligned\} \mbox\{原式\}=&2\left|\begin\{array\}\{cccccccccc\}\frac\{a\}\{2\}+n-1&2&2&2&\cdots&2\\\\ 0&a-2&&&&\\\\ 0&&a-2&&&\\\\ 0&&&a-2&&\\\\ \vdots&&&&&\\\\ 0&&&&&a-2\end\{array\}\right|\\\\ =&2\left(\frac\{a\}\{2\}+n-1\right)(a-2)^\{n-1\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
跟锦数学微信公众号. [在线资料](
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---
992、 2、 (16 分) 计算 $\displaystyle n+1$ 阶行列式
\begin\{aligned\} D\_\{n+1\}=\left|\begin\{array\}\{cccccccccc\}a&-1&0&0&\cdots&0\\\\ ax&a&-1&0&\cdots&0\\\\ ax^2&ax&a&-1&\cdots&0\\\\ \vdots&\vdots&\vdots&\vdots&&\vdots\\\\ ax^\{n-1\}&ax^\{n-2\}&ax^\{n-3\}&ax^\{n-4\}&\cdots&-1\\\\ ax^n&ax^\{n-1\}&ax^\{n-2\}&ax^\{n-3\}&\cdots&a\end\{array\}\right|. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(中南大学2023年高等代数考研试题) [行列式 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 第 $\displaystyle i$ 列 $\displaystyle \cdot(-x)$ 加到第 $\displaystyle i-1$ 列, $\displaystyle i=2,3,\cdots,n$ 得
\begin\{aligned\} D\_\{n+1\}=\left|\begin\{array\}\{cccccccccc\}a+x&-1&&&\\\\ 0&a+x&\ddots&&\\\\ 0&0&\ddots&-1&\\\\ \vdots&\vdots&&a+x&-1\\\\ 0&0&&0&a\end\{array\}\right|=a(a+x)^n. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
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---
993、 1、 (15 分) 已知行列式 $\displaystyle \left|\begin\{array\}\{cccccccccc\}1&9&2&4\\\\ 1&9&2&6\\\\ 2&0&0&1\\\\ 2&0&2&3\end\{array\}\right|$, $\displaystyle M\_\{ij\}$ 表示 $\displaystyle (ij)$ 元素的余子式. 求 $\displaystyle M\_\{21\}-2M\_\{22\}-3M\_\{23\}+4M\_\{24\}$. (中山大学2023年高等代数考研试题) [行列式 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) /
\begin\{aligned\} \mbox\{原式\}=&(-1)\cdot(-M\_\{21\})+(-2)\cdot M\_\{22\}+3\cdot(-M\_\{23\})+4\cdot M\_\{24\}\\\\ =&\left|\begin\{array\}\{cccccccccc\}1&9&2&4\\\\ -1&-2&3&4\\\\ 2&0&0&1\\\\ 2&0&2&3\end\{array\}\right|=-66. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
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---
994、 2、 (12 分) 已知行列式 $\displaystyle \left|\begin\{array\}\{cccccccccc\}1&-3&-4\\\\ 3&7&8\\\\ x&y&z\end\{array\}\right|=1$, 试求下列行列式: (1)、 $\displaystyle \left|\begin\{array\}\{cccccccccc\}1&1&1\\\\ 3&7&8\\\\ 1+x&1+y&1+z\end\{array\}\right|$; (2)、 $\displaystyle \left|\begin\{array\}\{cccccccccc\}x&2&1\\\\ y&2&-3\\\\ z&2&-4\end\{array\}\right|$; (3)、 $\displaystyle \left|\begin\{array\}\{cccccccccc\}1&-3&-4\\\\ 3-x&7-y&8-z\\\\ 2x&2y&2z\end\{array\}\right|$. (重庆大学2023年高等代数考研试题) [行列式 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / (1)、 原式 $\displaystyle =\left|\begin\{array\}\{cccccccccc\}1&1&1\\\\ 3&7&8\\\\ x&y&z\end\{array\}\right|$. 由
\begin\{aligned\} \left(\begin\{array\}\{cccccccccccccccccccc\}1&3&1\\\\ 1&7&-3\\\\ 1&8&-4\end\{array\}\right)\to\left(\begin\{array\}\{cccccccccccccccccccc\}1&0&4\\\\ 0&1&-1\\\\ 0&0&0\end\{array\}\right) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知
\begin\{aligned\} (1,-3,-4)=4(1,1,1)-(3,7,8). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
而
\begin\{aligned\} \mbox\{原式\}=&\frac\{1\}\{4\}\left|\begin\{array\}\{cccccccccc\}4&4&4\\\\ 3&7&8\\\\ x&y&z\end\{array\}\right|\\\\ =&\frac\{1\}\{4\}\left\[\left|\begin\{array\}\{cccccccccc\}1&-3&-4\\\\ 3&7&8\\\\ x&y&z\end\{array\}\right|+\left|\begin\{array\}\{cccccccccc\}3&7&8\\\\ 3&7&8\\\\ x&y&z\end\{array\}\right|\right\]=\frac\{1\}\{4\}(1+0)=\frac\{1\}\{4\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(2)、
\begin\{aligned\} \mbox\{原式\}=&\left|\begin\{array\}\{cccccccccc\}x&y&z\\\\ 2&2&2\\\\ 1&-3&-4\end\{array\}\right|=-2\left|\begin\{array\}\{cccccccccc\}1&-3&-4\\\\ 1&1&1\\\\ x&y&z\end\{array\}\right|\\\\ =&-2\left\[4 \left|\begin\{array\}\{cccccccccc\}1&1&1\\\\ 1&1&1\\\\ x&y&z\end\{array\}\right|-\left|\begin\{array\}\{cccccccccc\}3&7&8\\\\ 1&1&1\\\\ x&y&z\end\{array\}\right|\right\]=-2\left\[4\cdot 0+\frac\{1\}\{4\}\right\]=-\frac\{1\}\{2\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(3)、
\begin\{aligned\} \mbox\{原式\}=&2\left|\begin\{array\}\{cccccccccc\}1&-3&-4\\\\ 3-x&7-y&8-z\\\\ x&y&z\end\{array\}\right|=2\left|\begin\{array\}\{cccccccccc\}1&-3&-4\\\\ 3&7&8\\\\ x&y&z\end\{array\}\right|=2\cdot 1=2. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
跟锦数学微信公众号. [在线资料](
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---
995、 2、 (12 分) 设 $\displaystyle f(x)$ 是一个四阶 $\displaystyle \lambda$-矩阵的行列式, 问 $\displaystyle f(x)$ 的根如何? [题目不全, 跟锦数学微信公众号没法做哦.] (重庆师范大学2023年高等代数考研试题) [行列式 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / [题目不全, 跟锦数学微信公众号没法做哦.]跟锦数学微信公众号. [在线资料](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/
---
996、 3、 (12 分) 已知 $\displaystyle n$ 阶矩阵 $\displaystyle A,B$ 仅有第 $\displaystyle j$ 列不同, 证明:
\begin\{aligned\} |A|+|B|=\frac\{1\}\{2^\{n-1\}\}|A+B|. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(重庆师范大学2023年高等代数考研试题) [行列式 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 设
\begin\{aligned\} A=&(\gamma\_1,\cdots,\gamma\_\{j-1\},\alpha\_j, \gamma\_\{j+1\},\cdots,\gamma\_n),\\\\ B=&(\gamma\_1,\cdots,\gamma\_\{j-1\},\beta\_j, \gamma\_\{j+1\},\cdots,\gamma\_n), \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
则
\begin\{aligned\} &|A+B|=\left|2\gamma\_1,\cdots,2\gamma\_\{j-1\},\alpha\_j+\beta\_j, 2\gamma\_\{j+1\},\cdots,2\gamma\_n\right|\\\\ =&2^\{n-1\}\left|\gamma\_1,\cdots,\gamma\_\{j-1\},\alpha\_j+\beta\_j, \gamma\_\{j+1\},\cdots,\gamma\_n\right|\\\\ =&2^\{n-1\}\left\[\left|\gamma\_1,\cdots,\gamma\_\{j-1\},\alpha\_j, \gamma\_\{j+1\},\cdots,\gamma\_n\right| +\left|\gamma\_1,\cdots,\gamma\_\{j-1\},\beta\_j, \gamma\_\{j+1\},\cdots,\gamma\_n\right|\right\]\\\\ =&2^\{n-1\}(|A|+|B|). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
跟锦数学微信公众号. [在线资料](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
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---
997、 2、 (20 分) 设 $\displaystyle A=(a\_\{ij\})$ 是数域 $\displaystyle \mathbb\{F\}$ 上的 $\displaystyle n$ 阶方阵, $\displaystyle A\_k$ 是 $\displaystyle A$ 去掉第 $\displaystyle k$ 行剩下的 $\displaystyle n-1$ 行所组成的矩阵, $\displaystyle A\_\{ij\}$ 表示 $\displaystyle A$ 中元素 $\displaystyle a\_\{ij\}$ 的代数余子式. (1)、 若 $\displaystyle |A|\neq 0$, 证明: $\displaystyle (A\_\{k1\},\cdots,A\_\{kn\})^\mathrm\{T\}$ 是齐次线性方程组 $\displaystyle A\_kX=0$ 的一个基础解系. (2)、 若 $\displaystyle |A|=0$, 且元素 $\displaystyle a\_\{kl\}$ 的代数余子式 $\displaystyle A\_\{kl\}\neq 0$. 证明: $\displaystyle (A\_\{k1\},\cdots,A\_\{kn\})^\mathrm\{T\}$ 是齐次线性方程组 $\displaystyle AX=0$ 的一个基础解系. (北京工业大学2023年高等代数考研试题) [线性方程组 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / (1)、 由 $\displaystyle |A|\neq 0$ 知 $\displaystyle \mathrm\{rank\} A=n$, $\displaystyle A$ 的行向量组线性无关, $\displaystyle \mathrm\{rank\} A\_k=n-1$. 又由
\begin\{aligned\} &a\_\{i1\}A\_\{k1\}+\cdots+a\_\{in\}A\_\{kn\}=0, i\neq k\\\\ \Rightarrow&A\_k\left(\begin\{array\}\{cccccccccccccccccccc\}A\_\{k1\}\\\\\vdots\\\\A\_\{kn\}\end\{array\}\right)=\left(\begin\{array\}\{cccccccccccccccccccc\}0\\\\\vdots\\\\0\end\{array\}\right)\qquad(I) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知 $\displaystyle (A\_\{k1\},\cdots,A\_\{kn\})^\mathrm\{T\}$ 是 $\displaystyle A\_kX=0$ 的解. 注意到
\begin\{aligned\} 0\neq |A|=a\_\{k1\}A\_\{k1\}+\cdots+a\_\{kn\}A\_\{kn\} \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
蕴含 $\displaystyle (A\_\{k1\},\cdots,A\_\{kn\})^\mathrm\{T\}\neq 0$, 而是 $\displaystyle A\_kX=0$ 的一个基础解系. (2)、 由 $\displaystyle |A|=0$ 知 $\displaystyle \mathrm\{rank\} A\leq n-1$. 又由 $\displaystyle A\_\{kl\}\neq 0$ 知 $\displaystyle \mathrm\{rank\} (A^\star)\geq 1\Rightarrow \mathrm\{rank\} A\geq n-1$. 故 $\displaystyle \mathrm\{rank\} A=n-1$, $\displaystyle A\_kX=0$ 的基础解系只有一个线性无关的向量. 由 $\displaystyle (I)$ 知 $\displaystyle (A\_\{k1\},\cdots,A\_\{kn\})^\mathrm\{T\}$ 是 $\displaystyle A\_kX=0$ 的解. 又由 $\displaystyle A\_\{kl\}\neq 0$ 知 $\displaystyle (A\_\{k1\},\cdots,A\_\{kn\})^\mathrm\{T\}\neq 0$, 而是 $\displaystyle A\_kX=0$ 的一个基础解系.跟锦数学微信公众号. [在线资料](
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---
998、 2、 (15 分) 已知
\begin\{aligned\} A=\left(\begin\{array\}\{cccccccccccccccccccc\}2&3&a\\\\ 0&1&-1\\\\ 1&1&1\\\\ 3&5&1\end\{array\}\right),\quad \beta=\left(\begin\{array\}\{cccccccccccccccccccc\}4\\\\2\\\\1\\\\b\end\{array\}\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
问: $\displaystyle a,b$ 为何值时, $\displaystyle AX=\beta$ 有无穷多解, 有唯一解, 无解, 并求无穷多解时的通解. (北京科技大学2023年高等代数考研试题) [线性方程组 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 由
\begin\{aligned\} &(A,\beta)=\left(\begin\{array\}\{cccccccccccccccccccc\}2&3&a&4\\\\ 0&1&-1&2\\\\ 1&1&1&1\\\\ 3&5&1&b\end\{array\}\right)\to\left(\begin\{array\}\{cccccccccccccccccccc\}0&1&a-2&2\\\\ 0&1&-1&2\\\\ 1&1&1&1\\\\ 0&2&-2&b-3\end\{array\}\right)\\\\ \to&\left(\begin\{array\}\{cccccccccccccccccccc\}0&0&a-1&0\\\\ 0&1&-1&2\\\\ 1&0&2&-1\\\\ 0&0&0&b-7\end\{array\}\right)\to\left(\begin\{array\}\{cccccccccccccccccccc\}1&0&2&-1\\\\ 0&1&-1&2\\\\ 0&0&a-1&0\\\\ 0&0&0&b-7\end\{array\}\right) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知 (1)、 当 $\displaystyle b\neq 7$ 时, $\displaystyle AX=\beta$ 无解. (2)、 当 $\displaystyle b=7$ 时, (2-1)、 若 $\displaystyle a\neq 1$, 则 $\displaystyle |A|\neq 0$, 而 $\displaystyle AX=\beta$ 有唯一解. (2-2)、 若 $\displaystyle a=1$, 则
\begin\{aligned\} (A,\beta)\to\left(\begin\{array\}\{cccccccccccccccccccc\}1&0&2&-1\\\\ 0&1&-1&2\\\\ 0&0&0&0\end\{array\}\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
而 $\displaystyle AX=\beta$ 有无穷多解, 且取 $\displaystyle x\_3$ 为自由变量后通解为
\begin\{aligned\} k(-2,1,1)^\mathrm\{T\}+(-1,2,0)^\mathrm\{T\},\quad \forall\ k. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
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---
999、 (5)、 若
\begin\{aligned\} \alpha\_1=(1,2), \alpha\_2=(3,4), \beta\_1=(-1,3), \beta\_2=(5,7), \alpha\_iA^\mathrm\{T\} =\beta\_i\left(i=1,2\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
则 $\displaystyle A=\underline\{\ \ \ \ \ \ \ \ \ \ \}$. (北京理工大学2023年高等代数考研试题) [线性方程组 ]
[纸质资料](
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https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
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\begin\{aligned\} &\left(\begin\{array\}\{cccccccccccccccccccc\}\alpha\_1\\\\\alpha\_2\end\{array\}\right)A^\mathrm\{T\} =\left(\begin\{array\}\{cccccccccccccccccccc\}\beta\_1\\\\\beta\_2\end\{array\}\right) \Leftrightarrow A^\mathrm\{T\} =\left(\begin\{array\}\{cccccccccccccccccccc\}\alpha\_1\\\\\alpha\_2\end\{array\}\right)^\{-1\}\left(\begin\{array\}\{cccccccccccccccccccc\}\beta\_1\\\\\beta\_2\end\{array\}\right) =\left(\begin\{array\}\{cccccccccccccccccccc\}7&1\\\\ -4&1\end\{array\}\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 $\displaystyle A=\left(\begin\{array\}\{cccccccccccccccccccc\}7&-4\\\\ 1&1\end\{array\}\right)$.跟锦数学微信公众号. [在线资料](
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---
1000、 5、 已知非齐次线性方程组 $\displaystyle (I), (II)$:
\begin\{aligned\} (I): & \left\\{\begin\{array\}\{rrrrrrrrrrrrrrrr\}x\_1&+&x\_2&&&-&2x\_4&=&-6,\\\\ 4x\_1&-&x\_2&-&x\_3&-&x\_4&=&1,\\\\ 3x\_1&-&x\_2&-&x\_3&&&=&3;\end\{array\}\right.\\\\ (II): & \left\\{\begin\{array\}\{rrrrrrrrrrrrrrrr\} x\_1&+&mx\_2&-&x\_3&-&x\_4&=&-5,\\\\ &&nx\_2&-&x\_3&-&2x\_4&=&-11,\\\\ &&&&x\_3&-&2x\_4&=&-t+11.\end\{array\}\right. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(1)、 求方程组 $\displaystyle (I)$ 的通解; (2)、 当方程组 $\displaystyle (II)$ 中的参数 $\displaystyle m,n,t$ 为何值时, 方程组 $\displaystyle (I)$ 与 $\displaystyle (II)$ 同解? (北京邮电大学2023年高等代数考研试题) [线性方程组 ]
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https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
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https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / $\displaystyle (I)$ 的增广矩阵
\begin\{aligned\} \left(\begin\{array\}\{cccccccccccccccccccc\}1&1&0&-2&-6\\\\ 4&-1&-1&-1&1\\\\ 3&-1&-1&0&3\end\{array\}\right)\to\left(\begin\{array\}\{cccccccccccccccccccc\}1&0&0&-1&-2\\\\ 0&1&0&-1&-4\\\\ 0&0&1&-2&-5\end\{array\}\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
而 $\displaystyle (I)$ 的通解为
\begin\{aligned\} k(1,1,2,1)^\mathrm\{T\}+(-2,-4,-5,0)^\mathrm\{T\}, \quad \forall\ k. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
$\displaystyle (II)$ 的增广矩阵
\begin\{aligned\} \left(\begin\{array\}\{cccccccccccccccccccc\}1&m&-1&-1&-5\\\\ 0&n&-1&-2&-11\\\\ 0&0&1&-2&-t+11\end\{array\}\right)\to\left(\begin\{array\}\{cccccccccccccccccccc\}1&m&0&-3&6-t\\\\ 0&n&0&-4&-t\\\\ 0&0&1&-2&-t+11\end\{array\}\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
由 $\displaystyle (I), (II)$ 同解知
\begin\{aligned\} \left(\begin\{array\}\{cccccccccccccccccccc\}1&m&0&-3\\\\ 0&n&0&-4\\\\ 0&0&1&-2\end\{array\}\right)\left(\begin\{array\}\{cccccccccccccccccccc\}1\\\\1\\\\2\\\\1\end\{array\}\right)=\left(\begin\{array\}\{cccccccccccccccccccc\}0\\\\0\\\\0\end\{array\}\right)\Rightarrow m=2, n=4, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
\begin\{aligned\} \left(\begin\{array\}\{cccccccccccccccccccc\}1&m&0&-3\\\\ 0&n&0&-4\\\\ 0&0&1&-2\end\{array\}\right)\left(\begin\{array\}\{cccccccccccccccccccc\}-2\\\\-4\\\\-5\\\\0\end\{array\}\right)=\left(\begin\{array\}\{cccccccccccccccccccc\}6-t\\\\-t\\\\-t+11\end\{array\}\right)\Rightarrow t=16. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
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---
1001、 (3)、 当 $\displaystyle a,b$ 为何值时, 以下方程有唯一解, 无穷多解, 无解?
\begin\{aligned\} \left\\{\begin\{array\}\{rrrrrrrrrrrrrrrr\} x\_1&+&ax\_2&+&x\_3&=&3,\\\\ x\_1&+&x\_2&+&bx\_3&=&3,\\\\ x\_1&+&x\_2&+&2bx\_3&=&3. \end\{array\}\right. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(大连理工大学2023年高等代数考研试题) [线性方程组 ]
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https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 增广矩阵
\begin\{aligned\} (A,\beta)=&\left(\begin\{array\}\{cccccccccccccccccccc\}1&a&1&3\\\\ 1&1&b&3\\\\ 1&1&2b&3\end\{array\}\right)\to\left(\begin\{array\}\{cccccccccccccccccccc\}1&a&1&3\\\\ 0&1-a&b-1&0\\\\ 0&1-a&2b-1&0\end\{array\}\right)\\\\ \to&\left(\begin\{array\}\{cccccccccccccccccccc\}1&a&1&3\\\\ 0&1-a&b-1&0\\\\ 0&0&b&0\end\{array\}\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(3-1)、 若 $\displaystyle a\neq 1, b\neq 0$, 则 $\displaystyle |A|\neq 0$, 而 $\displaystyle Ax=\beta$ 有唯一解. (3-2)、 若 $\displaystyle a=1$, 则 $\displaystyle b-1,b$ 至少一个不为 $\displaystyle 0$, $\displaystyle \mathrm\{rank\} (A,\beta)=\mathrm\{rank\} A=2$, 而 $\displaystyle AX=\beta$ 有无穷多解. (3-3)、 若 $\displaystyle b=0$, 则 $\displaystyle b-1=-1\neq 0$, $\displaystyle \mathrm\{rank\} (A,\beta)=\mathrm\{rank\} A=2$, 而 $\displaystyle AX=\beta$ 有无穷多解.跟锦数学微信公众号. [在线资料](
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---
1002、 2、 证明题. (1)、 已知 $\displaystyle n$ 维向量组 $\displaystyle \alpha\_1,\cdots,\alpha\_r$ 为线性方程组 $\displaystyle AX=0$ 的一个基础解系, $\displaystyle n$ 维向量 $\displaystyle \beta\_0$ 不是 $\displaystyle AX=0$ 的解. 证明: 向量组
\begin\{aligned\} \beta,\beta+\alpha\_1,\cdots,\beta+\alpha\_r \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
线性无关. (大连理工大学2023年高等代数考研试题) [线性方程组 ]
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https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 由题设, $\displaystyle A\beta\_0\neq 0$. 于是
\begin\{aligned\} &x\beta+\sum\_i x\_i(\beta+\alpha\_i)=0\\\\ \stackrel\{A\cdot\}\{\Rightarrow\}&0=xA\beta+\sum\_i x\_iA\beta =\left(x+\sum\_i x\_i\right)A\beta \Rightarrow x+\sum\_i x\_i=0\\\\ \Rightarrow&s0=x\beta+\sum\_i x\_i(\beta+\alpha\_i)=\sum\_i x\_i\alpha\_i \Rightarrow x\_i=0\Rightarrow x=0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 $\displaystyle \beta,\beta+\alpha\_1,\cdots,\beta+\alpha\_r$ 线性无关.跟锦数学微信公众号. [在线资料](
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---
1003、 2、 计算题 (共 60 分, 每题 12 分). (1)、 当 $\displaystyle \lambda$ 为何值时, 齐次线性方程组
\begin\{aligned\} \left\\{\begin\{array\}\{rrrrrrrrrrrrrrrr\} (\lambda-2)x\_1&-&3x\_2&-&2x\_3&=&0,\\\\ -x\_1&+&(\lambda-8)x\_2&-&2x\_3&=&0,\\\\ 2x\_1&+&14x\_2&+&(\lambda+3)x\_3&=&0 \end\{array\}\right. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
有非零解? 并求出它的通解? (电子科技大学2023年高等代数考研试题) [线性方程组 ]
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https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 系数矩阵
\begin\{aligned\} A=&\left(\begin\{array\}\{cccccccccccccccccccc\}\lambda-2&-3&-2\\\\ -1&\lambda-8&-2\\\\ 2&14&\lambda+3\end\{array\}\right)\to\left(\begin\{array\}\{cccccccccccccccccccc\}1&13-10\lambda+\lambda^2&2-2\lambda\\\\ 1&8-\lambda&2\\\\ 0&2\lambda-2&\lambda-1\end\{array\}\right)\\\\ \to&\left(\begin\{array\}\{cccccccccccccccccccc\}0&(\lambda-3)^2&0\\\\ 1&8-\lambda&2\\\\ 0&2(\lambda-1)&\lambda-1\end\{array\}\right)\to\left(\begin\{array\}\{cccccccccccccccccccc\}1&8-\lambda&2\\\\ 0&(\lambda-3)^2&0\\\\ 0&2(\lambda-1)&\lambda-1\end\{array\}\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(1)、 当 $\displaystyle \lambda=3$ 时,
\begin\{aligned\} A\to\left(\begin\{array\}\{cccccccccccccccccccc\}1&5&2\\\\ 0&2&1\\\\ 0&0&0\end\{array\}\right)\to\left(\begin\{array\}\{cccccccccccccccccccc\}1&1&0\\\\ 0&2&1\\\\ 0&0&0\end\{array\}\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
而 $\displaystyle Ax=0$ 的通解为 $\displaystyle k\left(\begin\{array\}\{cccccccccccccccccccc\}-1\\\\1\\\\-2\end\{array\}\right),\forall\ k$. (2)、 若 $\displaystyle \lambda=1$, 则
\begin\{aligned\} A\to\left(\begin\{array\}\{cccccccccccccccccccc\}1&7&2\\\\ 0&1&0\\\\ 0&0&0\end\{array\}\right)\to\left(\begin\{array\}\{cccccccccccccccccccc\}1&0&2\\\\ 0&1&0\\\\ 0&0&0\end\{array\}\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
而 $\displaystyle Ax=0$ 的通解为 $\displaystyle k\left(\begin\{array\}\{cccccccccccccccccccc\}-2\\\\0\\\\1\end\{array\}\right),\forall\ k$.跟锦数学微信公众号. [在线资料](
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---
1004、 9、 证明非齐次线性方程组 $\displaystyle A\_\{s\times n\}X=\beta\_1, B\_\{l\times n\}X=\beta\_2$ 同解的充要条件是
\begin\{aligned\} \mathrm\{rank\}(A,\beta\_1)+\mathrm\{rank\}(B,\beta\_2)=\mathrm\{rank\} A+\mathrm\{rank\} B+2, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
或
\begin\{aligned\} &\mathrm\{rank\}\left(\begin\{array\}\{cccccccccccccccccccc\}A&\beta\_1\\\\ B&\beta\_2\end\{array\}\right)=\mathrm\{rank\}\left(\begin\{array\}\{cccccccccccccccccccc\}A\\\\B\end\{array\}\right)=\mathrm\{rank\} A=\mathrm\{rank\} B\\\\ =&\mathrm\{rank\}(A,\beta\_1)=\mathrm\{rank\}(B,\beta\_2). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(东北大学2023年高等代数考研试题) [线性方程组 ]
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https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / (1)、 $\displaystyle \Rightarrow$: (1-1)、 若一个方程组都无解, 则由它们同解知
\begin\{aligned\} &\mathrm\{rank\}(A,\beta\_1)=\mathrm\{rank\} A+1, \mathrm\{rank\}(B,\beta\_2)=\mathrm\{rank\} B+1\\\\ \Rightarrow&\mathrm\{rank\}(A,\beta\_1)+\mathrm\{rank\}(B,\beta\_2)=\mathrm\{rank\} A+\mathrm\{rank\} B+2. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(1-2)、 若一个方程组有解, 则由它们同解知
\begin\{aligned\} AX=\beta\_1\Leftrightarrow BX=\beta\_2\Leftrightarrow \left(\begin\{array\}\{cccccccccccccccccccc\}A\\\\B\end\{array\}\right)X=\left(\begin\{array\}\{cccccccccccccccccccc\}\beta\_1\\\\\beta\_2\end\{array\}\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
由线性方程组有解的充要条件知 $\displaystyle \mathrm\{rank\}\left(\begin\{array\}\{cccccccccccccccccccc\}A&\beta\_1\\\\ B&\beta\_2\end\{array\}\right)=\mathrm\{rank\}\left(\begin\{array\}\{cccccccccccccccccccc\}A\\\\B\end\{array\}\right)$,
\begin\{aligned\} \mathrm\{rank\} A=\mathrm\{rank\}(A,\beta\_1), \mathrm\{rank\} B=\mathrm\{rank\}(B,\beta\_2). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
又由
\begin\{aligned\} &\left\\{X\in\mathbb\{P\}^n; AX=\beta\_1\right\\}=\left\\{X\in\mathbb\{P\}^n; BX=\beta\_2\right\\}\\\\ \Rightarrow&\left\\{X\in\mathbb\{P\}^n; AX=0\right\\}=\left\\{X\in\mathbb\{P\}^n; BX=0\right\\}\\\\ \Rightarrow&\left\\{X\in\mathbb\{P\}^n; AX=0\right\\}=\left\\{X\in\mathbb\{P\}^n; \left(\begin\{array\}\{cccccccccccccccccccc\}A\\\\B\end\{array\}\right)X=0\right\\}\\\\ &=\left\\{X\in\mathbb\{P\}^n; BX=0\right\\}\\\\ \Rightarrow&\dim\left\\{X\in\mathbb\{P\}^n; AX=0\right\\}=\dim\left\\{X\in\mathbb\{P\}^n; \left(\begin\{array\}\{cccccccccccccccccccc\}A\\\\B\end\{array\}\right)X=0\right\\}\\\\ &=\dim\left\\{X\in\mathbb\{P\}^n; BX=0\right\\}\\\\ \Rightarrow&\mathrm\{rank\} A=\mathrm\{rank\}\left(\begin\{array\}\{cccccccccccccccccccc\}A\\\\B\end\{array\}\right)=\mathrm\{rank\} B \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
即知结论成立. (2)、 $\displaystyle \Leftarrow$: (2-1)、 若第一种情形成立, 则由
\begin\{aligned\} \mathrm\{rank\} A\leq \mathrm\{rank\}(A,\beta\_1)\leq \mathrm\{rank\} A+1,\\\\ \mathrm\{rank\} B\leq \mathrm\{rank\}(B,\beta\_2)\leq \mathrm\{rank\} B+1 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知 $\displaystyle \mathrm\{rank\} (A,\beta\_1)=\mathrm\{rank\} A+1, \mathrm\{rank\} (B,\beta\_2)=\mathrm\{rank\} B+1$. 而 $\displaystyle AX=\beta\_1, BX=\beta\_2$ 都无解, 而同解. (2-2)、 若第二种情形成立, 则
\begin\{aligned\} \left(\begin\{array\}\{cccccccccccccccccccc\}A\\\\ B\end\{array\}\right)X=\left(\begin\{array\}\{cccccccccccccccccccc\}\beta\_1\\\\\beta\_2\end\{array\}\right), AX=\beta\_1, BX=\beta\_2 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
都有解. 由
\begin\{aligned\} \mathrm\{rank\}\left(\begin\{array\}\{cccccccccccccccccccc\}A&\beta\_1\\\\ B&\beta\_2\end\{array\}\right)=\mathrm\{rank\}(A,\beta\_1)=\mathrm\{rank\}(B,\beta\_2) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知 $\displaystyle (B,\beta\_2)$ 的行向量组可由 $\displaystyle (A,\beta\_1)$ 的行向量组的一个极大无关组线性表出, 而可由 $\displaystyle (A,\beta\_1)$ 的行向量组线性表出, 从而 $\displaystyle AX=\beta\_1$ 的解都是 $\displaystyle BX=\beta\_2$ 的解; 同理, $\displaystyle (A,\beta\_1)$ 的行向量组可由 $\displaystyle (B,\beta\_2)$ 的行向量组的一个极大无关组线性表出, 而可由 $\displaystyle (B,\beta\_2)$ 的行向量组线性表出, 从而 $\displaystyle (B,\beta\_2)$ 的解都是 $\displaystyle AX=\beta\_1$ 的解. 故而两方程组同解.跟锦数学微信公众号. [在线资料](
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---
1005、 2、 设矩阵 $\displaystyle A=\left(\begin\{array\}\{cccccccccccccccccccc\}1&1&a\\\\ 1&a&1\\\\ a&1&1\end\{array\}\right), \beta=\left(\begin\{array\}\{cccccccccccccccccccc\}1\\\\1\\\\-2\end\{array\}\right)$, 若线性方程组 $\displaystyle AX=\beta$ 有解但不唯一. (1)、 (5 分) 求 $\displaystyle a$ 的值; (2)、 (10 分) 求一个正交矩阵 $\displaystyle Q$, 使得 $\displaystyle Q^\mathrm\{T\} AQ$ 为对角矩阵. (东北师范大学2023年高等代数与解析几何考研试题) [线性方程组 ]
[纸质资料](
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https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / (1)、
\begin\{aligned\} (A,\beta)=&\left(\begin\{array\}\{cccccccccccccccccccc\}1&1&a&1\\\\ 1&a&1&1\\\\ a&1&1&-2\end\{array\}\right)\to \left(\begin\{array\}\{cccccccccccccccccccc\}1&1&a&1\\\\ 0&a-1&1-a&0\\\\ 0&1-a&1-a^2&-2-a\end\{array\}\right) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
若 $\displaystyle a=1$, 则 $\displaystyle \mathrm\{rank\} A=1 < 2=\mathrm\{rank\}(A,\beta)$, 而 $\displaystyle Ax=\beta$ 无解, 与题设矛盾. 故 $\displaystyle a\neq 1$,
\begin\{aligned\} (A,\beta)\to\left(\begin\{array\}\{cccccccccccccccccccc\}1&0&a+1&1\\\\ 0&1&-1&0\\\\ 0&0&-(a-1)(a+2)&-(a+2)\end\{array\}\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
若 $\displaystyle a\neq -2$, 则 $\displaystyle |A|\neq 0$, $\displaystyle Ax=\beta$ 有唯一解, 与题设矛盾. 故 $\displaystyle a=-2$. (2)、 易知 $\displaystyle A$ 的特征值为 $\displaystyle 3,-3,0$. 由
\begin\{aligned\} &3E-A\to\left(\begin\{array\}\{cccccccccccccccccccc\}1&0&1\\\\ 0&1&0\\\\ 0&0&0 \end\{array\}\right), -3E-A\to\left(\begin\{array\}\{cccccccccccccccccccc\}1&0&-1\\\\ 0&1&2\\\\ 0&0&0 \end\{array\}\right),\\\\ &0 E-A\to\left(\begin\{array\}\{cccccccccccccccccccc\}1&0&-1\\\\ 0&1&-1\\\\ 0&0&0 \end\{array\}\right) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知 $\displaystyle A$ 的属于特征值 $\displaystyle 3,-3,0$ 的特征向量分别为
\begin\{aligned\} \xi\_1=\left(\begin\{array\}\{cccccccccccccccccccc\}-1\\\\0\\\\1 \end\{array\}\right), \xi\_2=\left(\begin\{array\}\{cccccccccccccccccccc\}1\\\\-2\\\\1 \end\{array\}\right), \xi\_3=\left(\begin\{array\}\{cccccccccccccccccccc\}1\\\\1\\\\1 \end\{array\}\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
将 $\displaystyle \xi\_1,\xi\_2,\xi\_3$ 标准正交化为 $\displaystyle \eta\_1,\eta\_2,\eta\_3$. 令
\begin\{aligned\} Q=(\eta\_1,\eta\_2,\eta\_3)=\left(\begin\{array\}\{cccccccccccccccccccc\} -\frac\{1\}\{\sqrt\{2\}\}&\frac\{1\}\{\sqrt\{6\}\}&\frac\{1\}\{\sqrt\{3\}\}\\\\ 0&-\frac\{2\}\{\sqrt\{6\}\}&\frac\{1\}\{\sqrt\{3\}\}\\\\ \frac\{1\}\{\sqrt\{2\}\}&\frac\{1\}\{\sqrt\{6\}\}&\frac\{1\}\{\sqrt\{3\}\} \end\{array\}\right), \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
则 $\displaystyle Q$ 正交, 且
\begin\{aligned\} Q^\mathrm\{T\} AQ=Q^\{-1\}AQ=\mathrm\{diag\}\left(3,-3,0\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
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---
1006、 1、 设 $\displaystyle A$ 是 $\displaystyle m\times n$ 矩阵, $\displaystyle \mathrm\{rank\} A=r$, 非齐次线性方程组 $\displaystyle AX=B$ 的解集合为 $\displaystyle S$. 证明: $\displaystyle S$ 中存在 $\displaystyle n-r+1$ 个线性无关的向量, 但任意 $\displaystyle n-r+2$ 个向量线性相关. (东南大学2023年高等代数考研试题) [线性方程组 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 由 $\displaystyle \mathrm\{rank\} A=r$ 知可设 $\displaystyle Ax=0$ 的基础解系为
\begin\{aligned\} \alpha\_1,\cdots,\alpha\_\{n-r\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
又设 $\displaystyle \gamma\neq 0$ 是 $\displaystyle Ax=B$ 的一个特解, 往证
\begin\{aligned\} \gamma, \alpha\_1+\gamma,\cdots, \alpha\_\{n-r\}+\gamma \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
是 $\displaystyle Ax=B$ 的 $\displaystyle n-r+1$ 个线性无关的解, 且 $\displaystyle Ax=B$ 的任一解 $\displaystyle \delta$ 都可由 $\displaystyle \gamma, \alpha\_1+\gamma,\cdots,\alpha\_\{n-r\}+\gamma$ 线性表出. 事实上, 由
\begin\{aligned\} &k\_0\gamma+\sum\_\{i=1\}^\{n-r\}k\_i(\alpha\_i+\gamma)=0\\\\ \Rightarrow&k\_0B+\sum\_\{i=1\}^\{n-r\}k\_iB=0\left(\mbox\{用 $\displaystyle A$ 作用\}\right)\\\\ \Rightarrow&k\_0+\sum\_\{i=1\}^\{n-r\}k\_i=0\left(B\neq 0\right)\\\\ \Rightarrow&\sum\_\{i=1\}^\{n-r\}k\_i\alpha\_i=0\left(\mbox\{代入本式第一行\}\right)\\\\ \Rightarrow&k\_1=\cdots=k\_\{n-r\}=0\left(\mbox\{$\alpha\_1,\cdots,\alpha\_\{n-r\}$ 是 $\displaystyle Ax=0$ 的基础解系\}\right)\\\\ \Rightarrow&k\_0=0\left(\mbox\{代入本式第一行\}\right) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知 $\displaystyle \gamma, \alpha\_1+\gamma,\cdots, \alpha\_\{n-r\}+\gamma$ 线性无关. 由
\begin\{aligned\} &\delta\mbox\{ 是 $\displaystyle Ax=B$ 的解\}\\\\ \Rightarrow&A\delta=B\Rightarrow A(\delta-\gamma)=0\\\\ \Rightarrow&\exists\ x\_i,\mathrm\{ s.t.\} \delta-\gamma=\sum\_\{i=1\}^\{n-r\}x\_i \alpha\_i\\\\ \Rightarrow&\delta=\left(1-\sum\_\{i=1\}^\{n-r\}x\_i\right)\gamma +\sum\_\{i=1\}^\{n-r\}x\_i(\alpha\_i+\gamma) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知 $\displaystyle Ax=\beta$ 的任一解 $\displaystyle \delta$ 都可由 $\displaystyle \gamma, \alpha\_1+\gamma,\cdots,\alpha\_\{n-r\}+\gamma$ 线性表出. 这就证明了方程组有 $\displaystyle n-r+1$ 个线性无关的解向量
\begin\{aligned\} \gamma, \alpha\_1+\gamma,\cdots, \alpha\_\{n-r\}+\gamma, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
但任意 $\displaystyle n-r+2$ 个解向量线性相关.跟锦数学微信公众号. [在线资料](
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---
1007、 (2)、 设齐次线性方程组 $\displaystyle \left\\{\begin\{array\}\{rrrrrrrrrrrrrrrr\}ax\_1&+&x\_2&+&x\_3&=&0,\\\\ x\_1&+&ax\_2&+&x\_3&=&0,\\\\ x\_1&+&x\_2&+&ax\_3&=&0\end\{array\}\right.$ 有非零解, 求 $\displaystyle a$. (福州大学2023年高等代数考研试题) [线性方程组 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 由题设, 方程组系数矩阵 $\displaystyle A$ 满足
\begin\{aligned\} 0=\left|\begin\{array\}\{cccccccccc\}a&1&1\\\\ 1&a&1\\\\ 1&1&a\end\{array\}\right|=(a-1)^2(a+2)\Rightarrow a=1\mbox\{或\} -2. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
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---
1008、 3、 解答题. (1)、 (12 分) 齐次线性方程组
\begin\{aligned\} \left\\{\begin\{array\}\{rrrrrrrrrrrrrrrr\}x\_1&+&x\_2&+&x\_3&=&0,\\\\ ax\_1&+&bx\_2&+&x\_3&=&0,\\\\ a^2x\_1&+&b^2x\_2&+&x\_3&=&0.\end\{array\}\right. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(1-1)、 当 $\displaystyle a,b$ 存在何种关系时, 齐次线性方程组没有非零解? (1-2)、 当 $\displaystyle a,b$ 存在何种关系时, 齐次线性方程组有非零解? 且求出全部解. (福州大学2023年高等代数考研试题) [线性方程组 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 方程组系数矩阵 $\displaystyle A$ 的行列式 $\displaystyle \det A=-(a-1)(a-b)(b-1)$. (1-1)、 当 $\displaystyle a\neq 1, a\neq b, b\neq 1$ 时, $\displaystyle \det A\neq 0$, 而 $\displaystyle Ax=0$ 只有零解, 没有非零解. (1-2)、 当 $\displaystyle a=b=1$ 时, $\displaystyle A\to \left(\begin\{array\}\{cccccccccccccccccccc\}1&1&1\\\\ 0&0&0\\\\ 0&0&0\end\{array\}\right)$, 而 $\displaystyle Ax=0$ 有非零解, 且通解为
\begin\{aligned\} k(-1,1,0)^\mathrm\{T\}+l(-1,0,1)^\mathrm\{T\}, \quad \forall\ k,l. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
当 $\displaystyle a=b\neq 1$ 时,
\begin\{aligned\} A=&\left(\begin\{array\}\{cccccccccccccccccccc\}1&1&1\\\\ a&a&1\\\\ a^2&a^2&1\end\{array\}\right)\to\left(\begin\{array\}\{cccccccccccccccccccc\}1&1&1\\\\ a-1&a-1&0\\\\ a^2-1&a^2-1&0\end\{array\}\right) \to\left(\begin\{array\}\{cccccccccccccccccccc\}0&0&1\\\\ 1&1&0\\\\ 0&0&0\end\{array\}\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
而 $\displaystyle Ax=0$ 有非零解, 且通解为
\begin\{aligned\} k(-1,1,0)^\mathrm\{T\}, \forall\ k. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
当 $\displaystyle 1=a\neq b$ 时,
\begin\{aligned\} A=\left(\begin\{array\}\{cccccccccccccccccccc\}1&1&1\\\\ 1&b&1\\\\ 1&b^2&1\end\{array\}\right)\to\left(\begin\{array\}\{cccccccccccccccccccc\}1&1&1\\\\ 0&b-1&0\\\\ 0&b^2-1&0\end\{array\}\right)\to\left(\begin\{array\}\{cccccccccccccccccccc\}1&0&1\\\\ 0&1&0\\\\ 0&0&0\end\{array\}\right), \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
而 $\displaystyle Ax=0$ 有非零解, 且通解为
\begin\{aligned\} k(-1,0,1)^\mathrm\{T\}, \quad \forall\ k. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
当 $\displaystyle a\neq b=1$ 时,
\begin\{aligned\} A=\left(\begin\{array\}\{cccccccccccccccccccc\}1&1&1\\\\ a&1&1\\\\ a^2&1&1\end\{array\}\right) \to\left(\begin\{array\}\{cccccccccccccccccccc\}1&1&1\\\\ a-1&0&0\\\\ a^2-1&0&0\end\{array\}\right)\to\left(\begin\{array\}\{cccccccccccccccccccc\}0&1&1\\\\ 1&0&0\\\\ 0&0&0\end\{array\}\right), \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
而 $\displaystyle Ax=0$ 有非零解, 且通解为
\begin\{aligned\} k(0,-1,1)^\mathrm\{T\}, \quad \forall\ k. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
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---
1009、 2、 设 $\displaystyle A$ 是 $\displaystyle n$ 阶实方阵, 证明以下两个命题等价: (1)、 齐次线性方程组 $\displaystyle Ax=0$ 与 $\displaystyle A^2x=0$ 在实 $\displaystyle n$ 维列向量空间中的解集相同. (2)、 齐次线性方程组 $\displaystyle Ax=0$ 与 $\displaystyle A^\{2022\}x=0$ 在实 $\displaystyle n$ 维列向量空间中的解集相同. [张祖锦注: 回忆的题目有问题, 反例见参考解解答. 没法做哦.] (复旦大学2023年代数(第4,6,7,8题没做)考研试题) [线性方程组 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 张祖锦注: 回忆的题目有问题, 比如 $\displaystyle A=\left(\begin\{array\}\{cccccccccccccccccccc\}0&1\\\\ 0&0\end\{array\}\right)\Rightarrow A^2=0$. 明显 $\displaystyle Ax=0$ 与 $\displaystyle A^2x=0$ 在实 $\displaystyle n$ 维列向量空间中的解集不同啊.跟锦数学微信公众号. [在线资料](
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---
1010、 (7)、 若方程组
\begin\{aligned\} \left\\{\begin\{array\}\{rrrrrrrrrrrrrrrr\} x\_1&+&x\_2&+&2x\_3&+&3x\_4&=&1,\\\\ x\_1&+&3x\_2&+&6x\_3&+&x\_4&=&3,\\\\ 3x\_1&-&x\_2&-&kx\_3&+&15x\_4&=&3,\\\\ x\_1&-&5x\_2&-&10x\_3&+&12x\_4&=&1 \end\{array\}\right. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
有唯一解, 则 $\displaystyle k\neq\underline\{\ \ \ \ \ \ \ \ \ \ \}$. (广西大学2023年高等代数考研试题) [线性方程组 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 系数矩阵为 $\displaystyle A=\left(\begin\{array\}\{cccccccccccccccccccc\}1&1&2&3\\\\ 1&3&6&1\\\\ 3&-1&-k&15\\\\ 1&-5&-10&12\end\{array\}\right)$, 而由题设, $\displaystyle 0\neq |A|=10(k-2)\Leftrightarrow k\neq 2$.跟锦数学微信公众号. [在线资料](
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---
1011、 5、 已知两组方程组如下:
\begin\{aligned\} (A): &\left\\{\begin\{array\}\{rrrrrrrrrrrrrrrr\} x\_1&+&x\_2&-&&-&2x\_4&=&-6,\\\\ 4x\_1&-&x\_2&-&x\_3&-&x\_4&=&1,\\\\ 3x\_1&-&x\_2&-&x\_3&&&=&3; \end\{array\}\right.\\\\ (B): &\left\\{\begin\{array\}\{rrrrrrrrrrrrrrrr\} x\_1&+&mx\_2&-&x\_3&-&x\_4&=&-5,\\\\ &&nx\_2&-&x\_3&+&2x\_4&=&-11,\\\\ &&&&x\_3&-&2x\_4&=&-t+1. \end\{array\}\right. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(1)、 求方程组 $\displaystyle (A)$ 的解, 用其导出组线性表示; (2)、 求 $\displaystyle m,n,t$ 的值, 使得 $\displaystyle (A)$ 与 $\displaystyle (B)$ 同解. [张祖锦算是算出来了, 但是题目有问题, 具体见参考解答.] (广西大学2023年高等代数考研试题) [线性方程组 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / (1)、 $\displaystyle (A)$ 的增广矩阵
\begin\{aligned\} (A,\beta)=\left(\begin\{array\}\{cccccccccccccccccccc\}1&1&0&-2&-6\\\\ 4&-1&-1&-1&1\\\\ 3&-1&-1&0&3\end\{array\}\right)\to \left(\begin\{array\}\{cccccccccccccccccccc\}1&0&0&-1&-2\\\\ 0&1&0&-1&-4\\\\ 0&0&1&-2&-5\end\{array\}\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故通解为 $\displaystyle k(1,1,2,1)^\mathrm\{T\}+(-2,-4,-5,0)^\mathrm\{T\}, \forall\ k$. (2)、 由两个方程组增广矩阵的秩相等知 $\displaystyle n\neq 0$, 而 $\displaystyle (B)$ 的增广矩阵
\begin\{aligned\} (B,\gamma)=\left(\begin\{array\}\{cccccccccccccccccccc\}1&m&-1&-1&-5\\\\ 0&n&-1&2&-11\\\\ 0&0&1&-2&-t+1\end\{array\}\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
由题设,
\begin\{aligned\} \gamma=B\left(\begin\{array\}\{cccccccccccccccccccc\}-2\\\\-4\\\\-5\\\\0\end\{array\}\right)\Rightarrow \left(\begin\{array\}\{cccccccccccccccccccc\}-5\\\\-11\\\\-t+1\end\{array\}\right)=\left(\begin\{array\}\{cccccccccccccccccccc\}3-4m\\\\5-4n\\\\-5\end\{array\}\right)\Rightarrow m=2, n=4, t=6. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
照这样子, 好像算出来了. 但是你会发现 $\displaystyle B$ 的通解根本不是第 1 步所求出来的. 所以题目有问题.跟锦数学微信公众号. [在线资料](
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---
1012、 (4)、 若齐次线性方程组
\begin\{aligned\} \left\\{\begin\{array\}\{rrrrrrrrrrrrrrrr\} a x\_1&+&x\_2&+&x\_3&=&0,\\\\ x\_1&+&a x\_2&+&x\_3&=&0,\\\\ x\_1&+&x\_2&+&a x\_3&=&0 \end\{array\}\right. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
的解空间维数为 $\displaystyle 1$, 则 $\displaystyle a=\underline\{\ \ \ \ \ \ \ \ \ \ \}$. (哈尔滨工程大学2023年高等代数考研试题) [线性方程组 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 由题设知系数矩阵 $\displaystyle A$ 的秩 $\displaystyle \mathrm\{rank\} A=3-1=2$. 由
\begin\{aligned\} A=&\left(\begin\{array\}\{cccccccccccccccccccc\}a&1&1\\\\ 1&a&1\\\\ 1&1&a\end\{array\}\right)\to \left(\begin\{array\}\{cccccccccccccccccccc\}0&1-a&1-a^2\\\\ 0&a-1&1-a\\\\ 1&1&a\end\{array\}\right) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知 $\displaystyle a\neq 1$, 进而
\begin\{aligned\} A\to \left(\begin\{array\}\{cccccccccccccccccccc\}0&1&a+1\\\\ 0&1&-1\\\\ 1&1&a\end\{array\}\right)\to\left(\begin\{array\}\{cccccccccccccccccccc\}0&0&a+2\\\\ 0&1&-1\\\\ 1&1&a\end\{array\}\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 $\displaystyle a=-2$.跟锦数学微信公众号. [在线资料](
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发表于 2023-3-5 09:17:17
