## 张祖锦2023年数学专业真题分类70天之第46天
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1036、 3、 方程组
\begin\{aligned\} (I): \left\\{\begin\{array\}\{rrrrrrrrrrrrrrrr\}x\_1&+&&&&&x\_4&=&0,\\\\ ax\_1&+&&+&a^2x\_3&&&=&0,\\\\ &&ax\_2&+&&&a^2x\_4&=&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\}
方程组 $\displaystyle (II): x\_1+x\_2+x\_3=0$. 若方程组 $\displaystyle (I)$ 的解都是方程组 $\displaystyle (II)$ 的解. (1)、 求 $\displaystyle a$. (2)、 求 $\displaystyle (I)$ 的通解. (武汉理工大学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) / 设 $\displaystyle (I),(II)$ 的系数矩阵分别为 $\displaystyle A,B$, 则
\begin\{aligned\} AX=0\Rightarrow BX=0, \mbox\{而\}AX=0\Leftrightarrow \left(\begin\{array\}\{cccccccccccccccccccc\}A\\\\B\end\{array\}\right)X=0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
于是
\begin\{aligned\} &\left\\{X; AX=0\right\\}\Leftrightarrow \left\\{X; \left(\begin\{array\}\{cccccccccccccccccccc\}A\\\\B\end\{array\}\right)X=0\right\\}\\\\ \Rightarrow&\dim\left\\{X; AX=0\right\\}=\dim \left\\{X; \left(\begin\{array\}\{cccccccccccccccccccc\}A\\\\B\end\{array\}\right)X=0\right\\}\\\\ \Rightarrow&4-\mathrm\{rank\} A=4-\mathrm\{rank\} \left(\begin\{array\}\{cccccccccccccccccccc\}A\\\\B\end\{array\}\right) \Rightarrow \mathrm\{rank\} A=\mathrm\{rank\}\left(\begin\{array\}\{cccccccccccccccccccc\}A\\\\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\}
这表明 $\displaystyle B$ 的行向量组可由 $\displaystyle A$ 的行向量线性表出. 由
\begin\{aligned\} &\left(\begin\{array\}\{cccccccccccccccccccc\}A\\\\B\end\{array\}\right)=\left(\begin\{array\}\{cccccccccccccccccccc\}1&0&0&1\\\\ a&0&a^2&0\\\\ 0&a&0&a^2\\\\ 1&1&1&0\end\{array\}\right)\to \left(\begin\{array\}\{cccccccccccccccccccc\}1&0&0&1\\\\ a&0&a^2&0\\\\ 0&a&0&a^2\\\\ 0&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\}
若 $\displaystyle a=0$, 则 $\displaystyle \mathrm\{rank\} A=1, \mathrm\{rank\}\left(\begin\{array\}\{cccccccccccccccccccc\}A\\\\B\end\{array\}\right)=2$. 这与 $\displaystyle \mathrm\{rank\} A=\mathrm\{rank\}\left(\begin\{array\}\{cccccccccccccccccccc\}A\\\\B\end\{array\}\right)$ 矛盾. 故 $\displaystyle a\neq 0$,
\begin\{aligned\} \left(\begin\{array\}\{cccccccccccccccccccc\}A\\\\B\end\{array\}\right)\to&\left(\begin\{array\}\{cccccccccccccccccccc\}1&0&0&1\\\\ 1&0&a&0\\\\ 0&1&0&a\\\\ 0&1&1&-1\end\{array\}\right)\to\left(\begin\{array\}\{cccccccccccccccccccc\}1&0&0&1\\\\ 0&0&a&-1\\\\ 0&1&0&a\\\\ 0&)&1&-1-a\end\{array\}\right)\\\\ \to&\left(\begin\{array\}\{cccccccccccccccccccc\}1&0&0&1\\\\ 0&1&0&a\\\\ 0&0&a&-1\\\\ 0&0&0&\frac\{1\}\{a\}-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
\begin\{aligned\} 3=\mathrm\{rank\} A=\mathrm\{rank\}\left(\begin\{array\}\{cccccccccccccccccccc\}A\\\\B\end\{array\}\right)\Rightarrow \frac\{1\}\{a\}-1-a=0\Rightarrow a=\frac\{-1\pm \sqrt\{5\}\}\{2\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(1)、 当 $\displaystyle a=\frac\{-1+ \sqrt\{5\}\}\{2\}$ 时,
\begin\{aligned\} A\to\left(\begin\{array\}\{cccccccccccccccccccc\}1&0&0&1\\\\ 0&1&0&\frac\{-1+\sqrt\{5\}\}\{2\}\\\\ 0&0&1&\frac\{-1-\sqrt\{5\}\}\{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 $\displaystyle (I)$ 的通解为
\begin\{aligned\} k\left(\begin\{array\}\{cccccccccccccccccccc\}-1,\frac\{1-\sqrt\{5\}\}\{2\},\frac\{1+\sqrt\{5\}\}\{2\},1\end\{array\}\right)^\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\}
(2)、 当 $\displaystyle a=\frac\{-1- \sqrt\{5\}\}\{2\}$ 时,
\begin\{aligned\} A\to\left(\begin\{array\}\{cccccccccccccccccccc\}1&0&0&1\\\\ 0&1&0&\frac\{-1-\sqrt\{5\}\}\{2\}\\\\ 0&0&1&\frac\{-1+\sqrt\{5\}\}\{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 $\displaystyle (I)$ 的通解为
\begin\{aligned\} k\left(\begin\{array\}\{cccccccccccccccccccc\}-1,\frac\{1+\sqrt\{5\}\}\{2\},\frac\{1-\sqrt\{5\}\}\{2\},1\end\{array\}\right)^\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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---
1037、 4、 (15 分) (可能有误) 已知三阶矩阵 $\displaystyle A$ 的第一行是 $\displaystyle (a,b,c)$, $\displaystyle a,b,c$ 不全为零, 矩阵 $\displaystyle B=\left(\begin\{array\}\{cccccccccccccccccccc\}1&2&3\\\\ 2&4&6\\\\ 3&6&k\end\{array\}\right)$, 其中 $\displaystyle k$ 为常数, 且 $\displaystyle AB=0$, 求线性方程组 $\displaystyle AX=0$ 的通解. [题目不全, 跟锦数学微信公众号没法做哦.] (西安电子科技大学2023年高等代数考研试题) [线性方程组 ]
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---
1038、 5、 已知齐次线性方程组
\begin\{aligned\} \left\\{\begin\{array\}\{rrrrrrrrrrrrrrrr\} &&x\_2&+&ax\_3&+&bx\_4&=&0,\\\\ -x\_1&&&+&cx\_3&+&dx\_4&=&0,\\\\ ax\_1&+&cx\_2&&&-&ex\_4&=&0,\\\\ bx\_1&+&dx\_2&-&ex\_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\}
的通解以 $\displaystyle x\_3,x\_4$ 为自由未知量. (1)、 求系数 $\displaystyle a,b,c,d,e$ 满足的条件; (2)、 求该齐次线性方程组的基础解系. (西北大学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)、 方程组的系数矩阵
\begin\{aligned\} &A=\left(\begin\{array\}\{cccccccccccccccccccc\}0&1&a&b\\\\ -1&0&c&d\\\\ a&c&0&-e\\\\ b&d&-e&0\end\{array\}\right)\to\left(\begin\{array\}\{cccccccccccccccccccc\}0&1&a&b\\\\ -1&0&c&d\\\\ 0&c&ac&ad-e\\\\ 0&d&bc-e&bd\end\{array\}\right)\\\\ \to&\left(\begin\{array\}\{cccccccccccccccccccc\}1&0&-c&-d\\\\ 0&1&a&b\\\\ 0&c&ac&ad-e\\\\ 0&d&bc-e&bd\end\{array\}\right)\to\left(\begin\{array\}\{cccccccccccccccccccc\}1&0&-c&-d\\\\ 0&1&a&b\\\\ 0&0&0&ad-bc-e\\\\ 0&0&bc-ad-e&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\}
故当且仅当
\begin\{aligned\} ad-bc-e=0, bc-ad-e=0\Leftrightarrow ad-bc=0, e=0 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
时, 方程组的通解以 $\displaystyle x\_3,x\_4$ 为自由未知量. (2)、 由第 1 步即知方程组的基础解系为 $\displaystyle (c,-a,1,0)^\mathrm\{T\}, (d,-b,0,1)^\mathrm\{T\}$.跟锦数学微信公众号. [在线资料](
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1039、 (4)、 设 $\displaystyle A$ 为 $\displaystyle 3$ 阶方阵, $\displaystyle \beta=(9,0,0)^\mathrm\{T\}$. 若非齐次线性方程组 $\displaystyle AX=\beta$ 的通解为
\begin\{aligned\} a\_1(2,1,-2)^\mathrm\{T\}+a\_2(-2,1,2)^\mathrm\{T\}+(2,-2,1)^\mathrm\{T\}, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
其中 $\displaystyle a\_1,a\_2$ 为任意常数. 求矩阵 $\displaystyle A$. (西南交通大学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](
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https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 由题设,
\begin\{aligned\} &A\left(\begin\{array\}\{cccccccccccccccccccc\}2&-2&2\\\\ 1&1&-2\\\\ -2&2&1\end\{array\}\right)=\left(\begin\{array\}\{cccccccccccccccccccc\}0&0&9\\\\ 0&0&0\\\\ 0&0&0\end\{array\}\right)\\\\ \Rightarrow& A=\left(\begin\{array\}\{cccccccccccccccccccc\}0&0&9\\\\ 0&0&0\\\\ 0&0&0\end\{array\}\right)\left(\begin\{array\}\{cccccccccccccccccccc\}2&-2&2\\\\ 1&1&-2\\\\ -2&2&1\end\{array\}\right)^\{-1\}=\left(\begin\{array\}\{cccccccccccccccccccc\}3&0&3\\\\ 0&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\}
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---
1040、 3、 设有齐次线性方程组
\begin\{aligned\} \left(\begin\{array\}\{cccccccccccccccccccc\}1+\lambda&2&3&\cdots&n\\\\ 1&2+\lambda&3&\cdots&n\\\\ 1&2&3+\lambda&\cdots&n\\\\ \vdots&\vdots&\vdots&&\vdots\\\\ 1&2&3&\cdots&n+\lambda\end\{array\}\right)\left(\begin\{array\}\{cccccccccccccccccccc\}x\_1\\\\x\_2\\\\x\_3\\\\\vdots\\\\x\_n\end\{array\}\right)=\left(\begin\{array\}\{cccccccccccccccccccc\}0\\\\0\\\\0\\\\\vdots\\\\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\}
讨论该方程组的解的情况. (湘潭大学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/dWvpeJFKnFr0WYPoidXXMA) / 系数矩阵
\begin\{aligned\} A=\left(\begin\{array\}\{cccccccccccccccccccc\}1+\lambda&2&\cdots&n\\\\ 1&2+\lambda&\cdots&n\\\\ \vdots&\vdots&&\vdots\\\\ 1&2&\cdots&n+\lambda\end\{array\}\right)\to\left(\begin\{array\}\{cccccccccccccccccccc\}1+\lambda&2&\cdots&n\\\\ -\lambda&\lambda&\cdots&0\\\\ \vdots&\vdots&&\vdots\\\\ -\lambda&0&\cdots&\lambda\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)、 若 $\displaystyle \lambda=0$, 则 $\displaystyle Ax=0$ 有无穷多解, 且通解为
\begin\{aligned\} k\_2(-2e\_1+e\_2)+\cdots+k\_n(-ne\_1+e\_n),\quad \forall\ k\_2,\cdots,k\_n. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(2)、 若 $\displaystyle \lambda\neq 0$, 则第 $\displaystyle i$ 行 $\displaystyle \cdot \left(\frac\{i\}\{\lambda\}\right)$ 加到第 $\displaystyle 1$ 行, $\displaystyle 2\leq i\leq n$, 并将第 $\displaystyle i$ 行乘以 $\displaystyle \frac\{1\}\{\lambda\}$, $\displaystyle 2\leq i\leq n$, 得
\begin\{aligned\} A\to\left(\begin\{array\}\{cccccccccccccccccccc\}\lambda+\frac\{n(n+1)\}\{2\}&0&\cdots&0\\\\ -1&1&\cdots&0\\\\ \vdots&\vdots&&\vdots\\\\ -1&0&\cdots&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\}
(2-1)、 若 $\displaystyle \lambda=-\frac\{n(n+1)\}\{2\}$, 则 $\displaystyle Ax=0$ 的通解为 $\displaystyle k(1,\cdots,1)^\mathrm\{T\}, \forall\ k$. (2-2)、 若 $\displaystyle \lambda\neq-\frac\{n(n+1)\}\{2\}$, 则 $\displaystyle Ax=0$ 只有零解.跟锦数学微信公众号. [在线资料](
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---
1041、 5、 (10 分) 证明: 与基础解系等价的线性无关向量组也是基础解系. (新疆大学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) / 设 $\displaystyle \eta\_1,\cdots,\eta\_r$ 是齐次线性方程组 $\displaystyle Ax=0$ 的基础解系, 而 $\displaystyle \xi\_1,\cdots,\xi\_r$ 也是 $\displaystyle Ax=0$ 的基础解系, 则 $\displaystyle \eta\_1,\cdots,\eta\_r$ 与 $\displaystyle \xi\_1,\cdots,\xi\_r$ 都是线性空间 $\displaystyle \left\\{x; Ax=0\right\\}$ 的基, 而可相互线性表出, 是等价的.跟锦数学微信公众号. [在线资料](
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---
1042、 2、 (15 分) 求方程组
\begin\{aligned\} \left\\{\begin\{array\}\{rrrrrrrrrrrrrrrr\} x\_1&-&x\_2&&&&&=&2022,\\\\ &&x\_2&-&x\_3&&&=&2023,\\\\ &&&&x\_3&-&x\_4&=&-2023,\\\\ -x\_1&&&&&+&x\_4&=&-2022 \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\}
的全部解. (云南大学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) / 增广矩阵
\begin\{aligned\} (A,\beta)\to\left(\begin\{array\}\{cccccccccccccccccccc\}1&-1&0&0&2022\\\\ 0&1&-1&0&2023\\\\ 0&0&1&-1&-2023\\\\ -1&0&0&1&-2022\end\{array\}\right)\to\left(\begin\{array\}\{cccccccccccccccccccc\}1&0&0&-1&2022\\\\ 0&1&0&-1&0\\\\ 0&0&1&-1&-2023\\\\ 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\}
而方程组的通解为
\begin\{aligned\} k\left(\begin\{array\}\{cccccccccccccccccccc\}1\\\\1\\\\1\\\\1\end\{array\}\right)+\left(\begin\{array\}\{cccccccccccccccccccc\}2022\\\\0\\\\-2023\\\\0\end\{array\}\right),\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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---
1043、 4、 (10 分) 给定线性方程组
\begin\{aligned\} (I)\left\\{\begin\{array\}\{rrrrrrrrrrrrrrrr\} a\_\{11\}x\_1&+&a\_\{12\}x\_2&+&\cdots&+&a\_\{1n\}x\_n&=&0,\\\\ a\_\{21\}x\_1&+&a\_\{22\}x\_2&+&\cdots&+&a\_\{2n\}x\_n&=&0,\\\\ \cdots&&\cdots&&\cdots&&\cdots&&\\\\ a\_\{n-1,1\}x\_1&+&a\_\{n-1,2\}x\_2&+&\cdots&+&a\_\{n-1,n\}x\_n&=&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\}
以 $\displaystyle M\_i$ 表示其系数矩阵划去第 $\displaystyle i$ 列后所剩 $\displaystyle n-1$ 阶方阵的行列式. 证明: (0-9)、 $\displaystyle (M\_1,-M\_2,\cdots,(-1)^\{n-1\}M\_n)^\mathrm\{T\}$ 是方程组的解; (0-10)、 若方程组系数矩阵的秩为 $\displaystyle n-1$, 则方程组的解全是
\begin\{aligned\} (M\_1,-M\_2,\cdots,(-1)^\{n-1\}M\_n)^\mathrm\{T\} \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
的倍数. (长安大学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) / (0-11)、
\begin\{aligned\} \sum\_\{j=1\}^n a\_\{ij\}(-1)^\{j-1\}M\_j =&(-1)^\{i-1\}\sum\_\{j=1\}^n (-1)^\{i+j\} a\_\{ij\}M\_j\\\\ =&(-1)^\{i-1\}\left|\begin\{array\}\{ccc\} a\_\{11\}&\cdots&a\_\{1n\}\\\\ \vdots&&\vdots\\\\ a\_\{n-1,1\}&\cdots&a\_\{n-1,n\}\\\\ a\_\{i1\}&\cdots&a\_\{in\} \end\{array\}\right| =0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(0-12)、 由 $\displaystyle \mathrm\{rank\}(A)=n-1$ 知 $\displaystyle A$ 的列秩为 $\displaystyle n-1$, 而有一个 $\displaystyle n-1$ 阶子式不等于 $\displaystyle 0$,
\begin\{aligned\} (M\_1,-M\_2,\cdots,(-1)^\{n-1\}M\_n)^\mathrm\{T\}\neq 0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
又 $\displaystyle Ax=0$ 的基础解系仅有一个线性无关的向量, 而
\begin\{aligned\} (M\_1,-M\_2,\cdots,(-1)^\{n-1\}M\_n)^\mathrm\{T\} \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
就是 $\displaystyle Ax=0$ 的一个基础解系, 方程组的解全是
\begin\{aligned\} (M\_1,-M\_2,\cdots,(-1)^\{n-1\}M\_n)^\mathrm\{T\} \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
的倍数.跟锦数学微信公众号. [在线资料](
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---
1044、 1、 填空题. (1)、 设 $\displaystyle \alpha\_1,\alpha\_2,\alpha\_3$ 是线性方程组 $\displaystyle \left\\{\begin\{array\}\{rrrrrrrrrrrrrrrr\}x\_1&+&x\_2&+&x\_3&+&x\_4&=&1,\\\\ &&&&&&&\end\{array\}\right.$ 的三个线性无关的向量, 则该线性方程组的系数矩阵的秩为 $\displaystyle \underline\{\ \ \ \ \ \ \ \ \ \ \}$. [虽然题目不全, 但还是可以做哦. 见参考解答.] (郑州大学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) / 设 $\displaystyle A\_\{s\times n\}, \beta\neq 0$, 则 $\displaystyle Ax=\beta$ 有 $\displaystyle n-\mathrm\{rank\} A+1$ 个线性无关的解向量. 故而 $\displaystyle 4-\mathrm\{rank\} A+1=3\Rightarrow \mathrm\{rank\} A=2$.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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---
1045、 1、 填空题 (每空 6 分, 共 60 分). (1)、 参数 $\displaystyle \mu=\underline\{\ \ \ \ \ \ \ \ \ \ \}$ 时, 方程组
\begin\{aligned\} \left\\{\begin\{array\}\{rrrrrrrrrrrrrrrr\} \mu x\_1&+&x\_2&+&x\_3&+&x\_4&=&0,\\\\ x\_1&+&\mu x\_2&+&x\_3&+&x\_4&=&0,\\\\ x\_1&+& x\_2&+&\mu x\_3&+&x\_4&=&0,\\\\ x\_1&+&x\_2&+&x\_3&+&\mu x\_4&=&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. (中国科学技术大学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) / 方程组有非零解 $\displaystyle \Leftrightarrow$ 系数矩阵行列式
\begin\{aligned\} 0=\left|\begin\{array\}\{cccccccccc\}\mu&1&1&1\\\\ 1&\mu&1&1\\\\ 1&1&\mu&1\\\\ 1&1&1&\mu\end\{array\}\right|=(\mu+3)(\mu-1)^3\Leftrightarrow \mu=-3\mbox\{或\} 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](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/
---
1046、 (2)、 克拉默法则仅适用于求解 $\displaystyle \underline\{\ \ \ \ \ \ \ \ \ \ \}$ 的线性方程组的解.
A. 非齐次
B. 齐次
C. 方程个数等于未知量个数且系数矩阵可逆
D. 任何有解 (中国矿业大学(北京)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) / $\displaystyle C$.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)/
---
1047、 4、 讨论非齐次线性方程组
\begin\{aligned\} \left\\{\begin\{array\}\{rrrrrrrrrrrrrrrr\} \lambda x\_1&+&\lambda x\_2&+&(2\lambda+1)x\_3&=&0,\\\\ x\_1&+&(\lambda^2+1)x\_2&+&2x\_3&=&\lambda,\\\\ x\_1&+&(2\lambda+1)x\_2&+&2x\_3&=&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\}
解的情况. (中国矿业大学(北京)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) / 增广矩阵
\begin\{aligned\} (A,\beta)=&\left(\begin\{array\}\{cccccccccccccccccccc\}\lambda&\lambda&2\lambda+1&0\\\\ 1&\lambda^2+1&2&\lambda\\\\ 1&2\lambda+1&2&2\end\{array\}\right)\to\left(\begin\{array\}\{cccccccccccccccccccc\}0&-2\lambda^2&1&-2\lambda\\\\ 0&\lambda^2-2\lambda&0&\lambda-2\\\\ 1&2\lambda+1&2&2\end\{array\}\right)\\\\ \to&\left(\begin\{array\}\{cccccccccccccccccccc\}0&-4\lambda&1&-4\\\\ 0&\lambda^2-2\lambda&0&\lambda-2\\\\ 1&2\lambda+1&2&2\end\{array\}\right)\to\left(\begin\{array\}\{cccccccccccccccccccc\}0&-4\lambda&1&-4\\\\ 0&0&\frac\{\lambda-2\}\{4\}&0\\\\ 1&2\lambda+1&2&2\end\{array\}\right)\\\\ \to&\left(\begin\{array\}\{cccccccccccccccccccc\}1&2\lambda+2&2&2\\\\ 0&-4\lambda&1&-4\\\\ 0&0&\frac\{\lambda-2\}\{4\}&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)、 若 $\displaystyle \lambda=0$, 则上式最后一行蕴含 $\displaystyle Ax=\beta$ 无解. (2)、 若 $\displaystyle \lambda\neq 0$, (2-1)、 当 $\displaystyle \lambda=2$ 时, $\displaystyle (A,\beta)\to\left(\begin\{array\}\{cccccccccccccccccccc\}1&0&\frac\{21\}\{8\}&-\frac\{1\}\{2\}\\\\ 0&1&-\frac\{1\}\{8\}&\frac\{1\}\{2\}\\\\ 0&0&0&0\end\{array\}\right)$, 而 $\displaystyle Ax=\beta$ 的通解为
\begin\{aligned\} \left(-\frac\{1\}\{2\},\frac\{1\}\{2\},0\right)^\mathrm\{T\}+k\left(-21,1,8\right)^\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\}
(2-2)、 当 $\displaystyle \lambda\neq 2$ 时, $\displaystyle |A|\neq 0$, $\displaystyle Ax=\beta$ 有唯一解, 经过计算知为 $\displaystyle \left(-\frac\{1\}\{\lambda\},\frac\{1\}\{\lambda\},0\right)^\mathrm\{T\}$.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)/
---
1048、 8、 $\displaystyle b$ 是一个非零向量, $\displaystyle \gamma\_0$ 是方程组 $\displaystyle AX=b$ 的一个解, 导出组的基础解系为 $\displaystyle \eta\_1,\cdots,\eta\_t$. 证明: 向量组
\begin\{aligned\} \gamma\_0,\gamma\_0+\eta\_1,\cdots,\gamma\_0+\eta\_t \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
是 $\displaystyle AX=b$ 解集的一个极大线性无关组. (中国矿业大学(北京)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\} &k\_0\gamma\_0+\sum\_\{i=1\}^t k\_i(\gamma\_0+\eta\_i)=0\qquad(I)\\\\ \stackrel\{A\cdot\}\{\Rightarrow\}&0=\sum\_\{i=0\}^t k\_ib\stackrel\{b\neq 0\}\{\Rightarrow\}\sum\_\{i=0\}^t k\_i=0\\\\ \stackrel\{(I)\}\{\Rightarrow\}&\sum\_\{i=1\}^t k\_i\eta\_i=0\Rightarrow k\_1=\cdots=k\_t=0 \stackrel\{(I)\}\{\Rightarrow\}k\_0=0 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知 $\displaystyle \gamma\_0,\gamma\_0+\eta\_1,\cdots,\gamma\_0+\eta\_t$ 线性无关. 再者, 对 $\displaystyle AX=b$ 的任一解 $\displaystyle X=\alpha$,
\begin\{aligned\} &A(\alpha-\gamma\_0)=b-b=0\Rightarrow \alpha-\gamma\_0=\sum\_\{i=1\}^t x\_i\eta\_i\\\\ \Rightarrow& \alpha=\left(1-\sum\_\{i=1\}^t x\_i\right)\gamma\_0+\sum\_\{i=1\}^t x\_i(\gamma\_0+\eta\_i). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
这就证明了题中向量组是 $\displaystyle AX=b$ 解集的一个极大线性无关组.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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---
1049、 (3)、 设 $\displaystyle A=(\alpha\_1,\alpha\_2,\alpha\_3,\alpha\_4)$ 为 $\displaystyle 4$ 阶方阵, 其中 $\displaystyle \alpha\_1,\alpha\_2,\alpha\_3$ 线性无关,
\begin\{aligned\} \alpha\_4=-\frac\{2\}\{3\}\alpha\_1-\frac\{2\}\{3\}\alpha\_2. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
若向量 $\displaystyle \beta=\alpha\_1+\alpha\_2+\alpha\_3+\alpha\_4$, 则 $\displaystyle AX=\beta$ 的通解可以写为 $\displaystyle X=\underline\{\ \ \ \ \ \ \ \ \ \ \}$. (中国矿业大学(徐州)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, $\displaystyle \mathrm\{rank\} A=3$, 而 $\displaystyle AX=0$ 的基础解系只有一个线性无关的向量 $\displaystyle \left(\frac\{2\}\{3\},\frac\{2\}\{3\},0,1\right)^\mathrm\{T\}$. 再者, $\displaystyle AX=\beta$ 有特解 $\displaystyle (1,1,1,1)^\mathrm\{T\}$. 故 $\displaystyle AX=\beta$ 的通解为
\begin\{aligned\} k\left(\begin\{array\}\{cccccccccccccccccccc\}2\\\\2\\\\0\\\\3\end\{array\}\right)+\left(\begin\{array\}\{cccccccccccccccccccc\}1\\\\1\\\\1\\\\1\end\{array\}\right),\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\}
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)/
---
1050、 3、 (16 分) 已知实线性方程组 $\displaystyle AX=b$ 有唯一解, 其中 $\displaystyle A$ 是 $\displaystyle m\times n$ 实矩阵, 且 $\displaystyle m\geq n$. 证明: $\displaystyle A^\mathrm\{T\} A$ 可逆, 且 $\displaystyle AX=b$ 的唯一解为 $\displaystyle (A^\mathrm\{T\} A)^\{-1\}A^\mathrm\{T\} b$. (中南大学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 $\displaystyle AX=b$ 有唯一解知 $\displaystyle AX=0$ 只有零解, 而 $\displaystyle \mathrm\{rank\} A=n$. 由
\begin\{aligned\} &A^\mathrm\{T\} AX=0\Rightarrow X^\mathrm\{T\} A^\mathrm\{T\} AX=0\\\\ \stackrel\{Y=AX\}\{\Rightarrow\}&Y^\mathrm\{T\} Y=0\Rightarrow 0=Y=AX\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\}
知 $\displaystyle A^\mathrm\{T\} AX=0$ 也只有零解, $\displaystyle \mathrm\{rank\}(A^\mathrm\{T\} A)=n$, 而 $\displaystyle A^\mathrm\{T\} A$ 可逆,
\begin\{aligned\} AX=b\Rightarrow A^\mathrm\{T\} AX=A^\mathrm\{T\} b\Rightarrow X=(A^\mathrm\{T\} A)^\{-1\}A^\mathrm\{T\} b. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
这就是 $\displaystyle AX=b$ 的唯一解.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)/
---
1051、 4、 (16 分) 设 $\displaystyle AX=0$ 的基础解系是 $\displaystyle \eta\_1,\cdots,\eta\_\{n-r\}$, 其中 $\displaystyle A$ 是 $\displaystyle m\times n$ 实矩阵. 设 $\displaystyle B=(\eta\_1,\cdots,\eta\_\{n-r\})$, 若 $\displaystyle n\times t$ 实矩阵 $\displaystyle C$ 满足 $\displaystyle AC=0$, 证明: 存在唯一的 $\displaystyle D$, 使得 $\displaystyle C=BD$. (中南大学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) / 设 $\displaystyle C=(\xi\_1,\cdots,\xi\_t)$, 则 $\displaystyle AC=0\Rightarrow A\xi\_i=0, 1\leq i\leq t$. 由基础解系的定义知
\begin\{aligned\} &\exists\ |\ d\_\{ij\},\mathrm\{ s.t.\} \xi\_i=\sum\_\{j=1\}^\{n-r\} d\_\{ji\}\eta\_j =(\eta\_1,\cdots,\eta\_\{n-r\})\left(\begin\{array\}\{cccccccccccccccccccc\}d\_\{1i\}\\\\ \vdots\\\\ d\_\{n-r,i\}\end\{array\}\right)=B\left(\begin\{array\}\{cccccccccccccccccccc\}d\_\{1i\}\\\\ \vdots\\\\ d\_\{n-r,i\}\end\{array\}\right)\\\\ \Rightarrow&C=(\xi\_1,\cdots,\xi\_t)=B\left(\begin\{array\}\{cccccccccccccccccccc\}d\_\{11\}&\cdots&d\_\{1t\}\\\\ \vdots&&\vdots\\\\ d\_\{n-r,1\}&\cdots&d\_\{n-r,t\}\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\}
令 $\displaystyle D=\left(\begin\{array\}\{cccccccccccccccccccc\}d\_\{11\}&\cdots&d\_\{1t\}\\\\ \vdots&&\vdots\\\\ d\_\{n-r,1\}&\cdots&d\_\{n-r,t\}\end\{array\}\right)$, 则 $\displaystyle C=BD$, 且 $\displaystyle D$ 是唯一的.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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---
1052、 8、 (18 分) 设 $\displaystyle n$ 阶实矩阵 $\displaystyle A,B,C,D$ 两两可换, 且 $\displaystyle E=AC+BD$, 其中 $\displaystyle E$ 是 $\displaystyle n$ 阶单位矩阵. 设齐次线性方程组 $\displaystyle ABX=0, AX=0, BX=0$ 的解空间分别为 $\displaystyle W,V\_1,V\_2$. 证明: $\displaystyle W=V\_1\oplus V\_2$. (中南大学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) / 易知 $\displaystyle V\_1,V\_2\subset W$. 由
\begin\{aligned\} &\alpha\in W\Rightarrow AB\alpha=0\\\\ \Rightarrow&\alpha=BD\alpha+AC\alpha\in V\_1+V\_2\left(\left\\{\begin\{array\}\{llllllllllll\} ABD\alpha=DAB\alpha=0\\\\ BAC\alpha=CAB\alpha=0 \end\{array\}\right.\right) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知 $\displaystyle W=V\_1+V\_2$. 又由
\begin\{aligned\} \alpha\in V\_1\cap V\_2\Rightarrow&A\alpha=0, B\alpha=0\\\\ \Rightarrow&\alpha=AC\alpha+BD\alpha=CA\alpha+DB\alpha=0 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知 $\displaystyle V\_1\cap V\_2=\left\\{0\right\\}$, 而 $\displaystyle W=V\_1\oplus V\_2$.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)/
---
1053、 3、 (14 分) 设 $\displaystyle A=(a\_\{ij\})\_\{n\times n\}$ 是线性方程组 $\displaystyle AX=B\neq 0$ 的系数矩阵, 且线性方程组有解, $\displaystyle \mathrm\{rank\} A=r$. (1)、 证明: 方程组有 $\displaystyle n-r+1$ 个线性无关的解向量; (2)、 用两种形式表示线性方程组 $\displaystyle AX=B\neq 0$ 的通解. (重庆大学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) / 设 $\displaystyle \overline\{A\}=(A,B)$. 由 $\displaystyle \mathrm\{rank\}(A)=\mathrm\{rank\}(\overline\{A\})=r$ 知 $\displaystyle AX=b$ 有特解 $\displaystyle \gamma\_0$, $\displaystyle AX=0$ 有基础解系 $\displaystyle \beta\_1,\cdots,\beta\_\{n-r\}$. 令
\begin\{aligned\} \gamma\_i=\beta\_i+\gamma\_0,\quad i=1,2,\cdots,n-r, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
则由
\begin\{aligned\} &k\_0\gamma\_0+k\_1\gamma\_1+\cdots+k\_\{n-r\}\gamma\_\{n-r\}=0\\\\ \Rightarrow&0=A(k\_0\gamma\_0+k\_1\gamma\_1+\cdots+k\_\{n-r\}\gamma\_\{n-r\}) =k\_0b+k\_1b+\cdots+k\_\{n-r\}b\\\\ \Rightarrow&k\_0+k\_1+\cdots+k\_\{n-r\}=0\\\\ \Rightarrow&0=k\_0\gamma\_0+\cdots+k\_\{n-r\}\gamma\_\{n-r\}\\\\ & =(k\_0+k\_1+\cdots+k\_\{n-r\})\gamma\_0+k\_1\beta\_1+\cdots+k\_\{n-r\}\beta\_\{n-r\}\\\\ & =k\_1\beta\_1+\cdots+k\_\{n-r\}\beta\_\{n-r\}\\\\ \Rightarrow&k\_1=\cdots=k\_\{n-r\}=0\Rightarrow k\_0=-\sum\_\{i=1\}^\{n-r\}k\_i=0 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知 $\displaystyle \gamma\_0,\cdots,\gamma\_\{n-r\}$ 线性无关, 都是 $\displaystyle AX=B$ 的解. 据此即知
\begin\{aligned\} \left\\{k\_0\gamma\_0+k\_1\gamma\_1+\cdots+k\_\{n-r\}\gamma\_\{n-r\}; k\_0,k\_1,\cdots,k\_\{n-r\}\in \mathbb\{K\}; \sum\_\{i=0\}^\{n-r\}k\_i=1\right\\} \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
中的元素满足 $\displaystyle AX=B$. 再若 $\displaystyle X$ 满足 $\displaystyle AX=B$, 则
\begin\{aligned\} X&=\gamma\_0+\sum\_\{i=1\}^\{n-r\} k\_i\beta\_i =\left(1-\sum\_\{i=1\}^\{n-r\} k\_i\right) \gamma\_0+\sum\_\{i=1\}^\{n-r\}k\_i\gamma\_i\\\\ &\in\left\\{k\_0\gamma\_0+k\_1\gamma\_1+\cdots+k\_\{n-r\}\gamma\_\{n-r\}; k\_0,k\_1,\cdots,k\_\{n-r\}\in \mathbb\{K\}; \sum\_\{i=0\}^\{n-r\}k\_i=1\right\\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
这就证明了题中集合就是 $\displaystyle AX=B$ 的解集. 于是 $\displaystyle AX=B$ 的通解为
\begin\{aligned\} k\_0\gamma\_0+k\_1\gamma\_1+\cdots+k\_\{n-r\}\gamma\_\{n-r\}; k\_0,k\_1,\cdots,k\_\{n-r\}\in \mathbb\{K\}; \sum\_\{i=0\}^\{n-r\}k\_i=1, \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\} \gamma\_0+k\_1\beta\_1+\cdots+k\_\{n-r\}\beta\_\{n-r\}. \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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---
1054、 11、 (13 分) 设 $\displaystyle A=(a\_\{ij\})\_\{n\times n\}$ 是实矩阵. 证明: 线性方程组 $\displaystyle AX=B$ 有解的充要条件是向量 $\displaystyle B$ 与齐次线性方程组 $\displaystyle A^\mathrm\{T\} X=0$ 的解空间正交. (重庆大学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) / 对矩阵 $\displaystyle B$, 设 $\displaystyle L(B)$ 表示由 $\displaystyle B$ 的列向量组生成的子空间. (0-3)、 设 $\displaystyle A^\mathrm\{T\}=(\beta\_1,\cdots, \beta\_n)$, 则
\begin\{aligned\} \alpha\in L(A^\mathrm\{T\})^\perp&\Leftrightarrow \beta\_i^\mathrm\{T\}\alpha=0, 1\leq i\leq n \Leftrightarrow \left(\begin\{array\}\{cccccccccccccccccccc\}\beta^\mathrm\{T\}\\\\\vdots\\\\\beta\_n^\mathrm\{T\}\end\{array\}\right)\alpha=0 \Leftrightarrow A\alpha=0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(0-4)、 设 $\displaystyle A=(\alpha\_1,\cdots,\alpha\_n)$, 则由第 1 步知 $\displaystyle A^\mathrm\{T\} Y=0$ 的解空间是 $\displaystyle L\left((A^\mathrm\{T\})^\mathrm\{T\}\right)=L(A)$ 的正交补子空间, 而
\begin\{aligned\} &\mbox\{ $\displaystyle B$ 与 $\displaystyle A^\mathrm\{T\} Y=0$ 的解空间正交\}\Leftrightarrow B\in L(A)=L(\alpha\_1,\cdots,\alpha\_n)\\\\ \Leftrightarrow& \exists\ X=(x\_1,\cdots, x\_n)^\mathrm\{T\},\mathrm\{ s.t.\} \sum\_\{i=1\}^n x\_i\alpha\_i=B \Leftrightarrow AX=B\mbox\{有解\}X. \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](
https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/
---
1055、 5、 (12 分) 设 $\displaystyle a\neq 0, b\neq ?$,
\begin\{aligned\} A=\left(\begin\{array\}\{cccccccccccccccccccc\}a&b&\cdots&b\\\\ b&a&\cdots&b\\\\ \vdots&\vdots&&\vdots\\\\ b&b&\cdots&a\end\{array\}\right), x=(x\_1,\cdots,x\_n)^\mathrm\{T\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
考虑齐次线性方程组 $\displaystyle AX=0$. (1)、 $\displaystyle a,b$ 为何值时, 方程组仅有零解; (2)、 $\displaystyle a,b$ 为何值时, 方程组有无穷多解, 并求其基础解系. [题目不全, 但跟锦数学微信公众号还是全部做了.] (重庆师范大学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) / 若 $\displaystyle a=b=0$, 则任一 $\displaystyle \mathbb\{P\}^n$ 中的向量都是解. 若 $\displaystyle a\neq 0, b=0$, 则方程组只有零解. 若 $\displaystyle b\neq 0$, 则设 $\displaystyle c=\frac\{a\}\{b\}$ 后有
\begin\{aligned\} A&=\left(\begin\{array\}\{cccccccccccccccccccc\}a&b&\cdots&b\\\\ b&a&\cdots&b\\\\ \vdots&\vdots&&\vdots\\\\ b&b&\cdots&a\end\{array\}\right)\to \left(\begin\{array\}\{cccccccccccccccccccc\}c&1&\cdots&1\\\\ 1&c&\cdots&1\\\\ \vdots&\vdots&&\vdots\\\\ 1&1&\cdots&c\end\{array\}\right) \to\left(\begin\{array\}\{cccccccccccccccccccc\}c-1&0&\cdots&0&1-c\\\\ 0&c-1&\cdots&0&1-c\\\\ \vdots&\vdots&&\vdots&\vdots\\\\ 0&0&\cdots&c-1&1-c\\\\ 1&1&\cdots&1&c\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)、 若 $\displaystyle c=1\Leftrightarrow a=b$, 则
\begin\{aligned\} A\to\left(\begin\{array\}\{cccccccccccccccccccc\}1&1&\cdots&1\\\\ 0&0&\cdots&0\\\\ \vdots&\vdots&&\vdots\\\\ 0&0&\cdots&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 $\displaystyle Ax=0$ 有无穷多解, 且通解为
\begin\{aligned\} \sum\_\{i=2\}^n k\_i(-e\_1+e\_i), \forall\ k\_2,\cdots,k\_n. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(2)、 若 $\displaystyle c\neq 1\Leftrightarrow a\neq b$, 则
\begin\{aligned\} A\to\left(\begin\{array\}\{cccccccccccccccccccc\}1&0&\cdots&0&-1\\\\ 0&1&\cdots&0&-1\\\\ \vdots&\vdots&&\vdots&\vdots\\\\ 0&0&\cdots&1&-1\\\\ 1&1&\cdots&1&c\end\{array\}\right) \to\left(\begin\{array\}\{cccccccccccccccccccc\}1&0&\cdots&0&-1\\\\ 0&1&\cdots&0&-1\\\\ \vdots&\vdots&&\vdots&\vdots\\\\ 0&0&\cdots&1&-1\\\\ 0&0&\cdots&0&c+n-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\}
(2-1)、 若 $\displaystyle c+n-1=0\Leftrightarrow a+(n-1)b=0$, 则 $\displaystyle Ax=0$ 有无穷多解, 且通解为
\begin\{aligned\} k(1,\cdots,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\}
(2-2)、 若 $\displaystyle c+n-1\neq 0\Leftrightarrow a+(n-1)b\neq 0$, 则 $\displaystyle |A|\neq 0$, $\displaystyle Ax=0$ 只有零解.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)/
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1056、 1、 填空题 (每题 5 分, 共 30 分). (1)、 已知 $\displaystyle 3$ 阶矩阵 $\displaystyle A$ 的伴随矩阵 $\displaystyle A^\star=\left(\begin\{array\}\{cccccccccccccccccccc\}1&0&0\\\\ 2&1&0\\\\ 1&2&1\end\{array\}\right)$, 则 $\displaystyle A=\underline\{\ \ \ \ \ \ \ \ \ \ \}$. (安徽大学2023年高等代数考研试题) [矩阵 ]
[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\} &AA^\star=|A|E\Rightarrow 1=|A^\star|=|A|^\{n-2\}=|A|\Rightarrow AA^\star=E\\\\ \Rightarrow& A=(A^\star)^\{-1\} =\left(\begin\{array\}\{cccccccccccccccccccc\}1&0&0\\\\ -2&1&0\\\\ 3&-2&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\}
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)/
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1057、 (2)、 矩阵 $\displaystyle A=\left(\begin\{array\}\{cccccccccccccccccccc\}0&a\_1&0&\cdots&0&0\\\\ 0&0&a\_2&\cdots&0&0\\\\ 0&0&0&\cdots&0&0\\\\ \vdots&\vdots&\vdots&&\vdots&\vdots\\\\ 0&0&0&\cdots&0&a\_\{n-1\}\\\\ a\_n&0&0&\cdots&0&0\end\{array\}\right)$, 其中 $\displaystyle a\_i\neq 0\ (i=1,\cdots,n)$, 则 $\displaystyle A^\{-1\}=\underline\{\ \ \ \ \ \ \ \ \ \ \}$. (安徽大学2023年高等代数考研试题) [矩阵 ]
[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\} \left(\begin\{array\}\{cccccccccccccccccccc\}&B\\\\ a&\end\{array\}\right)\left(\begin\{array\}\{cccccccccccccccccccc\}0&a^\{-1\}\\\\ B^\{-1\}&0\end\{array\}\right)=E \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知 $\displaystyle A^\{-1\}=\left(\begin\{array\}\{cccccccccccccccccccc\}&&&a\_n^\{-1\}\\\\ a\_1^\{-1\}&&&\\\\ &\ddots&&\\\\ &&a\_\{n-1\}^\{-1\}&\end\{array\}\right)$.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)/
---
1058、 (6)、 设 $\displaystyle A$ 的初等因子为 $\displaystyle \lambda,\lambda,\lambda^2,(\lambda-1)^2,\lambda+1,(\lambda+1)^2$, $\displaystyle E$ 与 $\displaystyle A$ 同阶, 则 $\displaystyle \lambda$-矩阵 $\displaystyle \lambda E-A$ 的所有不变因子为 $\displaystyle \underline\{\ \ \ \ \ \ \ \ \ \ \}$. (安徽大学2023年高等代数考研试题) [矩阵 ]
[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) / $\displaystyle A$ 是 $\displaystyle 1+1+2+2+1+2=9$ 阶方阵. 将 $\displaystyle A$ 的初等因子
\begin\{aligned\} \begin\{array\}\{rrr\} \lambda&\lambda&\lambda^2,\\\\ &&(\lambda-1)^2,\\\\ &\lambda+1&(\lambda+1)^2 \end\{array\} \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
乘起来, 并补上一些 $\displaystyle 1$ 即得 $\displaystyle \lambda E-A$ 的不变因子为
\begin\{aligned\} 1,1,1,1,1,1,\lambda,\lambda(\lambda+1),\lambda^2(\lambda+1)^2(\lambda-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](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/
发表于 2023-3-5 09:18:03
