## 张祖锦2023年数学专业真题分类70天之第43天
---
967、 7、 $\displaystyle E\_1,\cdots,E\_\{2^n-1\}$ 是 $\displaystyle \left\\{1,2,\cdots,n\right\\}$ 的非空子集的一个排序, 矩阵 $\displaystyle A\_\{(2^n-1)\times (2^n-1)\}$ 中元素
\begin\{aligned\} a\_\{ij\}=\left\\{\begin\{array\}\{llllllllllll\}0,&E\_i\cap E\_j=\varnothing,\\\\ 1,&E\_i\cap E\_j\neq \varnothing.\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\}
求 $\displaystyle \det A\_\{(2^n-1)\times (2^n-1)\}$. (南方科技大学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) / 往用数学归纳法证明 $\displaystyle \det A\_\{(2^n-1)\times (2^n-1)\}=\left\\{\begin\{array\}\{llllllllllll\}1,&n=1,\\\\ -1,&n\geq 2.\end\{array\}\right.$ 当 $\displaystyle n=1$ 时, $\displaystyle A=1$, 结论自明. 当 $\displaystyle n=2$ 时, 通过交换行与列, 不妨设
\begin\{aligned\} E\_1=\left\\{1\right\\}, E\_2=\left\\{1,2\right\\}, E\_3=\left\\{2\right\\}\Rightarrow A\_\{3\times 3\}=\left(\begin\{array\}\{cccccccccccccccccccc\}1&1&0\\\\ 1&1&1\\\\ 0&1&1\end\{array\}\right)\Rightarrow \det A=-1. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
设结论对 $\displaystyle n-1\ (n\geq 3)$ 时成立, 记 $\displaystyle E\_1,\cdots,E\_\{2^\{n-1\}-1\}$ 是 $\displaystyle \left\\{1,\cdots,n-1\right\\}$ 的非空子集的一个排列,
\begin\{aligned\} E\_\{2^\{n-1\}-1+j\}=E\_j\cup \left\\{n\right\\}, j=1,\cdots,2^\{n-1\}-1;\quad E\_\{2^n-1\}=\left\\{n\right\\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
则通过考查 $\displaystyle E\_i\cap E\_j$ 是否为空集知
\begin\{aligned\} A=\left(\begin\{array\}\{cccccccccccccccccccc\}B&B&0\\\\ B&ee^\mathrm\{T\}&e\\\\ 0&e^\mathrm\{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\}
其中
\begin\{aligned\} B\_\{(2^\{n-1\}-1)\times (2^\{n-1\}-1)\}, e=(\underbrace\{1,\cdots,1\}\_\{2^n-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\}
注意由 $\displaystyle A$ 的定义知 $\displaystyle A$ 是实对称矩阵! 通过初等变换知
\begin\{aligned\} A\to \left(\begin\{array\}\{cccccccccccccccccccc\}B&B&0\\\\ B&0&0\\\\ 0&e^\mathrm\{T\}&1\end\{array\}\right)\to\left(\begin\{array\}\{cccccccccccccccccccc\}B&B&0\\\\ B&0&0\\\\ 0&0&1\end\{array\}\right)\to\left(\begin\{array\}\{cccccccccccccccccccc\}0&B&0\\\\ B&0&0\\\\ 0&0&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\}
于是
\begin\{aligned\} |A|=(-1)^\{(2^\{n-1\}-1)^2\}\left|\begin\{array\}\{cccccccccc\}\mathrm\{diag\}(B,B,1)\end\{array\}\right| =(-1)|B|^2\xlongequal\{\tiny\mbox\{归纳假设\}\} -1^2=-1. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故有结论.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)/
---
968、 1、 (15 分) 求三维空间中由四点
\begin\{aligned\} A(1,1,1),\quad B(2,4,8),\quad C(3,9,27), \quad D(5,25,125) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
构成的三棱锥体积. (南京大学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) / 所求体积
\begin\{aligned\} =\frac\{1\}\{6\}\left|\det\left(\overrightarrow\{AB\},\overrightarrow\{AC\},\overrightarrow\{AD\}\right)\right| =\frac\{1\}\{6\}\left|\det\left(\begin\{array\}\{cccccccccccccccccccc\}1&3&7\\\\ 2&8&26\\\\ 4&24&124\end\{array\}\right)\right|=48. \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)/
---
969、 (2)、 排序 $\displaystyle 987654321$ 的逆序数为 $\displaystyle \underline\{\ \ \ \ \ \ \ \ \ \ \}$. (南京理工大学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) / $\displaystyle 8+7+\cdots+1=\frac\{8\cdot 9\}\{2\}=36$.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)/
---
970、 3、 (10 分) 求 $\displaystyle n$ 阶行列式
\begin\{aligned\} D\_n=\left|\begin\{array\}\{cccccccccc\}2&1&1&\cdots&1\\\\ 1&2&1&\cdots&1\\\\ 1&1&2&\cdots&1\\\\ \vdots&\vdots&\vdots&&\vdots\\\\ 1&1&1&\cdots&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\}
(南京理工大学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)、 我们有如下常用公式: 由
\begin\{aligned\} \left(\begin\{array\}\{cccccccccccccccccccc\}E&0\\\\ B&E\end\{array\}\right)\left(\begin\{array\}\{cccccccccccccccccccc\}E&A\\\\ -B&E\end\{array\}\right)\left(\begin\{array\}\{cccccccccccccccccccc\}E&-A\\\\ 0&E\end\{array\}\right)=&\left(\begin\{array\}\{cccccccccccccccccccc\}E&0\\\\ 0&E+BA\end\{array\}\right),\\\\ \left(\begin\{array\}\{cccccccccccccccccccc\}E&-A\\\\ 0&E\end\{array\}\right)\left(\begin\{array\}\{cccccccccccccccccccc\}E&A\\\\ -B&E\end\{array\}\right)\left(\begin\{array\}\{cccccccccccccccccccc\}E&0\\\\ B&E\end\{array\}\right)=&\left(\begin\{array\}\{cccccccccccccccccccc\}E+AB&0\\\\ 0&E\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\}
即知
\begin\{aligned\} |E+BA|=|E+AB|. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(2)、 设 $\displaystyle e=(1,\cdots,1)^\mathrm\{T\}$, 则
\begin\{aligned\} D\_n=|E\_n+ee^\mathrm\{T\}|\overset\{\tiny\mbox\{第1步\}\}\{=\} 1+e^\mathrm\{T\} e=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](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/
---
971、 3、 计算 $\displaystyle 2n$ 阶行列式
\begin\{aligned\} D\_\{2n\}=\left|\begin\{array\}\{cccccccccc\}a\_n&&&&&b\_n\\\\ &\ddots&&&\mathrm\{id\}dots&\\\\ &&a\_1&b\_1&&\\\\ &&c\_1&d\_1&&\\\\ &\mathrm\{id\}dots&&&\ddots&\\\\ c\_n&&&&&d\_n\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\}
(南京师范大学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) / 按第一行展开, $\displaystyle a\_n$ 的代数余子式按最后一列展开, $\displaystyle b\_n$ 的代数余子式按第一列展开, 得
\begin\{aligned\} D\_n=&a\_n\cdot(-1)^\{(2n-1)+(2n-1)\}d\_n\cdot D\_\{n-1\}\\\\ &+(-1)^\{1+2n\}b\_n\cdot(-1)^\{(2n-1)+1\}c\_n\cdot D\_\{n-1\}\\\\ =&(a\_nd\_n-b\_nc\_n)D\_\{n-1\}=\cdots=\prod\_\{k=1\}^n(a\_kd\_k-b\_kc\_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)/
---
972、 2、 (20 分) 证明:
\begin\{aligned\} \left|\begin\{array\}\{cccccccccc\}\cos t&1&0&0\\\\ 1&2\cos t&1&0\\\\ 0&1&2\cos t&1\\\\ 0&0&1&2\cos t\end\{array\}\right|=\cos 4t. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(南开大学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) / 我们有如下更为一般的结果, 而自然有题目的结论. $\displaystyle n$ 阶行列式
\begin\{aligned\} D\_n\equiv\left|\begin\{array\}\{cccccccccc\} \cos\alpha&1&0&\cdots&0&0\\\\ 1&2\cos\alpha&1&\cdots&0&0\\\\ 0&1&2\cos\alpha&\cdots&0&0\\\\ \vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\\\ 0&0&0&\cdots&2\cos \alpha&1\\\\ 0&0&0&\cdots&1&2\cos \alpha \end\{array\}\right|=\cos n\alpha. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
事实上, 按最后一列展开得
\begin\{aligned\} D\_n=2\cos\alpha \cdot D\_\{n-1\}-D\_\{n-2\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
其特征方程为
\begin\{aligned\} 0=\lambda^2-2\cos\alpha\cdot \lambda+1\Rightarrow \lambda=\mathrm\{e\}^\{\pm \mathrm\{ i\}\alpha\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故可设
\begin\{aligned\} &D\_n=A\mathrm\{e\}^\{\mathrm\{ i\} \alpha n\}+B\mathrm\{e\}^\{-\mathrm\{ i\} \alpha n\}\\\\ \Rightarrow&A=B=\frac\{1\}\{2\}\left(D\_1=\cos \alpha, D\_2=\cos2\alpha\right)\\\\ \Rightarrow&D\_n=\frac\{1\}\{2\}(\mathrm\{e\}^\{\mathrm\{ i\} \alpha n\}+\mathrm\{e\}^\{-\mathrm\{ i\} \alpha n\})=\cos n\alpha. \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)/
---
973、 1、 填空题. (1)、 设 $\displaystyle n$ 阶矩阵 $\displaystyle A$ 满足 $\displaystyle |A|=2$, 交换 $\displaystyle A$ 的第一行与第二行得到矩阵 $\displaystyle B$, 则 $\displaystyle |BA^\star|=\underline\{\ \ \ \ \ \ \ \ \ \ \}$. (厦门大学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) / $\displaystyle \mbox\{原式\}=|B|\cdot |A^\star|=-|A|\cdot |A|^\{n-1\}=-2^n$.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)/
---
974、 (2)、 已知 $\displaystyle 4$ 阶矩阵 $\displaystyle A=(a\_\{ij\})$ 满足
\begin\{aligned\} a\_\{1j\}=2023\left(j=1,2,3,4\right),\quad |A|=a, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
则 $\displaystyle \sum\_\{1\leq i,j\leq 4\}A\_\{ij\}=\underline\{\ \ \ \ \ \ \ \ \ \ \}$. (厦门大学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\}=&\sum\_\{j=1\}^4 A\_\{1j\}+\sum\_\{i=2\}^3 \sum\_\{j=1\}^4 A\_\{ij\} =\left|\begin\{array\}\{cccccccccc\}1&1&1&1\\\\ a\_\{21\}&a\_\{22\}&a\_\{23\}&a\_\{24\}\\\\ a\_\{31\}&a\_\{32\}&a\_\{33\}&a\_\{34\}\\\\ a\_\{41\}&a\_\{42\}&a\_\{43\}&a\_\{44\}\end\{array\}\right|+\sum\_\{i=2\}^3 0\\\\ =&\frac\{1\}\{2023\}\left|\begin\{array\}\{cccccccccc\}a\_\{11\}&a\_\{12\}&a\_\{12\}&a\_\{14\}\\\\ a\_\{21\}&a\_\{22\}&a\_\{23\}&a\_\{24\}\\\\ a\_\{31\}&a\_\{32\}&a\_\{33\}&a\_\{34\}\\\\ a\_\{41\}&a\_\{42\}&a\_\{43\}&a\_\{44\}\end\{array\}\right|=\frac\{a\}\{2023\}. \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)/
---
975、 1、 (15 分) 计算下列 $\displaystyle n$ 阶行列式 $\displaystyle \left|\begin\{array\}\{cccccccccc\}x-a&a&\cdots&a\\\\ a&x-a&\cdots&a\\\\ \vdots&\vdots&&\vdots\\\\ a&a&\cdots&x-a\end\{array\}\right|$. (山西大学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) / 当 $\displaystyle x\neq 2a$ 时,
\begin\{aligned\} \mbox\{原式\}=&\left|\begin\{array\}\{cccccccccc\}1&a&a&\cdots&a\\\\ 0&x-a&a&\cdots&a\\\\ 0&a&x-a&\cdots&a\\\\ \vdots&\vdots&\vdots&&\vdots\\\\ 0&a&a&\cdots&x-a\end\{array\}\right| =\left|\begin\{array\}\{cccccccccc\}1&a&a&\cdots&a\\\\ -1&x-2a&&&\\\\ -1&&x-2a&&\\\\ \vdots&&&\ddots&\\\\ -1&&&&x-2a\end\{array\}\right|\\\\ =&\left|\begin\{array\}\{cccccccccc\}1+\frac\{na\}\{x-2a\}&a&a&\cdots&a\\\\ 0&x-2a&&&\\\\ 0&&x-2a&&\\\\ \vdots&&&\ddots&\\\\ 0&&&&x-2a\end\{array\}\right|\\\\ =&\left(1+\frac\{na\}\{x-2a\}\right)(x-2a)^n =(x-2a)^\{n-1\}[x+(n-2)a]. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
上式两端都是关于 $\displaystyle x$ 的连续函数, 在 $\displaystyle x\neq 2a$ 时相等, 而恒等. 故原式 $\displaystyle =(x-2a)^\{n-1\}[x+(n-2)a]$.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)/
---
976、 2、 (15 分) 计算 $\displaystyle n$ 阶行列式 $\displaystyle D=\left|\begin\{array\}\{cccccccccc\}x\_1+a\_1^2&a\_1a\_2&\cdots&a\_1a\_n\\\\ a\_2a\_1&x\_2+a\_2^2&\cdots&a\_2a\_n\\\\ \vdots&\vdots&&\vdots\\\\ a\_na\_1&a\_na\_2&\cdots&x\_n+a\_n^2\end\{array\}\right|$. (陕西师范大学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) / 当 $\displaystyle x\_1\cdots x\_n\neq 0$ 时,
\begin\{aligned\} D\_n=&\left|\begin\{array\}\{ccccc\} 1&a\_1&a\_2&\cdots&a\_n\\\\ 0&x\_1+a\_1^2&a\_1a\_2&\cdots&a\_1a\_n\\\\ 0&a\_2a\_1&x\_2+a\_2^2&\cdots&a\_2a\_n\\\\ \vdots&\vdots&\vdots&\ddots&\vdots\\\\ 0&a\_na\_1&a\_na\_2&\cdots&x\_n+a\_n^2 \end\{array\}\right| =\left|\begin\{array\}\{ccccc\} 1&a\_1&a\_2&\cdots&a\_n\\\\ -a\_1&x\_1&\cdots&0\\\\ -a\_2&0&x\_2&\cdots&0\\\\ \vdots&\vdots&\vdots&\ddots&\vdots\\\\ -a\_n&0&0&\cdots&x\_n \end\{array\}\right|\\\\ =&\left|\begin\{array\}\{ccccc\} 1+\sum\_\{i=1\}^n \frac\{a\_i^2\}\{x\_i\}&a\_1&a\_2&\cdots&a\_n\\\\ 0&x\_1&0&\cdots&0\\\\ 0&0&x\_2&\cdots&0\\\\ \vdots&\vdots&\vdots&\ddots&\vdots\\\\ 0&0&0&\cdots&x\_n \end\{array\}\right| =\left(1+\sum\_\{i=1\}^n \frac\{a\_i^2\}\{x\_i\}\right)x\_1\cdots x\_n\\\\ =&x\_1\cdots x\_n+\sum\_\{i=1\}^n x\_1\cdots x\_\{i-1\}a\_i^2x\_\{i+1\}\cdots x\_n. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
左右两端都是 $\displaystyle x\_1,\cdots,x\_n$ 的连续函数, 而
\begin\{aligned\} D=x\_1\cdots x\_n+\sum\_\{i=1\}^n x\_1\cdots x\_\{i-1\}a\_i^2x\_\{i+1\}\cdots x\_n. \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)/
---
977、 1、 求行列式 $\displaystyle D\_n=\left|\begin\{array\}\{cccccccccc\}1&2&3&\cdots&n\\\\ n&1&2&\cdots&n-1\\\\ \vdots&\vdots&\vdots&&\vdots\\\\ 3&4&5&\cdots&2\\\\ 2&3&4&\cdots&1\end\{array\}\right|$. (上海财经大学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 $\displaystyle i$ 行 $\displaystyle \cdot(-1)$ 加到第 $\displaystyle i-1$ 行, $\displaystyle i=2,\cdots,n$, 得
\begin\{aligned\} D\_n=\left|\begin\{array\}\{cccccccccc\}1-n&1&1&\cdots&1\\\\ 1&1-n&1&\cdots&1\\\\ 1&1&1-n&\cdots&1\\\\ \vdots&\vdots&\vdots&&\vdots\\\\ 1&1&1&\cdots&1\\\\ 2&3&4&\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\}
第 $\displaystyle n$ 列 $\displaystyle \cdot(-1)$ 加到其它各列, 得
\begin\{aligned\} D\_n=\left|\begin\{array\}\{cccccccccc\}-n&0&0&\cdots&1\\\\ 0&-n&0&\cdots&1\\\\ 0&0&-n&\cdots&1\\\\ \vdots&\vdots&\vdots&&\vdots\\\\ 0&0&0&\cdots&1\\\\ 2&3&4&\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\}
第 $\displaystyle i$ 列 $\displaystyle \cdot\frac\{1\}\{n\}$ 加到第 $\displaystyle n$ 列, $\displaystyle 1\leq i\leq n-1$, 得
\begin\{aligned\} D\_n=&\left|\begin\{array\}\{cccccccccc\}-n&0&0&\cdots&0\\\\ 0&-n&0&\cdots&0\\\\ 0&0&-n&\cdots&0\\\\ \vdots&\vdots&\vdots&&\vdots\\\\ 0&0&0&\cdots&0\\\\ 2&3&4&\cdots&1+\sum\_\{i=1\}^\{n-1\}\frac\{i\}\{n\}\end\{array\}\right|\\\\ =&(-n)^\{n-1\}\left(1+\sum\_\{i=1\}^\{n-1\}\frac\{i\}\{n\}\right) =\frac\{(-n)^\{n-1\}(n+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)/
---
978、 1、 证明下面的等式
\begin\{aligned\} D\_n=\left|\begin\{array\}\{cccccccccc\}\frac\{1\}\{z\_1-c\_1\}&\frac\{1\}\{z\_1-c\_2\}&\cdots&\frac\{1\}\{z\_1-c\_n\}\\\\ \frac\{1\}\{z\_2-c\_1\}&\frac\{1\}\{z\_2-c\_2\}&\cdots&\frac\{1\}\{z\_2-c\_n\}\\\\ \vdots&\vdots&&\vdots\\\\ \frac\{1\}\{z\_n-c\_1\}&\frac\{1\}\{z\_n-c\_2\}&\cdots&\frac\{1\}\{z\_n-c\_n\}\end\{array\}\right| =\frac\{\displaystyle\prod\_\{1\leq i < j\leq n\}(z\_i-z\_j)(c\_j-c\_i)\}\{\displaystyle\prod\_\{1\leq i,j\leq n\}(z\_i-c\_j)\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(首都师范大学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) / 第 $\displaystyle n$ 行 $\displaystyle \cdot (-1)$ 加到第 $\displaystyle i$ 行, $\displaystyle 1\leq i\leq n-1$, 并注意到
\begin\{aligned\} \frac\{1\}\{z\_i-c\_j\}-\frac\{1\}\{z\_n-c\_j\} =\frac\{z\_n-c\_i\}\{(z\_i-c\_j)(z\_n-c\_j)\}, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
我们可以提出第 $\displaystyle j$ 列的公因子 $\displaystyle \frac\{1\}\{z\_n-c\_j\}\ (1\leq j\leq n)$, 提出第 $\displaystyle i$ 行的公因子 $\displaystyle z\_n-z\_i\ (1\leq ileq n-1)$, 得
\begin\{aligned\} D\_n=\frac\{\displaystyle\prod\_\{i=1\}^\{n-1\}(z\_n-z\_i)\}\{\displaystyle\prod\_\{j=1\}^n (z\_n-c\_j)\} \left|\begin\{array\}\{cccccccccc\}\frac\{1\}\{z\_1-c\_1\}&\frac\{1\}\{z\_1-c\_2\}&\cdots&\frac\{1\}\{z\_1-c\_n\}\\\\ \frac\{1\}\{z\_2-c\_1\}&\frac\{1\}\{z\_2-c\_2\}&\cdots&\frac\{1\}\{z\_2-c\_n\}\\\\ \vdots&\vdots&&\vdots\\\\ 1&1&\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\}
第 $\displaystyle n$ 列 $\displaystyle \cdot(-1)$ 加到第 $\displaystyle j$ 列, $\displaystyle 1\leq j\leq n-1$, 并注意到
\begin\{aligned\} \frac\{1\}\{z\_i-c\_j\}-\frac\{1\}\{z\_i-c\_n\} =\frac\{c\_j-c\_n\}\{(z\_i-c\_j)(z\_i-c\_n)\}, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
我们可以提出第 $\displaystyle i$ 行的公因子 $\displaystyle \frac\{1\}\{z\_i-c\_n\}\ (1\leq i\leq n-1)$, 提出第 $\displaystyle j$ 列的公因子 $\displaystyle c\_j-c\_n\ (1\leq j\leq n-1)$, 得
\begin\{aligned\} D\_n=&\frac\{\displaystyle\prod\_\{i=1\}^\{n-1\}(z\_n-z\_i)\}\{\displaystyle\prod\_\{j=1\}^n (z\_n-c\_j)\} \frac\{\displaystyle\prod\_\{j=1\}^\{n-1\}(c\_j-c\_n)\}\{\displaystyle\prod\_\{i=1\}^\{n-1\}(z\_i-c\_n)\}\\\\ &\left|\begin\{array\}\{cccccccccc\}\frac\{1\}\{z\_1-c\_1\}&\frac\{1\}\{z\_1-c\_2\}&\cdots&\frac\{1\}\{z\_1-c\_\{n-1\}\}&1\\\\ \frac\{1\}\{z\_2-c\_1\}&\frac\{1\}\{z\_2-c\_2\}&\cdots&\frac\{1\}\{z\_2-c\_\{n-1\}\}&1\\\\ \vdots&\vdots&&\vdots&\vdots\\\\ \frac\{1\}\{z\_\{n-1\}-c\_1\}&\frac\{1\}\{z\_\{n-1\}-c\_2\}&\cdots&\frac\{1\}\{z\_\{n-1\}-c\_\{n-1\}\}&1\\\\ 0&0&\cdots&0&1\end\{array\}\right|\\\\ =&\frac\{\displaystyle\prod\_\{i=1\}^\{n-1\}(z\_i-z\_n)\}\{\displaystyle\prod\_\{j=1\}^n (z\_n-c\_j)\} \frac\{\displaystyle\prod\_\{j=1\}^\{n-1\}(c\_n-c\_j)\}\{\displaystyle\prod\_\{i=1\}^\{n-1\}(z\_i-c\_n)\}D\_\{n-1\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
递推即得结论.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)/
---
979、 1、 求 $\displaystyle n$ 阶行列式 $\displaystyle D\_n=\left|\begin\{array\}\{cccccccccc\}2&\cdots&2&a\\\\ 2&\mathrm\{id\}dots&a&2\\\\ \vdots&\mathrm\{id\}dots&\vdots&\vdots\\\\ a&\cdots&2&2\end\{array\}\right|$. (太原理工大学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 $\displaystyle n$ 列与前 $\displaystyle n-1$ 列交换, 再第 $\displaystyle n$ 列与前 $\displaystyle n-2$ 列交换, 等等, 得
\begin\{aligned\} D\_n=&(-1)^\{(n-1)+(n-2)+\cdots+1\} \left|\begin\{array\}\{cccccccccc\}a&2&\cdots&2\\\\ 2&a&\cdots&2\\\\ \vdots&\vdots&&\vdots\\\\ 2&2&\cdots&a\end\{array\}\right|\\\\ =&(-1)^\frac\{(n-1)n\}\{2\} \left|\begin\{array\}\{cccccccccc\}a-2&0&\cdots&2-a\\\\ 0&a-2&\cdots&0&2-a\\\\ \vdots&\vdots&&\vdots&\vdots\\\\ 0&0&\cdots&a-2&2-a\\\\ 2&2&\cdots&2&a\end\{array\}\right|\\\\ =&(-1)^\frac\{(n-1)n\}\{2\} \left|\begin\{array\}\{cccccccccc\}a-2&0&\cdots&0&0\\\\ 0&a-2&\cdots&0&0\\\\ \vdots&\vdots&&\vdots&\vdots\\\\ 0&0&\cdots&a-2&0\\\\ 2&2&\cdots&2&a+2(n-1)\end\{array\}\right|\\\\ =&(-1)^\frac\{(n-1)n\}\{2\}[a+2(n-1)] (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](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/
---
980、 1、 设 $\displaystyle A=(a\_\{ij\})\_\{n\times n\}$, 并且 $\displaystyle a\_\{ij\}=\max\left\\{i,j\right\\}$, 求 $\displaystyle |A|$. (武汉理工大学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\} |A|=&\left|\begin\{array\}\{cccccccccc\}1&2&3&\cdots&n\\\\ 2&2&3&\cdots&n\\\\ 3&3&3&\cdots&n\\\\ \vdots&\vdots&\vdots&&\vdots\\\\ n&n&n&\cdots&n\end\{array\}\right|\\\\ =&\left|\begin\{array\}\{cccccccccc\}\left(\begin\{array\}\{cccccccccccccccccccc\}1&1&1&\cdots&1\\\\ &1&1&\cdots&1\\\\ &&1&\cdots&1\\\\ &&&\ddots&\vdots\\\\ &&&&1\end\{array\}\right)\left(\begin\{array\}\{cccccccccccccccccccc\}-1&&&&\\\\ -1&-1&&&\\\\ \vdots&\vdots&\ddots&\\\\ -1&-1&\cdots&-1&\\\\ n&n&\cdots&n&n\end\{array\}\right)\end\{array\}\right|\\\\ =&1\cdot(-1)^\{n-1\}n=(-1)^\{n-1\}n. \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)/
---
981、 2、 设 $\displaystyle A=\left|\begin\{array\}\{cccccccccc\}4&-3&1&d\\\\ a&b&c&d\\\\ a^2&b^2&c^2&d^2\\\\ a^4&b^4&c^4&d^4\end\{array\}\right|$, 其中 $\displaystyle A\_\{ij\}$ 为 $\displaystyle A$ 中 $\displaystyle (i,j)$ 元素的代数余子式, 求 $\displaystyle A\_\{11\}+A\_\{12\}+A\_\{13\}+A\_\{14\}$. (武汉理工大学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)、 我们给出一般结果:
\begin\{aligned\} D\_n=\left|\begin\{array\}\{cccc\} 1&1&\cdots&1\\\\ x\_1&x\_2&\cdots&x\_n\\\\ x\_1^2&x\_2^2&\cdots&x\_n^2\\\\ \cdots&\cdots&\cdots&\cdots\\\\ x\_1^\{n-2\}&x\_2^\{n-2\}&\cdots&x\_n^\{n-2\}\\\\ x\_1^n&x\_2^n&\cdots&x\_n^n \end\{array\}\right| =\prod\_\{1\leq i < j\leq n\} (x\_j-x\_i)\cdot \sum\_\{i=1\}^n x\_i. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
比较
\begin\{aligned\} \left|\begin\{array\}\{ccccc\} 1&1&\cdots&1&1\\\\ x\_1&x\_2&\cdots&x\_n&x\\\\ x\_1^2&x\_2^2&\cdots&x\_n^2&x\\\\ \cdots&\cdots&\cdots&\cdots&\cdots\\\\ x\_1^\{n-2\}&x\_2^\{n-2\}&\cdots&x\_n^\{n-2\}&x^\{n-2\}\\\\ x\_1^\{n-1\}&x\_2^\{n-1\}&\cdots&x\_n^\{n-1\}&x^\{n-1\}\\\\ x\_1^n&x\_2^n&\cdots&x\_n^n&x^n \end\{array\}\right|=\prod\_\{1\leq i < j\leq n\} (x\_j-x\_i)\cdot \prod\_\{i=1\}^n (x-x\_i) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
两端 $\displaystyle x^\{n-1\}$ 的系数有
\begin\{aligned\} -D\_n=\prod\_\{1\leq i < j\leq n\} (x\_j-x\_i)\cdot \left(-\sum\_\{i=1\}^n x\_i\right), \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
\begin\{aligned\} D\_n=\prod\_\{1\leq i < j\leq n\} (x\_j-x\_i)\cdot \sum\_\{i=1\}^n x\_i. \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\} &A\_\{11\}+A\_\{12\}+A\_\{13\}+A\_\{14\}=\left|\begin\{array\}\{cccccccccc\}1&1&1&1\\\\ a&b&c&d\\\\ a^2&b^2&c^2&d^2\\\\ a^4&b^4&c^4&d^4\end\{array\}\right|\\\\ \overset\{\tiny\mbox\{第1步\}\}\{=\}&(b-a)(c-a)(d-a)(c-b)(d-b)(d-c)(a+b+c+d). \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)/
---
982、 2、 (10 分) 计算 $\displaystyle n$ 阶行列式 $\displaystyle D\_n=\left|\begin\{array\}\{cccccccccc\}x&y&\cdots&y&y\\\\ z&x&\cdots&y&y\\\\ \vdots&\vdots&&\vdots&\vdots\\\\ z&z&\cdots&x&y\\\\ z&z&\cdots&z&x\end\{array\}\right|$. (西安电子科技大学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 $\displaystyle i$ 行 $\displaystyle \cdot(-1)$ 加到第 $\displaystyle i+1$ 行, $\displaystyle i=n-1,\cdots,1$, 得
\begin\{aligned\} D\_n=\left|\begin\{array\}\{cccccccccc\}x&y&y&\cdots&y&y\\\\ z-x&x-y&0&\cdots&0&0\\\\ 0&z-x&x-y&\cdots&0&0\\\\ \vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\\\ 0&0&0&\cdots&x-y&0\\\\ 0&0&0&\cdots&z-x&x-y\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\}
按第 $\displaystyle n$ 列展开得
\begin\{aligned\} D\_n&=(x-y)D\_\{n-1\}+(-1)^\{1+n\}y(z-x)^\{n-1\}\\\\ &=(x-y)D\_\{n-1\}+y(x-z)^\{n-1\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
交换 $\displaystyle y,z$ 的位置得
\begin\{aligned\} D\_n=(x-z)D\_\{n-1\}+z(x-y)^\{n-1\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
联合上述两式得
\begin\{aligned\} z\neq y\Rightarrow D\_n=\frac\{y(x-z)^n-z(x-y)^n\}\{y-z\}. \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)/
---
983、 1、 计算题. (1)、 计算 $\displaystyle n$ 阶行列式 $\displaystyle D\_n=\left|\begin\{array\}\{cccccccccc\}1&3&3&\cdots&3&3&1\\\\ 3&1&3&\cdots&3&3&1\\\\ 3&3&1&\cdots&3&3&1\\\\ \vdots&\vdots&\vdots&&\vdots&\vdots&\vdots\\\\ 3&3&3&\cdots&1&3&1\\\\ 3&3&3&\cdots&3&1&1\\\\ 1&1&1&\cdots&1&1&1\end\{array\}\right|$. (西安交通大学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 $\displaystyle i$ 行 $\displaystyle \cdot(-1)$ 加到其它各行, 得
\begin\{aligned\} D\_n=\left|\begin\{array\}\{cccccccccc\}1&3&3&\cdots&3&3&1\\\\ 2&-2&&&&&\\\\ 2&&-2&&&&\\\\ \vdots&&&\ddots&&&\\\\ 2&&&&-2&&\\\\ 2&&&&&-2&\\\\ 0&-2&-2&\cdots&-2&-2&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\}
按第 $\displaystyle n$ 列展开得
\begin\{aligned\} D\_n=(-1)^\{n+1\}\left|\begin\{array\}\{cccccccccc\}2&-2&&&&\\\\ 2&&-2&&&\\\\ \vdots&&&\ddots&&\\\\ 2&&&&-2&\\\\ 2&&&&&-2\\\\ 0&-2&-2&\cdots&-2&-2\end\{array\}\right|\_\{n-1\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
第 $\displaystyle i$ 列加到第 $\displaystyle 1$ 列, $\displaystyle 2\leq i\leq n-1$ 得
\begin\{aligned\} D\_n=&(-1)^\{n+1\}\left|\begin\{array\}\{cccccccccc\}0&-2&&&&\\\\ 0&&-2&&&\\\\ \vdots&&&\ddots&&\\\\ 0&&&&-2&\\\\ 0&&&&&-2\\\\ -2(n-2)&-2&-2&\cdots&-2&-2\end\{array\}\right|\_\{n-1\}\\\\ =&(-1)^\{n+1\}\cdot (-1)^\{(n-1)+1\}\left\[-2(n-2)\right\] (-2)^\{n-2\} =-(n-2)(-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](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/
---
984、 3、 求行列式 $\displaystyle \left|\begin\{array\}\{cccccccccc\}1+a\_1&1&\cdots&1\\\\ 2&2+a\_2&\cdots&2\\\\ \vdots&\vdots&&\vdots\\\\ n&n&\cdots&n+a\_n\end\{array\}\right|$, 其中 $\displaystyle a\_1\cdots a\_n\neq 0$. (西北大学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|\begin\{array\}\{cccccccccc\}1&1&\cdots&1\\\\ 0&1+a\_1&1&\cdots&1\\\\ 0&2&2+a\_2&\cdots&2\\\\ \vdots&\vdots&\vdots&&\vdots\\\\ 0&n&n&\cdots&n+a\_n\end\{array\}\right| =\left|\begin\{array\}\{cccccccccc\}1&1&\cdots&1\\\\ -1&a\_1&&&\\\\ \vdots&&\ddots&\\\\ -n&&&a\_n\end\{array\}\right|\\\\ =&\left|\begin\{array\}\{cccccccccc\}1+\sum\_\{i=1\}^n \frac\{i\}\{a\_i\}&1&\cdots&1\\\\ &a\_1&&\\\\ &&\ddots&\\\\ &&&a\_n\end\{array\}\right|=\left(1+\sum\_\{i=1\}^n \frac\{i\}\{a\_i\}\right)a\_1\cdots a\_n. \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)/
---
985、 1、 (15 分) 求行列式 $\displaystyle \left|\begin\{array\}\{cccccccccc\}a&b&c&d\\\\ d&a&b&c\\\\ c&d&a&b\\\\ b&c&d&a\end\{array\}\right|$. (西南财经大学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 $\displaystyle P=\left(\begin\{array\}\{cccccccccccccccccccc\}&E\_3\\\\ 1&\end\{array\}\right)$, 则 $\displaystyle P$ 的特征值为 $\displaystyle \pm 1, \pm \mathrm\{ i\}$. 而
\begin\{aligned\} \mbox\{原式\}=&|aE+bP+cP^2+dP^3| \stackrel\{f(x)=a+bx+cx^2+dx^3\}\{=\}f(1)f(-1)f(\mathrm\{ i\})f(-\mathrm\{ i\})\\\\ =&(a+b+c+d)(a-b+c-d)[(a-c)+(b-d)i][(a-c)-(b-d)]\\\\ =&(a+b+c+d)(a-b+c-d)[(a-c)^2+(b-d)^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)/
---
986、 2、 计算 $\displaystyle n\ (\geq 3)$ 阶行列式
\begin\{aligned\} D\_n=\left|\begin\{array\}\{cccccccccc\}x&a&a&\cdots&a&a\\\\ b&c&d&\cdots&d&d\\\\ b&d&c&\cdots&d&d\\\\ \vdots&\vdots&\vdots&&\vdots&\vdots\\\\ b&d&d&\cdots&d&d\\\\ b&d&d&\cdots&d&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\}
(西南大学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) / (1)、 设
\begin\{aligned\} e=(\underbrace\{1,\cdots,1\}\_\{n-1\})^\mathrm\{T\}, A=\left(\begin\{array\}\{cccccccccccccccccccc\}c&d&\cdots&d\\\\ d&c&\cdots&d\\\\ \vdots&\vdots&\ddots&\vdots\\\\ d&d&\cdots&c\end\{array\}\right)\_\{n-1\}, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
则不妨设$A$ 可逆 (否则取一列可逆矩阵, 做到最后结果取个极限). 由
\begin\{aligned\} \left(\begin\{array\}\{cccccccccccccccccccc\}1&-ae^\mathrm\{T\} A^\{-1\}\\\\ 0&E\_\{n-1\}\end\{array\}\right)\left(\begin\{array\}\{cccccccccccccccccccc\}x&ae^\mathrm\{T\}\\\\ be&A\end\{array\}\right)\left(\begin\{array\}\{cccccccccccccccccccc\}1&0\\\\ -be A^\{-1\}&E\_\{n-1\}\end\{array\}\right)=\left(\begin\{array\}\{cccccccccccccccccccc\}x-abe^\mathrm\{T\} A^\{-1\}e&\\\\ &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
\begin\{aligned\} D\_n=(x-ab e^\mathrm\{T\} A^\{-1\}e) |A|. \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\} Ae=[c+(n-2)d]e&\Rightarrow \frac\{1\}\{c+(n-2)d\} e=A^\{-1\}e\\\\ &\Rightarrow e^\mathrm\{T\} A^\{-1\}e=\frac\{n-1\}\{c+(n-2)d\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(3)、 容易算出
\begin\{aligned\} |A|=(c-d)^\{n-2\}[c+(n-2)d]. \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\} D\_n&=\left\[x-\frac\{(n-1)ab\}\{c+(n-2)d\}\right\] (c-d)^\{n-2\}[c+(n-2)d]\\\\ &=(c-d)^\{n-2\} \left\\{x[c+(n-2)d]-(n-1)ab\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)/
---
987、 2、 求行列式 $\displaystyle D=\left|\begin\{array\}\{cccccccccc\}1&a\_1&\cdots&a\_1^\{n-3\}&a\_1^\{n-1\}&a\_1^n\\\\ 1&a\_2&\cdots&a\_2^\{n-3\}&a\_2^\{n-1\}&a\_2^n\\\\ \vdots&\vdots&&\vdots&\vdots&\vdots\\\\ 1&a\_n&\cdots&a\_n^\{n-3\}&a\_n^\{n-1\}&a\_n^n\end\{array\}\right|$. (湘潭大学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) / 比较
\begin\{aligned\} \left|\begin\{array\}\{cccccccccc\}1&a\_1&\cdots&a\_1^\{n-3\}&a\_1^\{n-2\}&a\_1^\{n-1\}&a\_1^n\\\\ 1&a\_2&\cdots&a\_2^\{n-3\}&a\_2^\{n-2\}&a\_2^\{n-1\}&a\_2^n\\\\ \vdots&\vdots&&\vdots&\vdots&\vdots&\vdots\\\\ 1&a\_n&\cdots&a\_n^\{n-3\}&a\_n^\{n-2\}&a\_n^\{n-1\}&a\_n^n\\\\ 1&x&\cdots&x^\{n-3\}&x^\{n-2\}&x^\{n-1\}&x\_n\end\{array\}\right|=\prod\_\{k=1\}^n(x-a\_k)\cdot \prod\_\{1\leq i < j\leq n\}(a\_j-a\_i) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
$\displaystyle x^\{n-2\}$ 的次数即知
\begin\{aligned\} D=(-1)^\{(n+1)+(n-1)\}D=\prod\_\{1\leq i < j\leq n\}(a\_j-a\_i)\cdot \sum\_\{1\leq i < j\leq n\}a\_ia\_j. \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)/
---
988、 3、 (10 分) 计算 $\displaystyle n$ 阶行列式 $\displaystyle D=\left|\begin\{array\}\{cccccccccc\}x\_1-m&x\_2&\cdots&x\_n\\\\ x\_1&x\_2-m&\cdots&x\_n\\\\ \vdots&\vdots&&\vdots\\\\ x\_1&x\_2&\cdots&x\_n-m\end\{array\}\right|$. (新疆大学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) / 若 $\displaystyle n=1$, 则原式 $\displaystyle =x\_1-m$. 若 $\displaystyle n\geq 2$, 则当 $\displaystyle m=0$ 时, 原式 $\displaystyle =0$; 当 $\displaystyle m\neq 0$ 时,
\begin\{aligned\} \mbox\{原式\}&=\left|\begin\{array\}\{cccccccccc\}1&x\_1&x\_2&\cdots&x\_n\\\\ 0&x\_1-m&x\_2&\cdots&x\_n\\\\ 0&x\_1&x\_2-m&\cdots&x\_n\\\\ \vdots&\vdots&\vdots&\ddots&\vdots\\\\ 0&x\_1&x\_2&\cdots&x\_n-m\end\{array\}\right|\_\{n+1\}\\\\ &=\left|\begin\{array\}\{cccccccccc\}1&x\_1&x\_2&\cdots&x\_n\\\\ -1&-m&&\cdots&\\\\ -1&x\_1&-m&\cdots&\\\\ \vdots&\vdots&\vdots&\ddots&\vdots\\\\ -1&x\_1&&\cdots&-m\end\{array\}\right|\_\{n+1\}\\\\ &=\left|\begin\{array\}\{cccccccccc\}1-\frac\{1\}\{m\}\sum\_\{i=1\}^n x\_i&x\_1&x\_2&\cdots&x\_n\\\\ 0&-m&&\cdots&\\\\ 0&x\_1&-m&\cdots&\\\\ \vdots&\vdots&\vdots&\ddots&\vdots\\\\ 0&x\_1&&\cdots&-m\end\{array\}\right|\_\{n+1\}\\\\ &=(-m)^n \left(1-\frac\{1\}\{m\} \sum\_\{i=1\}^n x\_i\right) =(-m)^\{n-1\} \left(\sum\_\{i=1\}^n x\_i-m\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
不论何种情形, 总有原式 $\displaystyle =(-m)^\{n-1\} \left(\sum\_\{i=1\}^n x\_i-m\right)$.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)/
---
989、 3、 (15 分) 设 $\displaystyle f\_n(x)=\sum\_\{k=0\}^\{n-1\}\frac\{\mathrm\{e\}^\{k+1\}\}\{(n-k-1)!\}x^\{n-k-1\}$, $\displaystyle n=1,2,\cdots,2023$, 计算
\begin\{aligned\} \left(\begin\{array\}\{cccccccccccccccccccc\} f\_1(1)&f\_1(2)&\cdots&f\_1(2023)\\\\ f\_2(1)&f\_2(2)&\cdots&f\_2(2023)\\\\ \vdots&\vdots&&\vdots\\\\ f\_\{2023\}(1)&f\_\{2023\}(2)&\cdots&f\_\{2023\}(2023) \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\}
的行列式. (云南大学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) / 由
\begin\{aligned\} f\_n(x)=\mathrm\{e\}^n \sum\_\{k=0\}^\{n-1\} \frac\{1\}\{(n-k-1)!\} \left(\frac\{x\}\{\mathrm\{e\}\}\right)^\{n-k-1\} =\mathrm\{e\}^n\sum\_\{m=0\}^\{n-1\} \frac\{1\}\{m!\}\left(\frac\{x\}\{\mathrm\{e\}\}\right)^m \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\} f\_i(j)=\mathrm\{e\}^i \sum\_\{m=0\}^\{i-1\}\frac\{1\}\{m!\}\left(\frac\{j\}\{\mathrm\{e\}\}\right)^m. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
第 $\displaystyle i$ 行提出 $\displaystyle \mathrm\{e\}^i, i=1,\cdots, 2023$; 在第 $\displaystyle i$ 行 $\displaystyle \cdot(-1)$ 加到第 $\displaystyle i+1$ 行, $\displaystyle i=n-1,\cdots,1$, 得
\begin\{aligned\} \mbox\{原式\}=&\mathrm\{e\}^\{1+\cdots+2023\}\left|\begin\{array\}\{cccccccccc\}1&1&\cdots&1\\\\ \frac\{1\}\{\mathrm\{e\}\}&\frac\{2\}\{\mathrm\{e\}\}&\cdots&\frac\{2023\}\{\mathrm\{e\}\}\\\\ \vdots&\vdots&&\vdots\\\\ \frac\{1\}\{2022!\}\left(\frac\{1\}\{\mathrm\{e\}\}\right)^\{2022\}&\frac\{1\}\{2022!\}\left(\frac\{2\}\{\mathrm\{e\}\}\right)^\{2022\}&\cdots &\frac\{1\}\{2022!\}\left(\frac\{2023\}\{\mathrm\{e\}\}\right)^\{2022\}\end\{array\}\right|\\\\ =&\frac\{\mathrm\{e\}^\{2023\}\}\{2!\cdots 2022!\}\left|\begin\{array\}\{cccccccccc\}1&1&\cdots&1\\\\ 1&2&\cdots&2023\\\\ \vdots&\vdots&&\vdots\\\\ 1&2^\{2022\}&\cdots&2023^\{2022\}\end\{array\}\right|=\mathrm\{e\}^\{2023\}. \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 09:16:51
