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/答疑/pdf1/pdf2 / 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/公众号/
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/答疑/pdf1/pdf2 / 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/公众号/
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/答疑/pdf1/pdf2 / $\displaystyle 8+7+\cdots+1=\frac{8\cdot 9}{2}=36$.跟锦数学微信公众号. 在线资料/公众号/
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/答疑/pdf1/pdf2 / (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/公众号/
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/答疑/pdf1/pdf2 / 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/公众号/
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/答疑/pdf1/pdf2 / 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/公众号/
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/答疑/pdf1/pdf2 / $\displaystyle \mbox{原式}=|B|\cdot |A^\star|=-|A|\cdot |A|^{n-1}=-2^n$.跟锦数学微信公众号. 在线资料/公众号/
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/答疑/pdf1/pdf2 /
$$\begin{aligned} \mbox{原式}=&\sum_{j=1}^4 A_{1j}+\sum_{i=2}^3 \sum_{j=1}^4 A_{ij} =\left|\begin{array}{cccccccccc}1&1&1&1\\\\ a_{21}&a_{22}&a_{23}&a_{24}\\\\ a_{31}&a_{32}&a_{33}&a_{34}\\\\ a_{41}&a_{42}&a_{43}&a_{44}\end{array}\right|+\sum_{i=2}^3 0\\\\ =&\frac{1}{2023}\left|\begin{array}{cccccccccc}a_{11}&a_{12}&a_{12}&a_{14}\\\\ a_{21}&a_{22}&a_{23}&a_{24}\\\\ a_{31}&a_{32}&a_{33}&a_{34}\\\\ a_{41}&a_{42}&a_{43}&a_{44}\end{array}\right|=\frac{a}{2023}. \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
跟锦数学微信公众号. 在线资料/公众号/
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/答疑/pdf1/pdf2 / 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/公众号/
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/答疑/pdf1/pdf2 / 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/公众号/
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/答疑/pdf1/pdf2 / 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/公众号/
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/答疑/pdf1/pdf2 / 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/公众号/
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/答疑/pdf1/pdf2 / 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/公众号/
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/答疑/pdf1/pdf2 /
$$\begin{aligned} |A|=&\left|\begin{array}{cccccccccc}1&2&3&\cdots&n\\\\ 2&2&3&\cdots&n\\\\ 3&3&3&\cdots&n\\\\ \vdots&\vdots&\vdots&&\vdots\\\\ n&n&n&\cdots&n\end{array}\right|\\\\ =&\left|\begin{array}{cccccccccc}\left(\begin{array}{cccccccccccccccccccc}1&1&1&\cdots&1\\\\ &1&1&\cdots&1\\\\ &&1&\cdots&1\\\\ &&&\ddots&\vdots\\\\ &&&&1\end{array}\right)\left(\begin{array}{cccccccccccccccccccc}-1&&&&\\\\ -1&-1&&&\\\\ \vdots&\vdots&\ddots&\\\\ -1&-1&\cdots&-1&\\\\ n&n&\cdots&n&n\end{array}\right)\end{array}\right|\\\\ =&1\cdot(-1)^{n-1}n=(-1)^{n-1}n. \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
跟锦数学微信公众号. 在线资料/公众号/
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/答疑/pdf1/pdf2 / (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/公众号/
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/答疑/pdf1/pdf2 / 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/公众号/
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/答疑/pdf1/pdf2 / 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/公众号/
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/答疑/pdf1/pdf2 /
$$\begin{aligned} \mbox{原式}=&\left|\begin{array}{cccccccccc}1&1&\cdots&1\\\\ 0&1+a_1&1&\cdots&1\\\\ 0&2&2+a_2&\cdots&2\\\\ \vdots&\vdots&\vdots&&\vdots\\\\ 0&n&n&\cdots&n+a_n\end{array}\right| =\left|\begin{array}{cccccccccc}1&1&\cdots&1\\\\ -1&a_1&&&\\\\ \vdots&&\ddots&\\\\ -n&&&a_n\end{array}\right|\\\\ =&\left|\begin{array}{cccccccccc}1+\sum_{i=1}^n \frac{i}{a_i}&1&\cdots&1\\\\ &a_1&&\\\\ &&\ddots&\\\\ &&&a_n\end{array}\right|=\left(1+\sum_{i=1}^n \frac{i}{a_i}\right)a_1\cdots a_n. \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
跟锦数学微信公众号. 在线资料/公众号/
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/答疑/pdf1/pdf2 / 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/公众号/
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/答疑/pdf1/pdf2 / (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/公众号/
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/答疑/pdf1/pdf2 / 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/公众号/
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/答疑/pdf1/pdf2 / 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/公众号/
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/答疑/pdf1/pdf2 / 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/公众号/
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