## 张祖锦2023年数学专业真题分类70天之第42天
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944、 3、 计算题 (每题 10 分, 共 50 分). (1)、 设 $\displaystyle A=(a\_\{ij\})$ 为 $\displaystyle n$ 阶实方阵, $\displaystyle n\geq 2$, 其中 $\displaystyle a\_\{ij\}=\left\\{\begin\{array\}\{llllllllllll\}\frac\{i\}\{j\},&i\neq j,\\\\ 0,&i=j,\end\{array\}\right.$ $\displaystyle 1\leq i,j\leq n$, 求 $\displaystyle |A|$. (安徽大学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](
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https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / $\displaystyle |A|=\left|\begin\{array\}\{cccccccccc\}0&\frac\{1\}\{2\}&\frac\{1\}\{3\}&\cdots&\frac\{1\}\{n\}\\\\ 2&0&\frac\{2\}\{3\}&\cdots&\frac\{2\}\{n\}\\\\ 3&\frac\{3\}\{2\}&0&\cdots&\frac\{3\}\{n\}\\\\ \vdots&\vdots&\vdots&&\vdots\\\\ n&\frac\{n\}\{2\}&\frac\{n\}\{3\}&\cdots&0\end\{array\}\right|$. 第 $\displaystyle 1$ 行 $\displaystyle \cdot(-i)$ 加到第 $\displaystyle i$ 行, $\displaystyle 2\leq i\leq n$, 得
\begin\{aligned\} |A|=\left|\begin\{array\}\{cccccccccc\}0&\frac\{1\}\{2\}&\frac\{1\}\{3\}&\cdots&\frac\{1\}\{n\}\\\\ 2&-1&&&\\\\ 3&&-1&&\\\\ \vdots&&&\ddots&\\\\ 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\}
第 $\displaystyle i$ 列 $\displaystyle \cdot i$ 加到第 $\displaystyle 1$ 列, 得
\begin\{aligned\} |A|=\left|\begin\{array\}\{cccccccccc\}n-1&\frac\{1\}\{2\}&\frac\{1\}\{3\}&\cdots&\frac\{1\}\{n\}\\\\ 0&-1&&&\\\\ 0&&-1&&\\\\ \vdots&&&\ddots&\\\\ 0&&&&-1\end\{array\}\right|=(-1)^\{n-1\}(n-1). \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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945、 1、 (20 分) 设 $\displaystyle f(x)=x^n-x, n\geq 2$, 求下列 $\displaystyle n-1$ 阶行列式
\begin\{aligned\} \left|\begin\{array\}\{cccccccccc\}f\_n(x)&f\_\{n-1\}(x)&\cdots&f\_2(x)\\\\ f\_n(x+1)&f\_\{n-1\}(x+1)&\cdots&f\_2(x+1)\\\\ \vdots&\vdots&&\vdots\\\\ f\_n(x+n-2)&f\_\{n-1\}(x+n-2)&\cdots&f\_2(x+n-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\}
(北京工业大学2023年高等代数考研试题) [行列式 ]
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\begin\{aligned\} \mbox\{原式\}=&\left|\begin\{array\}\{cccccccccc\}1&x&x&\cdots&x\\\\ 0&f\_n(x)&f\_\{n-1\}(x)&\cdots&f\_2(x)\\\\ 0&f\_n(x+1)&f\_\{n-1\}(x+1)&\cdots&f\_2(x+1)\\\\ \vdots&\vdots&\vdots&&\vdots\\\\ 0&f\_n(x+n-2)&f\_\{n-1\}(x+n-2)&\cdots&f\_2(x+n-2)\end\{array\}\right|\\\\ =&\left|\begin\{array\}\{cccccccccc\}1&x&x&\cdots&x\\\\ 1&x^n&x^\{n-1\}&\cdots&x^2\\\\ 1&(x+1)^n&(x+1)^\{n-1\}&\cdots&(x+1)^2\\\\ \vdots&\vdots&\vdots&&\vdots\\\\ 1&(x+n-2)^n&(x+n-2)^\{n-1\}&\cdots&(x+n-2)^2\end\{array\}\right|\ \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
第 $\displaystyle n$ 列与前面 $\displaystyle n-2$ 列交换后, 第 $\displaystyle n$ 列与前面 $\displaystyle n-3$ 列交换, 等等,
\begin\{aligned\} \mbox\{原式\}=&(-1)^\{\sum\_\{i=1\}^\{n-2\}i\} \left|\begin\{array\}\{cccccccccc\}1&1&1&\cdots&1\\\\ x&x^2&x^3&\cdots&x^n\\\\ x+1&(x+1)^2&(x+1)^3&\cdots&(x+1)^n\\\\ \vdots&\vdots&\vdots&&\vdots\\\\ x+n-2&(x+n-2)^2&(x+n-2)^3&\cdots&(x+n-2)^n\end\{array\}\right|\\\\ =&(-1)^\frac\{(n-1)(n-2)\}\{2\} x(x+1)\cdots(x+n-2)\\\\ &\left|\begin\{array\}\{cccccccccc\}1&1&1&\cdots&1\\\\ 1&x&x^2&\cdots&x^\{n-1\}\\\\ 1&(x+1)&(x+1)^2&\cdots&(x+1)^\{n-1\}\\\\ \vdots&\vdots&\vdots&&\vdots\\\\ 1&(x+n-2)&(x+n-2)^2&\cdots&(x+n-2)^\{n-1\}\end\{array\}\right|\\\\ =&(-1)^\frac\{(n-1)(n-2)\}\{2\}\prod\_\{k=0\}^\{n-2\}(x+k) \left\\{\prod\_\{k=0\}^\{n-2\}(x+k-1)\cdot \prod\_\{0\leq i < j\leq n-2\}[(x+j)-(x-i)]\right\\}\\\\ =&(-1)^\frac\{(n-1)(n-2)\}\{2\}\prod\_\{k=0\}^\{n-2\} [(x+k)(x+k-1)]\cdot \prod\_\{j=1\}^\{n-2\}j!. \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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946、 1、 (15 分) 计算行列式
\begin\{aligned\} \left|\begin\{array\}\{cccccccccc\}2&2&2&\cdots&2\\\\ x\_1(x\_1-2)&x\_2(x\_2-2)&x\_3(x\_3-2)&\cdots&x\_n(x\_n-2)\\\\ x\_1^2(x\_1-2)&x\_2^2(x\_2-2)&x\_3^2(x\_3-2)&\cdots&x\_n^2(x\_n-2)\\\\ \vdots&\vdots&\vdots&&\vdots\\\\ x\_1^\{n-1\}(x\_1-2)&x\_2^\{n-2\}(x\_2-2)&x\_3^\{n-1\}(x\_3-2)&\cdots&x\_n^\{n-1\}(x\_n-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\}
(北京科技大学2023年高等代数考研试题) [行列式 ]
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https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 提出第一行的公因子 $\displaystyle 2$, 并加边得
\begin\{aligned\} \mbox\{原式\}=-2\left|\begin\{array\}\{cccccccccc\}0&1&1&1&\cdots&1\\\\ 1&x\_1&x\_2&x\_3&\cdots&x\_n\\\\ 0&x\_1(x\_1-2)&x\_2(x\_2-2)&x\_3(x\_3-2)&\cdots&x\_n(x\_n-2)\\\\ 0&x\_1^2(x\_1-2)&x\_2^2(x\_2-2)&x\_3^2(x\_3-2)&\cdots&x\_n^2(x\_n-2)\\\\ \vdots&\vdots&\vdots&\vdots&&\vdots\\\\ 0&x\_1^\{n-1\}(x\_1-2)&x\_2^\{n-2\}(x\_2-2)&x\_3^\{n-1\}(x\_3-2)&\cdots&x\_n^\{n-1\}(x\_n-2)\end\{array\}\right| \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
第 $\displaystyle i$ 行 $\displaystyle \cdot 2$ 加到第 $\displaystyle i+1$ 行, $\displaystyle i=2,\cdots,n-1$, 得
\begin\{aligned\} \mbox\{原式\}=-2\left|\begin\{array\}\{cccccccccc\}0&1&1&1&\cdots&1\\\\ 1&x\_1&x\_2&x\_3&\cdots&x\_n\\\\ 2&x\_1^2&x\_2^2&x\_3^2&\cdots&x\_n^2\\\\ 2^2&x\_1^3&x\_2^3&x\_3^3&\cdots&x\_n^3\\\\ \vdots&\vdots&\vdots&\vdots&&\vdots\\\\ 2^\{n-1\}&x\_1^n&x\_2^n&x\_3^n&\cdots&x\_n^n\end\{array\}\right| =-\left|\begin\{array\}\{cccccccccc\}0&1&1&1&\cdots&1\\\\ 2&x\_1&x\_2&x\_3&\cdots&x\_n\\\\ 2^2&x\_1^2&x\_2^2&x\_3^2&\cdots&x\_n^2\\\\ 2^3&x\_1^3&x\_2^3&x\_3^3&\cdots&x\_n^3\\\\ \vdots&\vdots&\vdots&\vdots&&\vdots\\\\ 2^n&x\_1^n&x\_2^n&x\_3^n&\cdots&x\_n^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\}
将第一列拆分得
\begin\{aligned\} \mbox\{原式\}=&-\left|\begin\{array\}\{cccccccccc\}1&1&1&1&\cdots&1\\\\ 2&x\_1&x\_2&x\_3&\cdots&x\_n\\\\ 2^2&x\_1^2&x\_2^2&x\_3^2&\cdots&x\_n^2\\\\ 2^3&x\_1^3&x\_2^3&x\_3^3&\cdots&x\_n^3\\\\ \vdots&\vdots&\vdots&\vdots&&\vdots\\\\ 2^n&x\_1^n&x\_2^n&x\_3^n&\cdots&x\_n^n\end\{array\}\right| -\left|\begin\{array\}\{cccccccccc\}-1&1&1&1&\cdots&1\\\\ 0&x\_1&x\_2&x\_3&\cdots&x\_n\\\\ 0&x\_1^2&x\_2^2&x\_3^2&\cdots&x\_n^2\\\\ 0&x\_1^3&x\_2^3&x\_3^3&\cdots&x\_n^3\\\\ \vdots&\vdots&\vdots&\vdots&&\vdots\\\\ 0&x\_1^n&x\_2^n&x\_3^n&\cdots&x\_n^n\end\{array\}\right|\\\\ =&-\prod\_\{i=1\}^n (x\_i-2)\cdot \prod\_\{1\leq i < j\leq n\}(x\_j-x\_i) +\prod\_\{i=1\}^n x\_i \cdot \prod\_\{1\leq i < j\leq n\}(x\_j-x\_i)\\\\ =&\left\[\prod\_\{i=1\}^n x\_i-\prod\_\{i=1\}^n (x\_i-2)\right\]\cdot \prod\_\{1\leq i < j\leq n\}(x\_j-x\_i). \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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947、 1、 填空题. (1)、 计算行列式: $\displaystyle \left|\begin\{array\}\{cccccccccc\}a\_1&b\_2&b\_3&b\_4\\\\ a\_2&-1&0&0\\\\ a\_3&0&-1&0\\\\ a\_4&0&0&-1\end\{array\}\right|=\underline\{\ \ \ \ \ \ \ \ \ \ \}$. (北京理工大学2023年高等代数考研试题) [行列式 ]
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https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
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https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 第 $\displaystyle i$ 列 $\displaystyle \cdot a\_i$ 加到第 $\displaystyle 1$ 列, $\displaystyle i=2,3,4$, 得
\begin\{aligned\} \mbox\{原式\}=&\left|\begin\{array\}\{cccccccccc\}a\_1+a\_2b\_2+a\_3b\_3+a\_4b\_4&b\_2&b\_3&b\_4\\\\ 0&-1&0&0\\\\ 0&0&-1&0\\\\ 0&0&0&-1\end\{array\}\right|\\\\ =&-(a\_1+a\_2b\_2+a\_3b\_3+a\_4b\_4). \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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---
948、 1、 求 $\displaystyle \left|\begin\{array\}\{cccccccccc\}1+a&2&\cdots&n\\\\ 1&2+a&\cdots&n\\\\ \vdots&\vdots&&\vdots\\\\ 1&2&\cdots&n+a\end\{array\}\right|$. (北京师范大学2023年高等代数与解析几何考研试题) [行列式 ]
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\begin\{aligned\} \mbox\{原式\}=&\left|\begin\{array\}\{cccccccccc\}1+a&2&\cdots&n\\\\ -a&a&\cdots&0\\\\ \vdots&\vdots&&\vdots\\\\ -a&0&\cdots&a\end\{array\}\right|=a^\{n-1\}\left|\begin\{array\}\{cccccccccc\}1+a&2&\cdots&n\\\\ -1&1&\cdots&0\\\\ \vdots&\vdots&&\vdots\\\\ -1&0&\cdots&1\end\{array\}\right|\\\\ =&a^\{n-1\}\left|\begin\{array\}\{cccccccccc\}a+\frac\{n(n+1)\}\{2\}&2&\cdots&n\\\\ 0&1&\cdots&0\\\\ \vdots&\vdots&&\vdots\\\\ 0&0&\cdots&1\end\{array\}\right|=a^\{n-1\}\left\[a+\frac\{n(n+1)\}\{2\}\right\]. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
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---
949、 1、 计算题. (1)、 计算行列式
\begin\{aligned\} \det(|i-j|)=\left|\begin\{array\}\{cccccccccc\}0&1&2&\cdots&n-2&n-1\\\\ 1&0&1&\cdots&n-3&n-2\\\\ 2&1&0&\cdots&n-4&n-3\\\\ \vdots&\vdots&\vdots&&\vdots&\vdots\\\\ n-2&n-3&n-4&\cdots&0&1\\\\ n-1&n-2&n-3&\cdots&1&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\}
(大连理工大学2023年高等代数考研试题) [行列式 ]
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\begin\{aligned\} \mbox\{原式\}& =\left|\begin\{array\}\{cccccccccc\} 0&1&2&\cdots&n-1\\\\ 1&-1&-1&\cdots&-1\\\\ \vdots&\vdots&\vdots&\ddots&\vdots\\\\ 1&1&1&\cdots&-1 \end\{array\}\right|\\\\ &\quad \left(\mbox\{第 $\displaystyle i$ 行乘以 $\displaystyle -1$ 加到第 $\displaystyle i+1$ 行, $\displaystyle i=n-1,\cdots,1$\}\right)\\\\ &=\left|\begin\{array\}\{cccccccccc\} n-1&n&n+1&\cdots&n-1\\\\ 0&-2&-2&\cdots&-1\\\\ 0&0&-2&\cdots&-1\\\\ \vdots&\vdots&\vdots&\ddots&\vdots\\\\ 0&0&0&\cdots&-1 \end\{array\}\right|\left(\begin\{array\}\{c\}\mbox\{第 $\displaystyle n$ 列加到第 $\displaystyle i$ 列,\}\\\\ i=1,2,\cdots,n-1\end\{array\}\right)\\\\ &=(n-1)(-2)^\{n-2\}(-1) =(-1)^\{n-3\}(n-1)2^\{n-2\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
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---
950、 1、 填空题 (共 30 分, 每题 5 分). (1)、 行列式
\begin\{aligned\} D\_5=\left|\begin\{array\}\{cccccccccc\}1&2&3&4&5\\\\ 2&1&2&3&4\\\\ 3&2&1&2&3\\\\ 4&3&2&1&2\\\\ 5&4&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\}
的值为 $\displaystyle \underline\{\ \ \ \ \ \ \ \ \ \ \}$. (电子科技大学2023年高等代数考研试题) [行列式 ]
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https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
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https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 第 $\displaystyle i$ 行 $\displaystyle \cdot(-1)$ 加到第 $\displaystyle i+1$ 行, $\displaystyle i=4,3,2,1$; 第 $\displaystyle i$ 列 $\displaystyle \cdot (-1)$ 加到第 $\displaystyle i+1$ 列, $\displaystyle i=4,3,2,1$; 第 $\displaystyle i$ 行 $\displaystyle \cdot \frac\{1\}\{2\}$ 加到第 $\displaystyle 1$ 列, $\displaystyle i=2,3,4,5$ 得
\begin\{aligned\} D\_5=&\left|\begin\{array\}\{cccccccccc\}1&2&3&4&5\\\\ 1&-1&-1&-1&-1\\\\ 1&1&-1&-1&-1\\\\ 1&1&1&-1&-1\\\\ 1&1&1&1&-1\end\{array\}\right|=\left|\begin\{array\}\{cccccccccc\}1&1&1&1&1\\\\ 1&-2&&&\\\\ 1&&-2&&\\\\ 1&&&-2&\\\\ 1&&&&-2\end\{array\}\right|\\\\ =&\left(1+\frac\{4\}\{2\}\right)\cdot (-2)^4=48. \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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---
951、 2、 设行列式
\begin\{aligned\} D=\left|\begin\{array\}\{cccccccccc\}a&b&c&d\\\\ -b&a&-d&c\\\\ -c&d&a&-b\\\\ -d&-c&b&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\}
(1)、 求 $\displaystyle D$ 中第一行第一列元素的余子式; (2)、 求 $\displaystyle A\_\{11\}+A\_\{12\}+A\_\{13\}+A\_\{14\}$; (3)、 求行列式 $\displaystyle D$. (东北大学2023年高等代数考研试题) [行列式 ]
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https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
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https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / (1)、 $\displaystyle D$ 中第一行第一列元素的余子式为
\begin\{aligned\} \left|\begin\{array\}\{cccccccccc\}a&-d&c\\\\ d&a&-b\\\\ -c&b&a\end\{array\}\right| =&a^2(a^2+b^2)+d(ad-bc)+c(bd+ac)\\\\ =&a(a^2+b^2+c^2+d^2). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(2)、 $\displaystyle A\_\{11\}+A\_\{12\}+A\_\{13\}+A\_\{14\}$
\begin\{aligned\} =&\left|\begin\{array\}\{cccccccccc\}1&1&1&1\\\\ -b&a&-d&c\\\\ -c&d&a&-b\\\\ -d&-c&b&a\end\{array\}\right|\\\\ =&\left|\begin\{array\}\{cccccccccc\}a&-d&c\\\\ d&a&-b\\\\ -c&b&a\end\{array\}\right|+\left|\begin\{array\}\{cccccccccc\}b&-d&c\\\\ c&a&-b\\\\ d&b&a\end\{array\}\right|+\left|\begin\{array\}\{cccccccccc\}-b&a&c\\\\ -c&d&-b\\\\ -d&-c&a\end\{array\}\right|+\left|\begin\{array\}\{cccccccccc\}b&a&-d\\\\ c&d&a\\\\ d&-c&b\end\{array\}\right|\\\\ =&a(a^2+b^2+c^2+d^2)+b(a^2+b^2+c^2+d^2)\\\\ &+c(a^2+b^2+c^2+d^2)+d(a^2+b^2+c^2+d^2)\\\\ =&(a+b+c+d)(a^2+b^2+c^2+d^2). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(3)、 为求行列式 $\displaystyle D$, 我们给出一个常用的行列式计算公式. 设 $\displaystyle A,B,C,D$ 为同级方阵且 $\displaystyle AC=CA$, 则
\begin\{aligned\} \left|\begin\{array\}\{cccccccccc\}A&B\\\\ C&D\end\{array\}\right|=|AD-CB|. \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, (3-1)、 若 $\displaystyle A$ 可逆, 则由
\begin\{aligned\} \left(\begin\{array\}\{cccccccccccccccccccc\} E&0\\\\ -CA^\{-1\}&E\end\{array\}\right)\left(\begin\{array\}\{cccccccccccccccccccc\} A&B\\\\ C&D\end\{array\}\right)\left(\begin\{array\}\{cccccccccccccccccccc\} E&-A^\{-1\}B\\\\ 0&E\end\{array\}\right)=\left(\begin\{array\}\{cccccccccccccccccccc\} A&0\\\\ 0&D-CA^\{-1\}B\end\{array\}\right) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知
\begin\{aligned\} \left|\begin\{array\}\{cccccccccc\} A&B\\\\ C&D\end\{array\}\right|&=|A|\cdot |D-CA^\{-1\}B|\\\\ &=|A|\cdot |D-A^\{-1\}CB|\left(AC=CA\Rightarrow CA^\{-1\}=A^\{-1\}C\right)\\\\ &=|A(D-A^\{-1\}CB|\left(|AB|=|A|\cdot |B|\right)\\\\ &=|AD-CB|. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(3-2)、 若 $\displaystyle A$ 奇异, 则关于 $\displaystyle \lambda$ 的多项式 $\displaystyle |\lambda E+A|=0$ 至多有 $\displaystyle n$ 个复根 $\displaystyle \lambda\_1,\cdots,\lambda\_n$. 而 \begin\{aligned\} &\forall\ \lambda\not\in \left\\{\lambda\_1,\cdots,\lambda\_n\right\\}, |\lambda E+A|\neq 0\\\\ \Rightarrow&\tilde\{A\}=\lambda E+A\mbox\{可逆, 满足\}\tilde\{A\} C=C\tilde\{A\}\\\\ \Rightarrow&\left|\begin\{array\}\{cccccccccc\} \tilde\{A\}&B\\\\ C&D\end\{array\}\right|=|\tilde\{A\} D-CB|\left(\mbox\{由 (1)\}\right)\\\\ \Rightarrow& \left|\begin\{array\}\{cccccccccc\} \lambda E+A&B\\\\ C&D\end\{array\}\right|=|(\lambda E+A)D-CB|. \qquad(210226: eq)\tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
设 $\displaystyle f(\lambda)\equiv \left|\begin\{array\}\{cccccccccc\} \lambda E+A&B\\\\ C&D\end\{array\}\right|-|(\lambda E+A)D-CB|$. 若 $\displaystyle f(\lambda)\not\equiv 0$, 则 $\displaystyle f(\lambda)$ 是次数 $\displaystyle \leq n$ 的多项式, 至多有 $\displaystyle n$ 个复根. 这与 (210226: eq) 矛盾. 故
\begin\{aligned\} &f(\lambda)\equiv 0, \forall\ \lambda\in\mathbb\{C\}\\\\ \Rightarrow&f(0)=0\Rightarrow \left|\begin\{array\}\{cccccccccc\}A&B\\\\ C&D\end\{array\}\right|=|AD-CB|. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(4)、 由第 3 步知
\begin\{aligned\} D=&\left|\left(\begin\{array\}\{cccccccccccccccccccc\}a&b\\\\-b&a\end\{array\}\right)\left(\begin\{array\}\{cccccccccccccccccccc\}a&-b\\\\ b&a\end\{array\}\right)-\left(\begin\{array\}\{cccccccccccccccccccc\}c&d\\\\ -d&c\end\{array\}\right)\left(\begin\{array\}\{cccccccccccccccccccc\}-c&d\\\\ -d&-c\end\{array\}\right)\right|\\\\ =&\left|\begin\{array\}\{cccccccccc\}a^2+b^2+c^2+d^2&\\\\ &a^2+b^2+c^2+d^2\end\{array\}\right|=(a^2+b^2+c^2+d^2)^2. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
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---
952、 1、 (15 分) 求下列行列式的值: $\displaystyle D=\left|\begin\{array\}\{cccccccccc\}1&1&1&\cdots&1&1\\\\ x\_1&x\_2&x\_3&\cdots&x\_\{n-1\}&x\_n\\\\ x\_1^2&x\_2^2&x\_3^2&\cdots&x\_\{n-1\}^2&x\_n^2\\\\ \vdots&\vdots&\vdots&&\vdots&\vdots\\\\ x\_1^\{n-2\}&x\_2^\{n-2\}&x\_3^\{n-2\}&\cdots&x\_\{n-1\}^\{n-2\}&x\_n^\{n-2\}\\\\ x\_1^\{n-1\}&x\_2^\{n-1\}&x\_3^\{n-1\}&\cdots&x\_\{n-1\}^\{n-1\}&x\_n^\{n-1\}\end\{array\}\right|$. (东北师范大学2023年高等代数与解析几何考研试题) [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\}\{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\}
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---
953、 (2)、 若 $\displaystyle f(x)=\left|\begin\{array\}\{cccccccccc\}5x&x&1&x\\\\ 1&x&1&-1\\\\ 3&2&x&1\\\\ 3&1&1&x\end\{array\}\right|$, 则 $\displaystyle x^4$ 的系数为 $\displaystyle \underline\{\ \ \ \ \ \ \ \ \ \ \}$, $\displaystyle x^3$ 的系数为 $\displaystyle \underline\{\ \ \ \ \ \ \ \ \ \ \}$. (广西大学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](
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https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / $\displaystyle 5,-4$. 直接计算即可.跟锦数学微信公众号. [在线资料](
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---
954、 2、 计算 $\displaystyle n$ 阶行列式
\begin\{aligned\} D=\left|\begin\{array\}\{cccccccccc\}1&2&3&\cdots&n-1&n\\\\ 2&3&4&\cdots&n&1\\\\ 3&4&5&\cdots&1&2\\\\ \vdots&\vdots&\vdots&&\vdots&\vdots\\\\ n&1&2&\cdots&n-2&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\}
的值. (广西大学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](
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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=\left|\begin\{array\}\{cccccccccc\}1&2&3&\cdots&n-1&n\\\\ 1&1&1&\cdots&1&1-n\\\\ 1&1&1&\cdots&1-n&1\\\\ \vdots&\vdots&\vdots&&\vdots&\vdots\\\\ 1-n&1&1&\cdots&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\}
第 $\displaystyle i$ 列加到第 $\displaystyle 1$ 列, $\displaystyle 2\leq i\leq n$, 得
\begin\{aligned\} D=&\left|\begin\{array\}\{cccccccccc\}\frac\{n(n+1)\}\{2\}&2&3&\cdots&n-1&n\\\\ 0&1&1&\cdots&1&1-n\\\\ 0&1&1&\cdots&1-n&1\\\\ \vdots&\vdots&\vdots&&\vdots&\vdots\\\\ 0&1&1&\cdots&1&1\end\{array\}\right|\\\\ =&\frac\{n(n+1)\}\{2\}\left|\begin\{array\}\{cccccccccc\}2&3&\cdots&n-1&n\\\\ 1&1&\cdots&1&1-n\\\\ 1&1&\cdots&1-n&1\\\\ \vdots&\vdots&&\vdots&\vdots\\\\ 1&1&\cdots&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\}
第 $\displaystyle n$ 行 $\displaystyle \cdot(-1)$ 加到第 $\displaystyle i$ 行, $\displaystyle 1\leq i\leq n-1$, 得
\begin\{aligned\} D=&\frac\{n(n+1)\}\{2\}\left|\begin\{array\}\{cccccccccc\}0&0&\cdots&0&-n\\\\ 0&0&\cdots&-n&0\\\\ \vdots&\vdots&&\vdots&\vdots\\\\ 0&-n&\cdots&0&0\\\\ 1&1&\cdots&1&1\end\{array\}\right|\\\\ =&\frac\{n(n+1)\}\{2\} (-1)^\{n-2\}(-1)^\{\tau(n,n-1,\cdots,1)\} =(-1)^\frac\{n(n-1)\}\{2\} \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\}
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---
955、 (2)、 行列式 $\displaystyle \left|\begin\{array\}\{cccccccccc\}2&0&0&0&-6\\\\ 0&3&0&0&9\\\\ 0&0&1&0&2\\\\ 0&0&0&5&-10\\\\ 5&5&6&6&1\end\{array\}\right|$ 的值为 $\displaystyle \underline\{\ \ \ \ \ \ \ \ \ \ \}$. (哈尔滨工程大学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 $\displaystyle 1$ 列乘以 $\displaystyle 3$ 加到第 $\displaystyle 5$ 列, 第 $\displaystyle 2$ 列乘以 $\displaystyle -3$ 加到第 $\displaystyle 5$ 列, 第 $\displaystyle 3$ 列乘以 $\displaystyle -2$ 加到第 $\displaystyle 5$ 列, 第 $\displaystyle 4$ 列乘以 $\displaystyle 2$ 加到第 $\displaystyle 5$ 列后就化为了下三角矩阵的行列式, 最终结果为
\begin\{aligned\} 2\cdot 3\cdot 1\cdot 5\cdot (1+3\cdot 5-3\cdot 5-2\cdot 6+2\cdot 6)=30. \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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---
956、 4、 若
\begin\{aligned\} A=\left(\begin\{array\}\{cccccccccccccccccccc\} a\_\{11\}&a\_\{12\}&\cdots&a\_\{1n\}\\\\ a\_\{21\}&a\_\{22\}&\cdots&a\_\{2n\}\\\\ \vdots&\vdots&&\vdots\\\\ a\_\{n1\}&a\_\{n2\}&\cdots&a\_\{nn\}\end\{array\}\right), F=\left(\begin\{array\}\{cccccccccccccccccccc\}1&1&\cdot&1\\\\ 1&1&\cdots&1\\\\ \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\}
(1)、 证明: $\displaystyle |A+xF|=|A|+x\sum\_\{j=1\}^n \sum\_\{i=1\}^n A\_\{ij\}$. (2)、 证明:
\begin\{aligned\} \sum\_\{j=1\}^n \sum\_\{i=1\}^n A\_\{ij\} =\left|\begin\{array\}\{cccccccccc\}1&a\_\{12\}-a\_\{11\}&\cdots&a\_\{1n\}-a\_\{1,n-1\}\\\\ 1&a\_\{22\}-a\_\{21\}&\cdots&a\_\{2n\}-a\_\{2,n-1\}\\\\ \vdots&\vdots&&\vdots\\\\ 1&a\_\{n2\}-a\_\{n1\}&\cdots&a\_\{nn\}-a\_\{n,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\}
(哈尔滨工业大学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](
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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 $\displaystyle e=(1,\cdots,1)^\mathrm\{T\}$, 则当 $\displaystyle A$ 可逆时,
\begin\{aligned\} &|A+xee^\mathrm\{T\}|=|A|\cdot |E+xA^\{-1\}e\cdot \mathrm\{e\}^\mathrm\{T\}|\\\\ \overset\{\tiny\mbox\{第1步\}\}\{=\}& |A|\left(1+xe^\mathrm\{T\} \cdot A^\{-1\}e\right) =|A|+xe^\mathrm\{T\} A^\star e. \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 $\displaystyle a\_\{ij\}$ 的连续函数, 而对一般的 $\displaystyle A$, 还有
\begin\{aligned\} |A+xee^\mathrm\{T\}|=|A|+xe^\mathrm\{T\} A^\star e=|A|+\sum\_\{j=1\}^n \sum\_\{i=1\}^n A\_\{ij\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(3)、 取 $\displaystyle x=1$, 由第 1 问结果知
\begin\{aligned\} \sum\_\{j=1\}^n \sum\_\{i=1\}^n A\_\{ij\}=|A+F|-|A|. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
利用行列式公式 $\displaystyle |\alpha+\beta,\cdots|=|\alpha,\cdots|+|\beta,\cdots|$ 知
\begin\{aligned\} \mbox\{左端\}=&\left|\begin\{array\}\{cccccccccc\}1&a\_\{12\}&\cdots&a\_\{1n\}\\\\ 1&a\_\{22\}&\cdots&a\_\{2n\}\\\\ \vdots&\vdots&&\vdots\\\\ 1&a\_\{n2\}&\cdots&a\_\{nn\}\end\{array\}\right|+\cdots+\left|\begin\{array\}\{cccccccccc\}a\_\{11\}&\cdots&a\_\{1,n-1\}&1\\\\ a\_\{21\}&\cdots&a\_\{2,n-1\}&1\\\\ \vdots&\vdots&&\vdots\\\\ a\_\{n1\}&\cdots&a\_\{n,n-1\}&1\end\{array\}\right|\\\\ \equiv&A\_1+\cdots+A\_n.\qquad(I) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
此外,
\begin\{aligned\} \mbox\{右端\}=&\left|\begin\{array\}\{cccccccccc\}1&0&0&\cdots&0\\\\ a\_\{11\}&1&a\_\{12\}-a\_\{11\}&\cdots&a\_\{1n\}-a\_\{1,n-1\}\\\\ a\_\{21\}&1&a\_\{22\}-a\_\{21\}&\cdots&a\_\{2n\}-a\_\{2,n-1\}\\\\ \vdots&\vdots&\vdots&&\vdots\\\\ a\_\{n1\}&1&a\_\{n2\}-a\_\{n1\}&\cdots&a\_\{nn\}-a\_\{n,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\}
第 $\displaystyle 1$ 列加到第 $\displaystyle 3$ 列, 第 $\displaystyle i$ 列加到第 $\displaystyle i+1$ 列, $\displaystyle i=3,\cdots,n-1$, 得
\begin\{aligned\} \mbox\{右端\}=\left|\begin\{array\}\{cccccccccc\}1&0&1&\cdots&1\\\\ a\_\{11\}&1&a\_\{12\}&\cdots&a\_\{1n\}\\\\ a\_\{21\}&1&a\_\{22\}&\cdots&a\_\{2n\}\\\\ \vdots&\vdots&\vdots&&\vdots\\\\ a\_\{n1\}&1&a\_\{n2\}&\cdots&a\_\{nn\}\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 $\displaystyle 1$ 行展开知 [回忆 $\displaystyle (I)$ 的记号]
\begin\{aligned\} \mbox\{右端\}=&A\_1+\sum\_\{i=2\}^n(-1)^\{1+(i+1)\} \left|\begin\{array\}\{cccccccccc\}a\_\{11\}&1&a\_\{12\}&\cdots&a\_\{1,i-1\}&a\_\{1,i+1\}&\cdots&a\_\{1n\}\\\\ a\_\{21\}&1&a\_\{22\}&\cdots&a\_\{1,i-1\}&a\_\{2,i+1\}&\cdots&a\_\{2n\}\\\\ \vdots&\vdots&\vdots&&\vdots&\vdots&&\vdots\\\\ a\_\{n1\}&1&a\_\{n2\}&\cdots&a\_\{n,i-1\}&a\_\{n,i+1\}&\cdots&a\_\{nn\}\end\{array\}\right|\\\\ =&A\_1+\sum\_\{i=2\}^n A\_i\stackrel\{(I)\}\{=\}\mbox\{左端\}. \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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957、 2、 设 $\displaystyle D\_n=\left|\begin\{array\}\{cccccccccc\}1&1&1&\cdots&1\\\\ a\_1&a\_2&a\_3&\cdots&a\_n\\\\ a\_1^2&a\_2^2&a\_3^2&\cdots&a\_n^2\\\\ \vdots&\vdots&\vdots&&\vdots\\\\ a\_1^\{n-1\}&a\_2^\{n-1\}&a\_3^\{n-1\}&\cdots&a\_n^\{n-1\}\end\{array\}\right|$. 求 $\displaystyle \sum\_\{i=1\}^n \sum\_\{j=1\}^n A\_\{ij\}$, 其中 $\displaystyle A\_\{ij\}$ 是 $\displaystyle D\_n$ 中 $\displaystyle (i,j)$ 元的代数余子式. (河北工业大学2023年高等代数考研试题) [行列式 ]
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\begin\{aligned\} \mbox\{原式\}=&\sum\_\{j=1\}^n A\_\{1j\}+\sum\_\{j=1\}^n A\_\{2j\}+\cdots+\sum\_\{j=1\}^n A\_\{nj\}\\\\ =&\sum\_\{j=1\}^n A\_\{1j\}+0+\cdots+0\left(\mbox\{后面的 $\displaystyle n-1$ 个行列式第 $\displaystyle 1, i$ 行的元素全为 $\displaystyle 1$\}\right)\\\\ =&D\_n=\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\}
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---
958、 1、 计算 $\displaystyle n$ 阶行列式
\begin\{aligned\} D\_n=\left|\begin\{array\}\{cccccccccc\}1&1&\cdots&1\\\\ 1&2&\cdots&n\\\\ \vdots&\vdots&&\vdots\\\\ 1&2^\{n-2\}&\cdots&n^\{n-2\}\\\\ 1&2^n&\cdots&n^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\}
(湖南大学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)、 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)、 回到题目. 由第 1 步知
\begin\{aligned\} &D\_n=\prod\_\{1\leq i < j\leq n\}(j-i)\cdot \sum\_\{i=1\}^n i\\\\ =&\prod\_\{j=2\}^n (j-1)! \cdot \frac\{n(n+1)\}\{2\} =\frac\{n(n+1)\}\{2\}\prod\_\{i=1\}^\{n-1\}i!. \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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---
959、 (4)、 行列式 $\displaystyle \left|\begin\{array\}\{cccccccccc\}6&1&8\\\\ 7&5&3\\\\ 2&9&4\end\{array\}\right|$ 的值为 $\displaystyle \underline\{\ \ \ \ \ \ \ \ \ \ \}$. (华东师范大学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) / 直接按第一行展开知原式 $\displaystyle =360$.跟锦数学微信公众号. [在线资料](
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---
960、 2、 若 $\displaystyle n$ 是奇数, 证明行列式
\begin\{aligned\} D=\left|\begin\{array\}\{cccccccccc\} 1&2&3&\cdots&n-1&n\\\\ 2^2&3^2&4^2&\cdots&n^2&(n+1)^2\\\\ 3^3&4^3&5^3&\cdots&(n+1)^3&(n+1)^3\\\\ \vdots&\vdots&\vdots&&\vdots&\vdots\\\\ (n-1)^\{n-1\}&n^\{n-1\}&(n+1)^\{n-1\}&\cdots&(n+1)^\{n-1\}&(n+1)^\{n-1\}\\\\ n^n&(n+1)^n&(n+1)^n&\cdots&(n+1)^n&(n+1)^n \end\{array\}\right|\neq 0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(华南理工大学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) / 由 $\displaystyle n$ 是奇数知 $\displaystyle n+1$ 是偶数. 设题中矩阵为 $\displaystyle A=(a\_\{ij\})$, 则
\begin\{aligned\} a\_\{i,n+1-i\}=2b\_\{i,n+1-i\}+1, a\_\{i,n+2-i\}=2b\_\{i,n+2-i\}, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
而
\begin\{aligned\} D=&\sum\_\{j\_1,\cdots,j\_n\}(-1)^\{\tau(j\_1,\cdots,j\_n)\}a\_\{1j\_1\}\cdots a\_\{nj\_n\}\\\\ =&\sum\_\{j\_1,\cdots,j\_n\}(-1)^\{\tau(j\_1,\cdots,j\_n)\}a\_\{1j\_1\}\cdots a\_\{i,n-i\} \cdot (2b\_\{i,n+1-i\}+1) \cdot 2b\_\{i,n+2-i\}\cdots 2b\_\{i,n+2-i\}\\\\ \equiv&\sum\_\{j\_1,\cdots,j\_n\}(-1)^\{\tau(j\_1,\cdots,j\_n)\}a\_\{1j\_1\}\cdots a\_\{i,n-i\} \cdot 1\cdot 0\cdots 0\left(\mod 2\right)\\\\ \equiv& \left|\begin\{array\}\{cccccccccc\} 1&2&3&\cdots&n-1&1\\\\ 2^2&3^2&4^2&\cdots&1&0\\\\ 3^3&4^3&5^3&\cdots&0&0\\\\ \vdots&\vdots&\vdots&&\vdots&\vdots\\\\ (n-1)^\{n-1\}&1&0&\cdots&0&0\\\\ 1&0&0&\cdots&0&0 \end\{array\}\right|\left(\mod 2\right)\\\\ \equiv& (-1)^\{\tau(n,n-1,\cdots,2,1)\}=(-1)^\frac\{n(n-1)\}\{2\}\left(\mod 2\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 $\displaystyle D\neq 0$.跟锦数学微信公众号. [在线资料](
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---
961、 2、 计算行列式 $\displaystyle D=\left|\begin\{array\}\{cccccccccc\}x\_1y\_1&x\_1y\_2&x\_1y\_3&x\_1y\_4\\\\ x\_2y\_1&x\_2y\_2&x\_2y\_3&x\_2y\_4\\\\ x\_3y\_1&x\_3y\_2&x\_3y\_3&x\_3y\_4\\\\ x\_4y\_1&x\_4y\_2&x\_4y\_3&x\_4y\_4\end\{array\}\right|$. (华南师范大学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) / $\displaystyle D$ 对应的矩阵 $\displaystyle A=\left(\begin\{array\}\{cccccccccccccccccccc\}x\_1\\\\\vdots\\\\x\_4\end\{array\}\right)(y\_1,\cdots,y\_4)$ 满足 $\displaystyle \mathrm\{rank\} A\leq 1$, 而 $\displaystyle |D|=\left\\{\begin\{array\}\{llllllllllll\}x\_1y\_1,&n=1,\\\\ 0,&n\geq 2.\end\{array\}\right.$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](
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---
962、 1、 (15 分) 计算行列式
\begin\{aligned\} \left|\begin\{array\}\{cccccccccc\}2^2-2&2^3-2&\cdots&2^\{2023\}-2\\\\ 3^2-3&3^3-3&\cdots&3^\{2023\}-3\\\\ \vdots&\vdots&&\vdots\\\\ 2023^2-2023&2023^3-2023&\cdots&2023^\{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\}
(华中科技大学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) / 加边:
\begin\{aligned\} \mbox\{原式\}=\left|\begin\{array\}\{cccccccccc\}1&1&1&\cdots&1\\\\ 0&2^2-2&2^3-2&\cdots&2^\{2023\}-2\\\\ 0&3^2-3&3^3-3&\cdots&3^\{2023\}-3\\\\ \vdots&\vdots&\vdots&&\vdots\\\\ 0&2023^2-2023&2023^3-2023&\cdots&2023^\{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\}
第 $\displaystyle 1$ 行加到其它各行:
\begin\{aligned\} \mbox\{原式\}=\left|\begin\{array\}\{cccccccccc\}1&1^2&1^3&\cdots&1^\{2023\}\\\\ 2&2^2&2^3&\cdots&2^\{2023\}\\\\ 3&3^2&3^3&\cdots&3^\{2023\}\\\\ \vdots&\vdots&\vdots&&\vdots\\\\ 2023&2023^2&2023^3&\cdots&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\}
第 $\displaystyle i$ 行提出公因子 $\displaystyle i$, $\displaystyle 2\leq i\leq 2023$ 后化为范德蒙行列式
\begin\{aligned\} \mbox\{原式\}=2023!\prod\_\{1\leq i < j\leq 2023\} =2023!\prod\_\{j=2\}^\{2023\}(j-1)!=\prod\_\{k=2\}^\{2023\}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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---
963、 1、 填空题. (1)、 设三阶行列式 $\displaystyle D=\left|\begin\{array\}\{cccccccccc\}2&1&1\\\\ 1&2&1\\\\ 1&1&2\end\{array\}\right|$, $\displaystyle A\_\{ij\}$ 表示 $\displaystyle D$ 的 $\displaystyle (ij)$ 位置的代数余子式, 则
\begin\{aligned\} A\_\{11\}+2A\_\{12\}+3A\_\{13\}=\underline\{\ \ \ \ \ \ \ \ \ \ \}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(华中师范大学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) / $\displaystyle \mbox\{原式\}=\left|\begin\{array\}\{cccccccccc\}1&2&3\\\\ 1&2&1\\\\ 1&1&2\end\{array\}\right|=-2$.跟锦数学微信公众号. [在线资料](
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---
964、 2、 计算与证明. (1)、 计算 $\displaystyle n$ 阶行列式 $\displaystyle \left|\begin\{array\}\{cccccccccc\}2&-4&&\\\\ 2&\ddots&\ddots&\\\\ &\ddots&\ddots&-4\\\\ &&2&2\end\{array\}\right|$. (华中师范大学2023年高等代数考研试题) [行列式 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 设原式 $\displaystyle =D\_n$, 则 $\displaystyle D\_1=2, D\_2=12$, $\displaystyle D\_n=2D\_\{n-1\}+8D\_\{n-2\}$. 特征方程为 $\displaystyle \lambda^2-2\lambda-8=0$, 而可设
\begin\{aligned\} D\_n=c\_14^n+c\_2(-2)^n\stackrel\{D\_1=2, D\_2=12\}\{\Rightarrow\}c\_1=\frac\{2\}\{3\},c\_2=\frac\{1\}\{3\}\Rightarrow D\_n=\frac\{2\cdot 4^n+(-2)^n\}\{3\}. \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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---
965、 1、 计算行列式 $\displaystyle D\_n=\left|\begin\{array\}\{cccccccccc\}a\_0&x&x&\cdots&x\\\\ x&a\_1&0&\cdots&0\\\\ x&0&a\_2&\cdots&0\\\\ \vdots&\vdots&\vdots&\ddots&\vdots\\\\ x&0&0&\cdots&a\_\{n-1\}\end\{array\}\right|$, 其中 $\displaystyle a\_1a\_2\cdots a\_\{n-1\}\neq 0$. (暨南大学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) / 当 $\displaystyle n=1$ 时, $\displaystyle D\_1=a\_0$. 当 $\displaystyle n\geq 2$ 时, 第 $\displaystyle i$ 列 $\displaystyle \cdot\frac\{-x\}\{a\_\{i-1\}\}$ 加到第 $\displaystyle 1$ 列, $\displaystyle 2\leq i\leq n$, 得
\begin\{aligned\} D\_n=\left(a\_0-\sum\_\{i=1\}^\{n-1\} \frac\{x^2\}\{a\_i\}\right)a\_1\cdots a\_\{n-1\}. \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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---
966、 2、 设
\begin\{aligned\} g(y)=\left|\begin\{array\}\{cccccccccc\}y&0&0&0&\cdots&0&1\\\\ y^2&2&0&0&\cdots&0&1\\\\ y^3&3&3&0&\cdots&0&1\\\\ \vdots&\vdots&\vdots&\vdots&&\vdots&\vdots\\\\ y^n&n&C\_n^2&C\_n^3&\cdots&C\_n^\{n-1\}&1\\\\ y^\{n+1\}&n+1&C\_\{n+1\}^2&C\_\{n+1\}^3&\cdots&C\_\{n+1\}^\{n-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\}
求 $\displaystyle g(y+1)-g(y)$. (南昌大学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&0&0&0&\cdots&0&1\\\\ 2y+1&2&0&0&\cdots&0&1\\\\ 3y^2+3y+1&3&3&0&\cdots&0&1\\\\ \vdots&\vdots&\vdots&\vdots&&\vdots&\vdots\\\\ ny^\{n-1\}+\cdots+ny+1&n&C\_n^2&C\_n^3&\cdots&C\_n^\{n-1\}&1\\\\ (n+1)y^n+\cdots+(n+1)y+1&n+1&C\_\{n+1\}^2&C\_\{n+1\}^3&\cdots&C\_\{n+1\}^\{n-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\}
第 $\displaystyle 2$ 列 $\displaystyle \cdot(-y)$, 第 $\displaystyle 3$ 列 $\displaystyle \cdot(-y^2)$, $\displaystyle \cdots$, 第 $\displaystyle n$ 列 $\displaystyle \cdot(-y^\{n-1\})$, 第 $\displaystyle n$ 列 $\displaystyle \cdot(-1)$ 加到第 $\displaystyle 1$ 列得
\begin\{aligned\} \mbox\{原式\}= \left|\begin\{array\}\{cccccccccc\}0&0&0&0&\cdots&0&1\\\\ 0&2&0&0&\cdots&0&1\\\\ 0&3&3&0&\cdots&0&1\\\\ \vdots&\vdots&\vdots&\vdots&&\vdots&\vdots\\\\ 0&n&C\_n^2&C\_n^3&\cdots&C\_n^\{n-1\}&1\\\\ (n+1)y^n&n+1&C\_\{n+1\}^2&C\_\{n+1\}^3&\cdots&C\_\{n+1\}^\{n-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\}
按第 $\displaystyle 1$ 列展开后, 再按第 1 行展开知
\begin\{aligned\} \mbox\{原式\}=(-1)^\{(n+1)+1\}(n+1)y^n\cdot (-1)^\{1+n\}1\cdot 2\cdot 3\cdots n =-(n+1)!y^n. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
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发表于 2023-3-5 09:16:30
