Questions in category: 行列式 (Determinant)

<[1] [2] >

## 1. 柯西行列式(Cauchy determinant)

Posted by haifeng on 2022-04-05 23:25:25 last update 2022-04-05 23:43:27 | Answers (1) | 收藏

$\det(A_n)= \begin{vmatrix} \frac{1}{x_1+y_1} & \frac{1}{x_1+y_2} & \cdots & \frac{1}{x_1+y_n}\\ \frac{1}{x_2+y_1} & \frac{1}{x_2+y_2} & \cdots & \frac{1}{x_2+y_n}\\ \vdots & \vdots & &\vdots\\ \frac{1}{x_n+y_1} & \frac{1}{x_n+y_2} & \cdots & \frac{1}{x_n+y_n}\\ \end{vmatrix}$

$\det(A)=\dfrac{\prod\limits_{1\leqslant i < j\leqslant n}(x_i-x_j)(y_i-y_j)}{\prod\limits_{1\leqslant i,j\leqslant n}(x_i+y_j)}$

## 2. 求行列式

Posted by haifeng on 2017-05-04 17:14:36 last update 2017-05-04 17:37:57 | Answers (2) | 收藏

$D=\begin{vmatrix} \lambda-1 & -1 & -1 &\cdots & -1\\ -1 & \lambda-1 & -1 &\cdots & -1\\ -1 & -1 &\lambda-1 &\cdots & -1\\ \vdots &\vdots &\vdots &\ddots & \vdots\\ -1 & -1 & -1 & \cdots &\lambda-1 \end{vmatrix}_{n}$

$D=\begin{vmatrix} \lambda-1 & -\frac{c_1}{c_2} & -\frac{c_1}{c_3} &\cdots & -\frac{c_1}{c_n}\\ -\frac{c_2}{c_1} & \lambda-1 & -\frac{c_2}{c_3} &\cdots & -\frac{c_2}{c_n}\\ -\frac{c_3}{c_1} & -\frac{c_3}{c_2} &\lambda-1 &\cdots & -\frac{c_3}{c_n}\\ \vdots &\vdots &\vdots &\ddots & \vdots\\ -\frac{c_n}{c_1} & -\frac{c_n}{c_2} & -\frac{c_n}{c_3} & \cdots &\lambda-1 \end{vmatrix}_{n}$

## 3. 设 $a,b,c$ 是已知常数, $A_n$ 为 $n$ 阶方阵, $\beta\in\mathbb{R}^n$. 设 $|A_n|=a$, $\begin{vmatrix}A &\beta\\ \beta^T & b\end{vmatrix}=0$, 求 $\begin{vmatrix}A &\beta\\ \beta^T & c\end{vmatrix}$ 的值.

Posted by haifeng on 2016-10-07 08:42:15 last update 2016-10-07 08:42:15 | Answers (1) | 收藏

## 4. 设 $A_{m\times n}$, $B_{n\times m}$ 为实矩阵, 证明 $|I_m-AB|=|I_n-BA|$.

Posted by haifeng on 2016-04-04 22:25:47 last update 2021-05-18 19:04:43 | Answers (0) | 收藏

$\det(I_m-AB)=\det(I_n-BA)$

$\bigl|\lambda I_m-AB\bigr|=\lambda^{m-n}\bigl|\lambda I_n-BA\bigr|$

References:

[1] 王品超  编著  《高等代数新方法》,  山东教育出版社.

## 5. 求行列式

Posted by haifeng on 2016-04-03 21:20:32 last update 2016-04-03 21:20:32 | Answers (0) | 收藏

$\begin{vmatrix} a_1-b_1 & a_1-b_2 & \cdots & a_1-b_n\\ a_2-b_1 & a_2-b_2 & \cdots & a_2-b_n\\ \vdots & \vdots & \ddots & \vdots \\ a_n-b_1 & a_n-b_2 & \cdots & a_n-b_n\\ \end{vmatrix}$

[Hint]

$A= \begin{pmatrix} a_1-b_1 & a_1-b_2 & \cdots & a_1-b_n\\ a_2-b_1 & a_2-b_2 & \cdots & a_2-b_n\\ \vdots & \vdots & \ddots & \vdots \\ a_n-b_1 & a_n-b_2 & \cdots & a_n-b_n\\ \end{pmatrix}$

## 6. 设 $A_1,A_2,B_1,B_2$ 都是 $3\times 1$ 矩阵. 令 $A=(A_1,A_2,B_1)$, $B=(A_1,A_2,B_2)$. 假设 $|A|=2$, $|B|=3$, 求 $|A+2B|$.

Posted by haifeng on 2016-04-03 17:06:06 last update 2016-04-03 17:06:06 | Answers (1) | 收藏

## 7. 计算行列式

Posted by haifeng on 2015-08-28 17:02:38 last update 2015-08-29 10:02:22 | Answers (1) | 收藏

$\begin{vmatrix} 0 & a & a & \cdots & a\\ b & 0 & a & \cdots & a\\ b & b & 0 & \cdots & a\\ \vdots & \vdots & \ddots & \ddots & a\\ b & b & b & \cdots & 0\\ \end{vmatrix}_n$

Hint

$a^n\cdot \begin{vmatrix} 0 & 1 & 1 & \cdots & 1\\ t & 0 & 1 & \cdots & 1\\ t & t & 0 & \cdots & 1\\ \vdots & \vdots & \ddots & \ddots & 1\\ t & t & t & \cdots & 0\\ \end{vmatrix}_n=:a^n\cdot D_n(t).$

$D_n(t)$ 是关于 $t$ 的多项式, 我们证明

$D_n(t)=(-1)^{n-1}(t+t^2+\cdots+t^{n-1}).$

## 8. 将 $\det(I_n+tA)$ 展开

Posted by haifeng on 2015-07-27 09:37:56 last update 2015-07-27 09:37:56 | Answers (1) | 收藏

## 9. $\text{det}:\mathcal{M}(n,\mathbb{R})\rightarrow\mathbb{R}$ 是开映射或闭映射?

Posted by haifeng on 2015-06-05 14:31:52 last update 2015-06-05 15:28:41 | Answers (0) | 收藏

[讨论]

$|A|=ad-bc,$

$-\varepsilon^2 < bc < \varepsilon^2.$

$-\varepsilon^2 < ad < \varepsilon^2.$

$-2\varepsilon^2 < ad-bc < 2\varepsilon^2.$

$A=\begin{pmatrix} a & b & c\\ u & v & w\\ x & y & z\\ \end{pmatrix}$

$|A|=a\begin{vmatrix} v & w\\ y & z\\ \end{vmatrix}-b\begin{vmatrix} u & w\\ x & z\\ \end{vmatrix}+c\begin{vmatrix} u & v\\ x & y\\ \end{vmatrix}$

$|A|\in(-2\sqrt{3}\varepsilon^3, 2\sqrt{3}\varepsilon^3).$

$f$ 为连续的闭映射当且仅当 $\forall A\subset X$, $\overline{f(A)}=f(\bar{A})$.

## 10. 用归纳法证明下面的行列式等式

Posted by haifeng on 2014-12-11 10:54:42 last update 2014-12-11 10:55:38 | Answers (1) | 收藏

$D_n=\begin{vmatrix} \cos\alpha & 1 & & & & \\ 1 & 2\cos\alpha & 1 & & & \\ & 1 & 2\cos\alpha & 1 & & \\ & & \ddots & \ddots & \ddots & \\ & & & \ddots & \ddots & 1\\ & & & & 1 & 2\cos\alpha \end{vmatrix}_n=\cos n\alpha$

<[1] [2] >