行列式典型例题

计算三阶行列式

$$D = \begin{vmatrix} 2 & 0 & 1 \\ 1 & -4 & -1 \\ -1 & 8 & 3 \\ \end{vmatrix}$$

$D = -4$

$$D = \begin{vmatrix} 1 & 1 & 1 \\ a & b & c \\ a^2 & b^2 & c^2 \\ \end{vmatrix}$$

$D = (b-a)(c-b)(c-a)$

逆序数

求下列排列的逆序数

1. $\tau(4,1,3,2)$
2. $\tau(2,5,6,3,4,1)$
3. $\tau(1,3,5,\cdots,(2n-1),2,4,6,\cdots,(2n))$

$4, 9, \frac{1}{2}n(n-1)$

已知$a_{3j} a_{12} a_{41} a_{2k}$在 4 阶行列式带负号，那$j$和$k$分别为？

$j=4, k=3$

四阶行列式中含有因子$a_{11}a_{23}$的项是？

$-a_{11}a_{23}a_{32}a_{44},a_{11}a_{23}a_{34}a_{42}$

$$f(x) = \begin{vmatrix} 2x & x & x & 2 \\ 1 & x & 1 & -1 \\ 3 & 2 & x & 1 \\ 1 & 1 & 0 & x \\ \end{vmatrix}$$

其中$x^4$和$x^3$的系数分别是？

2, -4

n 阶行列式的计算

$$D = \begin{vmatrix} 2 & -1 & 1 & 6 \\ 4 & -1 & 5 & 0 \\ -1 & 2 & 0 & -5 \\ 1 & 4 & -2 & -2 \\ \end{vmatrix}$$

D = 120

$$D = \begin{vmatrix} 4 & 1 & 2 & 4 \\ 1 & 2 & 0 &2 \\ 10 & 5 & 2 & 0 \\ 0 & 1 & 1 & 7 \\ \end{vmatrix}$$

D = 0

$$\begin{vmatrix} x+1 & 2 & -1 \\ 2 & x+1 & 1 \\ -1 & 1 & x+1 \\ \end{vmatrix} = 0$$

$x_1=-3,x_2=\sqrt{3},x_3=-\sqrt{3}$

$$D = \begin{vmatrix} 1 & \frac{1}{2} & \frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & 1 & \frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & \frac{1}{2} & 1 & \frac{1}{2} \\ \frac{1}{2} & \frac{1}{2} & \frac{1}{2} & 1 \\ \end{vmatrix}$$

$D = \frac{5}{16}$

$$D = \begin{vmatrix} 1 & b_1 & 0 & 0 \\ -1 & 1-b_1 & b_2 & 0 \\ 0 & -1 & 1-b_2 & b_3 \\ 0 & 0 & -1 & 1-b_3 \\ \end{vmatrix}$$

$D = 1$

$$D = \begin{vmatrix} 1 & -1 & 1 & x-1 \\ 1 & -1 & x+1 & -1 \\ 1 & x-1 & 1 & -1 \\ x+1 & -1 & 1 & -1 \\ \end{vmatrix}$$

$D=x^4$

$D_n=\text{det}(a_{ij})$，其中$a_{ij}=|i-j|$，求 D。

$D=(-1)^{n-1}(n-1)2^{n-2}$

$$D = \begin{vmatrix} b+c & c+a & a+b \\ a & b & c \\ a^2 & b^2 & c^2 \\ \end{vmatrix}$$

$D=(a+b+c)(b-a)(c-a)(c-b)$

$$D = \begin{vmatrix} 1 & 1 & 1 & 1 \\ a & b & c & d \\ a^2 & b^2 & c^2 & d^2 \\ a^4 & b^4 & c^4 & d^4 \\ \end{vmatrix}$$

$D=(a-b)(a-c)(a-d)(b-c)(b-d)(c-d)(a+b+c+d)$

$$D = \begin{vmatrix} a_1 & 0 & a_2 & 0 \\ 0 & b_1 & 0 & b_2 \\ c_1 & 0 & c_2 & 0 \\ 0 & d_1 & 0 & d_2 \\ \end{vmatrix}$$

$D = (a_1c_2 - a_2c_1)(b_1d_2 - b_2d_1)$

$$D_{2n} = \begin{vmatrix} a_n & & & & & b_n \\ & \ddots & & & \ddots \\ & & a_1 & b_1 & & \\ & & c_1 & d_1 & & \\ & \ddots & & & \ddots & \\ c_n & & & & & d_n \end{vmatrix}$$

$D = \prod_{i=1}^n(a_id_i - b_ic_i)$

三对角线行列式

$$D_5 = \begin{vmatrix} 3 & 2 & 0 & 0 & 0 \\ 1 & 3 & 2 & 0 & 0 \\ 0 & 1 & 3 & 2 & 0 \\ 0 & 0 & 1 & 3 & 2 \\ 0 & 0 & 0 & 1 & 3 \\ \end{vmatrix}$$

$D=63$

证明

$$\begin{vmatrix} x & -1 & 0 & 0 \\ 0 & x & -1 & 0 \\ 0 & 0 & x & -1 \\ a_0 & a_1 & a_2 & a_3 \\ \end{vmatrix} = a_3x^3 + a_2x^2 + a_1x + a_0$$ $$D_n = \begin{vmatrix} x & a & \cdots & a \\ a & x & \cdots & a \\ \vdots & \vdots & & \vdots \\ a & a & \cdots & x \\ \end{vmatrix}$$

$D = [x + (n-1)a](x-a)^{n-1}$

$$D_n = \begin{vmatrix} 1+a_1 & 1 & \cdots & 1 \\ 1 & 1+a_2 & \cdots & 1 \\ \vdots & \vdots & & \vdots \\ 1 & 1 & \cdots & 1+a_n \end{vmatrix}$$

$D_n=\prod_{i=1}^n a_i (1 + \sum_{i=1}^n\frac{1}{a_i})$

$$D = \begin{vmatrix} 3 & 1 & -1 & 2 \\ -5 & 1 & 3 & -4 \\ 2 & 0 & 1 & -1 \\ 1 & -5 & 3 & -3 \\ \end{vmatrix}$$

$D$ 的 $(i,j)$ 元的代数余子式记作 $A_{ij}$, 求$A_{31} + 3A_{32} - 2A_{33} + 2A_{34}$

24

$$\begin{vmatrix} 0 & a & b & 0 \\ a & 0 & 0 & b \\ 0 & c & d & 0 \\ c & 0 & 0 & d \\ \end{vmatrix}$$

$-(ad-bc)^2$

$$\begin{vmatrix} \lambda & -1 & 0 & 0 \\ 0 & \lambda & -1 & 0 \\ 0 & 0 & \lambda & -1 \\ 4 & 3 & 2 & \lambda + 1 \\ \end{vmatrix}$$

$\lambda^4 + \lambda^3 + 2\lambda^2 + 3\lambda + 4$

$$A = \begin{pmatrix} 0 & 1 & 1 & \cdots & 1 & 1 \\ 1 & 0 & 1 & \cdots & 1 & 1 \\ \vdots & \vdots & \vdots & & \vdots & \vdots \\ 1 & 1 & 1 & \cdots & 0 & 1 \\ 1 & 1 & 1 & \cdots & 1 & 0 \\ \end{pmatrix}$$

$|A|$ = ?.

$(-1)^{n-1}(n-1)$

$$D = \begin{vmatrix} 3 & 0 & 4 & 0 \\ 2 & 2 & 2 & 2 \\ 0 & -7 & 0 & 0 \\ 5 & 3 & -2 & 2 \\ \end{vmatrix}$$

求第四行各元素余子式之和

-28