行列式典型例题

计算三阶行列式

D=∣2011−4−1−183∣D = \begin{vmatrix} 2 & 0 & 1 \\ 1 & -4 & -1 \\ -1 & 8 & 3 \\ \end{vmatrix}

D=−4D = -4

D=∣111abca2b2c2∣D = \begin{vmatrix} 1 & 1 & 1 \\ a & b & c \\ a^2 & b^2 & c^2 \\ \end{vmatrix}

D=(b−a)(c−b)(c−a)D = (b-a)(c-b)(c-a)

逆序数

求下列排列的逆序数

  1. τ(4,1,3,2)\tau(4,1,3,2)
  2. τ(2,5,6,3,4,1)\tau(2,5,6,3,4,1)
  3. τ(1,3,5,⋯ ,(2n−1),2,4,6,⋯ ,(2n))\tau(1,3,5,\cdots,(2n-1),2,4,6,\cdots,(2n))

4,9,12n(n−1)4, 9, \frac{1}{2}n(n-1)

已知a3ja12a41a2ka_{3j} a_{12} a_{41} a_{2k}在 4 阶行列式带负号，那jj和kk分别为？

j=4,k=3j=4, k=3

四阶行列式中含有因子a11a23a_{11}a_{23}的项是？

−a11a23a32a44,a11a23a34a42-a_{11}a_{23}a_{32}a_{44},a_{11}a_{23}a_{34}a_{42}

f(x)=∣2xxx21x1−132x1110x∣f(x) = \begin{vmatrix} 2x & x & x & 2 \\ 1 & x & 1 & -1 \\ 3 & 2 & x & 1 \\ 1 & 1 & 0 & x \\ \end{vmatrix}

其中x4x^4和x3x^3的系数分别是？

2, -4

n 阶行列式的计算

D=∣2−1164−150−120−514−2−2∣D = \begin{vmatrix} 2 & -1 & 1 & 6 \\ 4 & -1 & 5 & 0 \\ -1 & 2 & 0 & -5 \\ 1 & 4 & -2 & -2 \\ \end{vmatrix}

D = 120

D=∣41241202105200117∣D = \begin{vmatrix} 4 & 1 & 2 & 4 \\ 1 & 2 & 0 &2 \\ 10 & 5 & 2 & 0 \\ 0 & 1 & 1 & 7 \\ \end{vmatrix}

D = 0

∣x+12−12x+11−11x+1∣=0\begin{vmatrix} x+1 & 2 & -1 \\ 2 & x+1 & 1 \\ -1 & 1 & x+1 \\ \end{vmatrix} = 0

x1=−3,x2=3,x3=−3x_1=-3,x_2=\sqrt{3},x_3=-\sqrt{3}

D=∣1121212121121212121121212121∣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=516D = \frac{5}{16}

D=∣1b100−11−b1b200−11−b2b300−11−b3∣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=1D = 1

D=∣1−11x−11−1x+1−11x−11−1x+1−11−1∣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=x4D=x^4

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

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

D=∣b+cc+aa+babca2b2c2∣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=(a+b+c)(b-a)(c-a)(c-b)

D=∣1111abcda2b2c2d2a4b4c4d4∣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=(a-b)(a-c)(a-d)(b-c)(b-d)(c-d)(a+b+c+d)

D=∣a10a200b10b2c10c200d10d2∣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=(a1c2−a2c1)(b1d2−b2d1)D = (a_1c_2 - a_2c_1)(b_1d_2 - b_2d_1)

D2n=∣anbn⋱⋱a1b1c1d1⋱⋱cndn∣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=∏i=1n(aidi−bici)D = \prod_{i=1}^n(a_id_i - b_ic_i)

三对角线行列式

D5=∣3200013200013200013200013∣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=63D=63

证明

∣x−1000x−1000x−1a0a1a2a3∣=a3x3+a2x2+a1x+a0\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 Dn=∣xa⋯aax⋯a⋮⋮⋮aa⋯x∣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−1D = [x + (n-1)a](x-a)^{n-1}

Dn=∣1+a11⋯111+a2⋯1⋮⋮⋮11⋯1+an∣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}

Dn=∏i=1nai(1+∑i=1n1ai)D_n=\prod_{i=1}^n a_i (1 + \sum_{i=1}^n\frac{1}{a_i})

D=∣31−12−513−4201−11−53−3∣D = \begin{vmatrix} 3 & 1 & -1 & 2 \\ -5 & 1 & 3 & -4 \\ 2 & 0 & 1 & -1 \\ 1 & -5 & 3 & -3 \\ \end{vmatrix}

DD 的 (i,j)(i,j) 元的代数余子式记作 AijA_{ij}, 求A31+3A32−2A33+2A34A_{31} + 3A_{32} - 2A_{33} + 2A_{34}

24

∣0ab0a00b0cd0c00d∣\begin{vmatrix} 0 & a & b & 0 \\ a & 0 & 0 & b \\ 0 & c & d & 0 \\ c & 0 & 0 & d \\ \end{vmatrix}

−(ad−bc)2-(ad-bc)^2

∣λ−1000λ−1000λ−1432λ+1∣\begin{vmatrix} \lambda & -1 & 0 & 0 \\ 0 & \lambda & -1 & 0 \\ 0 & 0 & \lambda & -1 \\ 4 & 3 & 2 & \lambda + 1 \\ \end{vmatrix}

λ4+λ3+2λ2+3λ+4\lambda^4 + \lambda^3 + 2\lambda^2 + 3\lambda + 4

A=(011⋯11101⋯11⋮⋮⋮⋮⋮111⋯01111⋯10)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∣|A| = ?.

(−1)n−1(n−1)(-1)^{n-1}(n-1)

D=∣304022220−70053−22∣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