文章目录
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- [一 n阶行列式](#一 n阶行列式)
- [二 三阶行列式](#二 三阶行列式)
- [三 特殊行列式](#三 特殊行列式)
- 结语
一 n阶行列式
∣ a 11 a 12 ⋯ a 1 n a 21 a 22 ⋯ a 2 n ⋯ ⋯ ⋯ ⋯ a n 1 a n 2 ⋯ a n n ∣ \begin{vmatrix}a_{11}&a_{12}&\cdots&a_{1n}\\a_{21}&a_{22}&\cdots&a_{2n}\\\cdots&\cdots&\cdots&\cdots\\a_{n1}&a_{n2}&\cdots&a_{nn}\\\end{vmatrix} a11a21⋯an1a12a22⋯an2⋯⋯⋯⋯a1na2n⋯ann n 2 n^2 n2个数,取不同行不同列数乘积的和或差,成为n阶行列式,记作 D n D_n Dn
D n = ∑ ( − 1 ) τ ( p 1 p 2 ⋯ p n ) a a 1 p 1 a a 2 p 3 ⋯ a a n p n ( n ! 项的和或差 ) D_n=\sum(-1)^{\tau(p_1p_2\cdots p_n)}a_{a_1p_1}a_{a_2p_3}\cdots a_{a_np_n}(n!项的和或差) Dn=∑(−1)τ(p1p2⋯pn)aa1p1aa2p3⋯aanpn(n!项的和或差)
p 1 p 2 ⋯ p n 为 12 ⋅ n p_1p_2\cdots p_n为1 2 \cdot n p1p2⋯pn为12⋅n的全排列
D n = ∑ ( − 1 ) τ ( q 1 q 2 ⋯ q n ) a q 1 1 a q 2 2 ⋯ a q n n ( n ! 项的和或差 ) D_n=\sum(-1)^{\tau(q_1q_2\cdots q_n)}a_{q_11}a_{q_22}\cdots a_{q_nn}(n!项的和或差) Dn=∑(−1)τ(q1q2⋯qn)aq11aq22⋯aqnn(n!项的和或差)
q 1 q 2 ⋯ q n 为 12 ⋅ n q_1q_2\cdots q_n为1 2 \cdot n q1q2⋯qn为12⋅n的全排列
二 三阶行列式
三阶行列式 ∣ a 11 a 12 a 13 a 21 a 22 a 23 a 31 a 32 a 33 ∣ \begin{vmatrix}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33}\\\end{vmatrix} a11a21a31a12a22a32a13a23a33
∣ a 11 a 12 a 13 a 21 a 22 a 23 a 31 a 32 a 33 ∣ = a 11 a 22 a 33 − a 11 a 23 a 32 + a 12 a 23 a 31 − a 12 a 21 a 33 + a 13 a 21 a 32 − a 12 a 22 a 31 \begin{vmatrix}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33}\\\end{vmatrix}=\\ a_{11}a_{22}a_{33}-a_{11}a_{23}a_{32}+a_{12}a_{23}a_{31}-a_{12}a_{21}a_{33}+a_{13}a_{21}a_{32}-a_{12}a_{22}a_{31} a11a21a31a12a22a32a13a23a33 =a11a22a33−a11a23a32+a12a23a31−a12a21a33+a13a21a32−a12a22a31
三阶行列式对角线法则:三条实线看做是平行于主对角线的连线,三条虚线看做是平行于副对角线的连线,实线上三元素的乘积为正好,虚线上三元素的乘积为负号。
例 计算 D = ∣ 1 2 − 4 − 2 2 1 − 3 4 − 2 ∣ D=\begin{vmatrix}1&2&-4\\-2&2&1\\-3&4&-2\\\end{vmatrix} D= 1−2−3224−41−2
∣ 1 2 − 4 − 2 2 1 − 3 4 − 2 ∣ = − 4 − 4 − 6 − 8 + 32 − 24 = − 14 \begin{vmatrix}1&2&-4\\-2&2&1\\-3&4&-2\\\end{vmatrix}=\\ -4-4-6-8+32-24\\ =-14 1−2−3224−41−2 =−4−4−6−8+32−24=−14
例 解方程 ∣ 1 1 1 2 3 x 4 9 x 2 ∣ = 0 \begin{vmatrix}1&1&1\\2&3&x\\4&9&x^2\\\end{vmatrix}=0 1241391xx2 =0
D = 3 x 2 − 9 x + 4 x − 2 x 2 + 18 − 12 = x 2 − 5 x + 6 = 0 x = 2 或者 x = 3 D=3x^2-9x+4x-2x^2+18-12=\\ x^2-5x+6=0\\ x=2或者x=3 D=3x2−9x+4x−2x2+18−12=x2−5x+6=0x=2或者x=3
注:
- 四阶行列式总共 4 ! = 24 项 4!=24项 4!=24项;五阶行列式总共120项......因此对角线法则只适用于二、三阶行列式的计算。
- 一节行列式 ∣ − 2 ∣ = − 2 \begin{vmatrix}-2\end{vmatrix}=-2 −2 =−2
三 特殊行列式
例 ∣ a 11 a 12 a 13 a 14 0 a 22 a 23 a 24 0 0 a 33 a 34 0 0 0 a 44 ∣ \begin{vmatrix}a_{11}&a_{12}&a_{13}&a_{14}\\0&a_{22}&a_{23}&a_{24}\\0&0&a_{33}&a_{34}\\0&0&0&a_{44}\end{vmatrix} a11000a12a2200a13a23a330a14a24a34a44
D = a 11 a 22 a 33 a 44 D=a_{11}a_{22}a_{33}a_{44} D=a11a22a33a44
推广: ∣ a 11 a 12 ⋯ a 1 n 0 a 22 ⋯ a 2 n ⋯ ⋯ ⋯ ⋯ 0 0 0 a n n ∣ = a 11 a 22 ⋯ a n n \begin{vmatrix}a_{11}&a_{12}&\cdots&a_{1n}\\0&a_{22}&\cdots&a_{2n}\\\cdots&\cdots&\cdots&\cdots\\0&0&0&a_{nn}\end{vmatrix}=a_{11}a_{22}\cdots a_{nn} a110⋯0a12a22⋯0⋯⋯⋯0a1na2n⋯ann =a11a22⋯ann,称为上山角行列式。
推广: ∣ a 11 0 ⋯ 0 a 21 a 22 ⋯ 0 ⋯ ⋯ ⋯ ⋯ a n 1 a n 2 ⋯ a n n ∣ = a 11 a 22 ⋯ a n n \begin{vmatrix}a_{11}&0&\cdots&0\\a_{21}&a_{22}&\cdots&0\\\cdots&\cdots&\cdots&\cdots\\a_{n1}&a_{n2}&\cdots&a_{nn}\end{vmatrix}=a_{11}a_{22}\cdots a_{nn} a11a21⋯an10a22⋯an2⋯⋯⋯⋯00⋯ann =a11a22⋯ann,称为下山角行列式。
特别地: ∣ a 11 a 22 ⋯ a n n ∣ = a 11 a 22 ⋯ a n n \begin{vmatrix}a_{11}\\&a_{22}\\&&\cdots\\&&&a_{nn}\end{vmatrix}=a_{11}a_{22}\cdots a_{nn} a11a22⋯ann =a11a22⋯ann称为对角行列式。
例: f ( x ) = ∣ x 1 1 2 1 x 1 − 1 3 2 x 1 1 1 2 x 1 ∣ 中 x 3 f(x)=\begin{vmatrix}x&1&1&2\\1&x&1&-1\\3&2&x&1\\1&1&2x&1\end{vmatrix}中x^3 f(x)= x1311x2111x2x2−111 中x3的系数
通过观察有 x 3 的项只有两项 x ⋅ x ⋅ x ⋅ 1 − x ⋅ x ⋅ 1 ⋅ 2 x = − x 3 所以 x 3 的系数为 − 1 通过观察有x^3的项只有两项\\ x\cdot x\cdot x\cdot1-x\cdot x\cdot1\cdot 2x=-x^3\\ 所以x^3的系数为-1 通过观察有x3的项只有两项x⋅x⋅x⋅1−x⋅x⋅1⋅2x=−x3所以x3的系数为−1
推广: ∣ 0 0 ⋯ a 1 n 0 ⋯ a 2 n − 1 a 2 n ⋯ ⋯ ⋯ ⋯ a n 1 a n 2 ⋯ a n n ∣ = ( − 1 ) n ( n − 1 ) 2 a 1 n a 2 n − 1 ⋯ a n 1 \begin{vmatrix}0&0&\cdots&a_{1n}\\0&\cdots&a_{2n-1}&a_{2n}\\\cdots&\cdots&\cdots&\cdots\\a_{n1}&a_{n2}&\cdots&a_{nn}\end{vmatrix}=(-1)^{\frac{n(n-1)}{2}}a_{1n}a_{2n-1}\cdots a_{n1} 00⋯an10⋯⋯an2⋯a2n−1⋯⋯a1na2n⋯ann =(−1)2n(n−1)a1na2n−1⋯an1称为反下山角行列式。
推广: ∣ a 11 a 12 ⋯ a 1 n a 21 ⋯ a 2 n − 1 a 0 ⋯ ⋯ ⋯ ⋯ a n 1 0 ⋯ 0 ∣ = ( − 1 ) n ( n − 1 ) 2 a 1 n a 2 n − 1 ⋯ a n 1 \begin{vmatrix}a_{11}&a_{12}&\cdots&a_{1n}\\a_{21}&\cdots&a_{2n-1}&a_{0}\\\cdots&\cdots&\cdots&\cdots\\a_{n1}&0&\cdots&0\end{vmatrix}=(-1)^{\frac{n(n-1)}{2}}a_{1n}a_{2n-1}\cdots a_{n1} a11a21⋯an1a12⋯⋯0⋯a2n−1⋯⋯a1na0⋯0 =(−1)2n(n−1)a1na2n−1⋯an1称为反上三角行列式。
推广: ∣ a 1 n a 2 n − 1 ⋯ a n 1 ∣ = ( − 1 ) n ( n − 1 ) 2 a 1 n a 2 n − 1 ⋯ a n 1 \begin{vmatrix}&&&a_{1n}\\&&a_{2n-1}\\&&\cdots\\a_{n1}&&&\end{vmatrix}=(-1)^{\frac{n(n-1)}{2}}a_{1n}a_{2n-1}\cdots a_{n1} an1a2n−1⋯a1n =(−1)2n(n−1)a1na2n−1⋯an1称为反对角行列式。
结语
❓QQ:806797785
⭐️文档笔记地址:https://gitee.com/gaogzhen/math
参考:
[1]同济大学数学系.工程数学.线性代数 第6版 [M].北京:高等教育出版社,2014.6.p4-5.
[2]同济六版《线性代数》全程教学视频[CP/OL].2020-02-07.p3.