线代强化NO4|行列式的计算
数值型
- 定义
n 阶行列式的定义:阵中所有取自不同行不同列的 n 个元素乘积的代数和。它的数学表达式:
∣a11a12⋯a1na21a22⋯a2n⋮⋮⋮an1an2⋯ann∣=∑i1,i2,⋯,in(−1)τ(i1,i2,⋯,in)a1i1a2i2⋯anin\begin{vmatrix} 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{vmatrix} = \sum_{i_1,i_2,\cdots,i_n} (-1)^{\tau(i_1,i_2,\cdots,i_n)} a_{1i_1}a_{2i_2}\cdots a_{ni_n} a11a21⋮an1a12a22⋮an2⋯⋯⋯a1na2n⋮ann=i1,i2,⋯,in∑(−1)τ(i1,i2,⋯,in)a1i1a2i2⋯anin
- 注:可推出低阶行列式计算公式
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二阶行列式
∣abcd∣=ad−bc\begin{vmatrix} a & b \\ c & d \end{vmatrix} = ad - bc acbd=ad−bc -
三阶行列式
∣a1a2a3b1b2b3c1c2c3∣=a1b2c3+a2b3c1+a3b1c2−a3b2c1−a2b1c3−a1b3c2\begin{vmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{vmatrix} = a_1b_2c_3 + a_2b_3c_1 + a_3b_1c_2 - a_3b_2c_1 - a_2b_1c_3 - a_1b_3c_2 a1b1c1a2b2c2a3b3c3=a1b2c3+a2b3c1+a3b1c2−a3b2c1−a2b1c3−a1b3c2
展开定理:行列式的值等于其任何一行(或列)所有元素与其代数余子式乘积之和,即
∣A∣=ai1Ai1+ai2Ai2+⋯+ainAin=a1jA1j+a2jA2j+⋯+anjAnj(i=j=1,2,⋯,n)|A|=a_{i1}A_{i1}+a_{i2}A_{i2}+\cdots+a_{in}A_{in}=a_{1j}A_{1j}+a_{2j}A_{2j}+\cdots+a_{nj}A_{nj}(i = j = 1,2,\cdots,n)∣A∣=ai1Ai1+ai2Ai2+⋯+ainAin=a1jA1j+a2jA2j+⋯+anjAnj(i=j=1,2,⋯,n)
【注】为了充分简化计算,一般要求所展开的行或列仅有一到两个非零元,所以运用展开定理的关键在于展开之前的准备工作,要先借助行列式的性质 “化零”。 -
非降阶性质
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降阶性质
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公式:
- 上、下三角行列式
∣a11a12⋯a1n0a22⋯a2n⋮⋮⋮00⋯ann∣=∣a110⋯0a21a22⋯0⋮⋮⋮an1an2⋯ann∣=a11a22⋯ann\begin{vmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ 0 & a_{22} & \cdots & a_{2n} \\ \vdots & \vdots & & \vdots \\ 0 & 0 & \cdots & a_{nn} \end{vmatrix} = \begin{vmatrix} a_{11} & 0 & \cdots & 0 \\ a_{21} & a_{22} & \cdots & 0 \\ \vdots & \vdots & & \vdots \\ a_{n1} & a_{n2} & \cdots & a_{nn} \end{vmatrix} = a_{11}a_{22}\cdots a_{nn} a110⋮0a12a22⋮0⋯⋯⋯a1na2n⋮ann=a11a21⋮an10a22⋮an2⋯⋯⋯00⋮ann=a11a22⋯ann
- 上、下三角行列式
∣a11⋯a1,n−1a1na21⋯a2,n−10⋮⋮⋮an1⋯00∣=∣0⋯0a1n0⋯a2,n−1a2n⋮⋮⋮an1⋯an,n−1ann∣=(−1)n(n−1)2a1na2,n−1⋯an1\begin{vmatrix} a_{11} & \cdots & a_{1,n-1} & a_{1n} \\ a_{21} & \cdots & a_{2,n-1} & 0 \\ \vdots & & \vdots & \vdots \\ a_{n1} & \cdots & 0 & 0 \end{vmatrix}= \begin{vmatrix} 0 & \cdots & 0 & a_{1n} \\ 0 & \cdots & a_{2,n-1} & a_{2n} \\ \vdots & & \vdots & \vdots \\ a_{n1} & \cdots & a_{n,n-1} & a_{nn} \end{vmatrix} = (-1)^{\frac{n(n-1)}{2}} a_{1n}a_{2,n-1}\cdots a_{n1} a11a21⋮an1⋯⋯⋯a1,n−1a2,n−1⋮0a1n0⋮0=00⋮an1⋯⋯⋯0a2,n−1⋮an,n−1a1na2n⋮ann=(−1)2n(n−1)a1na2,n−1⋯an1
利用基本性质将行列式化为上、下三角行列式是计算数值型行列式的一种重要的方法。
- 范德蒙
∣111⋯1a1a2a3⋯ana12a22a32⋯an2⋮⋮⋮⋮a1n−1a2n−1a3n−1⋯ann−1∣=∣1a1a12⋯a1n−11a2a22⋯a2n−11a3a32⋯a3n−1⋮⋮⋮⋮1anan2⋯ann−1∣=∏1≤i<j≤n(aj−ai)\begin{vmatrix} 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{vmatrix} = \begin{vmatrix} 1 & a_1 & a_1^2 & \cdots & a_1^{n-1} \\ 1 & a_2 & a_2^2 & \cdots & a_2^{n-1} \\ 1 & a_3 & a_3^2 & \cdots & a_3^{n-1} \\ \vdots & \vdots & \vdots & & \vdots \\ 1 & a_n & a_n^2 & \cdots & a_n^{n-1} \end{vmatrix} = \prod_{\substack{1 \leq i < j \leq n}} (a_j - a_i) 1a1a12⋮a1n−11a2a22⋮a2n−11a3a32⋮a3n−1⋯⋯⋯⋯1anan2⋮ann−1=111⋮1a1a2a3⋮ana12a22a32⋮an2⋯⋯⋯⋯a1n−1a2n−1a3n−1⋮ann−1=1≤i<j≤n∏(aj−ai)
3. 拉普拉斯:∣AB∣=∣A∣∣B∣|AB| = |A||B|∣AB∣=∣A∣∣B∣
抽象型
假设 A,B 分别为 n 阶和 m 阶方阵,则:
- ∣A∣=∣AT∣|\boldsymbol{A}| = |\boldsymbol{A}^T|∣A∣=∣AT∣
- ∣kA∣=kn∣A∣|k\boldsymbol{A}| = k^n |\boldsymbol{A}|∣kA∣=kn∣A∣
- 当 m = n 时,∣AB∣=∣A∣∣B∣=∣BA∣|\boldsymbol{AB}| = |\boldsymbol{A}||\boldsymbol{B}| = |\boldsymbol{BA}|∣AB∣=∣A∣∣B∣=∣BA∣
- 当 A 可逆时,∣A−1∣=∣A∣−1|\boldsymbol{A}^{-1}| = |\boldsymbol{A}|^{-1}∣A−1∣=∣A∣−1
- ∣A∗∣=∣A∣n−1|\boldsymbol{A}^*| = |\boldsymbol{A}|^{n-1}∣A∗∣=∣A∣n−1
- ∣Ak∣=∣A∣k|\boldsymbol{A}^k| = |\boldsymbol{A}|^{k}∣Ak∣=∣A∣k
- ∣A00B∣=∣AC0B∣=∣A0CB∣=∣A∣∣B∣\begin{vmatrix} \boldsymbol{A} & \boldsymbol{0} \\ \boldsymbol{0} & \boldsymbol{B} \end{vmatrix} =\begin{vmatrix} \boldsymbol{A} & \boldsymbol{C} \\ \boldsymbol{0} & \boldsymbol{B} \end{vmatrix} = \begin{vmatrix} \boldsymbol{A} & \boldsymbol{0} \\ \boldsymbol{C} & \boldsymbol{B} \end{vmatrix} = |\boldsymbol{A}||\boldsymbol{B}|A00B=A0CB=AC0B=∣A∣∣B∣
- ∣0BA0∣=∣CBA0∣=∣0BAC∣=(−1)mn∣A∣∣B∣\begin{vmatrix} \boldsymbol{0} & \boldsymbol{B} \\ \boldsymbol{A} & \boldsymbol{0} \end{vmatrix} =\begin{vmatrix} \boldsymbol{C} & \boldsymbol{B} \\ \boldsymbol{A} & \boldsymbol{0} \end{vmatrix} = \begin{vmatrix} \boldsymbol{0} & \boldsymbol{B} \\ \boldsymbol{A} & \boldsymbol{C} \end{vmatrix} = (-1)^{mn} |\boldsymbol{A}||\boldsymbol{B}|0AB0=CAB0=0ABC=(−1)mn∣A∣∣B∣
- 假设矩阵 A 的特征值为λ1,…,λn\lambda_{1},\dots, \lambda_{n}λ1,…,λn,则有∣A∣=∏i=1nλi|\boldsymbol{A}| = \prod_{i=1}^n \lambda_i∣A∣=∏i=1nλi
![![[Pasted image 20251109161108.png]]](https://i-blog.csdnimg.cn/direct/f645b4479c4c4d91bab10351cce586f3.png)
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解:
∣a100b1b400a40b3a300a2b20∣=∣a1b100b4a40000a3b300b2a2∣=(a1a4−b1b4)(a3a2−b2b3)\begin{vmatrix} a_1 & 0 & 0 & b_1 \\ b_4 & 0 & 0 & a_4 \\ 0 & b_3 & a_3 & 0 \\ 0 & a_2 & b_2 & 0 \end{vmatrix} = \begin{vmatrix} a_1 & b_1 & 0 & 0 \\ b_4 & a_4 & 0 & 0 \\ 0 & 0 & a_3 & b_3 \\ 0 & 0 & b_2 & a_2 \end{vmatrix} = (a_1a_4 - b_1b_4)(a_3a_2 - b_2b_3)a1b40000b3a200a3b2b1a400=a1b400b1a40000a3b200b3a2=(a1a4−b1b4)(a3a2−b2b3)
选D -
拓:
∣1020011001233101∣\begin{vmatrix} 1 & 0 & 2 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 1 & 2 & 3 \\ 3 & 1 & 0 & 1 \end{vmatrix}1003011121200031
不含有二阶零块,无法用拉普拉斯展开。 -
改:
∣1001012310021211∣=∣1001012301231211∣=∣1100120003211112∣=∣1112∣∣2112∣=1×3=3\begin{vmatrix} 1 & 0 & 0 & 1 \\ 0 & 1 & 2 & 3 \\ 1 & 0 & 0 & 2 \\ 1 & 2 & 1 & 1 \end{vmatrix} = \begin{vmatrix} 1 & 0 & 0 & 1 \\ 0 & 1 & 2 & 3 \\ 0 & 1 & 2 & 3 \\ 1 & 2 & 1 & 1 \end{vmatrix} = \begin{vmatrix} 1 & 1 & 0 & 0 \\ 1 & 2 & 0 & 0 \\ 0 & 3 & 2 & 1 \\ 1 & 1 & 1 & 2 \end{vmatrix} = \begin{vmatrix} 1 & 1 \\ 1 & 2 \end{vmatrix} \begin{vmatrix} 2 & 1 \\ 1 & 2 \end{vmatrix} = 1 \times 3 = 3 1011010202011321=1001011202211331=1101123100210012=11122112=1×3=3
【小结】 -
二阶零子块:矩阵的某个二阶子式的所有元素均为 0;
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若四阶行列式含有二阶零子块,则可以利用性质 2 化成
∣ACOB∣=∣AOCB∣=∣A∣∣B∣\begin{vmatrix} \boldsymbol{A} & \boldsymbol{C} \\ \boldsymbol{O} & \boldsymbol{B} \end{vmatrix} = \begin{vmatrix} \boldsymbol{A} & \boldsymbol{O} \\ \boldsymbol{C} & \boldsymbol{B} \end{vmatrix} = |\boldsymbol{A}||\boldsymbol{B}|AOCB=ACOB=∣A∣∣B∣
![![[Pasted image 20251109163001.png]]](https://i-blog.csdnimg.cn/direct/d40f542afb6546cba693b69ba6fb17d3.png)
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法1
特殊值法。从考试规律来看,∣B+A∣∣B+A∣∣B+A∣一定为一个常数,故α1,α2,α3\alpha_{1},\alpha_{2},\alpha_{3}α1,α2,α3的取值对∣B+A∣∣B+A∣∣B+A∣不影响。可对α1,α2,α3\alpha_{1},\alpha_{2},\alpha_{3}α1,α2,α3采用特殊值法令A=E,即
∣B+A∣=∣α1+α2+α3,α1+2α2+4α3,α1+3α2+9α3∣=∣111123149∣=(2−1)(3−1)(3−2)=2|B+A|=\begin{vmatrix}\alpha_1+\alpha_2+\alpha_3, \alpha_1+2\alpha_2+4\alpha_3, \alpha_1+3\alpha_2+9\alpha_3\end{vmatrix}=\begin{vmatrix}1&1&1\\1&2&3\\1&4&9\end{vmatrix}=(2-1)(3-1)(3-2)=2∣B+A∣=α1+α2+α3,α1+2α2+4α3,α1+3α2+9α3=111124139=(2−1)(3−1)(3−2)=2 -
法2
∣B+A∣=∣α1+α2+α3,α1+2α2+4α3,α1+3α2+9α3∣|B+A|=\begin{vmatrix}\alpha_1+\alpha_2+\alpha_3, \alpha_1+2\alpha_2+4\alpha_3, \alpha_1+3\alpha_2+9\alpha_3\end{vmatrix}∣B+A∣=α1+α2+α3,α1+2α2+4α3,α1+3α2+9α3
=∣(α1,α2,α3)(111123149)∣=∣α1α2α3∣∣111123149∣=1⋅2=2=\left|(\alpha_1,\alpha_2,\alpha_3)\begin{pmatrix}1&1&1\\1&2&3\\1&4&9\end{pmatrix}\right|=|\alpha_1\alpha_2\alpha_3|\begin{vmatrix}1&1&1\\1&2&3\\1&4&9\end{vmatrix}=1\cdot2=2=(α1,α2,α3)111124139=∣α1α2α3∣111124139=1⋅2=2
【小结】如果矩阵是按列分块的,计算其行列式的第一种基本思路就是利用行列式性质对行列式做列变换,将行列式化简,第二种思路是利用矩阵按列分块的矩阵的运算法则:
[α1,α2,⋯,αn][k1k2⋮kn]=k1α1+k2α2+⋯+knαn[\alpha_1,\alpha_2,\cdots,\alpha_n]\begin{bmatrix}k_1\\k_2\\\vdots\\k_n\end{bmatrix}=k_1\alpha_1+k_2\alpha_2+\cdots+k_n\alpha_n[α1,α2,⋯,αn]k1k2⋮kn=k1α1+k2α2+⋯+knαn
,对该公式要从正反两方面来掌握,一方面,对公式本身的运用要熟练;另一方面,当我们看到向量组α1,α2,⋯,αnα_{1},α_{2},⋯,α_{n}α1,α2,⋯,αn的线性组合k1α1+k2α2+⋯+knαnk_{1}α_{1}+k_{2}α_{2}+⋯+k_{n}α_{n}k1α1+k2α2+⋯+knαn时,要能够反向运用该公式,将其分解为
[α1,α2,⋯,αn][k1k2⋮kn][\alpha_1,\alpha_2,\cdots,\alpha_n]\begin{bmatrix}k_1\\k_2\\\vdots\\k_n\end{bmatrix}[α1,α2,⋯,αn]k1k2⋮kn
![![[Pasted image 20251109180207.png]]](https://i-blog.csdnimg.cn/direct/1577e9214aff4603a3e274ec937d7264.png)
解:
∣2A∗B−1∣=2n∣A∗B−1∣=2n∣A∗∣∣B−1∣=2n⋅∣A∣n−1⋅1∣B∣|2A^*B^{-1}|=2^n|A^*B^{-1}|=2^n|A^*||B^{-1}|=2^n\cdot|A|^{n-1}\cdot\frac{1}{|B|}∣2A∗B−1∣=2n∣A∗B−1∣=2n∣A∗∣∣B−1∣=2n⋅∣A∣n−1⋅∣B∣1
=2n⋅2n−1⋅1−3=−22n−13=2^n\cdot2^{n-1}\cdot\frac{1}{-3}=-\frac{2^{2n-1}}{3}=2n⋅2n−1⋅−31=−322n−1
![![[Pasted image 20251109184624.png]]](https://i-blog.csdnimg.cn/direct/a7f6604264634a399dfaf9fa185398d3.png)
(−1)m⋅n∣A∣∣B∣=(−1)m⋅nab(-1)^{m\cdot n}|A||B|=(-1)^{m\cdot n}ab(−1)m⋅n∣A∣∣B∣=(−1)m⋅nab
![![[Pasted image 20251109184839.png]]](https://i-blog.csdnimg.cn/direct/689610c62ad84cc7b3129a73e3f08f41.png)
∣AAT∣=∣A∣∣AT∣=∣A∣∣A∣=∣A∣2=∣E∣=1⟹∣A∣=−1|AA^T|=|A||A^T|=|A||A|=|A|^2=|E|=1\implies |A|=-1∣AAT∣=∣A∣∣AT∣=∣A∣∣A∣=∣A∣2=∣E∣=1⟹∣A∣=−1
∣A+E∣=∣A+AAT∣=∣A(E+AT)∣=∣A∣∣E+AT∣=∣A∣∣(E+AT)T∣|A+E|=|A+AA^T|=|A(E+A^T)|=|A||E+A^T|=|A||(E+A^T)^T|∣A+E∣=∣A+AAT∣=∣A(E+AT)∣=∣A∣∣E+AT∣=∣A∣∣(E+AT)T∣
=∣A∣∣ET+(AT)T∣=∣A∣∣E+A∣=−∣E+A∣=|A||E^T+(A^T)^T|=|A||E+A|=-|E+A|=∣A∣∣ET+(AT)T∣=∣A∣∣E+A∣=−∣E+A∣
故∣A+E∣=0|A+E|=0∣A+E∣=0
![![[Pasted image 20251109190353.png]]](https://i-blog.csdnimg.cn/direct/06710d67f1124330beb4bfd015026709.png)
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法1:
由相似矩阵特征值相同,可得B的特征值也为2,3,4,5
B−E的特征值为1,2,3,4,∣B−E∣=1⋅2⋅3⋅4=24∣B−E∣=1⋅2⋅3⋅4=24∣B−E∣=1⋅2⋅3⋅4=24 -
法2:
A相似于B,B−E相似于A−E, ∣B−E∣=∣A−E∣=1⋅2⋅3⋅4=24∣B−E∣=∣A−E∣=1⋅2⋅3⋅4=24∣B−E∣=∣A−E∣=1⋅2⋅3⋅4=24
【小结】行列式计算的三个基本思路:
- 运用行列式的定义、性质和公式;
- 结合矩阵的运算及公式;
- 结合特征值。
