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Cramer_rule克莱姆法则讲解

發布時間:2023/12/10 编程问答 46 豆豆
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克萊姆法則講解

最近因準備保研在復習線性代數,克萊姆法則(Cramer’s rule)老是記混,故此在博客整理一下克萊姆法則。

1.考慮方程組

{a11x1+a12x2+???+a1nxn=b1a21x1+a22x2+???+a2nxn=b2???????????an1x1+an2x2+???+annxn=bn\begin{cases} a_{11}x_1+a_{12}x_2+···+a_{1n}x_n=b_1\\ a_{21}x_1+a_{22}x_2+···+a_{2n}x_n=b_2\\ ···········\\ a_{n1}x_1+a_{n2}x_2+···+a_{nn}x_n=b_n\\ \end{cases} ??????????a11?x1?+a12?x2?+???+a1n?xn?=b1?a21?x1?+a22?x2?+???+a2n?xn?=b2????????????an1?x1?+an2?x2?+???+ann?xn?=bn??
與二、三元方程組類似,n元方程組的解也可用行列式表示。

2.方程組的矩陣表示

AXβ[a11a12...a1na21a22...a2n…………an1an2...ann][x1x2...xn]=[b1b2...bn]\begin{matrix} A & X& & \beta \\ \left[\begin{array}{rr} a_{11} & a_{12} & ... & a_{1n}\\ a_{21} & a_{22} & ... & a_{2n}\\ \dots &\dots &\dots &\dots \\ a_{n1} & a_{n2} & ... & a_{nn}\\ \end{array}\right] & \left[\begin{array}{rr} x_1 \\ x_2 \\ ... \\ x_n \\ \end{array}\right] & = & \left[\begin{array}{rr} b_1 \\ b_2 \\ ... \\ b_n \\ \end{array}\right] \end{matrix} A?????a11?a21?an1??a12?a22?an2??.........?a1n?a2n?ann????????X?????x1?x2?...xn????????=?β?????b1?b2?...bn????????

3.解過程

分別記錄A,Aj如下:A,A_{j}如下:A,Aj?:
A=∣a11a12...a1na21a22...a2n…………an1an2...ann∣A= \left |\begin{array}{cccc} a_{11} & a_{12} & ... & a_{1n}\\ a_{21} & a_{22} & ... & a_{2n}\\ \dots &\dots &\dots &\dots \\ a_{n1} & a_{n2} & ... & a_{nn}\\ \end{array}\right| A=?a11?a21?an1??a12?a22?an2??.........?a1n?a2n?ann???

Aj=∣a11...a1(j?1)b1a1(j+1)...a1na21...a2(j?1)b2a2(j+1)...a2n…………………an1...an(j?1)bnan(j+1)...ann∣A_{j}=\left |\begin{array}{cccc} a_{11} & ... & a_{1(j-1)}&b_1&a_{1(j+1)}&... & a_{1n}\\ a_{21} & ... & a_{2(j-1)}&b_2&a_{2(j+1)}&... & a_{2n}\\ \dots &\dots &\dots&\dots&\dots &\dots &\dots\\ a_{n1} & ... & a_{n(j-1)}&b_n&a_{n(j+1)}&... & a_{nn}\\ \end{array}\right| Aj?=?a11?a21?an1??.........?a1(j?1)?a2(j?1)?an(j?1)??b1?b2?bn??a1(j+1)?a2(j+1)?an(j+1)??.........?a1n?a2n?ann???

若系數行列式不等于0,即:
A=∣a11a12...a1na21a22...a2n…………an1an2...ann∣≠0A= \left |\begin{array}{cccc} a_{11} & a_{12} & ... & a_{1n}\\ a_{21} & a_{22} & ... & a_{2n}\\ \dots &\dots &\dots &\dots \\ a_{n1} & a_{n2} & ... & a_{nn}\\ \end{array}\right| \neq 0 A=?a11?a21?an1??a12?a22?an2??.........?a1n?a2n?ann????=0

則解為:
xj=AjA=∣a11...a1(j?1)b1a1(j+1)...a1na21...a2(j?1)b2a2(j+1)...a2n…………………an1...an(j?1)bnan(j+1)...ann∣∣a11a12...a1na21a22...a2n…………an1an2...ann∣x_j=\frac{A_j}{A}=\frac{\left |\begin{array}{cccc} a_{11} & ... & a_{1(j-1)}&b_1&a_{1(j+1)}&... & a_{1n}\\ a_{21} & ... & a_{2(j-1)}&b_2&a_{2(j+1)}&... & a_{2n}\\ \dots &\dots &\dots&\dots&\dots &\dots &\dots\\ a_{n1} & ... & a_{n(j-1)}&b_n&a_{n(j+1)}&... & a_{nn}\\ \end{array}\right|}{\left |\begin{array}{cccc} a_{11} & a_{12} & ... & a_{1n}\\ a_{21} & a_{22} & ... & a_{2n}\\ \dots &\dots &\dots &\dots \\ a_{n1} & a_{n2} & ... & a_{nn}\\ \end{array}\right|} xj?=AAj??=?a11?a21?an1??a12?a22?an2??.........?a1n?a2n?ann????a11?a21?an1??.........?a1(j?1)?a2(j?1)?an(j?1)??b1?b2?bn??a1(j+1)?a2(j+1)?an(j+1)??.........?a1n?a2n?ann????

克萊姆法則:

  • 定理一:若系數行列A≠0A\neq0A?=0,則方程組有為一解xj=AjAx_j=\frac{A_j}{A}xj?=AAj??,即:
    KaTeX parse error: Expected '}', got 'EOF' at end of input: …{array}\right|
    定理一的個人理解: 如果要證明定理一就必須分三步,即
    Step1:證明方程組有解;
    Step2:證明方程組的解唯一;
    Step3:證明方程組解的公式。
  • 定理二:若齊次方程組的系數行列式 A≠0A\neq0A?=0 則方程組有惟一零解。
    補充知識:齊次線性方程組為方程組中常數項為000,即在本題中為β\betaβ中每一項bi=0b_i=0bi?=0都成立,即:
    {a11x1+a12x2+???+a1nxn=0a21x1+a22x2+???+a2nxn=0???????????an1x1+an2x2+???+annxn=0\begin{cases} a_{11}x_1+a_{12}x_2+···+a_{1n}x_n=0\\ a_{21}x_1+a_{22}x_2+···+a_{2n}x_n=0\\ ···········\\ a_{n1}x_1+a_{n2}x_2+···+a_{nn}x_n=0\\ \end{cases} ??????????a11?x1?+a12?x2?+???+a1n?xn?=0a21?x1?+a22?x2?+???+a2n?xn?=0???????????an1?x1?+an2?x2?+???+ann?xn?=0?
    齊次線性方程組的矩陣表示為:
    AXβ[a11a12...a1na21a22...a2n…………an1an2...ann][x1x2...xn]=[00...0]\begin{matrix} A & X& & \beta \\ \left[\begin{array}{rr} a_{11} & a_{12} & ... & a_{1n}\\ a_{21} & a_{22} & ... & a_{2n}\\ \dots &\dots &\dots &\dots \\ a_{n1} & a_{n2} & ... & a_{nn}\\ \end{array}\right] & \left[\begin{array}{rr} x_1 \\ x_2 \\ ... \\ x_n \\ \end{array}\right] & = & \left[\begin{array}{rr} 0 \\ 0 \\ ... \\ 0 \\ \end{array}\right] \end{matrix} A?????a11?a21?an1??a12?a22?an2??.........?a1n?a2n?ann????????X?????x1?x2?...xn????????=?β?????00...0???????
    定理二的個人理解: 因為當齊次方程組的系數行列式 A≠0A\neq0A?=0 時方程組有惟一零解,而此時又因為齊次線性方程組中常數項為000,所以唯一的解只能為000解。
  • 補充知識:若要證明齊次線性方程組有非零解,則系數行列式A=0A=0A=0
    補充知識的個人理解: 若要保證齊次線性方程組有非零解,則系數行列式A=0A=0A=0,此時因為系數行列式A=0A=0A=0,則方程組解不唯一,就有除了唯一零解之外的非零解。

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