Cramer_rule克莱姆法则讲解

克莱姆法则讲解

最近因准备保研在复习线性代数,克莱姆法则(Cramer’s rule)老是记混,故此在博客整理一下克莱姆法则。

1.考虑方程组

{ a 11 x 1 + a 12 x 2 + ⋅ ⋅ ⋅ + a 1 n x n = b 1 a 21 x 1 + a 22 x 2 + ⋅ ⋅ ⋅ + a 2 n x n = b 2 ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ a n 1 x 1 + a n 2 x 2 + ⋅ ⋅ ⋅ + a n n x n = b n \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} a11x1+a12x2++a1nxn=b1a21x1+a22x2++a2nxn=b2an1x1+an2x2++annxn=bn
与二、三元方程组类似,n元方程组的解也可用行列式表示。

2.方程组的矩阵表示

A X β [ a 11 a 12 . . . a 1 n a 21 a 22 . . . a 2 n … … … … a n 1 a n 2 . . . a n n ] [ x 1 x 2 . . . x n ] = [ b 1 b 2 . . . b n ] \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} Aa11a21an1a12a22an2.........a1na2nannXx1x2...xn=βb1b2...bn

3.解过程

分别记录 A , A j 如 下 : A,A_{j}如下: A,Aj:
A = ∣ a 11 a 12 . . . a 1 n a 21 a 22 . . . a 2 n … … … … a n 1 a n 2 . . . a n n ∣ 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=a11a21an1a12a22an2.........a1na2nann

A j = ∣ a 11 . . . a 1 ( j − 1 ) b 1 a 1 ( j + 1 ) . . . a 1 n a 21 . . . a 2 ( j − 1 ) b 2 a 2 ( j + 1 ) . . . a 2 n … … … … … … … a n 1 . . . a n ( j − 1 ) b n a n ( j + 1 ) . . . a n n ∣ 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=a11a21an1.........a1(j1)a2(j1)an(j1)b1b2bna1(j+1)a2(j+1)an(j+1).........a1na2nann

若系数行列式不等于0,即:
A = ∣ a 11 a 12 . . . a 1 n a 21 a 22 . . . a 2 n … … … … a n 1 a n 2 . . . a n n ∣ ≠ 0 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| \neq 0 A=a11a21an1a12a22an2.........a1na2nann=0

则解为:
x j = A j A = ∣ a 11 . . . a 1 ( j − 1 ) b 1 a 1 ( j + 1 ) . . . a 1 n a 21 . . . a 2 ( j − 1 ) b 2 a 2 ( j + 1 ) . . . a 2 n … … … … … … … a n 1 . . . a n ( j − 1 ) b n a n ( j + 1 ) . . . a n n ∣ ∣ a 11 a 12 . . . a 1 n a 21 a 22 . . . a 2 n … … … … a n 1 a n 2 . . . a n n ∣ 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=a11a21an1a12a22an2.........a1na2nanna11a21an1.........a1(j1)a2(j1)an(j1)b1b2bna1(j+1)a2(j+1)an(j+1).........a1na2nann

克莱姆法则:

  • 定理一:若系数行列 A ≠ 0 A\neq0 A=0,则方程组有为一解 x j = A j A x_j=\frac{A_j}{A} xj=AAj,即:
    KaTeX parse error: Expected '}', got 'EOF' at end of input: …{array}\right|
    定理一的个人理解: 如果要证明定理一就必须分三步,即
    Step1:证明方程组有解;
    Step2:证明方程组的解唯一;
    Step3:证明方程组解的公式。
  • 定理二:若齐次方程组的系数行列式 A ≠ 0 A\neq0 A=0 则方程组有惟一零解。
    补充知识:齐次线性方程组为方程组中常数项为 0 0 0,即在本题中为 β \beta β中每一项 b i = 0 b_i=0 bi=0都成立,即:
    { a 11 x 1 + a 12 x 2 + ⋅ ⋅ ⋅ + a 1 n x n = 0 a 21 x 1 + a 22 x 2 + ⋅ ⋅ ⋅ + a 2 n x n = 0 ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ a n 1 x 1 + a n 2 x 2 + ⋅ ⋅ ⋅ + a n n x n = 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} a11x1+a12x2++a1nxn=0a21x1+a22x2++a2nxn=0an1x1+an2x2++annxn=0
    齐次线性方程组的矩阵表示为:
    A X β [ a 11 a 12 . . . a 1 n a 21 a 22 . . . a 2 n … … … … a n 1 a n 2 . . . a n n ] [ x 1 x 2 . . . x n ] = [ 0 0 . . . 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} Aa11a21an1a12a22an2.........a1na2nannXx1x2...xn=β00...0
    定理二的个人理解: 因为当齐次方程组的系数行列式 A ≠ 0 A\neq0 A=0 时方程组有惟一零解,而此时又因为齐次线性方程组中常数项为 0 0 0,所以唯一的解只能为 0 0 0解。
  • 补充知识:若要证明齐次线性方程组有非零解,则系数行列式 A = 0 A=0 A=0
    补充知识的个人理解: 若要保证齐次线性方程组有非零解,则系数行列式 A = 0 A=0 A=0,此时因为系数行列式 A = 0 A=0 A=0,则方程组解不唯一,就有除了唯一零解之外的非零解。

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