线性代数 01.04 克拉默法则

第一章第四节克拉默法则 \color{blue}{第一章第四节克拉默法则}

一、非齐次线性方程组的克拉默法则 \color{blue}{一、非齐次线性方程组的克拉默法则}

设非其次线性方程组,(b 1 ,b 2 ,⋯,b n ,不全为0) 设非其次线性方程组,(b_1, b_2, \cdots, b_n,不全为0)
⎧ ⎩ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ 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 a n =b n (1) \left \{ \begin{array}{l} a_{11}x_1 + a_{12}x_2 + \cdots + a_{1n}x_n = b_1 \\ a_{21}x_1 + a_{22}x_2 + \cdots + a_{2n}x_n = b_2 \\ \cdots \quad \cdots \quad \cdots \quad \cdots \\ a_{n1}x_1 + a_{n2}x_2 + \cdots + a_{nn}a_n = b_n \end{array} \right. \tag 1
若(1)的系数行列式若(1)的系数行列式
D=∣ ∣ ∣ ∣ ∣ ∣ a 11 a 21 ⋯a n1 a 12 a 22 ⋯a n2 ⋯⋯⋯⋯ a 1n a 2n ⋯a nn ∣ ∣ ∣ ∣ ∣ ∣ ≠0 (2) D = \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} \neq 0 \tag 2
则线性方程组(1)有唯一解则线性方程组(1)有唯一解
x j =D j D ,(j=1,2,⋯,n) (3) x_j = \dfrac{D_j}{D}, (j = 1, 2, \cdots, n) \tag 3

证明：先证x j =D j D ,j=1,2,⋯,n是(1)的解,要证x j =D j D ,j=1,2,⋯,n是(1)的解,只需证明(3)满足(1)即可,为此把(1)改写成:a i1 x 1 +a i2 x 2 +⋯+a in x n =b i ,i=1,2,⋯,n.即证明：a i1 D 1 D +a i2 D 2 D +⋯+a in D n D =b i ,i=1,2,⋯,n. 证明：先证x_j = \dfrac{D_j}{D},j = 1, 2, \cdots, n是(1)的解,\\ 要证x_j = \dfrac{D_j}{D},j = 1, 2, \cdots, n是(1)的解,只需证明\\ (3)满足(1)即可,为此把(1)改写成:\\ a_{i1}x_1 + a_{i2}x_2 + \cdots + a_{in}x_n = b_i, i = 1, 2, \cdots, n. \\ 即证明：\\ a_{i1}\dfrac{D_1}{D} + a_{i2}\dfrac{D_2}{D} + \cdots + a_{in}\dfrac{D_n}{D} = b_i, i = 1, 2, \cdots, n.
做n + 1阶行列式
D n+1 =∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ b i b 1 ⋯b i ⋯b n a i1 a 11 ⋯a i1 ⋯a n1 ⋯⋯⋯⋯⋯⋯ a ij a 1j ⋯a ij ⋯a nj ⋯⋯⋯⋯⋯⋯ a in a 1n ⋯a in ⋯a nn ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ D_{n + 1} = \begin{vmatrix} b_i & a_{i1} & \cdots & a_{ij} & \cdots & a_{in} \\ b_1 & a_{11} & \cdots & a_{1j} & \cdots & a_{1n} \\ \cdots & \cdots & \cdots & \cdots & \cdots & \cdots \\ b_i & a_{i1} & \cdots & a_{ij} & \cdots & a_{in} \\ \cdots & \cdots & \cdots & \cdots & \cdots & \cdots \\ b_n & a_{n1} & \cdots & a_{nj} & \cdots & a_{nn} \\ \end{vmatrix}
显然D n+1 =0.把D n+1 按第一行展开.需要求出第一行每个元素的代数余子式.第一行元素a ij 的代数余子式为: 显然D_{n + 1} = 0.把D_{n + 1} 按第一行展开.需要求出第一行\\ 每个元素的代数余子式.第一行元素a_{ij}的代数余子式为:
A ij =(−1) 1+j+1 ∣ ∣ ∣ ∣ ∣ b 1 ⋮b n a 11 ⋮a n1 ⋯⋯ a 1(j−1) ⋮a n(j−1) a 1(j+1) ⋮a n(j+1) ⋯⋯ a 1n ⋮a nn ∣ ∣ ∣ ∣ ∣ =(−1) 1+j+1 (−1) j−1 ∣ ∣ ∣ ∣ ∣ a 11 ⋮a n1 ⋯⋯ a 1(j−1) ⋮a n(j−1) b 1 ⋮b n a 1(j+1) ⋮a n(j+1) ⋯⋯ a 1n ⋮a nn ∣ ∣ ∣ ∣ ∣ =(−1) 2j+1 D j =−D j (j=1,2,⋯,n). A_{ij} = (-1)^{1 + j + 1} \begin{vmatrix} b_1 & a_{11} & \cdots & a_{1(j-1)} & a_{1(j+1)} & \cdots & a_{1n} \\ \vdots & \vdots & & \vdots & \vdots & & \vdots \\ b_n & a_{n1} & \cdots & a_{n(j-1)} & a_{n(j+1)} & \cdots & a_{nn} \\ \end{vmatrix} \\ = (-1)^{1 + j + 1}(-1)^{j-1} \begin{vmatrix} a_{11} & \cdots & a_{1(j-1)} & b_1 & a_{1(j+1)} & \cdots & a_{1n} \\ \vdots & & \vdots & \vdots & \vdots & & \vdots \\ a_{n1} & \cdots & a_{n(j-1)} & b_n & a_{n(j+1)} & \cdots & a_{nn} \\ \end{vmatrix} \\ =(-1)^{2j + 1}D_j \\ = -D_j(j = 1, 2, \cdots, n).
所以,D n+1 =b i D−a i1 D 1 −a i2 D 2 −⋯−a in D n =0即:a i1 D 1 D +a i2 D 2 D +⋯+a in D n D =b i (i=1,2,⋯,n)这说明x j =D j D ,j=1,2,⋯,n是(1)的解所以, D_{n + 1} = b_iD - a_{i1}D_1 - a_{i2}D_2 - \cdots - a_{in}D_n = 0 \\ 即:a_{i1}\dfrac{D_1}{D} + a_{i2}\dfrac{D_2}{D} + \cdots + a_{in}\dfrac{D_n}{D} = b_i (i = 1, 2, \cdots, n) \\ 这说明x_j = \dfrac{D_j}{D}, j = 1, 2, \cdots, n是(1)的解

再证唯一性,假设x j =c j ,j=1,2,⋯,n也是(1)的解,在(2)两端同时乘以c j 再证唯一性,假设x_j = c_j, j = 1, 2, \cdots, n\\ 也是(1)的解,在(2)两端同时乘以c_j
c j D=∣ ∣ ∣ ∣ ∣ a 11 ⋮a n1 ⋯⋯ a 1j c j ⋮a nj c j ⋯⋯ a 1n ⋮a nn ∣ ∣ ∣ ∣ ∣ =∣ ∣ ∣ ∣ ∣ a 11 ⋮a n1 ⋯⋯ (a 11 c 1 +⋯+a 1j c j +⋯+a 1n c n )⋮(a n1 c 1 +⋯+a nj c j +⋯+a nn c n ) ⋯⋯ a 1n ⋮a nn ∣ ∣ ∣ ∣ ∣ =∣ ∣ ∣ ∣ ∣ a 11 ⋮a n1 ⋯⋯ b 1 ⋮b n ⋯⋯ a 1n ⋮a nn ∣ ∣ ∣ ∣ ∣ =D j c_jD = \begin{vmatrix} a_{11} & \cdots & a_{1j}c_j & \cdots & a_{1n} \\ \vdots & & \vdots & & \vdots \\ a_{n1} & \cdots & a_{nj}c_j & \cdots & a_{nn} \\ \end{vmatrix} \\ = \begin{vmatrix} a_{11} & \cdots & (a_{11}c_1 + \cdots + a_{1j}c_j + \cdots + a_{1n}c_n) & \cdots & a_{1n} \\ \vdots & & \vdots & & \vdots \\ a_{n1} & \cdots & (a_{n1}c_1 + \cdots + a_{nj}c_j + \cdots + a_{nn}c_n) & \cdots & a_{nn} \\ \end{vmatrix} \\ = \begin{vmatrix} a_{11} & \cdots & b_1 & \cdots & a_{1n} \\ \vdots & & \vdots & & \vdots \\ a_{n1} & \cdots & b_n & \cdots & a_{nn} \\ \end{vmatrix} \\ = D_j
由于D≠0,所以由于 D \neq 0,所以
c j =D j D j=1,2,⋯,n.故线性方程组(1)有唯一解(3). c_j = \dfrac{D_j}{D} j = 1, 2, \cdots, n.\\ 故线性方程组(1)有唯一解(3).

例1.解线性方程组
⎧ ⎩ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ 2x 1 +x 2 −5x 3 +x 4 =8x 1 −3x 2 −6x 4 =9 2x 2 −x 3 +2x 4 =−5x 1 +4x 2 −7x 3 +6x 4 =0 \left \{ \begin{array}{l} 2x_1 + x_2 -5x_3 + x_4 = 8 \\ x_1 -3x_2 \qquad \ -6x_4 = 9 \\ \qquad \ 2x_2 - x_3 + 2x_4 = -5 \\ x_1 + 4x_2 - 7x_3 + 6x_4 = 0 \end{array} \right.
解：D=∣ ∣ ∣ ∣ ∣ ∣ 2101 1−324 −50−1−7 1−626 ∣ ∣ ∣ ∣ ∣ ∣ =∣ ∣ ∣ ∣ ∣ ∣ 0100 7−327 −50−1−7 13−6212 ∣ ∣ ∣ ∣ ∣ ∣ =−∣ ∣ ∣ ∣ 727 −5−1−7 13212 ∣ ∣ ∣ ∣ =∣ ∣ ∣ −3−7 3−2 ∣ ∣ ∣ =27D 1 =∣ ∣ ∣ ∣ ∣ ∣ 89−50 1−324 −50−1−7 1−626 ∣ ∣ ∣ ∣ ∣ ∣ =81D 2 =∣ ∣ ∣ ∣ ∣ ∣ 2101 89−50 −50−1−7 1−626 ∣ ∣ ∣ ∣ ∣ ∣ =−108D 3 =∣ ∣ ∣ ∣ ∣ ∣ 2101 1−324 89−50 1−626 ∣ ∣ ∣ ∣ ∣ ∣ =−27D 4 =∣ ∣ ∣ ∣ ∣ ∣ 2101 1−324 −50−1−7 89−50 ∣ ∣ ∣ ∣ ∣ ∣ =27 解：\\ D = \begin{vmatrix} 2 & 1 & -5 & 1 \\ 1 & -3 & 0 & -6 \\ 0 & 2 & -1 & 2 \\ 1 & 4 & -7 & 6 \\ \end{vmatrix} \\ = \begin{vmatrix} 0 & 7 & -5 & 13 \\ 1 & -3 & 0 & -6 \\ 0 & 2 & -1 & 2 \\ 0 & 7 & -7 & 12 \\ \end{vmatrix} \\ = -\begin{vmatrix} 7 & -5 & 13 \\ 2 & -1 & 2 \\ 7 & -7 & 12 \\ \end{vmatrix} \\ = \begin{vmatrix} -3 & 3 \\ -7 & -2 \\ \end{vmatrix} = 27 \\ D_1 = \begin{vmatrix} 8 & 1 & -5 & 1 \\ 9 & -3 & 0 & -6 \\ -5 & 2 & -1 & 2 \\ 0 & 4 & -7 & 6 \\ \end{vmatrix} = 81 \\ D_2 = \begin{vmatrix} 2 & 8 & -5 & 1 \\ 1 & 9 & 0 & -6 \\ 0 & -5 & -1 & 2 \\ 1 & 0 & -7 & 6 \\ \end{vmatrix} = -108 \\ D_3 = \begin{vmatrix} 2 & 1 & 8 & 1 \\ 1 & -3 & 9 & -6 \\ 0 & 2 & -5 & 2 \\ 1 & 4 & 0 & 6 \\ \end{vmatrix} = -27 \\ D_4 = \begin{vmatrix} 2 & 1 & -5 & 8 \\ 1 & -3 & 0 & 9 \\ 0 & 2 & -1 & -5 \\ 1 & 4 & -7 & 0 \\ \end{vmatrix} = 27
于是得原方程组的解⎧ ⎩ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ x 1 =D 1 D =8127 =3x 2 =D 2 D =−10827 =−4x 3 =D 3 D =−2727 =−1x 4 =D 4 D =2727 =1 于是得原方程组的解 \left \{ \begin{array}{l}x_1 = \dfrac{D_1}{D} = \dfrac{81}{27} = 3 \\ x_2 = \dfrac{D_2}{D} = \dfrac{-108}{27} = -4 \\ x_3 = \dfrac{D_3}{D} = \dfrac{-27}{27} = -1 \\ x_4 = \dfrac{D_4}{D} = \dfrac{27}{27} = 1 \end{array} \right.

定理2.如果线性方程组(1)的系数行列式D不等于0,则(1)有唯一的解. 定理2.如果线性方程组(1)的系数行列式D不等于0,则(1)有唯一的解.
定理2 ′ .如果线性方程组(1)的无解或有多个解,则它的系数行列式必为0. 定理2^{\prime}.如果线性方程组(1)的无解或有多个解,则它的系数行列式必为0.

二、齐次线性方程组的克拉默法则 \color{blue}{二、齐次线性方程组的克拉默法则}

设齐次线性方程组,(b 1 ,b 2 ,⋯,b n ,不全为0) 设齐次线性方程组,(b_1, b_2, \cdots, b_n,不全为0)
⎧ ⎩ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ a 11 x 1 +a 12 x 2 +⋯+a 1n x n =0a 21 x 1 +a 22 x 2 +⋯+a 2n x n =0⋯⋯⋯⋯a n1 x 1 +a n2 x 2 +⋯+a nn a n =0 (4) \left \{ \begin{array}{l} a_{11}x_1 + a_{12}x_2 + \cdots + a_{1n}x_n = 0 \\ a_{21}x_1 + a_{22}x_2 + \cdots + a_{2n}x_n = 0 \\ \cdots \quad \cdots \quad \cdots \quad \cdots \\ a_{n1}x_1 + a_{n2}x_2 + \cdots + a_{nn}a_n = 0 \end{array} \right. \tag 4

若(4)的系数行列式若(4)的系数行列式
D=∣ ∣ ∣ ∣ ∣ ∣ a 11 a 21 ⋯a n1 a 12 a 22 ⋯a n2 ⋯⋯⋯⋯ a 1n a 2n ⋯a nn ∣ ∣ ∣ ∣ ∣ ∣ ≠0 (5) D = \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} \neq 0 \tag 5
则线性方程组(4)没有非零解则线性方程组(4)没有非零解

定理3.如果齐次线性方程组(4)的系数行列式D不等于0，则齐次线性方程组(4)没有非零解. 定理3.如果齐次线性方程组(4)的系数行列式D不等于0，则齐次线性方程组(4)没有非零解.
定理3 ′ .如果齐次线性方程组(4)有非零解,则它的系数行列式必为0. 定理3^{\prime}.如果齐次线性方程组(4)有非零解,则它的系数行列式必为0.

例2.问λ在什么条件下,方程组{λx 1 +x 2 =0x 1 +λx 2 =0 有非零解？例2.问\lambda在什么条件下,方程组\left \{ \begin{array}{l} \lambda x_1 + x_2 = 0 \\ x_1 + \lambda x_2 = 0 \end{array} \right. 有非零解？
解:由定理3 ′ 知，若方程组有非零解,则其系数行列式必为零D=∣ ∣ ∣ λ1 1λ ∣ ∣ ∣ =λ 2 −1=0λ=±1 解: 由定理3^{\prime}知，若方程组有非零解,则其系数行列式必为零\\ D = \begin{vmatrix} \lambda & 1 \\ 1 & \lambda \\ \end{vmatrix} = \lambda^2 - 1 = 0 \\ \lambda = \pm 1

例3.设非齐次线性方程组例3.设非齐次线性方程组
⎧ ⎩ ⎨ ⎪ ⎪ λx 1 +x 2 +x 3 =1x 1 +λx 2 +x 3 =λx 1 +x 2 +λx 3 =λ 2    \left \{ \begin{array}{l} \lambda x_1 + x_2 + x_3 = 1 \\ x_1 + \lambda x_2 + x_3 = \lambda \\ x_1 + x_2 + \lambda x_3 = \lambda^2 \end{array} \right.
问λ为何值时,该方程组有唯一解,并求其解. 问\lambda为何值时,该方程组有唯一解,并求其解.
解:方程组系数行列式为D=∣ ∣ ∣ ∣ λ11 1λ1 11λ ∣ ∣ ∣ ∣ =(λ+2)(λ−1) 2 当λ≠−2,λ≠1时，方程组有唯一解。解:方程组系数行列式为\\ D= \begin{vmatrix} \lambda & 1 & 1 \\ 1 & \lambda & 1 \\ 1 & 1 & \lambda \\ \end{vmatrix} = (\lambda + 2)(\lambda - 1)^2 \\ 当\lambda \neq -2, \lambda \neq 1时，方程组有唯一解。
D 1 =∣ ∣ ∣ ∣ 1λλ 2  1λ1 11λ ∣ ∣ ∣ ∣ =∣ ∣ ∣ ∣ 00λ 2 −1 1λ1 11λ ∣ ∣ ∣ ∣ =(λ 2 −1)∣ ∣ ∣ 1λ 11 ∣ ∣ ∣ =−(λ+1)(λ−1) 2   D_1= \begin{vmatrix} 1 & 1 & 1 \\ \lambda & \lambda & 1 \\ \lambda^2 & 1 & \lambda \\ \end{vmatrix} = \begin{vmatrix} 0 & 1 & 1 \\ 0 & \lambda & 1 \\ \lambda^2 - 1 & 1 & \lambda \\ \end{vmatrix} \\ = (\lambda^2 - 1) \begin{vmatrix} 1 & 1 \\ \lambda & 1 \\ \end{vmatrix} = -(\lambda + 1)(\lambda - 1)^2
D 2 =∣ ∣ ∣ ∣ λ11 1λλ 2  11λ ∣ ∣ ∣ ∣ =∣ ∣ ∣ ∣ ∣ 010 1−λ 2 λλ 2 −λ 1−λ1λ−1 ∣ ∣ ∣ ∣ ∣ =−∣ ∣ ∣ 1−λ 2 λ 2 −λ 1−λλ−1 ∣ ∣ ∣ =−(1−λ 2 )(λ−1)+(λ 2 −λ)(1−λ)=(λ−1) 2   D_2= \begin{vmatrix} \lambda & 1 & 1 \\ 1 & \lambda & 1 \\ 1 & \lambda^2 & \lambda \\ \end{vmatrix} = \begin{vmatrix} 0 & 1-\lambda^2 & 1- \lambda \\ 1 & \lambda & 1 \\ 0 & \lambda^2 -\lambda & \lambda -1\\ \end{vmatrix} \\ = - \begin{vmatrix} 1-\lambda^2 & 1- \lambda \\ \lambda^2 -\lambda & \lambda -1\\ \end{vmatrix} \\ = -(1 - \lambda^2)(\lambda - 1) + (\lambda^2 - \lambda)(1 - \lambda) \\ = (\lambda - 1)^2

D 3 =∣ ∣ ∣ ∣ λ11 1λ1 1λλ 2 ∣ ∣ ∣ ∣ =∣ ∣ ∣ ∣ ∣ λ1−λ 2 1 101 10λ 2 ∣ ∣ ∣ ∣ ∣ =−(1−λ 2 )∣ ∣ ∣ 11 1λ 2 ∣ ∣ ∣ =−(1−λ 2 )(λ 2 −1)=(λ+1) 2 (λ−1) 2 D_3= \begin{vmatrix} \lambda & 1 & 1 \\ 1 & \lambda & \lambda \\ 1 & 1 & \lambda^2 \\ \end{vmatrix} = \begin{vmatrix} \lambda & 1 & 1 \\ 1-\lambda^2 & 0 & 0 \\ 1 & 1 & \lambda^2 \\ \end{vmatrix} \\ = - (1-\lambda^2) \begin{vmatrix} 1 & 1 \\ 1 & \lambda^2 \\ \end{vmatrix} \\ = - (1-\lambda^2)(\lambda^2 - 1) \\ = (\lambda + 1)^2(\lambda - 1)^2
⎧ ⎩ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ x 1 =D 1 D =−(λ+1)(λ−1) 2 (λ+2)(λ−1) 2 =−λ+1λ+2 x 2 =D 2 D =−(λ−1) 2 (λ+2)(λ−1) 2 =1λ+2 x 3 =D 3 D =(λ+1) 2 (λ−1) 2 (λ+2)(λ−1) 2 =(λ+1) 2 λ+2 \left \{ \begin{array}{l} x_1 = \dfrac{D_1}{D} = \dfrac{-(\lambda + 1)(\lambda - 1)^2}{(\lambda + 2)(\lambda - 1)^2} = -\dfrac{\lambda + 1}{\lambda + 2} \\ x_2 = \dfrac{D_2}{D} = \dfrac{- (\lambda - 1)^2}{(\lambda + 2)(\lambda - 1)^2} = \dfrac{1}{\lambda + 2} \\ x_3 = \dfrac{D_3}{D} = \dfrac{ (\lambda + 1)^2(\lambda - 1)^2}{(\lambda + 2)(\lambda - 1)^2} = \dfrac{(\lambda + 1)^2}{\lambda + 2} \end{array} \right.