含有n个未知数
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n
x_1,x_2,\cdots,x_n
x1,x2,⋯,xn的n个线性方程的方程组
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\begin{cases} 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\cdots,\\ a_{n1}x_1+a_{n2}x_2+\cdots+a_{nn}x_n=b_n,\\ \end{cases}
⎩
⎨
⎧a11x1+a12x2+⋯+a1nxn=b1,a21x1+a22x2+⋯+a2nxn=b2,⋯⋯,an1x1+an2x2+⋯+annxn=bn,
克拉默法则 如果线性方程组的系数矩阵A的行列式不等于零,即
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|A|=\begin{vmatrix} a_{11}&\cdots&a_{1n}\\ \vdots&&\vdots\\ a_{n1}&\cdots&a_{nn} \end{vmatrix} \not=0
∣A∣=
a11⋮an1⋯⋯a1n⋮ann
=0 那么方程组有唯一解
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x_1=\frac{|A_1|}{|A|},x_2=\frac{|A_2|}{|A|},\cdots,x_n=\frac{|A_n|}{|A|}
x1=∣A∣∣A1∣,x2=∣A∣∣A2∣,⋯,xn=∣A∣∣An∣
其中
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A_j(j=1,2,\cdots,n)
Aj(j=1,2,⋯,n)是吧系数矩阵A中第
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j列的元素哟好难过方程组右端常数项代替后所得到的n阶矩阵,即
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A_j=\begin{pmatrix} a_{11}&\cdots&a_{1j-1}&b_1&a_{1j+1}&\cdots&a_{1n}\\ \vdots&&\vdots&\vdots&\vdots&&\vdots\\ a_{n1}&\cdots&a_{nj-1}&b_n&a_{nj+1}&\cdots&a_{nn}\\ \end{pmatrix}
Aj=
a11⋮an1⋯⋯a1j−1⋮anj−1b1⋮bna1j+1⋮anj+1⋯⋯a1n⋮ann
证明:
把方程组写成矩阵方程
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阶矩阵,因
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,故
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存在
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根据逆矩阵的唯一性,知
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是方程组的唯一解向量
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即
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证明:\\ 把方程组写成矩阵方程 Ax=b\\ A=(a_{ij})_{n\times n}位n阶矩阵,因|A|\not=0,故A^{-1}存在\\ x=A^{-1}b\\ 根据逆矩阵的唯一性,知x=A^{-1}b是方程组的唯一解向量\\ x=\frac{A^*}{|A|}b=\frac{1}{|A|}\begin{pmatrix} A_{11}&A_{21}&\cdots&A_{n1}\\ A_{12}&A_{22}&\cdots&A_{n2}\\ \vdots&\vdots&&\vdots\\ A_{1n}&A_{2n}&\cdots&A_{nn}\\ \end{pmatrix} \begin{pmatrix} b_1\\ b_2\\ \vdots\\ b_n\\ \end{pmatrix}\\ =\frac{1}{|A|}\begin{pmatrix} b_1A_{11}+b_2A_{21}+\cdots+b_nA_{n1}\\ b_1A_{12}+b_2A_{22}+\cdots+b_nA_{n2}\\ \vdots\\ b_1A_{1n}+b_2A_{2n}+\cdots+b_nA_{nn}\\ \end{pmatrix}\\ 即x_j=\frac{1}{|A|}(b_1A_{1j}+b_2A_{2j}+\cdots+b_nA_{nj})=\frac{1}{|A|}|A_j|(j=1,2,\cdots,n)
证明:把方程组写成矩阵方程Ax=bA=(aij)n×n位n阶矩阵,因∣A∣=0,故A−1存在x=A−1b根据逆矩阵的唯一性,知x=A−1b是方程组的唯一解向量x=∣A∣A∗b=∣A∣1
A11A12⋮A1nA21A22⋮A2n⋯⋯⋯An1An2⋮Ann
b1b2⋮bn
=∣A∣1
b1A11+b2A21+⋯+bnAn1b1A12+b2A22+⋯+bnAn2⋮b1A1n+b2A2n+⋯+bnAnn
即xj=∣A∣1(b1A1j+b2A2j+⋯+bnAnj)=∣A∣1∣Aj∣(j=1,2,⋯,n)
例16 用克拉默法则求解线性方程组
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\begin{cases} x_1-x_2-x_3=2\\ 2x_1-x_2-3x_3=1\\ 3x_1+2x_2-5x_3=0\\ \end{cases}
⎩
⎨
⎧x1−x2−x3=22x1−x2−3x3=13x1+2x2−5x3=0
解
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根据克拉默法则,有
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3
解\\ |A|=\begin{vmatrix} 1&-1&-1\\ 2&-1&-3\\ 3&2&-5 \end{vmatrix} =5+9-4-(3+10-6)=3\not=0\\ 根据克拉默法则,有\\ x_1=\frac{A_1}{|A|}=\frac{1}{3}\begin{vmatrix} 2&-1&-1\\ 1&-1&-3\\ 0&2&-5 \end{vmatrix} =\frac{1}{3}(10-2-5+12)=5\\ x_2=\frac{A_1}{|A|}=\frac{1}{3}\begin{vmatrix} 1&2&-1\\ 2&1&-3\\ 3&0&-5 \end{vmatrix} =\frac{1}{3}(-5-18+3+20)=0\\ x_3=\frac{A_1}{|A|}=\frac{1}{3}\begin{vmatrix} 1&-1&2\\ 2&-1&1\\ 3&2&0 \end{vmatrix} =\frac{1}{3}(-3+8+6-2)=3\\
解∣A∣=
123−1−12−1−3−5
=5+9−4−(3+10−6)=3=0根据克拉默法则,有x1=∣A∣A1=31
210−1−12−1−3−5
=31(10−2−5+12)=5x2=∣A∣A1=31
123210−1−3−5
=31(−5−18+3+20)=0x3=∣A∣A1=31
123−1−12210
=31(−3+8+6−2)=3
结语
❓QQ:806797785
⭐️文档笔记地址 https://github.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.p44-46.
[2]同济六版《线性代数》全程教学视频[CP/OL].2020-02-07.p11.
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