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一、二元线性方程组与二阶行列式

用消元法解二元线性方程组

{ a 11 x 1 + a 12 x 2 = b 1 a 21 x 1 + a 22 x 2 = b 2 \left\{ \begin{array}{r} a_{11}x_{1} + a_{12}x_{2} = b_{1} \\ a_{21}x_{1} + a_{22}x_{2} = b_{2} \end{array} \right.\

为消去未知数x2 , 以a22与a12分别乘上列两方程的两端

即 { a 22 a 11 x 1 + a 22 a 12 x 2 = a 22 b 1 a 12 a 21 x 1 + a 12 a 22 x 2 = a 12 b 2 \left\{ \begin{array}{r} a_{22}a_{11}x_{1} + a_{22}a_{12}x_{2} = a_{22}b_{1} \\ a_{12}a_{21}x_{1} + a_{12}a_{22}x_{2} = a_{12}b_{2} \end{array} \right.\

然后两个方程相减 , 得(a11a22‒a12a21)x1=b1a22‒a12b2

为消去未知数x1 , 以a21与a11分别乘上列两方程的两端

即 { a 21 a 11 x 1 + a 21 a 12 x 2 = a 21 b 1 a 11 a 21 x 1 + a 11 a 22 x 2 = a 11 b 2 \left\{ \begin{array}{r} a_{21}a_{11}x_{1} + a_{21}a_{12}x_{2} = a_{21}b_{1} \\ a_{11}a_{21}x_{1} + a_{11}a_{22}x_{2} = a_{11}b_{2} \end{array} \right.\

然后两个方程相减 , 得(a11a22‒a12a21)x2=a11b2‒b1a21

当a11a22‒a12a21≠0时 , 求得方程组 { a 11 x 1 + a 12 x 2 = b 1 a 21 x 1 + a 22 x 2 = b 2 \left\{ \begin{array}{r} a_{11}x_{1} + a_{12}x_{2} = b_{1} \\ a_{21}x_{1} + a_{22}x_{2} = b_{2} \end{array} \right.\ 的解为

x 1 = b 1 a 22 − a 12 b 2 a 11 a 22 − a 12 a 21 , x 2 = a 11 b 2 − b 1 a 21 a 11 a 22 − a 12 a 21 x_{1} = \frac{b_{1}a_{22} - a_{12}b_{2}}{a_{11}a_{22} - a_{12}a_{21}}\ ,\ x_{2} = \frac{a_{11}b_{2} - b_{1}a_{21}}{a_{11}a_{22} - a_{12}a_{21}}\

式中的分子、分母都是四个数分两对相乘再相减而得

其中分母a11a22‒a12a21是由方程组 { a 11 x 1 + a 12 x 2 = b 1 a 21 x 1 + a 22 x 2 = b 2 \left\{ \begin{array}{r} a_{11}x_{1} + a_{12}x_{2} = b_{1} \\ a_{21}x_{1} + a_{22}x_{2} = b_{2} \end{array} \right.\ 的四个系数确定的

把这四个数按它们在方程组 { a 11 x 1 + a 12 x 2 = b 1 a 21 x 1 + a 22 x 2 = b 2 \left\{ \begin{array}{r} a_{11}x_{1} + a_{12}x_{2} = b_{1} \\ a_{21}x_{1} + a_{22}x_{2} = b_{2} \end{array} \right.\ 中的位置

排成二行二列(横排称行、竖排称列)的数表 a 11 a 12 a 21 a 22 \begin{matrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{matrix} ,

表达式a11a22‒a12a21称为数表 a 11 a 12 a 21 a 22 \begin{matrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{matrix} 所确定的二阶行列式,并记作 | a 11 a 12 a 21 a 22 | \left| \begin{matrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{matrix} \right|

数aij(i=1 , 2 ; j=1,2)称为行列式 | a 11 a 12 a 21 a 22 | \left| \begin{matrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{matrix} \right| 的元素或元，

元素aij的第一个下标i称为行标 , 表明该元素位于第i行;

第二个下标j称为列标 , 表明该元素位于第j列

位于第i行第j列的元素称为行列式 | a 11 a 12 a 21 a 22 | \left| \begin{matrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{matrix} \right| 的(i , j)元，

上述二阶行列式的定义 , 可用对角线法则来记忆 , 参看图1.1

把a11到 a22的实连线称为主对角线 , a12到a21的虚连线称为副对角线

于是二阶行列式

就是主对角线上的两元素之积减去副对角线上两元素之积所得的差

利用二阶行列式的概念 ,

x 1 = b 1 a 22 − a 12 b 2 a 11 a 22 − a 12 a 21 , x 2 = a 11 b 2 − b 1 a 21 a 11 a 22 − a 12 a 21 x_{1} = \frac{b_{1}a_{22} - a_{12}b_{2}}{a_{11}a_{22} - a_{12}a_{21}}\ ,\ x_{2} = \frac{a_{11}b_{2} - b_{1}a_{21}}{a_{11}a_{22} - a_{12}a_{21}} 式中x1 , x2的分子也可写成二阶行列式,

即b1a22‒a12b2= | b 1 a 12 b 2 a 22 | \left| \begin{matrix} b_{1} & a_{12} \\ b_{2} & a_{22} \end{matrix} \right| , a11b2‒b1a21= | a 11 b 1 a 21 b 2 | \left| \begin{matrix} a_{11} & b_{1} \\ a_{21} & b_{2} \end{matrix} \right|

若记D= | a 11 a 12 a 21 a 22 | \left| \begin{matrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{matrix} \right| , D1= | b 1 a 12 b 2 a 22 | \left| \begin{matrix} b_{1} & a_{12} \\ b_{2} & a_{22} \end{matrix} \right| , D2= | a 11 b 1 a 21 b 2 | \left| \begin{matrix} a_{11} & b_{1} \\ a_{21} & b_{2} \end{matrix} \right|

那么 x 1 = b 1 a 22 − a 12 b 2 a 11 a 22 − a 12 a 21 , x 2 = a 11 b 2 − b 1 a 21 a 11 a 22 − a 12 a 21 x_{1} = \frac{b_{1}a_{22} - a_{12}b_{2}}{a_{11}a_{22} - a_{12}a_{21}}\ ,\ x_{2} = \frac{a_{11}b_{2} - b_{1}a_{21}}{a_{11}a_{22} - a_{12}a_{21}}

可写成x1= D 1 D \frac{D_{1}}{D} = | b 1 a 12 b 2 a 22 | | a 11 a 12 a 21 a 22 | \frac{\left| \begin{matrix} b_{1} & a_{12} \\ b_{2} & a_{22} \end{matrix} \right|}{\left| \begin{matrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{matrix} \right|} , x2= D 2 D \frac{D_{2}}{D} = | a 11 b 1 a 21 b 2 | | a 11 a 12 a 21 a 22 | \frac{\left| \begin{matrix} a_{11} & b_{1} \\ a_{21} & b_{2} \end{matrix} \right|}{\left| \begin{matrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{matrix} \right|}

注意这里的分母D是由方程组 { a 11 x 1 + a 12 x 2 = b 1 a 21 x 1 + a 22 x 2 = b 2 \left\{ \begin{array}{r} a_{11}x_{1} + a_{12}x_{2} = b_{1} \\ a_{21}x_{1} + a_{22}x_{2} = b_{2} \end{array} \right.\ 的系数所确定的二阶行列式(称系数行列式),

x1的分子D1是用常数项b1 , b2替换D中第1列的元素a11 , a21所得的二阶行列式

x2的分子D2是用常数项b1 , b2替换D中第2列的元素a12 , a22所得的二阶行列式

例1求解二元线性方程组 { 3 x 1 − 2 x 2 = 12 2 x 1 + x 2 = 1 \left\{ \begin{array}{r} 3x_{1} - 2x_{2} = 12 \\ 2x_{1} + x_{2} = 1\ \ \ \ \ \end{array} \right.\

解: 由于D= | 3 − 2 2 1 | \left| \begin{matrix} 3 & - 2 \\ 2 & 1 \end{matrix} \right| =3‒(‒4)=7≠0

D1= | 12 − 2 1 1 | \left| \begin{matrix} 12 & - 2 \\ 1 & 1 \end{matrix} \right| =12‒(‒2)=14 , D2= | 3 − 2 2 1 | \left| \begin{matrix} 3 & - 2 \\ 2 & 1 \end{matrix} \right| =3‒24=‒21

因此x1= D 1 D \frac{D_{1}}{D} = 14 7 \frac{14}{7} =2 , x2= D 2 D \frac{D_{2}}{D} = − 21 7 \frac{- 21}{7} =‒3

二、三阶行列式

定义1设有9个数排成3行3列的数表 a 11 a 12 a 13 a 21 a 22 a 23 a 31 a 32 a 33 \begin{matrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{matrix}

记 | a 11 a 12 a 13 a 21 a 22 a 23 a 31 a 32 a 33 | \left| \begin{matrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{matrix} \right|

=a11a22a33+a12a23a31+a13a21a32‒a11a23a32‒a12a21a33‒a13a22a31

上式称为数表 a 11 a 12 a 13 a 21 a 22 a 23 a 31 a 32 a 33 \begin{matrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{matrix} 所确定的三阶行列式

上述定义表明三阶行列式含6项

每项均为不同行不同列的三个元素的乘积再冠以正负号

其规律遵循图1.2所示的对角线法则:

图中有三条实线看做是平行于主对角线的连线

三条虚线看做是平行于副对角线的连线

实线上三元素的乘积冠正号 , 虚线上三元素的乘积冠负号

例2计算三阶行列式D= | 1 2 − 4 − 2 2 1 − 3 4 − 2 | \left| \begin{matrix} 1 & 2 & - 4 \\ - 2 & 2 & 1 \\ - 3 & 4 & - 2 \end{matrix} \right|

解: 按对角线法则 , 有

D=1×2×(‒2)+2×1×(‒3)+(‒4)×(‒2)×4‒1×1×4

‒2×(‒2)×(‒2)‒(‒4)×2×(‒3)=‒4‒6+32‒4‒8‒24=‒14.

例3求解方程 | 1 1 1 2 3 x 4 9 x 2 | \left| \begin{matrix} 1 & 1 & 1 \\ 2 & 3 & x \\ 4 & 9 & x^{2} \end{matrix} \right| =0

解: 方程左端的三阶行列式D=3x2+4x+18‒9x‒2x2‒12=x2‒5x+6

由x2‒5x+6=0解得x=2或x=3.

对角线法则只适用于二阶与三阶行列式 , 为研究四阶及更高阶行列式

需要用到全排列的知识 , 然后引出n阶行列式的概念