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含有n个未知数x1 , x2 , ⋯ , xn的n个线性方程的方程组

{ 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 \left\{ \begin{array}{r} \begin{matrix} 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} \end{matrix} \\ \begin{matrix} \cdots\cdots\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \cdots\cdots \\ a_{n1}x_{1} + a_{n2}x_{2} + \cdots + a_{nn}x_{n} = b_{n} \end{matrix} \end{array} \right.\

它的解可以用n阶行列式表示 , 即有下列法则

克拉默法则 如果线性方程组 { 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 \left\{ \begin{array}{r} \begin{matrix} 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} \end{matrix} \\ \begin{matrix} \cdots\cdots\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \cdots\cdots \\ a_{n1}x_{1} + a_{n2}x_{2} + \cdots + a_{nn}x_{n} = b_{n} \end{matrix} \end{array} \right.\

的系数矩阵A的行列式不等于零，即|A|= | a 11 ⋯ a 1 n ⋮ ⋮ a n 1 ⋯ a n n | ≠ 0 \left| \begin{matrix} a_{11} & \cdots & a_{1n} \\ \vdots & & \vdots \\ a_{n1} & \cdots & a_{nn} \end{matrix} \right| \neq 0

那么 , 方程组有惟一解x1= | A 1 | | A | \frac{\left| A_{1} \right|}{|A|} , x2= | A 2 | | A | \frac{\left| A_{2} \right|}{|A|} , ⋯ , xn= | A n | | A | \frac{\left| A_{n} \right|}{|A|}

其中Aj(j=1 , 2 , ⋯ , n)是把系数矩阵A中第j列的元素

用方程组右端的常数项代替后所得到的n阶矩阵

即Aj= [ a 11 ⋮ a n 1 ⋯ ⋯ a 1 , j − 1 ⋮ a n , j − 1 b 1 ⋮ b n a 1 , j − 1 ⋮ a n , j + 1 ⋯ ⋯ a 1 n ⋮ a n n ] \left\lbrack \begin{matrix} a_{11} \\ \vdots \\ a_{n1} \end{matrix}\ \ \ \ \ \ \ \ \ \begin{matrix} \cdots \\ \\ \cdots \end{matrix}\ \ \ \ \ \ \ \ \ \ \begin{matrix} a_{1\ ,\ j - 1} \\ \vdots \\ a_{n\ ,\ j - 1} \end{matrix}\ \ \ \ \ \ \ \ \ \ \begin{matrix} b_{1} \\ \vdots \\ b_{n} \end{matrix}\ \ \ \ \ \ \ \ \ \ \begin{matrix} a_{1\ ,\ j - 1} \\ \vdots \\ a_{n\ ,\ j + 1} \end{matrix}\ \ \ \ \ \ \ \ \ \ \begin{matrix} \cdots \\ \\ \cdots \end{matrix}\ \ \ \ \ \ \ \ \ \begin{matrix} a_{1n} \\ \vdots \\ a_{nn} \end{matrix} \right\rbrack

证: 把方程组写成矩阵方程Ax=b

这里A=(aij)n×n为n阶矩阵 , 因|A|≠0 , 故A-1存在

由Ax=b , 有A-1Ax=A-1b , 即x=A-1b

根据逆矩阵的惟一性 , 知x=A-1b是方程组的惟一的解向量

由逆矩阵公式A-1= 1 | A | \frac{1}{|A|}\ A* , 有x=A-1b= 1 | A | \frac{1}{|A|}\ A*b

即 [ x 1 x 2 ⋮ x n ] = 1 | A | [ A 11 A 12 ⋮ A 1 n A 21 A 21 ⋮ A 2 n ⋯ ⋯ ⋯ A n 1 A n 2 ⋮ A n n ] [ b 1 b 2 ⋮ b n ] = 1 | A | [ b 1 A 11 + b 2 A 21 + ⋯ + b n A n 1 b 1 A 12 + b 2 A 22 + ⋯ + b n A n 2 ⋮ b 1 A 1 n + b 2 A 2 n + ⋯ + b n A n n ] \left\lbrack \begin{array}{r} \begin{matrix} x_{1} \\ x_{2} \end{matrix} \\ \begin{matrix} \vdots \\ x_{n} \end{matrix} \end{array} \right\rbrack = \frac{1}{|A|}\ \left\lbrack \begin{array}{r} \begin{matrix} A_{11} \\ A_{12} \end{matrix} \\ \begin{matrix} \vdots \\ A_{1n} \end{matrix} \end{array}\ \ \ \begin{array}{r} \begin{matrix} A_{21} \\ A_{21} \end{matrix} \\ \begin{matrix} \vdots \\ A_{2n} \end{matrix} \end{array}\ \ \ \begin{array}{r} \begin{matrix} \cdots \\ \cdots \end{matrix} \\ \begin{matrix} \\ \cdots \end{matrix} \end{array}\ \ \ \begin{array}{r} \begin{matrix} A_{n1} \\ A_{n2} \end{matrix} \\ \begin{matrix} \vdots \\ A_{nn} \end{matrix} \end{array} \right\rbrack\left\lbrack \begin{array}{r} \begin{matrix} b_{1} \\ b_{2} \end{matrix} \\ \begin{matrix} \vdots \\ b_{n} \end{matrix} \end{array} \right\rbrack = \frac{1}{|A|}\ \left\lbrack \begin{array}{r} \begin{matrix} b_{1}A_{11} + b_{2}A_{21} + \cdots + b_{n}A_{n1} \\ b_{1}A_{12} + b_{2}A_{22} + \cdots + b_{n}A_{n2} \end{matrix} \\ \begin{matrix} \vdots \\ b_{1}A_{1n} + b_{2}A_{2n} + \cdots + b_{n}A_{nn} \end{matrix} \end{array} \right\rbrack

亦即xj= 1 | A | ( b 1 A 1 j + b 2 A 2 j + ⋯ + b n A n j ) \frac{1}{|A|}\ \left( b_{1}A_{1j} + b_{2}A_{2j} + \cdots + b_{n}A_{nj} \right) = 1 | A | \frac{1}{|A|}\ |Aj| (j=1 , 2 , ⋯ , n)

克拉默法则可视为行列式的一个应用

而所给出的证明又可看作逆矩阵的一个应用

它解决的是方程个数与未知数个数相等并且系数行列式不等于零的线性方程组，

是求解一般线性方程组的一个特殊的情形

例16分别用克拉默法则和逆矩阵方法求解线性方程组

{ x 1 − x 2 − x 3 = 2 2 x 1 − x 2 − 3 x 3 = 1 3 x 1 − 2 x 2 − 5 x 3 = 0 \left\{ \begin{matrix} x_{1} - x_{2} - x_{3} = 2 \\ {2x}_{1} - x_{2} - 3x_{3} = 1 \\ {3x}_{1} - {2x}_{2} - 5x_{3} = 0 \end{matrix} \right.\

解: (1)用克拉默法则

因方程组的系数矩阵的行列式|A|= | 1 − 1 − 1 2 − 1 − 3 3 2 − 5 | \left| \begin{matrix} 1 & - 1 & - 1 \\ 2 & - 1 & - 3 \\ 3 & 2 & - 5 \end{matrix} \right| =3≠0

由克拉默法则 , 知它有惟一解 , 并且

x1= 1 | A | | 2 − 1 − 1 1 − 1 − 3 0 2 − 5 | ⇒ r 1 ↔ r 2 − 1 3 | 1 − 1 − 3 2 − 1 − 1 0 2 − 5 | ⇒ r 2 − 2 r 1 r 3 − 2 r 2 − 1 3 | 1 − 1 − 3 0 1 5 0 0 − 15 | \frac{1}{|A|}\left| \begin{matrix} 2 & - 1 & - 1 \\ 1 & - 1 & - 3 \\ 0 & 2 & - 5 \end{matrix} \right|\overset{\ \ r_{1} \leftrightarrow r_{2}\ \ \ }{\Rightarrow}\frac{- 1}{3}\left| \begin{matrix} 1 & - 1 & - 3 \\ 2 & - 1 & - 1 \\ 0 & 2 & - 5 \end{matrix} \right|\overset{\ \ \begin{matrix} r_{2} - 2r_{1} \\ r_{3} - 2r_{2} \end{matrix}\ \ \ \ }{\Rightarrow}\frac{- 1}{3}\left| \begin{matrix} 1 & - 1 & - 3 \\ 0 & 1 & 5 \\ 0 & 0 & - 15 \end{matrix} \right| =5

x2= 1 | A | | 1 2 − 1 2 1 − 3 3 0 − 5 | ⇒ r 2 − 2 r 1 r 3 − 3 r 1 1 3 | 1 2 − 1 0 − 3 − 1 0 − 6 − 2 | \frac{1}{|A|}\left| \begin{matrix} 1 & 2 & - 1 \\ 2 & 1 & - 3 \\ 3 & 0 & - 5 \end{matrix} \right|\overset{\ \ \begin{matrix} r_{2} - 2r_{1} \\ r_{3} - 3r_{1} \end{matrix}\ \ \ \ }{\Rightarrow}\frac{1}{3}\left| \begin{matrix} 1 & 2 & - 1 \\ 0 & - 3 & - 1 \\ 0 & - 6 & - 2 \end{matrix} \right| =0

x3= 1 | A | | 1 − 1 2 2 − 1 1 3 2 0 | ⇒ r 2 − 2 r 1 r 3 − 2 r 2 1 3 | 1 − 1 2 0 1 − 3 0 5 − 6 | \frac{1}{|A|}\left| \begin{matrix} 1 & - 1 & 2 \\ 2 & - 1 & 1 \\ 3 & 2 & 0 \end{matrix} \right|\overset{\ \ \begin{matrix} r_{2} - 2r_{1} \\ r_{3} - 2r_{2} \end{matrix}\ \ \ \ }{\Rightarrow}\frac{1}{3}\left| \begin{matrix} 1 & - 1 & 2 \\ 0 & 1 & - 3 \\ 0 & 5 & - 6 \end{matrix} \right| = 1 3 | 1 − 3 5 − 6 | \frac{1}{3}\ \left| \begin{matrix} 1 & - 3 \\ 5 & - 6 \end{matrix} \right| =3

(2)用逆矩阵方法

因|A|=3≠0 , 故A可逆 , 于是

x=A-1b= [ 1 − 1 − 1 2 − 1 − 3 3 2 − 5 ] − 1 [ 2 1 0 ] = 1 3 [ 11 − 7 2 1 − 2 1 7 − 5 1 ] [ 2 1 0 ] = 1 3 [ 15 0 9 ] \begin{bmatrix} 1 & - 1 & - 1 \\ 2 & - 1 & - 3 \\ 3 & 2 & - 5 \end{bmatrix}^{- 1}\begin{bmatrix} 2 \\ 1 \\ 0 \end{bmatrix} = \frac{1}{3}\ \begin{bmatrix} 11 & - 7 & 2 \\ 1 & - 2 & 1 \\ 7 & - 5 & 1 \end{bmatrix}\begin{bmatrix} 2 \\ 1 \\ 0 \end{bmatrix} = \frac{1}{3}\ \begin{bmatrix} 15 \\ 0 \\ 9 \end{bmatrix} = [ 5 0 3 ] \begin{bmatrix} 5 \\ 0 \\ 3 \end{bmatrix}

即有 { x 1 = 5 x 2 = 0 x 3 = 3 \left\{ \begin{matrix} x_{1} = 5 \\ x_{2} = 0 \\ x_{3} = 3 \end{matrix} \right.\