返回

一般说来 , 低阶行列式的计算比高阶行列式的计算要简便

于是 , 我们自然地考虑用低阶行列式来表示高阶行列式的问题

为此 , 先引进余子式和代数余子式的概念

在n阶行列式中 , 把(i , j)元a所在的第i行和第j列划去后

留下来的n‒1阶行列式叫做(i , j)元aij的余子式 , 记作Mij

记Aij=(‒1)i+jMij , Aij叫做(i , j)元aij的代数余子式

例如四阶行列式D=|a11a21a31a41a12a22a32a42a13a23a33a43a14a24a34a44|\left| \begin{array}{r} \begin{matrix} a_{11} \\ a_{21} \end{matrix} \\ \begin{matrix} a_{31} \\ a_{41} \end{matrix} \end{array}\ \ \ \begin{array}{r} \begin{matrix} a_{12} \\ a_{22} \end{matrix} \\ \begin{matrix} a_{32} \\ a_{42} \end{matrix} \end{array}\ \ \ \begin{array}{r} \begin{matrix} a_{13} \\ a_{23} \end{matrix} \\ \begin{matrix} a_{33} \\ a_{43} \end{matrix} \end{array}\ \ \ \begin{array}{r} \begin{matrix} a_{14} \\ a_{24} \end{matrix} \\ \begin{matrix} a_{34} \\ a_{44} \end{matrix} \end{array} \right|

中(3 , 2)元a32的余子式为M32=|a11a21a31a41a12a22a32a42a13a23a33a43a14a24a34a44|\left| \begin{array}{r} \begin{matrix} a_{11} \\ a_{21} \end{matrix} \\ \begin{matrix} a_{31} \\ a_{41} \end{matrix} \end{array}\ \ \ \begin{array}{r} \begin{matrix} a_{12} \\ a_{22} \end{matrix} \\ \begin{matrix} a_{32} \\ a_{42} \end{matrix} \end{array}\ \ \ \begin{array}{r} \begin{matrix} a_{13} \\ a_{23} \end{matrix} \\ \begin{matrix} a_{33} \\ a_{43} \end{matrix} \end{array}\ \ \ \begin{array}{r} \begin{matrix} a_{14} \\ a_{24} \end{matrix} \\ \begin{matrix} a_{34} \\ a_{44} \end{matrix} \end{array} \right|=|a11a13a14a21a23a24a41a43a44|\left| \begin{matrix} a_{11} & a_{13} & a_{14} \\ a_{21} & a_{23} & a_{24} \\ a_{41} & a_{43} & a_{44} \end{matrix} \right|

代数余子式为A32=(‒1)3+2M32=‒M32

引理 一个n阶行列式 , 如果其中第i行所有元素除(i , j)元aij外都为零

那么这行列式等于aij与它的代数余子式的乘积 , 即D=aijAij

也就是说当一行中只有一个非零元素时

行列式的值就是这个非零元素与它的代数余子式的乘积

例如

|a1100a21a22a23a31a32a33|\left| \begin{matrix} a_{11} & 0 & 0 \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{matrix} \right|=a11|a22a23a32a33|\left| \begin{matrix} a_{22} & a_{23} \\ a_{32} & a_{33} \end{matrix} \right|

|a11a12a130a220a31a32a33|=|0a220a11a12a13a31a32a33|=|a2200a12a11a13a32a31a33|=\left| \begin{matrix} a_{11} & a_{12} & a_{13} \\ 0 & a_{22} & 0 \\ a_{31} & a_{32} & a_{33} \end{matrix} \right| = \left| \begin{matrix} 0 & a_{22} & 0 \\ a_{11} & a_{12} & a_{13} \\ a_{31} & a_{32} & a_{33} \end{matrix} \right| = \left| \begin{matrix} a_{22} & 0 & 0 \\ a_{12} & a_{11} & a_{13} \\ a_{32} & a_{31} & a_{33} \end{matrix} \right| =a22|a11a13a31a33|\left| \begin{matrix} a_{11} & a_{13} \\ a_{31} & a_{33} \end{matrix} \right|

定理2行列式等于它的任一行(列)的各元素与其对应的代数余子式乘积之和

即D=ai1Ai1+ai2Ai2+⋯+ainAin (i=1 , 2 , ⋯ , n)

或D=a1jA1j+a2jA2j+⋯+anjAnj(j=1 , 2 , ⋯ , n)

这个定理叫做行列式按行(列)展开法则

利用这一法则并结合行列式的性质 , 可以简化行列式的计算

例如：|a11a12a13a21a22a23a31a32a33|\left| \begin{matrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{matrix}\ \ \right|=|a11a12a13a21+0+00+a22+00+0+a23a31a32a33|\left| \begin{matrix} a_{11} & a_{12} & a_{13} \\ a_{21} + 0 + 0 & 0 + a_{22} + 0 & 0 + 0 + a_{23} \\ a_{31} & a_{32} & a_{33} \end{matrix}\ \ \right|

=|a11a12a13a2100a31a32a33|+|a11a12a130a220a31a32a33|+|a11a12a1300a23a31a32a33|\left| \begin{matrix} a_{11} & a_{12} & a_{13} \\ a_{21} & 0 & 0 \\ a_{31} & a_{32} & a_{33} \end{matrix}\ \ \right| + \left| \begin{matrix} a_{11} & a_{12} & a_{13} \\ 0 & a_{22} & 0 \\ a_{31} & a_{32} & a_{33} \end{matrix}\ \ \right| + \left| \begin{matrix} a_{11} & a_{12} & a_{13} \\ 0 & 0 & a_{23} \\ a_{31} & a_{32} & a_{33} \end{matrix}\ \ \right|

=a21|a12a13a32a33|+a22|a11a13a31a33|+a23|a11a12a31a32|a_{21}\left| \begin{matrix} a_{12} & a_{13} \\ a_{32} & a_{33} \end{matrix} \right| + a_{22}\left| \begin{matrix} a_{11} & a_{13} \\ a_{31} & a_{33} \end{matrix} \right| + a_{23}\left| \begin{matrix} a_{11} & a_{12} \\ a_{31} & a_{32} \end{matrix} \right|

例12 D=|31−12−513−4201−11−53−3|D = \left| \begin{matrix} 3 & 1 & - 1 & 2 \\ - 5 & 1 & 3 & - 4 \\ 2 & 0 & 1 & - 1 \\ 1 & - 5 & 3 & - 3 \end{matrix} \right|

按第一行展开：

1. A11=(−1)1+1M11=M11A_{11} = ( - 1)^{1 + 1}M_{11} = M_{11}

划掉第 1 行第 1 列：M11=|13−401−1−53−3|M_{11} = \left| \begin{matrix} 1 & 3 & - 4 \\ 0 & 1 & - 1 \\ - 5 & 3 & - 3 \end{matrix} \right|

2. A12=(−1)1+2M12=−M12A_{12} = ( - 1)^{1 + 2}M_{12} = - M_{12}

划掉第 1 行第 2 列：M12=|−53−421−113−3|M_{12} = \left| \begin{matrix} - 5 & 3 & - 4 \\ 2 & 1 & - 1 \\ 1 & 3 & - 3 \end{matrix} \right|

3. A13=(−1)1+3M13=M13A_{13} = ( - 1)^{1 + 3}M_{13} = M_{13}

划掉第 1 行第 3 列：M13=|−51−420−11−5−3|M_{13} = \left| \begin{matrix} - 5 & 1 & - 4 \\ 2 & 0 & - 1 \\ 1 & - 5 & - 3 \end{matrix} \right|

4. A14=(−1)1+4M14=−M14A_{14} = ( - 1)^{1 + 4}M_{14} = - M_{14}

划掉第 1 行第 4 列：M14=|−5132011−53|M_{14} = \left| \begin{matrix} - 5 & 1 & 3 \\ 2 & 0 & 1 \\ 1 & - 5 & 3 \end{matrix} \right|

展开式：D=3×M11−1×M12+(−1)×M13−2×M14D = 3 \times M_{11} - 1 \times M_{12} + ( - 1) \times M_{13} - 2 \times M_{14}

即：D=3|13−401−1−53−3|−|−53−421−113−3|−|−51−420−11−5−3|−2|−5132011−53|D = 3\left| \begin{matrix} 1 & 3 & - 4 \\ 0 & 1 & - 1 \\ - 5 & 3 & - 3 \end{matrix} \right| - \left| \begin{matrix} - 5 & 3 & - 4 \\ 2 & 1 & - 1 \\ 1 & 3 & - 3 \end{matrix} \right| - \left| \begin{matrix} - 5 & 1 & - 4 \\ 2 & 0 & - 1 \\ 1 & - 5 & - 3 \end{matrix} \right| - 2\left| \begin{matrix} - 5 & 1 & 3 \\ 2 & 0 & 1 \\ 1 & - 5 & 3 \end{matrix} \right|

例15证明范德蒙德行列式

Dn=|1x1x12⋮x1n−11x2x22⋮x2n−1⋯⋯⋯⋯1xnxn2⋮xnn−1|=∏1≤j<i≤n(xi−xj)D_{n} = \left| \begin{array}{r} \begin{matrix} 1 \\ x_{1} \end{matrix} \\ \begin{matrix} x_{1}^{2} \\ \vdots \\ x_{1}^{n - 1} \end{matrix} \end{array}\ \ \ \begin{array}{r} \begin{matrix} 1 \\ x_{2} \end{matrix} \\ \begin{matrix} x_{2}^{2} \\ \vdots \\ x_{2}^{n - 1} \end{matrix} \end{array}\ \ \ \begin{array}{r} \begin{matrix} \cdots \\ \cdots \end{matrix} \\ \begin{matrix} \cdots \\ \\ \cdots \end{matrix} \end{array}\ \ \ \begin{array}{r} \begin{matrix} 1 \\ x_{n} \end{matrix} \\ \begin{matrix} x_{n}^{2} \\ \vdots \\ x_{n}^{n - 1} \end{matrix} \end{array} \right| = \prod_{1 \leq j < i \leq n}^{}(x_{i} - x_{j})\

其结论为：n阶范德蒙德行列式等于所有满足“行标大于列标”的(xi−xj)(x_{i} - x_{j})的乘积，

（注：∏\prod为乘积符号，展开后包含(x2−x1)、(x3−x1)、(x3−x2)、…、(xn−xn−1)(x_{2} - x_{1})、(x_{3} - x_{1})、(x_{3} - x_{2})、\ldots 、(x_{n} - x_{n - 1})等所有两两差的因子）

证:

1.基础步：验证n=2时成立当n=2时，2阶范德蒙德行列式为：D2=|11x1x2|D_{2} = \left| \begin{matrix} 1 & 1 \\ x_{1} & x_{2} \end{matrix} \right|

根据2阶行列式的计算规则（主对角线乘积减副对角线乘积）：

D2=1⋅x2−1⋅x1=x2−x1D_{2} = 1 \cdot x_{2} - 1 \cdot x_{1} = x_{2} - x_{1}

而按结论公式，n=2时∏1≤j<i≤2(xi−xj)=x2−x1\prod_{1 \leq j < i \leq 2}^{}(x_{i} - x_{j}) = x_{2} - x_{1}，与计算结果一致，

故n=2时成立。

2.归纳假设：假设n-1阶范德蒙德行列式成立

假设对任意n-1阶范德蒙德行列式Dn−1D_{n - 1}，其结论成立，即：

Dn−1=|11…1x2x3…xnx22x32…xn2⋮⋮⋱⋮x2n−2x3n−2…xnn−2|=∏2≤j<i≤n(xi−xj)D_{n - 1} = \left| \begin{matrix} 1 & 1 & \ldots & 1 \\ x_{2} & x_{3} & \ldots & x_{n} \\ x_{2}^{2} & x_{3}^{2} & \ldots & x_{n}^{2} \\ \vdots & \vdots & \ddots & \vdots \\ x_{2}^{n - 2} & x_{3}^{n - 2} & \ldots & x_{n}^{n - 2} \end{matrix} \right| = \prod_{2 \leq j < i \leq n}^{}(x_{i} - x_{j})

3.归纳递推：证明n阶范德蒙德行列式成立

对n阶行列式DnD_{n}进行行变换（行列式行变换不改变行列式值），

消去最后一行的高次项：

将第n-1行乘以(−x1)( - x_{1})后加到第n行；

将第n-2行乘以(−x1)( - x_{1})后加到第n-1行；

...

将第1行乘以(−x1)( - x_{1})后加到第2行。

变换后的行列式为：

Dn=|111…10x2−x1x3−x1…xn−x10x2(x2−x1)x3(x3−x1)…xn(xn−x1)⋮⋮⋮⋱⋮0x2n−2(x2−x1)x3n−2(x3−x1)…xnn−2(xn−x1)|D_{n} = \left| \begin{matrix} 1 & 1 & 1 & \ldots & 1 \\ 0 & x_{2} - x_{1} & x_{3} - x_{1} & \ldots & x_{n} - x_{1} \\ 0 & x_{2}(x_{2} - x_{1}) & x_{3}(x_{3} - x_{1}) & \ldots & x_{n}(x_{n} - x_{1}) \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & x_{2}^{n - 2}(x_{2} - x_{1}) & x_{3}^{n - 2}(x_{3} - x_{1}) & \ldots & x_{n}^{n - 2}(x_{n} - x_{1}) \end{matrix} \right|

根据行列式按第一列展开法则

（第一列除第一个元素外均为0，

展开后仅保留第一个元素与对应代数余子式的乘积）：

Dn=1⋅(−1)1+1⋅M11=M11D_{n} = 1 \cdot ( - 1)^{1 + 1} \cdot M_{11} = M_{11} ，其中M11M_{11}是第一行第一列元素的余子式，

即划去第一行第一列后剩余的n-1阶行列式：

M11=|x2−x1x3−x1…xn−x1x2(x2−x1)x3(x3−x1)…xn(xn−x1)⋮⋮⋱⋮x2n−2(x2−x1)x3n−2(x3−x1)…xnn−2(xn−x1)|M_{11} = \left| \begin{matrix} x_{2} - x_{1} & x_{3} - x_{1} & \ldots & x_{n} - x_{1} \\ x_{2}(x_{2} - x_{1}) & x_{3}(x_{3} - x_{1}) & \ldots & x_{n}(x_{n} - x_{1}) \\ \vdots & \vdots & \ddots & \vdots \\ x_{2}^{n - 2}(x_{2} - x_{1}) & x_{3}^{n - 2}(x_{3} - x_{1}) & \ldots & x_{n}^{n - 2}(x_{n} - x_{1}) \end{matrix} \right|

将每一列的公因子(xi−x1)(x_{i} - x_{1})（i=2,3,...,n）提出

（行列式每列提公因子不改变行列式值）：

M11=(x2−x1)(x3−x1)…(xn−x1)⋅|11…1x2x3…xnx22x32…xn2⋮⋮⋱⋮x2n−2x3n−2…xnn−2|M_{11} = (x_{2} - x_{1})(x_{3} - x_{1})\ldots(x_{n} - x_{1}) \cdot \left| \begin{matrix} 1 & 1 & \ldots & 1 \\ x_{2} & x_{3} & \ldots & x_{n} \\ x_{2}^{2} & x_{3}^{2} & \ldots & x_{n}^{2} \\ \vdots & \vdots & \ddots & \vdots \\ x_{2}^{n - 2} & x_{3}^{n - 2} & \ldots & x_{n}^{n - 2} \end{matrix} \right|

注意到上式中的n-1阶行列式正是归纳假设中的Dn−1D_{n - 1}，代入归纳假设的结论：

M11=[∏i=2n(xi−x1)]⋅[∏2≤j<i≤n(xi−xj)]M_{11} = \left\lbrack \prod_{i = 2}^{n}(x_{i} - x_{1}) \right\rbrack \cdot \left\lbrack \prod_{2 \leq j < i \leq n}^{}(x_{i} - x_{j}) \right\rbrack

合并两个乘积项，恰好覆盖所有1≤j<i≤n1 \leq j < i \leq n的(xi−xj)(x_{i} - x_{j})，即：

Dn=∏1≤j<i≤n(xi−xj)D_{n} = \prod_{1 \leq j < i \leq n}^{}(x_{i} - x_{j})

因此，n阶范德蒙德行列式成立。

推论 行列式某一行(列)的元素与另一行(列)的对应元素的代数余子式

乘积之和等于零.

即ai1Aj1+ai2Aj2+⋯+ainAjn=0 (i≠j)或a1iA1j+a2iA2j+⋯+aniAnj=0 (i≠j)

例13：3阶行列式D=|102314211|D = \left| \begin{matrix} 1 & 0 & 2 \\ 3 & 1 & 4 \\ 2 & 1 & 1 \end{matrix} \right|

1.选定j=2j = 2，计算代数余子式A21,A22,A23A_{21},A_{22},A_{23}

A21=(−1)2+1|0211|=−(0⋅1−2⋅1)=−(−2)=2A_{21} = ( - 1)^{2 + 1}\left| \begin{matrix} 0 & 2 \\ 1 & 1 \end{matrix} \right| = - (0 \cdot 1 - 2 \cdot 1) = - ( - 2) = 2

A22=(−1)2+2|1221|=1⋅1−2⋅2=1−4=−3A_{22} = ( - 1)^{2 + 2}\left| \begin{matrix} 1 & 2 \\ 2 & 1 \end{matrix} \right| = 1 \cdot 1 - 2 \cdot 2 = 1 - 4 = - 3

A23=(−1)2+3|1021|=−(1⋅1−0⋅2)=−1A_{23} = ( - 1)^{2 + 3}\left| \begin{matrix} 1 & 0 \\ 2 & 1 \end{matrix} \right| = - (1 \cdot 1 - 0 \cdot 2) = - 1

2.取i=1i = 1，j=2j = 2，验证不同行的代数余子式乘积之和：

a11A21+a12A22+a13A23=1⋅2+0⋅(−3)+2⋅(−1)=2+0−2=0a_{11}A_{21} + a_{12}A_{22} + a_{13}A_{23} = 1 \cdot 2 + 0 \cdot ( - 3) + 2 \cdot ( - 1) = 2 + 0 - 2 = 0

说明第1行的元素与第2行的代数余子式乘积之和为0。

综合定理2及其推论 , 有关于代数余子式的重要性质:

当i=j时,∑k=1nakiAkj=D,当i≠j时,∑k=1nakiAkj=0当i = j时\ ,\ \sum_{k = 1}^{n}{a_{ki}A_{kj} = D}\ ,\ 当i \neq j时\ \ ,\ \sum_{k = 1}^{n}{a_{ki}A_{kj} = 0}

或

当i=j时,∑k=1naikAjk=D,当i≠j时,∑k=1naikAjk=0当i = j时\ ,\ \sum_{k = 1}^{n}{a_{ik}A_{jk} = D}\ ,\ 当i \neq j时\ \ ,\ \sum_{k = 1}^{n}{a_{ik}A_{jk} = 0}

例14：3阶行列式D=|102314211|D = \left| \begin{matrix} 1 & 0 & 2 \\ 3 & 1 & 4 \\ 2 & 1 & 1 \end{matrix} \right|

1.选定j=2j = 2，计算代数余子式A21,A22,A23A_{21},A_{22},A_{23}

A21=(−1)2+1|0211|=−(0⋅1−2⋅1)=−(−2)=2A_{21} = ( - 1)^{2 + 1}\left| \begin{matrix} 0 & 2 \\ 1 & 1 \end{matrix} \right| = - (0 \cdot 1 - 2 \cdot 1) = - ( - 2) = 2

A22=(−1)2+2|1221|=1⋅1−2⋅2=1−4=−3A_{22} = ( - 1)^{2 + 2}\left| \begin{matrix} 1 & 2 \\ 2 & 1 \end{matrix} \right| = 1 \cdot 1 - 2 \cdot 2 = 1 - 4 = - 3

A23=(−1)2+3|1021|=−(1⋅1−0⋅2)=−1A_{23} = ( - 1)^{2 + 3}\left| \begin{matrix} 1 & 0 \\ 2 & 1 \end{matrix} \right| = - (1 \cdot 1 - 0 \cdot 2) = - 1

2当i=j时当i = j时\

按第2行展开求DD：

D=3⋅A21+1⋅A22+4⋅A23D = 3 \cdot A_{21} + 1 \cdot A_{22} + 4 \cdot A_{23}

=3⋅2+1⋅(−3)+4⋅(−1)=6−3−4=−1= 3 \cdot 2 + 1 \cdot ( - 3) + 4 \cdot ( - 1) = 6 - 3 - 4 = - 1\quad

3.当i≠j时当i \neq j时\

取i=1i = 1，j=2j = 2，验证不同行的代数余子式乘积之和：

a11A21+a12A22+a13A23=1⋅2+0⋅(−3)+2⋅(−1)=2+0−2=0a_{11}A_{21} + a_{12}A_{22} + a_{13}A_{23} = 1 \cdot 2 + 0 \cdot ( - 3) + 2 \cdot ( - 1) = 2 + 0 - 2 = 0

说明第1行的元素与第2行的代数余子式乘积之和为0。