线性代数 01.03 行列式按行(列)展开

$\color{blue}{第一章 第三节 行列式按行(列)展开}$

$\color{blue}{一、余子式和代数余子式}$

$在n阶行列式中，把元素a_{ij}所在的第i行和第j列划去后，\\ 留下的n-1阶行列式叫做元素a_{ij}的余子式.记作M_{ij}.即$
$a_{ij}的余子式记作M_{ij}.$
$a_{ij}的代数余子式A_{ij} = (-1)^{i + j}M_{ij}$

例如，四阶行列式
$D = \begin{vmatrix} a_{11} & a_{12} & a_{13} & a_{14} \\ a_{21} & a_{22} & a_{23} & a_{24} \\ a_{31} & a_{32} & a_{33} & a_{34} \\ a_{41} & a_{42} & a_{43} & a_{44} \\ \end{vmatrix}$
$其中元素a_{32}的余子式和代数余子式分别为$
$M_{32} = \begin{vmatrix} a_{11} & a_{13} & a_{14} \\ a_{21} & a_{23} & a_{24} \\ a_{41} & a_{43} & a_{44} \\ \end{vmatrix}$

$A_{32} = (-1)^{3 + 2} \begin{vmatrix} a_{11} & a_{13} & a_{14} \\ a_{21} & a_{23} & a_{24} \\ a_{41} & a_{43} & a_{44} \\ \end{vmatrix} = -\begin{vmatrix} a_{11} & a_{13} & a_{14} \\ a_{21} & a_{23} & a_{24} \\ a_{41} & a_{43} & a_{44} \\ \end{vmatrix} = -M_{32}$

$\color{blue}{二、行列式按行(列)展开定理}$

$引理:设D为n阶行列式,如果D的第i行所有元素除a_{ij}外,\\ 其余元素均为零,那么行列式D等于a_{ij}与其代数余子式的\\ 乘积,即$
$D = a_{ij}A_{ij}$

$证:设 D = \begin{vmatrix} a_{11} & \cdots & a_{1j} & \cdots & a_{1n} \\ \vdots & & \vdots & & \vdots \\ 0 & \cdots & a_{ij} & \cdots & 0 \\ \vdots & & \vdots & & \vdots \\ a_{n1} & \cdots & a_{nj} & \cdots & a_{nn} \\ \end{vmatrix}$
$把a_{i}行依次上移至第一行,在把a_{j}列依次前移至第一列$
$D = (-1)^{i - 1} \begin{vmatrix} 0 & \cdots & a_{ij} & \cdots & 0 \\ a_{11} & \cdots & a_{1j} & \cdots & a_{1n} \\ \cdots & \cdots & \cdots & \cdots & \cdots \\ a_{(i-1)1} & \cdots & a_{(i -1)j} & \cdots & a_{(i -1)n} \\ a_{(i+1)1} & \cdots & a_{(i +1)j} & \cdots & a_{(i +1)n} \\ \cdots & \cdots & \cdots & \cdots & \cdots \\ a_{n1} & \cdots & a_{nj} & \cdots & a_{nn} \\ \end{vmatrix} \\ = (-1)^{i - 1}(-1)^{j - 1} \begin{vmatrix} a_{ij} & 0 & \cdots & 0 & 0 & \cdots & 0 \\ a_{1j} & a_{11} & \cdots & a_{1(j-1)} & a_{1(j+1)} & \cdots & a_{1n} \\ \cdots & \cdots & \cdots & \cdots & \cdots & \cdots & \cdots \\ a_{(i -1)j} & a_{(i-1)1} & \cdots & a_{(i -1)(j-1)} & a_{(i - 1)(j+1)} & \cdots & a_{(i -1)n} \\ a_{(i +1)j} & a_{(i+1)1} & \cdots & a_{(i +1)(j-1)} & a_{(i + 1)(j+1)} & \cdots & a_{(i +1)n} \\ \cdots & \cdots & \cdots & \cdots & \cdots & \cdots & \cdots \\ a_{nj} & a_{n1} & \cdots & a_{n(j-1)} & a_{n(j+1)} & \cdots & a_{nn} \\ \end{vmatrix} \\ = (-1)^{i + j}a_{ij}M_{ij} = a_{ij}A_{ij}$

$定理1.行列式等于它的任一行(列)的各个元素与其对应的代数余子式乘积之和,即$
$D = a_{i1}A_{i1} + a_{i2}A_{i2} + \cdots + a_{in}A_{in} (i = 1, 2, \cdots, n)$
$D = a_{1j}A_{1j} + a_{2j}A_{2j} + \cdots + a_{nj}A_{nj} (j = 1, 2, \cdots, n)$

证:设
$D = \begin{vmatrix} a_{11} & a_{12} \cdots & \cdots & a_{1n} \\ \cdots & \cdots & \cdots & \cdots \\ a_{i1} & a_{i2} & \cdots & a_{in} \\ \cdots & \cdots & \cdots & \cdots \\ a_{n1} & a_{n2} & \cdots & a_{nn} \\ \end{vmatrix} \\ = \begin{vmatrix} a_{11} & a_{12} \cdots & \cdots & a_{1n} \\ \cdots & \cdots & \cdots & \cdots \\ a_{i1} + 0 + \cdots + 0 & 0 + a_{i2} + 0 + \cdots + 0 & \cdots & 0 + \cdots + 0 + a_{in} \\ \cdots & \cdots & \cdots & \cdots \\ a_{n1} & a_{n2} & \cdots & a_{nn} \\ \end{vmatrix} \\ = \begin{vmatrix} a_{11} & a_{12} \cdots & \cdots & a_{1n} \\ \cdots & \cdots & \cdots & \cdots \\ a_{i1} & 0 & \cdots & 0 \\ \cdots & \cdots & \cdots & \cdots \\ a_{n1} & a_{n2} & \cdots & a_{nn} \\ \end{vmatrix} + \begin{vmatrix} a_{11} & a_{12} \cdots & \cdots & a_{1n} \\ \cdots & \cdots & \cdots & \cdots \\ 0 & a_{i2} & \cdots & 0 \\ \cdots & \cdots & \cdots & \cdots \\ a_{n1} & a_{n2} & \cdots & a_{nn} \\ \end{vmatrix} \\ + \begin{vmatrix} a_{11} & a_{12} \cdots & \cdots & a_{1n} \\ \cdots & \cdots & \cdots & \cdots \\ 0 & 0 & \cdots & a_{in} \\ \cdots & \cdots & \cdots & \cdots \\ a_{n1} & a_{n2} & \cdots & a_{nn} \\ \end{vmatrix} \\ = a_{i1}A_{i1} + a_{i2}A_{i2} + \cdots + a_{in}A_{in}(i = 1, 2, \cdots, n)$

类似地、若按列证明，可得
$D = a_{1j}A_{1j} + a_{2j}A_{2j} + \cdots + a_{nj}A_{nj}(j = 1, 2, \cdots, n)$

例1.计算
$D = \begin{vmatrix} 3 & 1 & -1 & 2 \\ -5 & 1 & 3 & -4 \\ 2 & 0 & 1 & -1 \\ 1 & -5 & 3 & -3 \\ \end{vmatrix}$

解：
$D = \begin{vmatrix} 5 & 1 & -1 & 1 \\ -11 & 1 & 3 & -1 \\ 0 & 0 & 1 & 0 \\ -5 & -5 & 3 & 0 \\ \end{vmatrix} \\ = 1 \times (-1)^{3+3} \begin{vmatrix} 5 & 1 & 1 \\ -11 & 1 & -1 \\ -5 & -5 & 0 \\ \end{vmatrix} = \begin{vmatrix} 5 & 1 & 1 \\ -6 & 2 & 0 \\ -5 & -5 & 0 \\ \end{vmatrix} \\ = 1 \times (-1)^{1 + 3}\begin{vmatrix} -6 & 2 \\ -5 & -5 \\ \end{vmatrix} = 40$

例2.计算
$D_{2n} = \begin{vmatrix} a & & & & & & & & & & b \\ & & & & & & & & & & \\ & & a & & & 0 & & & b & & \\ & & & \cdots & & & & \cdots & & & \\ & & & & a & & b & & & & \\ & & 0 & & & & & & 0 & & \\ & & & & c & & d & & & & \\ & & & \cdots & & & & \cdots & & & \\ & & & & & & & & & & \\ & & c & & & 0 & & & d & & \\ & & & & & & & & & & \\ c & & & & & & & & & & d \\ \end{vmatrix}$

解:按第一行展开
$D_{2n} = a \begin{vmatrix} a & & & & & & & b & 0 \\ & & \cdots & & & & \cdots & & \\ & & & a & & b & & & \\ & & & & & & & & \\ & & & c & & d & & & \\ & & \cdots & & & & \cdots & & \\ c & & & & & & & d & \\ 0 & & & & & & & & d \\ \end{vmatrix} + b(-1)^{2n + 1} \begin{vmatrix} 0 & a & & & & & & & b \\ & & \cdots & & & & \cdots & & \\ & & & a & & b & & & \\ & & & & & & & & \\ & & & c & & d & & & \\ & & \cdots & & & & \cdots & & \\ & c & & & & & & & d \\ c & & & & & & & & 0 \\ \end{vmatrix} \\ = adD_{2(n-1)} - bc(-1)^{2n - 1 + 1}D_{2(n-1)} \\ = (ad - bc)D_{2(n-1)} \\ 依次类推公式，得\\ D_{2n} = (ad - bc)^{n}$

例3.证明范蒙得(Vandermonde)行列式
$D_n = \begin{vmatrix} 1 & 1 & \cdots & 1 \\ x_1 & x_2 & \cdots & x_n \\ x_1^2 & x_2^2 & \cdots & x_n^2 \\ \vdots & \vdots & & \vdots \\ x_1^{n - 1} & x_2^{n - 1} & \cdots & x_n^{n - 1} \\ \end{vmatrix} = \prod _{n \geq i > j \geq 1}(x_i - x_j)$
证:用数学归纳法.因为
$D_2 = \begin{vmatrix} 1 & 1 \\ x_1 & x_2 \\ \end{vmatrix} = x_2 - x_1 = \prod_{2 \geq i > j \geq 1}(x_i - x_j)$
$所以，当n = 2时,(1)成立.$
$现在假设(1)对于n-1阶Vandermonde行列式，即$
$D_{n - 1} = \begin{vmatrix} 1 & 1 & \cdots & 1 \\ x_1 & x_2 & \cdots & x_n \\ x_1^2 & x_2^2 & \cdots & x_n^2 \\ \vdots & \vdots & & \vdots \\ x_1^{n - 2} & x_2^{n - 2} & \cdots & x_n^{n - 2} \\ \end{vmatrix} = \prod _{n \geq i > j \geq 2}(x_i - x_j)$

$我们来证明对n阶Vandermonde行列式也成立.$
$D_n = \begin{vmatrix} 1 & 1 & 1& \cdots & 1 \\ 0 & x_2 - x_1 & x_3 - x_1 & \cdots & x_n - x_1 \\ 0 & x_2(x_2 - x_1) & x_3(x_3 - x_1) & \cdots & x_n(x_n - x_1) \\ \vdots & \vdots & \vdots& & \vdots \\ 0 & x_2^{n - 2}(x_2 - x_1) & x_3^{n - 2}(x_3 - x_1) & \cdots & x_n^{n - 2}(x_n - x_1) \\ \end{vmatrix} \\ = (x_2 - x_1)(x_3 - x_1) \cdots (x_n - x_1) \begin{vmatrix} 1 & 1 & \cdots & 1 \\ x_2 & x_3 & \cdots & x_n \\ \vdots & \vdots & & \vdots \\ x_2^{n - 2} & x_3^{n - 2} & \cdots & x_n^{n - 2} \\ \end{vmatrix} \\ = (x_2 - x_1)(x_3 - x_1) \cdots (x_n - x_1) \prod_{n \geq i > j \geq 2}(x_i - x_j) \\ = \prod_{n \geq i > j \geq 1}(x_i - x_j)$

例4.计算
$D = \begin{vmatrix} 1 & 1 & 1 & 1 \\ 2 & 3 & 4 & 5 \\ 2^2 & 3^2 & 4 ^2 & 5^2 \\ 2^3 & 3^3 & 4^3 & 5^3 \\ \end{vmatrix} = (3 - 2)(4-2)(5-2)(4-3)(5-3)(5-4) = 12$

$\color{blue}{三、行列式展开定理的推论}$

$推论:行列式某一行(列)的元素与另一行(列)的对应元素的代数余子式乘积之和等于零.即$
$a_{i1}A_{j1} + a_{i2}A_{j2} + \cdots + a_{in}A_{jn} = 0, i \neq j,$
$或a_{1i}A_{1j} + a_{2i}A_{2j} + \cdots + a_{ni}A_{nj} = 0, i \neq j$
证：设
$D = \begin{vmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ \cdots & \cdots & \cdots & \cdots \\ a_{i1} & a_{i2} & \cdots & a_{in} \\ \cdots & \cdots & \cdots & \cdots \\ a_{j1} & a_{j2} & \cdots & a_{jn} \\ \cdots & \cdots & \cdots & \cdots \\ a_{n1} & a_{n2} & \cdots & a_{nn} \\ \end{vmatrix}$
把D按照第j行展开，有
$\begin{vmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ \cdots & \cdots & \cdots & \cdots \\ a_{i1} & a_{i2} & \cdots & a_{in} \\ \cdots & \cdots & \cdots & \cdots \\ a_{j1} & a_{j2} & \cdots & a_{jn} \\ \cdots & \cdots & \cdots & \cdots \\ a_{n1} & a_{n2} & \cdots & a_{nn} \\ \end{vmatrix}=a_{j1}A_{j1} + a_{j2}A_{j2} + \cdots + a_{jn}A_{jn}$
$上式两端用a_{i1},a_{i2}, \cdots, a_{in}代替\\ a_{j1},a_{j2}, \cdots, a_{jn},得$
$\begin{vmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ \cdots & \cdots & \cdots & \cdots \\ a_{i1} & a_{i2} & \cdots & a_{in} \\ \cdots & \cdots & \cdots & \cdots \\ a_{i1} & a_{i2} & \cdots & a_{in} \\ \cdots & \cdots & \cdots & \cdots \\ a_{n1} & a_{n2} & \cdots & a_{nn} \\ \end{vmatrix}=a_{i1}A_{j1} + a_{i2}A_{j2} + \cdots + a_{in}A_{jn}$
$等式左端有两个相同，故行列式等于0$
$即 a_{i1}A_{j1} + a_{i2}A_{j2} + \cdots + a_{in}A_{jn} = 0(i \neq j)$
$同理可证$
$a_{1i}A_{1j} + a_{2i}A_{2j} + \cdots + a_{ni}A_{nj} = 0, i \neq j)$

综合定理1和推论有
$\sum_ {k = 1}^{n} a_{ki}A_{kj} = D\delta_{ij} = \left \{ \begin{array}{l} D \quad 当i = j, \\ 0 \quad 当i \neq j; \end{array} \right.$
$或\sum_ {k = 1}^{n} a_{ik}A_{jk} = D\delta_{ij} = \left \{ \begin{array}{l} D \quad 当i = j, \\ 0 \quad 当i \neq j; \end{array} \right.$
$其中 \delta_{ij} = \left \{ \begin{array}{l} 1, 当i = j, \\ 0, 当i \neq j. \end{array} \right.$

例5.已知行列式
$D = \begin{vmatrix} 1 & 2 & 3 & 4 \\ 1 & 1 & 1 & 1 \\ 2 & 3 & 4 & 5 \\ 5 & 4 & 2 & 3 \\ \end{vmatrix}$
$求A_{41} + A_{42} + A_{43} + A_{44}, 其中A_{41}, A_{42}, A_{43}, A_{44},是D的第四行元素的代数余子式.$
解:
$\because a_{21}A_{41} + a_{22}A_{42} + a_{23}A_{43} + a_{24}A_{44} = 0 \\ 即 1 \times A_{41} + 1 \times A_{42} + 1 \times A_{43} + 1 \times A_{44} = 0 \\ \therefore A_{41} + A_{42} + A_{43} + A_{44} = 0$

行列式主要研究的问题：
$\left \{ \begin{array}{l} a_{11}x_1 + a_{12}x_2 + \cdots + a_{1n}x_n = b_1 \\ a_{21}x_1 + a_{22}x_2 + \cdots + a_{2n}x_n = b_2 \\ \cdots \quad \cdots \quad \cdots \quad \cdots \\ a_{n1}x_1 + a_{n2}x_2 + \cdots + a_{nn}a_n = b_n \end{array} \right.$
$①D = \begin{vmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ a_{21} & a_{22} & \cdots & a_{2n} \\ \cdots & \cdots & & \cdots \\ a_{n1} & a_{n2} & \cdots & a_{nn} \\ \end{vmatrix} = ?$
$②D \neq 0,方程组是否有唯一解?$
$③若方程组有唯一解,那么\\ x_j = \dfrac{D_j}{D} (j = 1, 2, \cdots, n)是否是方程组的解?$