线性代数6.矩阵的行列式-代数余子式
6.矩阵的行列式-代数余子式
6.1 余子式和代数余子式
设存在n阶行列式\(|A|\),并存在\(|A|\)中的元素\(a_{ij}\)
则\(|A|\)中,除去元素\(a_{ij}\)所在的第\(i\)行和第\(j\)列所有元素后,剩下元素所形成的行列式称为\(a_{ij}\)的\(余子式\),记为\(M_{ij}\)
且存在\(A_{ij}=(-1)^{i+j}\cdot M_{ij} \Rightarrow A_{ij}\)称为\(a_{ij}\)的\(代数余子式\)
6.2 代数余子式在行列式求值中的应用
6.2.1 通过某元素的代数余子式求行列式的值
设存在如下行列式
\[|A|=
\begin{vmatrix}
a_{11} & 0 & 0 &...& 0\\
a_{21} & a_{22} & a_{23} &...& a_{2n}\\
a_{31} & a_{32} & a_{33} &...& a_{3n}\\
& & ......\\
a_{n1} & a_{n2} & a_{n3} &...& a_{nn}\\
\end{vmatrix}
\]
若|A|中,第1行元素除\(a_{11}\)外均为0,则根据下三角元素行列式求值方法及后续分块矩阵知识点,可得:
\[\tag {1}
|A|=
\begin{vmatrix}
a_{11} & \cancel{0} & \cancel{0} &...& \cancel{0}\\
\cancel{a_{21}} & a_{22} & a_{23} &...& a_{2n}\\
\cancel{a_{31}} & a_{32} & a_{33} &...& a_{3n}\\
& & ......\\
\cancel{a_{n1}} & a_{n2} & a_{n3} &...& a_{nn}\\
\end{vmatrix}
=a_{11}\cdot
\begin{vmatrix}
\cancel{a_{11}} & \cancel{0} & \cancel{0} &...& \cancel{0}\\
\cancel{a_{21}} & a_{22} & a_{23} &...& a_{2n}\\
\cancel{a_{31}} & a_{32} & a_{33} &...& a_{3n}\\
& & ......\\
\cancel{a_{n1}} & a_{n2} & a_{n3} &...& a_{nn}\\
\end{vmatrix}
=a_{11}\cdot M_{11}=a_{11}\cdot A_{11}
\]
根据以上推论,若存在以下包含元素\(a_{ij}\)的行列式:
\[|A|'=
\begin{vmatrix}
a_{11} & a_{12}... & a_{1j} &...& a_{1n}\\
a_{21} & a_{22}... & a_{2j} &...& a_{2n}\\
a_{31} & a_{32}... & a_{3j} &...& a_{3n}\\
& & ......\\
0 & 0... & a_{ij} &0...& 0\\
& & ......\\
a_{n1} & a_{n2}... & a_{nj} &...& a_{nn}\\
\end{vmatrix}
\]
若|A|'中,第i行元素除\(a_{ij}\)外均为0
则根据行列互换的性质,可先将第i行分别与第\((i-1),(i-2),(i-3),...,1\)行互换:
\[|A|'=(-1) \cdot
\begin{vmatrix}
a_{11} & a_{12}... & a_{1j} &...& a_{1n}\\
a_{21} & a_{22}... & a_{2j} &...& a_{2n}\\
a_{31} & a_{32}... & a_{3j} &...& a_{3n}\\
& & ......\\
0 & 0... & a_{ij} &0...& 0\\
a_{(i-1)1} & a_{(i-1)2} & ...a_{(i-1)j} &...& a_{(i-1)n}\\
a_{(i+1)1} & a_{(i+1)2} & ...a_{(i+1)j} &...& a_{(i+1)n}\\
& & ......\\
a_{n1} & a_{n2}... & a_{nj} &...& a_{nn}\\
\end{vmatrix}\\
\]
\[......
\]
\[=(-1)^{i-1} \cdot
\begin{vmatrix}
0 & 0... & a_{ij} &0...& 0\\
a_{11} & a_{12} & a_{1j} &...& a_{1n}\\
a_{21} & a_{22} & a_{2j} &...& a_{2n}\\
& & ......\\
a_{(i-1)1} & a_{(i-1)2} & ...a_{(i-1)j} &...& a_{(i-1)n}\\
a_{(i+1)1} & a_{(i+1)2} & ...a_{(i+1)j} &...& a_{(i+1)n}\\
& & ......\\
a_{n1} & a_{n2} & a_{nj} &...& a_{nn}\\
\end{vmatrix}\\
\]
再将第j列分别与第\((j-1),(j-2),...,1\)列互换:
\[|A|'=(-1)^{(i-1)+ 1}
\begin{vmatrix}
0 & 0... & a_{ij} &0 &0...& 0\\
a_{11} & a_{12}... & a_{1j} &a_{1(j-1)} &...& a_{1n}\\
a_{21} & a_{22}... & a_{2j} &a_{2(j-1)} &...& a_{2n}\\
& & ......\\
a_{(i-1)1} & a_{(i-1)2}... & a_{(i-1)j} &a_{(i-1)(j-1)} &...& a_{(i-1)n}\\
a_{(i+1)1} & a_{(i+1)2}... & a_{(i+1)j} &a_{(i+1)(j-1)} &...& a_{(i+1)n}\\
& & ......\\
a_{n1} & a_{n2}... & a_{nj} &a_{n(j-1)} &...& a_{nn}\\
\end{vmatrix}
\]
\[...
\]
\[=(-1)^{(i-1)+ (j-1)}
\begin{vmatrix}
a_{ij} &0 & 0... & 0 &0...& 0\\
a_{1j} &a_{11} & a_{12}... & a_{1(j-1)} &...& a_{1n}\\
a_{2j} &a_{21} & a_{22}... & a_{2(j-1)} &...& a_{2n}\\
& & ......\\
a_{(i-1)j} &a_{(i-1)1} & a_{(i-1)2}... & a_{(i-1)(j-1)} &...& a_{(i-1)n}\\
a_{(i+1)j} &a_{(i+1)1} & a_{(i+1)2}... & a_{(i+1)(j-1)} &...& a_{(i+1)n}\\
& & ......\\
a_{nj} &a_{n1} & a_{n2}... & a_{n(j-1)} &...& a_{nn}\\
\end{vmatrix}
\]
根据公式(1)可得:
\[\qquad\qquad\qquad|A|'=a_{ij}\cdot (-1)^{(i-1)+ (j-1)}\cdot M_{ij}\\
\qquad\qquad\quad=a_{ij}\cdot (-1)^{(i+j)}\cdot M_{ij}\\
=a_{ij}\cdot A_{ij}
\]
\(\Rightarrow\) 对任意行列式\(|A|\)及其中任意元素\(a_{ij}\),存在:
\[\tag{2}
|A|= a_{ij}\cdot A_{ij}
\]
**注意区分:\(|A|\)为矩阵A的行列式,\(A_{ij}\)为元素\(a_{ij}\)的代数余子式
6.2.2 行列式的按行/按列展开(行列式的降阶)
设存在以下包含第i行的行列式\(|A|\):
\[|A|=
\begin{vmatrix}
a_{11} & a_{12}... & a_{1j} &...& a_{1n}\\
a_{21} & a_{22}... & a_{2j} &...& a_{2n}\\
a_{31} & a_{32}... & a_{3j} &...& a_{3n}\\
& & ......\\
a_{i1} & a_{i2}... & a_{ij} &...& a_{in}\\
& & ......\\
a_{n1} & a_{n2}... & a_{nj} &...& a_{nn}\\
\end{vmatrix}
\]
则根据行列式的行相加规则,有:
\[|A|=
\begin{vmatrix}
a_{11} & a_{12}... & a_{1j} &...& a_{1n}\\
a_{21} & a_{22}... & a_{2j} &...& a_{2n}\\
a_{31} & a_{32}... & a_{3j} &...& a_{3n}\\
& & ......\\
a_{i1} &0... & 0 &...& 0\\
& & ......\\
a_{n1} & a_{n2}... & a_{nj} &...& a_{nn}\\
\end{vmatrix}
\]
\[+
\begin{vmatrix}
a_{11} & a_{12}... & a_{1j} &...& a_{1n}\\
a_{21} & a_{22}... & a_{2j} &...& a_{2n}\\
a_{31} & a_{32}... & a_{3j} &...& a_{3n}\\
& & ......\\
0 &a_{i2}... & 0 &...& 0\\
& & ......\\
a_{n1} & a_{n2}... & a_{nj} &...& a_{nn}\\
\end{vmatrix}
\]
\[+......\\
+
\begin{vmatrix}
a_{11} & a_{12}... & a_{1j} &...& a_{1n}\\
a_{21} & a_{22}... & a_{2j} &...& a_{2n}\\
a_{31} & a_{32}... & a_{3j} &...& a_{3n}\\
& & ......\\
0 &0... & a_{ij} &...& 0\\
& & ......\\
a_{n1} & a_{n2}... & a_{nj} &...& a_{nn}\\
\end{vmatrix}
\]
\[+......\\
+
\begin{vmatrix}
a_{11} & a_{12}... & a_{1j} &...& a_{1n}\\
a_{21} & a_{22}... & a_{2j} &...& a_{2n}\\
a_{31} & a_{32}... & a_{3j} &...& a_{3n}\\
& & ......\\
0 &0... & 0 &...& a_{in}\\
& & ......\\
a_{n1} & a_{n2}... & a_{nj} &...& a_{nn}\\
\end{vmatrix}
\]
则根据下三角元素行列式及分块矩阵求值方法得:
\[\tag{3}
|A|=a_{i1}\cdot A_{i1}+a_{i2}\cdot A_{i2}+...+a_{in}\cdot A_{in}
=\sum_{x=1}^n a_{ix}\cdot A_{ix}
\]
同理可得同一列元素也具有:
\[\tag{4}
|A|=a_{1i}\cdot A_{1i}+a_{2i}\cdot A_{2i}+...+a_{ni}\cdot A_{ni}
=\sum_{x=1}^n a_{xi}\cdot A_{xi}
\]
6.2.3 行列式的【异乘变零】定理
设存在以下包含第\(i\)行和第\(j\)行的行列式
\[|A|=
\begin{vmatrix}
a_{11} & a_{12} & a_{13} &...& a_{1n}\\
a_{21} & a_{22} & a_{23} &...& a_{2n}\\
a_{31} & a_{32} & a_{33} &...& a_{3n}\\
& & ......\\
a_{i1} & a_{i2} & a_{i3} &...& a_{in}\\
& & ......\\
a_{j1} & a_{j2} & a_{j3} &...& a_{jn}\\
& & ......\\
a_{n1} & a_{n2} & a_{n3} &...& a_{nn}\\
\end{vmatrix}
\]
根据行列式的【异乘变零】定理,有:
\[\tag{5}
\sum_{x=1}^na_{ix}\cdot A_{jx}=0
\]
**【异乘变零】定理的证明过程略(题主不认同牵强附会的证明方法)

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