2.4* 矩阵公式总结

操作顺序可交换

$$({\mathbf{A}}^{?}{)}^{?}=({\mathbf{A}}^{?}{)}^{?}⟹\{\begin{array}{l}({\mathbf{A}}^{\ast }{)}^{-1}=({\mathbf{A}}^{-1}{)}^{\ast }\\ ({\mathbf{A}}^{\mathrm{T}}{)}^{\ast }=({\mathbf{A}}^{\ast }{)}^{\mathrm{T}}\\ ({\mathbf{A}}^{-1}{)}^{\mathrm{T}}=({\mathbf{A}}^{\mathrm{T}}{)}^{-1}\end{array}$$

整体操作需对调

$$(\mathbf{AB}{)}^{?}={\mathbf{B}}^{?}{\mathbf{A}}^{?}⟹\{\begin{array}{l}(\mathbf{AB}{)}^{-1}={\mathbf{B}}^{-1}{\mathbf{A}}^{-1}\\ (\mathbf{AB}{)}^{\mathrm{T}}={\mathbf{B}}^{\mathrm{T}}{\mathbf{A}}^{\mathrm{T}}\\ (\mathbf{AB}{)}^{\ast }={\mathbf{B}}^{\ast }{\mathbf{A}}^{\ast }\end{array}$$

拓展：$(\mathbf{ABC}{)}^{?}={\mathbf{C}}^{?}{\mathbf{B}}^{?}{\mathbf{A}}^{?}$

重复操作会还原

$$({\mathbf{A}}^{?}{)}^{?}=\mathbf{A}⟹\{\begin{array}{l}({\mathbf{A}}^{\mathrm{T}}{)}^{\mathrm{T}}=\mathbf{A}\\ ({\mathbf{A}}^{-1}{)}^{-1}=\mathbf{A}\end{array}$$

WARNING

两次取伴随矩阵不会还原：$({\mathbf{A}}^{\ast }{)}^{\ast }\ne \mathbf{A}$

转置的优良性

$${\mathbf{A}}^{\mathrm{T}}⟹\{\begin{array}{l}|{\mathbf{A}}^{\mathrm{T}}|=|\mathbf{A}|\\ (k\mathbf{A}{)}^{\mathrm{T}}=k{\mathbf{A}}^{\mathrm{T}}\\ (\mathbf{A}+\mathbf{B}{)}^{\mathrm{T}}={\mathbf{A}}^{\mathrm{T}}+{\mathbf{B}}^{\mathrm{T}}\end{array}$$

WARNING

$$\{\begin{array}{l}(\mathbf{A}+\mathbf{B}{)}^{-1}\ne {\mathbf{A}}^{-1}+{\mathbf{B}}^{-1}\\ (\mathbf{A}+\mathbf{B}{)}^{\ast }\ne {\mathbf{A}}^{\ast }+{\mathbf{B}}^{\ast }\end{array}$$

逆矩阵的定义

$${\mathbf{AA}}^{-1}=\mathbf{E}⟹\{\begin{array}{rl}& |{\mathbf{A}}^{-1}|=\frac{1}{|\mathbf{A}|}\\ & (k\mathbf{A}{)}^{-1}=\frac{1}{k}{\mathbf{A}}^{-1}\end{array}$$

伴随矩阵的推导

$${\mathbf{A}}^{\ast }=|\mathbf{A}|{\mathbf{A}}^{-1}⟹\{\begin{array}{l}(k\mathbf{A}{)}^{\ast }=k^{n-1}{\mathbf{A}}^{\ast }\\ |{\mathbf{A}}^{\ast }|=|\mathbf{A}{|}^{n-1}\\ ({\mathbf{A}}^{\ast }{)}^{\ast }=|\mathbf{A}{|}^{n-2}\mathbf{A}\end{array}$$

${\mathbf{A}}^{\ast }=|\mathbf{A}|{\mathbf{A}}^{-1}$ 这一点要牢牢记住。