矩阵求逆的方法总结
1. 矩阵的逆
定义
A→A−1AB=BA=E⇒A−1=B
A\to A^{-1} \\ AB=BA=E\Rightarrow A^{-1}=B
A→A−1AB=BA=E⇒A−1=B
性质
adjA=A∗AA∗=A∗A=∣A∣EAA∗∣A∣=EA−1=adjAdetA
adjA = A^*\\ AA^*=A^*A=|A|E\\\frac{AA^*}{|A|}=E\\A^{-1} = \frac{adjA}{\det A}
adjA=A∗AA∗=A∗A=∣A∣E∣A∣AA∗=EA−1=detAadjA
2. 初等变换法：
[AE]⟶初等行变换[EA−1]
\left[\begin{array}{c|c}A&E\end{array}\right]\overset{初等行变换}{\longrightarrow }\left[\begin{array}{c|c}E & A^{-1}\end{array}\right]
[AE]⟶初等行变换[EA−1]
[AE]⟶初等列变换[EA−1]
\left[\begin{array}{}A\\\hline E\end{array}\right]\overset{初等列变换}{\longrightarrow} \left[\begin{array}{}E\\\hline A^{-1}\end{array}\right]
[AE]⟶初等列变换[EA−1]
3. 特征多项式法
∣A−λE∣=det(A−λE)⟶关于λ的n次多项式anλn+an−1λn−1+⋯+a1λ+a0
|A-\lambda E| = \det(A-\lambda E)\overset{关于\lambda的n次多项式}{\longrightarrow}a_n\lambda^n+a_{n-1}\lambda^{n-1}+\cdots+a_1\lambda+a_0
∣A−λE∣=det(A−λE)⟶关于λ的n次多项式anλn+an−1λn−1+⋯+a1λ+a0
AAA可逆则：
AAA可逆⇒\Rightarrow⇒无0特征值λ≠0\lambda\ne0λ=0⇒a0≠0\Rightarrow a_0\ne0⇒a0=0；
特征多项式是AAA的化零多项式；
anAn+an−1An−1+⋯+a1A+a0E=0
a_nA^n+a_{n-1}A^{n-1}+\cdots+a_1A+a_0E = 0
anAn+an−1An−1+⋯+a1A+a0E=0
因此有
an−a0An+an−1−a0An−1+⋯+a1−a0A=E⇒A(an−a0An−1+an−1−a0An−2+⋯+a1−a0E)=E⇒A−1=an−a0An−1+an−1−a0An−2+⋯+a1−a0E
\frac{a_n}{-a_0}A^n+\frac{a_{n-1}}{-a_0}A^{n-1}+\cdots+\frac{a_1}{-a_0}A=E\Rightarrow \\A(\frac{a_n}{-a_0}A^{n-1}+\frac{a_{n-1}}{-a_0}A^{n-2}+\cdots+\frac{a_1}{-a_0}E)=E\Rightarrow \\A^{-1}=\frac{a_n}{-a_0}A^{n-1}+\frac{a_{n-1}}{-a_0}A^{n-2}+\cdots+\frac{a_1}{-a_0}E
−a0anAn+−a0an−1An−1+⋯+−a0a1A=E⇒A(−a0anAn−1+−a0an−1An−2+⋯+−a0a1E)=E⇒A−1=−a0anAn−1+−a0an−1An−2+⋯+−a0a1E
例：
A=(1−102)A = \begin{pmatrix}1&-1\\0&2\end{pmatrix}A=(10−12)，求A−1A^{-1}A−1。
解：
∣A−λE∣=(λ−1)(λ−2)=λ2−3λ+2
|A-\lambda E| = (\lambda-1)(\lambda-2) = \lambda^2-3\lambda+2
∣A−λE∣=(λ−1)(λ−2)=λ2−3λ+2
A−1=1−2A−3−2E=−12(1−102)+32(1001)=(112012)
A^{-1} = \frac{1}{-2}A -\frac{3}{-2}E = -\frac{1}{2}\begin{pmatrix}1&-1\\0&2\end{pmatrix}+\frac{3}{2}\begin{pmatrix}1&0\\0&1\end{pmatrix} = \begin{pmatrix}1&\frac{1}{2}\\0&\frac{1}{2}\end{pmatrix}
A−1=−21A−−23E=−21(10−12)+23(1001)=(102121)
4. 最小多项式
A→Tordm标准型→Smith矩阵→dn(λ)→最小多项式akλk+ak−1λk−1+⋅+a0=0(k≤n)akAk+ak−1Ak−1+⋯+a0E=0A(ak−a0Ak−1+ak−1−a0Ak−2+⋯+a1−a0E)=EA−1=ak−a0Ak−1+ak−1−a0Ak−2+⋯+a1−a0E
A\to Tordm标准型\to Smith矩阵\\\to d_n(\lambda)\to最小多项式\\a_k\lambda^k+a_{k-1}\lambda^{k-1}+\cdot+a_0=0(k\le n)\\a_kA^k+a_{k-1}A^{k-1}+\cdots+a_0E = 0\\A(\frac{a_k}{-a_0}A^{k-1}+\frac{a_{k-1}}{-a_0}A^{k-2}+\cdots+\frac{a_1}{-a_0}E)=E\\A^{-1} = \frac{a_k}{-a_0}A^{k-1}+\frac{a_{k-1}}{-a_0}A^{k-2}+\cdots+\frac{a_1}{-a_0}E
A→Tordm标准型→Smith矩阵→dn(λ)→最小多项式akλk+ak−1λk−1+⋅+a0=0(k≤n)akAk+ak−1Ak−1+⋯+a0E=0A(−a0akAk−1+−a0ak−1Ak−2+⋯+−a0a1E)=EA−1=−a0akAk−1+−a0ak−1Ak−2+⋯+−a0a1E
例：
若矩阵AAA的SmithSmithSmith矩阵为A=(11111)A = \begin{pmatrix}1&&&\\&1&&\\&&1&1\\&&&1\end{pmatrix}A=11111，求A−1A^{-1}A−1。
解：
由题意得
A−λE⟶Smith(1λ−1λ−1(λ−1)2)
A-\lambda E\overset{Smith}{\longrightarrow}\begin{pmatrix}1&&&\\&\lambda-1&&\\&&\lambda-1&\\&&&(\lambda-1)^2\end{pmatrix}
A−λE⟶Smith1λ−1λ−1(λ−1)2
所以AAA的最小多项式为
dn(λ)=λ2−2λ+1
d_{n}(\lambda) = \lambda^2-2\lambda+1
dn(λ)=λ2−2λ+1
于是有
A2−2A+E=0
A^2-2A+E = 0
A2−2A+E=0
所以
A−1=1−1A+−2−1E=(111−11)
A^{-1} = \frac{1}{-1}A+\frac{-2}{-1}E = \begin{pmatrix}1&&&\\&1&&\\&&1&-1\\&&&1\end{pmatrix}
A−1=−11A+−1−2E=111−11
5. LeVerrier-Faddeev\text{LeVerrier-Faddeev}LeVerrier-Faddeev算法
B0=0an=1B1=AB0+anEan−1=−tr(AB1)B2=AB1+an−1Ean−2=−12tr(AB2)⋯Bn=ABn−1+a1Ea0=−1ntr(ABn)0=ABn+a0E⇒A−1=−Bna0
B_0=0\quad a_n = 1\\B_1 = AB_0+a_nE\quad a_{n-1} = -tr(AB_1)\\B_2 = AB_1+a_{n-1}E\quad a_{n-2} = -\frac{1}{2}tr(AB_2)\\\cdots\\B_n = AB_{n-1}+a_{1}E\quad a_0 = -\frac{1}{n}tr(AB_{n})\\0=AB_{n}+a_{0}E\\\Rightarrow A^{-1}=-\frac{B_n}{a_0}
B0=0an=1B1=AB0+anEan−1=−tr(AB1)B2=AB1+an−1Ean−2=−21tr(AB2)⋯Bn=ABn−1+a1Ea0=−n1tr(ABn)0=ABn+a0E⇒A−1=−a0Bn
例：
设A=(315331464)A = \begin{pmatrix}3&1&5\\3&3&1\\4&6&4\end{pmatrix}A=334136514，求A−1A^{-1}A−1。
解：
利用LeVerrier-Faddeev\text{LeVerrier-Faddeev}LeVerrier-Faddeev算法，先确定n=3n=3n=3
B0=(000000000)a3=1B1=AB0+a3E=(100010001)a2=−tr(AB1)=−10B2=AB1+a2E=(−7153−7146−6)a1=−12tr(AB2)=4B3=AB2+a1E=(626−14−8−8126−146)a0=−13tr(AB3)=−40AB3+a0E=(000000000)⇒A−1=−B3a0=(0.150.65−0.35−0.2−0.20.30.15−0.350.15)
B_0 = \begin{pmatrix}0&0&0\\0&0&0\\0&0&0\end{pmatrix}\quad a_3=1\\B_1=AB_0+a_3E=\begin{pmatrix}1&0&0\\0&1&0\\0&0&1\end{pmatrix}\quad a_2=-tr(AB_1)=-10\\B_2 = AB_1+a_2E=\begin{pmatrix}-7&1&5\\3&-7&1\\4&6&-6\end{pmatrix}\quad a_1 =-\frac{1}{2}tr(AB_2)=4\\B_3=AB_2+a_1E=\begin{pmatrix}6&26&-14\\-8&-8&12\\6&-14&6\end{pmatrix} \quad a_0=-\frac{1}{3}tr(AB_3)=-40\\AB_3+a_0E=\begin{pmatrix}0&0&0\\0&0&0\\0&0&0\end{pmatrix}\\\Rightarrow A^{-1}=-\frac{B_3}{a_0}=\begin{pmatrix}0.15&0.65&-0.35\\-0.2&-0.2&0.3\\0.15&-0.35&0.15\end{pmatrix}
B0=000000000a3=1B1=AB0+a3E=100010001a2=−tr(AB1)=−10B2=AB1+a2E=−7341−7651−6a1=−21tr(AB2)=4B3=AB2+a1E=6−8626−8−14−14126a0=−31tr(AB3)=−40AB3+a0E=000000000⇒A−1=−a0B3=0.15−0.20.150.65−0.2−0.35−0.350.30.15
6. 分块求逆
分块矩阵求逆公式如下
(A11A12A21A22)−1=((A11−A12A22−1A21)−1−(A11−A12A22−1A21)−1A12A22−1−A22−1A21(A11−A12A22−1A21)−1A22−1+A22−1A21(A11−A12A22−1A21)−1A12A22−1)
\begin{array}{c}\left(\begin{array}{ll}A_{11} & A_{12} \\A_{21} & A_{22}\end{array}\right)^{-1}= \\\left(\begin{array}{cc}\left(A_{11}-A_{12} A_{22}^{-1} A_{21}\right)^{-1} & -\left(A_{11}-A_{12} A_{22}^{-1} A_{21}\right)^{-1} A_{12} A_{22}^{-1} \\-A_{22}^{-1} A_{21}\left(A_{11}-A_{12} A_{22}^{-1} A_{21}\right)^{-1} & A_{22}^{-1}+A_{22}^{-1} A_{21}\left(A_{11}-A_{12} A_{22}^{-1} A_{21}\right)^{-1} A_{12} A_{22}^{-1}\end{array}\right)\end{array}
(A11A21A12A22)−1=((A11−A12A22−1A21)−1−A22−1A21(A11−A12A22−1A21)−1−(A11−A12A22−1A21)−1A12A22−1A22−1+A22−1A21(A11−A12A22−1A21)−1A12A22−1)
7. Sherman-Morrison-Woodbury\text{Sherman-Morrison-Woodbury}Sherman-Morrison-Woodbury公式
Am×mA_{m\times m}Am×m、Dn×nD_{n\times n}Dn×n非奇异，Bm×nB_{m\times n}Bm×n、Cn×mC_{n\times m}Cn×m，若CA−1B+D−1CA^{-1}B+D^{-1}CA−1B+D−1非奇异，则矩阵A+BDCA+BDCA+BDC非奇异，且
(A+BDC)−1=A−1−A−1B(D−1+CA−1B)−1CA−1
(A+BDC)^{-1} = A^{-1}-A^{-1}B(D^{-1}+CA^{-1}B)^{-1}CA^{-1}
(A+BDC)−1=A−1−A−1B(D−1+CA−1B)−1CA−1