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\begin_body

\begin_layout Section
逆矩阵
\end_layout

\begin_layout Subsection
逆矩阵的概念
\end_layout

\begin_layout Frame
\begin_inset Argument 3
status open

\begin_layout Plain Layout
allowframebreaks
\end_layout

\end_inset


\begin_inset Argument 4
status open

\begin_layout Plain Layout
逆矩阵的概念
\end_layout

\end_inset


\end_layout

\begin_deeper
\begin_layout Standard
在数的运算中, 对于数 
\begin_inset Formula $a\neq0$
\end_inset

, 总存在唯一一个数 
\begin_inset Formula $a^{-1}$
\end_inset

, 使得
\begin_inset Formula 
\[
a\cdot a^{-1}=a^{-1}\cdot a=1.
\]

\end_inset


\end_layout

\begin_layout Standard
数的逆在解方程中起着重要作用, 例如, 解一元线性方程
\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
ax=b.
\]

\end_inset


\end_layout

\begin_layout Standard
当 
\begin_inset Formula $a\neq0$
\end_inset

 时, 其解为
\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
x=a^{-1}b.
\]

\end_inset


\end_layout

\begin_layout Standard
对一个矩阵 
\begin_inset Formula $A$
\end_inset

, 是否也存在类似的运算? 在回答这个问题之前, 我们先引入可逆矩阵与逆矩阵的概念.
\end_layout

\begin_layout Definition
逆矩阵的运算性质对于 
\begin_inset Formula $n$
\end_inset

 阶矩阵 
\begin_inset Formula $A$
\end_inset

, 如果存在一个 
\begin_inset Formula $n$
\end_inset

 阶矩阵 
\begin_inset Formula $B$
\end_inset

, 使得
\begin_inset Formula 
\[
AB=BA=E,
\]

\end_inset

则称矩阵 
\begin_inset Formula $A$
\end_inset

 为
\series bold
可逆矩阵
\series default
, 而矩阵 
\begin_inset Formula $B$
\end_inset

 称为 
\series bold

\begin_inset Formula $A$
\end_inset

 的逆矩阵
\series default
.
 
\end_layout

\begin_layout Proposition
若矩阵 
\begin_inset Formula $A$
\end_inset

 是可逆的, 则 
\begin_inset Formula $A$
\end_inset

 的逆矩阵是唯一的.
\end_layout

\begin_layout Definition
如果 
\begin_inset Formula $n$
\end_inset

 阶矩阵 
\begin_inset Formula $A$
\end_inset

 的行列式 
\begin_inset Formula $|A|\ne0$
\end_inset

, 则称 
\begin_inset Formula $A$
\end_inset

 为
\series bold
非奇异的
\series default
, 否则称为
\series bold
奇异的
\series default
.
\end_layout

\end_deeper
\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Frame
\begin_inset Argument 4
status open

\begin_layout Plain Layout
逆矩阵的概念
\end_layout

\end_inset


\end_layout

\begin_deeper
\begin_layout Example
设 
\begin_inset Formula $A=\begin{bmatrix}2 & 1\\
-1 & 0
\end{bmatrix}$
\end_inset

, 求 
\begin_inset Formula $A$
\end_inset

 的逆矩阵.
 
\end_layout

\begin_layout Solution*
利用待定系数法, 设 
\begin_inset Formula $A$
\end_inset

 的逆矩阵 
\begin_inset Formula $B=\begin{bmatrix}a & b\\
c & d
\end{bmatrix}$
\end_inset

, 则
\begin_inset ERT
status open

\begin_layout Plain Layout


\backslash
vspace{-5mm}
\end_layout

\end_inset


\begin_inset Formula 
\[
AB=\begin{bmatrix}2 & 1\\
-1 & 0
\end{bmatrix}\begin{bmatrix}a & b\\
c & d
\end{bmatrix}=\begin{bmatrix}1 & 0\\
0 & 1
\end{bmatrix}\Longrightarrow\begin{bmatrix}2a+c & 2b+d\\
-a & -b
\end{bmatrix}=\begin{bmatrix}1 & 0\\
0 & 1
\end{bmatrix}\Longrightarrow\begin{cases}
2a+c=1\\
2b+d=0\\
-a=0\\
-b=1
\end{cases}\hspace{-2em}\Longrightarrow\begin{cases}
a=0\\
b=-1\\
c=1\\
d=2
\end{cases}
\]

\end_inset

又因为 
\begin_inset Formula $\begin{bmatrix}2 & 1\\
-1 & 0
\end{bmatrix}\begin{bmatrix}0 & -1\\
1 & 2
\end{bmatrix}=\begin{bmatrix}0 & -1\\
1 & 2
\end{bmatrix}\begin{bmatrix}2 & 1\\
-1 & 0
\end{bmatrix}=\begin{bmatrix}1 & 0\\
0 & 1
\end{bmatrix}$
\end_inset

, 所以 
\begin_inset Formula $A^{-1}=\begin{bmatrix}0 & -1\\
1 & 2
\end{bmatrix}$
\end_inset

.
\end_layout

\end_deeper
\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Frame
\begin_inset Argument 4
status open

\begin_layout Plain Layout
逆矩阵的概念
\end_layout

\end_inset


\end_layout

\begin_deeper
\begin_layout Example
证明矩阵 
\begin_inset Formula $A$
\end_inset

 无逆矩阵: 
\begin_inset Formula $A=\begin{bmatrix}1 & 0\\
0 & 0
\end{bmatrix}$
\end_inset

.
\end_layout

\begin_layout Proof
假定 
\begin_inset Formula $A$
\end_inset

 有逆矩阵 
\begin_inset Formula $B=\left(b_{ij}\right)_{2\times2}$
\end_inset

 使 
\begin_inset Formula $AB=BA=E_{2}$
\end_inset

, 则
\begin_inset Formula 
\[
\begin{bmatrix}1 & 0\\
0 & 0
\end{bmatrix}\begin{bmatrix}b_{11} & b_{12}\\
b_{21} & b_{22}
\end{bmatrix}=\begin{bmatrix}b_{11} & b_{12}\\
0 & 0
\end{bmatrix}=E_{2}=\begin{bmatrix}1 & 0\\
0 & 1
\end{bmatrix}\text{. }
\]

\end_inset

但这是不可能的, 因为由 
\begin_inset Formula $\begin{bmatrix}b_{11} & b_{12}\\
0 & 0
\end{bmatrix}=\begin{bmatrix}1 & 0\\
0 & 1
\end{bmatrix}$
\end_inset

, 将推出 
\begin_inset Formula $0=1$
\end_inset

 的谬论来.
 因此 
\begin_inset Formula $A$
\end_inset

 无逆矩阵.
\end_layout

\end_deeper
\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Frame
\begin_inset Argument 3
status open

\begin_layout Plain Layout
allowframebreaks
\end_layout

\end_inset


\begin_inset Argument 4
status open

\begin_layout Plain Layout
逆矩阵的概念
\end_layout

\end_inset


\end_layout

\begin_deeper
\begin_layout Example
如果 
\begin_inset Formula $A=\begin{bmatrix}a_{1} & 0 & \cdots & 0\\
0 & a_{2} & \cdots & 0\\
\vdots & \vdots & \ddots & \vdots\\
0 & 0 & \cdots & a_{n}
\end{bmatrix}$
\end_inset

, 其中 
\begin_inset Formula $a_{i}\neq0$
\end_inset

, (
\begin_inset Formula $i=1,2,\cdots,n$
\end_inset

).
 验证
\begin_inset Formula 
\[
A^{-1}=\begin{bmatrix}1/a_{1} & 0 & \cdots & 0\\
0 & 1/a_{2} & \cdots & 0\\
\vdots & \vdots & \ddots & \vdots\\
0 & 0 & \cdots & 1/a_{n}
\end{bmatrix}.
\]

\end_inset


\end_layout

\begin_layout Proof
由于
\begin_inset ERT
status open

\begin_layout Plain Layout


\backslash
vspace{-3mm}
\end_layout

\end_inset


\begin_inset Formula 
\begin{align*}
\begin{bmatrix}a_{1} & 0 & \cdots & 0\\
0 & a_{2} & \cdots & 0\\
\vdots & \vdots & \ddots & \vdots\\
0 & 0 & \cdots & a_{n}
\end{bmatrix}\begin{bmatrix}1/a_{1} & 0 & \cdots & 0\\
0 & 1/a_{2} & \cdots & 0\\
\vdots & \vdots & \ddots & \vdots\\
0 & 0 & \cdots & 1/a_{n}
\end{bmatrix} & =\begin{bmatrix}1 & 0 & \cdots & 0\\
0 & 1 & \cdots & 0\\
\vdots & \vdots & \ddots & \vdots\\
0 & 0 & \cdots & 1
\end{bmatrix}\\
 & \hspace{-5em}=\begin{bmatrix}1/a_{1} & 0 & \cdots & 0\\
0 & 1/a_{2} & \cdots & 0\\
\vdots & \vdots & \ddots & \vdots\\
0 & 0 & \cdots & 1/a_{n}
\end{bmatrix}\begin{bmatrix}a_{1} & 0 & \cdots & 0\\
0 & a_{2} & \cdots & 0\\
\vdots & \vdots & \ddots & \vdots\\
0 & 0 & \cdots & a_{n}
\end{bmatrix},
\end{align*}

\end_inset

所以
\begin_inset ERT
status open

\begin_layout Plain Layout


\backslash
vspace{-3mm}
\end_layout

\end_inset


\begin_inset Formula 
\[
A^{-1}=\begin{bmatrix}1/a_{1} & 0 & \cdots & 0\\
0 & 1/a_{2} & \cdots & 0\\
\vdots & \vdots & \ddots & \vdots\\
0 & 0 & \cdots & 1/a_{n}
\end{bmatrix}.
\]

\end_inset


\end_layout

\end_deeper
\begin_layout Subsection
伴随矩阵及其与逆矩阵的关系
\end_layout

\begin_layout Frame
\begin_inset Argument 3
status open

\begin_layout Plain Layout
allowframebreaks
\end_layout

\end_inset


\begin_inset Argument 4
status open

\begin_layout Plain Layout
伴随矩阵及其与逆矩阵的关系
\end_layout

\end_inset


\end_layout

\begin_deeper
\begin_layout Definition
行列式 
\begin_inset Formula $|A|$
\end_inset

 的各个元素 
\begin_inset Formula $a_{ij}$
\end_inset

 对应的代数余子式 
\begin_inset Formula $A_{ij}$
\end_inset

 所构成的矩阵
\begin_inset Formula 
\begin{equation}
A^{*}=\begin{bmatrix}A_{11} & A_{21} & \cdots & A_{n1}\\
A_{12} & A_{22} & \cdots & A_{n2}\\
\vdots & \vdots & \ddots & \vdots\\
A_{1n} & A_{2n} & \cdots & A_{nn}
\end{bmatrix}\label{eq:Aadj}
\end{equation}

\end_inset

称为矩阵 
\series bold

\begin_inset Formula $A$
\end_inset

 的伴随矩阵
\series default
.
\end_layout

\begin_layout Standard
由定义与代数余子式的性质, 可以证明伴随矩阵满足的一个基本性质
\begin_inset Formula 
\[
AA^{*}=A^{*}A=|A|E.
\]

\end_inset


\end_layout

\begin_layout Proof
\begin_inset Argument 1
status open

\begin_layout Plain Layout
简证
\end_layout

\end_inset

 设 
\begin_inset Formula $A=(a_{ij})_{n\times n}$
\end_inset

, 
\begin_inset Formula $A^{*}=(A_{ji})_{n\times n}$
\end_inset

, (或用 
\begin_inset Formula $(A^{*})_{ij}=A_{ji}$
\end_inset

 来表示矩阵 
\begin_inset Formula $A^{*}$
\end_inset

 的第 
\begin_inset Formula $i$
\end_inset

 行第 
\begin_inset Formula $j$
\end_inset

 列的元素), 则 
\begin_inset Formula $A\cdot A^{*}$
\end_inset

 的第 
\begin_inset Formula $i$
\end_inset

 行第 
\begin_inset Formula $j$
\end_inset

 列的元素为
\begin_inset Formula 
\[
\sum_{k=1}^{n}(A)_{ik}\cdot(A^{*})_{kj}=\sum_{k=1}^{n}a_{ik}\cdot A_{jk}=\left|A\right|\cdot\delta_{ij}=\left|A\right|(E)_{ij}.
\]

\end_inset


\end_layout

\begin_layout Theorem
\begin_inset Formula $n$
\end_inset

 阶矩阵 
\begin_inset Formula $A$
\end_inset

 可逆的充分必要条件是其行列式 
\begin_inset Formula $|A|\neq0$
\end_inset

.
 且当 
\begin_inset Formula $A$
\end_inset

 可逆时, 有
\end_layout

\begin_layout Theorem
\begin_inset Formula 
\[
A^{-1}=\frac{1}{|A|}A^{*},
\]

\end_inset

其中 
\begin_inset Formula $A^{*}$
\end_inset

 为 
\begin_inset Formula $A$
\end_inset

 的伴随矩阵.
\end_layout

\begin_layout Corollary
\begin_inset CommandInset label
LatexCommand label
name "cor:3.2-1"

\end_inset

 若 
\begin_inset Formula $AB=E$
\end_inset

 (或 
\begin_inset Formula $BA=E$
\end_inset

), 则 
\begin_inset Formula $B=A^{-1}$
\end_inset

.
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Corollary
若存在 
\begin_inset Formula $n$
\end_inset

 阶方阵 
\begin_inset Formula $B$
\end_inset

, 使得 
\begin_inset Formula $n$
\end_inset

 阶方阵 
\begin_inset Formula $A$
\end_inset

 满足 
\begin_inset Formula $AB=E$
\end_inset

, 则有 
\begin_inset Formula $BA=E$
\end_inset

.
 反之亦然, 也即两个 
\begin_inset Formula $n$
\end_inset

 阶方阵 
\begin_inset Formula $A,B$
\end_inset

 若满足 
\begin_inset Formula $BA=E$
\end_inset

, 则有 
\begin_inset Formula $AB=E$
\end_inset

.
 即对于满足 
\begin_inset Formula $AB=E$
\end_inset

 的两个 
\begin_inset Formula $n$
\end_inset

 阶方阵 
\begin_inset Formula $A,B$
\end_inset

 是可交换的.
\end_layout

\begin_layout Remark
用伴随矩阵求矩阵的逆一般不推荐, 但对于初学者应当经历几次用伴随矩阵求逆的繁琐计算, 然后记住前面的 (
\begin_inset CommandInset ref
LatexCommand ref
reference "eq:Aadj"
plural "false"
caps "false"
noprefix "false"

\end_inset

) 式及以上性质才是本小节的目标.
\end_layout

\end_deeper
\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Frame
\begin_inset Argument 4
status open

\begin_layout Plain Layout
逆矩阵的概念
\end_layout

\end_inset


\end_layout

\begin_deeper
\begin_layout Example
设 
\begin_inset Formula $A,B,C$
\end_inset

 均为 
\begin_inset Formula $n$
\end_inset

 阶矩阵, 且满足 
\begin_inset Formula $ABC=E$
\end_inset

, 则下式中哪些必定成立, 理由是什么? 
\end_layout

\begin_layout Example
(1) 
\begin_inset Formula $BCA=E$
\end_inset

; 
\end_layout

\begin_layout Example
(2) 
\begin_inset Formula $BAC=E$
\end_inset

; 
\end_layout

\begin_layout Example
(3) 
\begin_inset Formula $ACB=E$
\end_inset

; 
\end_layout

\begin_layout Example

\family roman
\series medium
\shape up
\size normal
\emph off
\bar no
\strikeout off
\xout off
\uuline off
\uwave off
\noun off
\color none
(4) 
\family default
\series default
\shape default
\size default
\emph default
\bar default
\strikeout default
\xout default
\uuline default
\uwave default
\noun default
\color inherit

\begin_inset Formula $CBA=E$
\end_inset

;
\end_layout

\begin_layout Example
(5) 
\begin_inset Formula $CAB=E$
\end_inset

.
\end_layout

\begin_layout Solution*
由 
\begin_inset Formula $ABC=E$
\end_inset

; 有 
\begin_inset Formula $(AB)C=E$
\end_inset

 或 
\begin_inset Formula $A(BC)=E$
\end_inset

.
 根据可逆矩阵的定义, 前者表明 
\begin_inset Formula $AB$
\end_inset

 与 
\begin_inset Formula $C$
\end_inset

 互为逆矩阵, 则有 
\begin_inset Formula $(AB)C=C(AB)=CAB=E$
\end_inset

; 
\end_layout

\begin_layout Solution*
后者表明 
\begin_inset Formula $A$
\end_inset

 与 
\begin_inset Formula $BC$
\end_inset

 互为逆矩阵, 可推出 
\begin_inset Formula $A(BC)=(BC)A=BCA=E$
\end_inset

.
 因此 (1) 与 (5) 必定成立.
\end_layout

\begin_layout Solution*
至于 (2), (3), (4), 当 
\begin_inset Formula $A=\begin{bmatrix}1 & 2\\
2 & 1
\end{bmatrix}$
\end_inset

, 
\begin_inset Formula $B=\begin{bmatrix}-1 & 2\\
2 & 1
\end{bmatrix}$
\end_inset

, 
\begin_inset Formula $C=\begin{bmatrix}\frac{7}{15} & -\frac{8}{15}\\
\frac{1}{15} & \frac{1}{15}
\end{bmatrix}$
\end_inset

 时, 可以验证 
\begin_inset Formula $ABC=E$
\end_inset

, 而 (2), (3), (4) 均不成立.
\end_layout

\end_deeper
\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Frame
\begin_inset Argument 3
status open

\begin_layout Plain Layout
allowframebreaks
\end_layout

\end_inset


\begin_inset Argument 4
status open

\begin_layout Plain Layout
求伴随矩阵
\end_layout

\end_inset


\end_layout

\begin_deeper
\begin_layout Example
\begin_inset CommandInset label
LatexCommand label
name "exa:3.2-4"

\end_inset

 矩阵 
\begin_inset Formula $A=\begin{bmatrix}1 & 0 & 1\\
2 & 1 & 0\\
-3 & 2 & -5
\end{bmatrix}$
\end_inset

, 求矩阵 
\begin_inset Formula $A$
\end_inset

 的伴随矩阵 
\begin_inset Formula $A^{*}$
\end_inset

.
\end_layout

\begin_layout Solution*
按定义, 因为
\begin_inset Formula 
\[
\begin{array}{ccc}
A_{11}=\begin{vmatrix}1 & 0\\
2 & -5
\end{vmatrix}=-5, & A_{12}=-\begin{vmatrix}2 & 0\\
-3 & -5
\end{vmatrix}=10, & A_{13}=\begin{vmatrix}2 & 1\\
-3 & 2
\end{vmatrix}=7,\\
A_{21}=-\begin{vmatrix}0 & 1\\
2 & -5
\end{vmatrix}=2, & A_{22}=\begin{vmatrix}1 & 1\\
-3 & -5
\end{vmatrix}=-2, & A_{23}=-\begin{vmatrix}1 & 0\\
-3 & 2
\end{vmatrix}=-2,\\
A_{31}=\begin{vmatrix}0 & 1\\
1 & 0
\end{vmatrix}=-1, & A_{32}=-\begin{vmatrix}1 & 1\\
2 & 0
\end{vmatrix}=2, & A_{33}=\begin{vmatrix}1 & 0\\
2 & 1
\end{vmatrix}=1.
\end{array}
\]

\end_inset

所以
\begin_inset Formula 
\[
\begin{gathered}A^{*}=\begin{bmatrix}A_{11} & A_{21} & A_{31}\\
A_{12} & A_{22} & A_{32}\\
A_{13} & A_{23} & A_{33}
\end{bmatrix}=\begin{bmatrix}-5 & 2 & -1\\
10 & -2 & 2\\
7 & -2 & 1
\end{bmatrix}.\end{gathered}
\]

\end_inset


\end_layout

\end_deeper
\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Frame
\begin_inset Argument 4
status open

\begin_layout Plain Layout
用伴随矩阵求逆矩阵
\end_layout

\end_inset


\end_layout

\begin_deeper
\begin_layout Example
求例 
\begin_inset CommandInset ref
LatexCommand ref
reference "exa:3.2-4"
plural "false"
caps "false"
noprefix "false"

\end_inset

 中矩阵 
\begin_inset Formula $A$
\end_inset

 的逆矩阵 
\begin_inset Formula $A^{-1}$
\end_inset

.
\end_layout

\begin_layout Solution*
由于 
\begin_inset Formula $|A|=\begin{vmatrix}1 & 0 & 1\\
2 & 1 & 0\\
-3 & 2 & -5
\end{vmatrix}=2\neq0$
\end_inset

, 利用例 
\begin_inset CommandInset ref
LatexCommand ref
reference "exa:3.2-4"
plural "false"
caps "false"
noprefix "false"

\end_inset

 的结果, 知 
\begin_inset Formula $A^{*}=\begin{bmatrix}-5 & 2 & -1\\
10 & -2 & 2\\
7 & -2 & 1
\end{bmatrix}$
\end_inset

, 所以
\begin_inset Formula 
\[
A^{-1}=\frac{1}{|A|}A^{*}=\frac{1}{2}\begin{bmatrix}-5 & 2 & -1\\
10 & -2 & 2\\
7 & -2 & 1
\end{bmatrix}=\begin{bmatrix}-\frac{5}{2} & 1 & -\frac{1}{2}\\
5 & -1 & 1\\
\frac{7}{2} & -1 & \frac{1}{2}
\end{bmatrix}.
\]

\end_inset


\end_layout

\end_deeper
\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Frame
\begin_inset Argument 3
status open

\begin_layout Plain Layout
allowframebreaks
\end_layout

\end_inset


\begin_inset Argument 4
status open

\begin_layout Plain Layout
用伴随矩阵求逆矩阵
\end_layout

\end_inset


\end_layout

\begin_deeper
\begin_layout Example
求 
\begin_inset Formula $A=\begin{bmatrix}1 & 1 & -1\\
1 & 2 & -3\\
0 & 1 & 1
\end{bmatrix}$
\end_inset

 的逆矩阵.
\end_layout

\begin_layout Solution*
由于 
\begin_inset Formula $|A|=\begin{vmatrix}1 & 1 & -1\\
1 & 2 & -3\\
0 & 1 & 1
\end{vmatrix}=3\neq0$
\end_inset

,
\begin_inset Formula 
\[
\begin{array}{ccc}
A_{11}=(-1)^{1+1}\begin{vmatrix}2 & -3\\
1 & 1
\end{vmatrix}, & A_{12}=(-1)^{1+2}\begin{vmatrix}1 & -3\\
0 & 1
\end{vmatrix}, & A_{13}=(-1)^{1+3}\begin{vmatrix}1 & 2\\
0 & 1
\end{vmatrix},\\
A_{21}=(-1)^{2+1}\begin{vmatrix}1 & -1\\
1 & 1
\end{vmatrix}, & A_{22}=(-1)^{2+2}\begin{vmatrix}1 & -1\\
0 & 1
\end{vmatrix}, & A_{23}=(-1)^{2+3}\begin{vmatrix}1 & 1\\
0 & 1
\end{vmatrix},\\
A_{31}=(-1)^{3+1}\begin{vmatrix}1 & -1\\
2 & -3
\end{vmatrix}, & A_{32}=(-1)^{3+2}\begin{vmatrix}1 & -1\\
1 & -3
\end{vmatrix}, & A_{33}=(-1)^{3+3}\begin{vmatrix}1 & 1\\
1 & 2
\end{vmatrix}.
\end{array}
\]

\end_inset


\begin_inset Formula 
\[
A_{11}=5,\ A_{12}=-1,\ A_{13}=1,\ A_{21}=-2,\ A_{22}=1,\ A_{23}=-1,\ A_{31}=-1,\ A_{32}=2,\ A_{33}=1.
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Solution*
于是 
\begin_inset Formula $A$
\end_inset

 的伴随矩阵 
\begin_inset Formula $A^{*}=\begin{bmatrix}5 & -2 & -1\\
-1 & 1 & 2\\
1 & -1 & 1
\end{bmatrix}$
\end_inset

, 
\begin_inset Formula $A$
\end_inset

 的逆矩阵为
\begin_inset Formula 
\[
A^{-1}=\frac{1}{|A|}A^{*}=\frac{1}{3}\begin{bmatrix}5 & -2 & -1\\
-1 & 1 & 2\\
1 & -1 & 1
\end{bmatrix}=\begin{bmatrix}\frac{5}{3} & -\frac{2}{3} & -\frac{1}{3}\\
-\frac{1}{3} & \frac{1}{3} & \frac{2}{3}\\
\frac{1}{3} & -\frac{1}{3} & \frac{1}{3}
\end{bmatrix}.
\]

\end_inset


\end_layout

\end_deeper
\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Frame
\begin_inset Argument 4
status open

\begin_layout Plain Layout
用伴随矩阵求逆矩阵
\end_layout

\end_inset


\end_layout

\begin_deeper
\begin_layout Example
已知 
\begin_inset Formula $A=\begin{bmatrix}1 & 0 & 0 & 0 & 0\\
0 & 2 & 0 & 0 & 0\\
0 & 0 & 3 & 0 & 0\\
0 & 0 & 0 & 4 & 0\\
0 & 0 & 0 & 0 & 5
\end{bmatrix}$
\end_inset

, 试用伴随矩阵法求 
\begin_inset Formula $A^{-1}$
\end_inset

.
\end_layout

\begin_layout Solution*
因 
\begin_inset Formula $|A|=5!\neq0$
\end_inset

, 故 
\begin_inset Formula $A^{-1}$
\end_inset

 存在.
 由伴随矩阵法得
\begin_inset Formula 
\[
\begin{aligned}A^{-1}= & \frac{A^{*}}{|A|}=\frac{1}{5!}\begin{bmatrix}2\cdot3\cdot4\cdot5 & 0 & 0 & 0 & 0\\
0 & 1\cdot3\cdot4\cdot5 & 0 & 0 & 0\\
0 & 0 & 1\cdot2\cdot4\cdot5 & 0 & 0\\
0 & 0 & 0 & 1\cdot2\cdot3\cdot5 & 0\\
0 & 0 & 0 & 0 & 1\cdot2\cdot3\cdot4
\end{bmatrix}\\
 & =\begin{bmatrix}1 & 0 & 0 & 0 & 0\\
0 & 1/2 & 0 & 0 & 0\\
0 & 0 & 1/3 & 0 & 0\\
0 & 0 & 0 & 1/4 & 0\\
0 & 0 & 0 & 0 & 1/5
\end{bmatrix}.
\end{aligned}
\]

\end_inset


\end_layout

\end_deeper
\begin_layout Subsection
逆矩阵的运算性质
\end_layout

\begin_layout Frame
\begin_inset Argument 4
status open

\begin_layout Plain Layout
逆矩阵的运算性质
\end_layout

\end_inset


\end_layout

\begin_deeper
\begin_layout Standard
(1) 若矩阵 
\begin_inset Formula $A$
\end_inset

 可逆, 则 
\begin_inset Formula $A^{-1}$
\end_inset

 也可逆, 且 
\begin_inset Formula $\left(A^{-1}\right)^{-1}=A$
\end_inset

;
\end_layout

\begin_layout Standard
(2) 若矩阵 
\begin_inset Formula $A$
\end_inset

 可逆, 数 
\begin_inset Formula $k\neq0$
\end_inset

, 则 
\begin_inset Formula $(kA)^{-1}=\frac{1}{k}A^{-1}$
\end_inset

;
\end_layout

\begin_layout Standard
(3) 两个同阶可逆矩阵 
\begin_inset Formula $A,B$
\end_inset

 的乘积是可逆矩阵, 且
\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
(AB)^{-1}=B^{-1}A^{-1}\text{; }
\]

\end_inset


\end_layout

\begin_layout Standard
(4) 若矩阵 
\begin_inset Formula $A$
\end_inset

 可逆, 则 
\begin_inset Formula $A^{T}$
\end_inset

 也可逆, 且有 
\begin_inset Formula $\left(A^{T}\right)^{-1}=\left(A^{-1}\right)^{T}$
\end_inset

;
\end_layout

\begin_layout Standard
(5) 若矩阵 
\begin_inset Formula $A$
\end_inset

 可逆, 则 
\begin_inset Formula $\left|A^{-1}\right|=|A|^{-1}$
\end_inset

.
\end_layout

\end_deeper
\begin_layout Subsection
矩阵方程
\end_layout

\begin_layout Frame
\begin_inset Argument 4
status open

\begin_layout Plain Layout
矩阵方程
\end_layout

\end_inset


\end_layout

\begin_deeper
\begin_layout Standard
对标准矩阵方程
\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\begin{aligned}AX & =B,\\
XA & =B,\\
AXB & =C,
\end{aligned}
\]

\end_inset


\end_layout

\begin_layout Standard
利用矩阵乘法的运算规律和逆矩阵的运算性质, 通过在方程两边左乘或右乘相应的矩阵的逆矩阵, 可求出其解分别为
\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\begin{aligned}X & =A^{-1}B,\\
X & =BA^{-1},\\
X & =A^{-1}CB^{-1},
\end{aligned}
\]

\end_inset


\end_layout

\begin_layout Standard
而其它形式的矩阵方程, 则可通过矩阵的有关运算性质转化为标准矩阵方程后进行求解.
\end_layout

\end_deeper
\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Frame
\begin_inset Argument 4
status open

\begin_layout Plain Layout
矩阵方程
\end_layout

\end_inset


\end_layout

\begin_deeper
\begin_layout Example
设有线性方程组:
\end_layout

\begin_layout Example
\begin_inset Formula 
\begin{equation}
\begin{cases}
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\cdots\cdots\cdots\cdots\cdots\cdots\cdots\\
a_{n1}x_{1}+a_{n2}x_{2}+\cdots+a_{nn}x_{n}=b_{n}
\end{cases}\label{eq:3.4-1}
\end{equation}

\end_inset


\end_layout

\begin_layout Example
假定这个方程组的系数矩阵为 
\begin_inset Formula $A$
\end_inset

, 则方程组可改写为 
\begin_inset Formula 
\begin{equation}
Ax=b,\label{eq:3.4-2}
\end{equation}

\end_inset

其中 
\begin_inset Formula $x=\begin{bmatrix}x_{1}\\
x_{2}\\
\vdots\\
x_{n}
\end{bmatrix}$
\end_inset

, 
\begin_inset Formula $b=\begin{bmatrix}b_{1}\\
b_{2}\\
\vdots\\
b_{n}
\end{bmatrix}$
\end_inset

.
\end_layout

\begin_layout Example
当 
\begin_inset Formula $|A|\neq0$
\end_inset

 时, 
\begin_inset Formula $A^{-1}$
\end_inset

 存在.
 用 
\begin_inset Formula $A^{-1}$
\end_inset

 左乘 (
\begin_inset CommandInset ref
LatexCommand ref
reference "eq:3.4-2"
plural "false"
caps "false"
noprefix "false"

\end_inset

) 式得 
\begin_inset Formula $A^{-1}(Ax)=A^{-1}b$
\end_inset

, 即 
\begin_inset Formula $x=A^{-1}b$
\end_inset

.
\end_layout

\end_deeper
\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Frame
\begin_inset Argument 4
status open

\begin_layout Plain Layout
矩阵方程
\end_layout

\end_inset


\end_layout

\begin_deeper
\begin_layout Example
设 
\begin_inset Formula $A,B,C$
\end_inset

 是同阶矩阵, 且 
\begin_inset Formula $A$
\end_inset

 可逆, 下列结论如果正确, 试证明之, 如果不正确, 试举反例说明之.
\end_layout

\begin_layout Example
(1) 若 
\begin_inset Formula $AB=AC$
\end_inset

, 则 
\begin_inset Formula $B=C$
\end_inset

;
\end_layout

\begin_layout Example
(2) 若 
\begin_inset Formula $AB=CB$
\end_inset

, 则 
\begin_inset Formula $A=C$
\end_inset

.
\end_layout

\begin_layout Solution*
(1) 正确.
 因为若 
\begin_inset Formula $AB=AC$
\end_inset

, 等式两边同时左乘以 
\begin_inset Formula $A^{-1}$
\end_inset

, 有
\begin_inset Formula 
\[
A^{-1}AB=A^{-1}AC\Longrightarrow EB=EC\Longrightarrow B=C.
\]

\end_inset

证毕.
\end_layout

\begin_layout Solution*
(2) 不正确.
 例如, 设
\begin_inset Formula 
\[
A=\begin{bmatrix}1 & 2\\
0 & 1
\end{bmatrix},\ B=\begin{bmatrix}1 & 1\\
1 & 1
\end{bmatrix},\ C=\begin{bmatrix}3 & 0\\
0 & 1
\end{bmatrix},
\]

\end_inset

则 
\begin_inset Formula $AB=\begin{bmatrix}1 & 2\\
0 & 1
\end{bmatrix}\begin{bmatrix}1 & 1\\
1 & 1
\end{bmatrix}=\begin{bmatrix}3 & 3\\
1 & 1
\end{bmatrix},$
\end_inset

 
\begin_inset Formula $CB=\begin{bmatrix}3 & 0\\
0 & 1
\end{bmatrix}\begin{bmatrix}1 & 1\\
1 & 1
\end{bmatrix}=\begin{bmatrix}3 & 3\\
1 & 1
\end{bmatrix}$
\end_inset

.
 显然有 
\begin_inset Formula $AB=AC$
\end_inset

, 但 
\begin_inset Formula $A\neq C$
\end_inset

.
\end_layout

\end_deeper
\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Frame
\begin_inset Argument 4
status open

\begin_layout Plain Layout
矩阵方程
\end_layout

\end_inset


\end_layout

\begin_deeper
\begin_layout Example
设 
\begin_inset Formula $A=\begin{bmatrix}1 & 2 & 3\\
2 & 2 & 1\\
3 & 4 & 3
\end{bmatrix}$
\end_inset

, 
\begin_inset Formula $B=\begin{bmatrix}2 & 1\\
5 & 3
\end{bmatrix}$
\end_inset

, 
\begin_inset Formula $C=\begin{bmatrix}1 & 3\\
2 & 0\\
3 & 1
\end{bmatrix}$
\end_inset

, 求矩阵 
\begin_inset Formula $X$
\end_inset

 使满足 
\begin_inset Formula $AXB=C$
\end_inset

.
\end_layout

\begin_layout Solution*
由于 
\begin_inset Formula $|A|=\begin{vmatrix}1 & 2 & 3\\
2 & 2 & 1\\
3 & 4 & 3
\end{vmatrix}=2\neq0$
\end_inset

, 
\begin_inset Formula $|B|=\begin{vmatrix}2 & 1\\
5 & 3
\end{vmatrix}=1\neq0$
\end_inset

, 所以 
\begin_inset Formula $A^{-1},B^{-1}$
\end_inset

 都存在.
 且 
\begin_inset Formula $A^{-1}=\begin{bmatrix}1 & 3 & -2\\
-3/2 & -3 & 5/2\\
1 & 1 & -1
\end{bmatrix}$
\end_inset

.
 同理 
\begin_inset Formula $B^{-1}=\begin{bmatrix}3 & -1\\
-5 & 2
\end{bmatrix}$
\end_inset

,
\end_layout

\begin_layout Solution*
又由 
\begin_inset Formula $AXB=C\Longrightarrow A^{-1}AXBB^{-1}=A^{-1}CB^{-1}$
\end_inset

.
\end_layout

\begin_layout Solution*
即 
\begin_inset Formula $X=A^{-1}CB^{-1}=\begin{bmatrix}1 & 3 & -2\\
-3/2 & -3 & 5/2\\
1 & 1 & -1
\end{bmatrix}\begin{bmatrix}1 & 3\\
2 & 0\\
3 & 1
\end{bmatrix}\begin{bmatrix}3 & -1\\
-5 & 2
\end{bmatrix}=\begin{bmatrix}-2 & 1\\
10 & -4\\
-10 & 4
\end{bmatrix}$
\end_inset

.
\end_layout

\end_deeper
\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Frame
\begin_inset Argument 3
status open

\begin_layout Plain Layout
allowframebreaks
\end_layout

\end_inset


\begin_inset Argument 4
status open

\begin_layout Plain Layout
矩阵方程
\end_layout

\end_inset


\end_layout

\begin_deeper
\begin_layout Example
设矩阵 
\begin_inset Formula $A,B$
\end_inset

 满足 
\begin_inset Formula $A^{*}BA=2BA-8E$
\end_inset

, 其中 
\begin_inset Formula $A=\begin{bmatrix}1\\
 & -2\\
 &  & 1
\end{bmatrix}$
\end_inset

, 
\begin_inset Formula $A^{*}$
\end_inset

 为 
\begin_inset Formula $A$
\end_inset

 的的伴随矩阵, 
\begin_inset Formula $E$
\end_inset

为单位矩阵, 求矩阵 
\begin_inset Formula $B$
\end_inset

.
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Solution*
由于 
\begin_inset Formula $|A|=-2\neq0$
\end_inset

, 故 
\begin_inset Formula $A$
\end_inset

 可逆, 从而 
\begin_inset Formula $A^{*}=|A|\cdot A^{-1}=-2A^{-1}$
\end_inset

.
 又 
\begin_inset Formula 
\[
A^{*}BA=2BA-8E\Longleftrightarrow A^{*}BA-2BA=-8E\Longleftrightarrow\left(A^{*}-2E\right)BA=-8E,
\]

\end_inset

其中, 
\begin_inset Formula $A^{*}-2E=-2A^{-1}-2E=-2\left(\begin{bmatrix}1\\
 & -1/2\\
 &  & 1
\end{bmatrix}+\begin{bmatrix}1\\
 & 1\\
 &  & 1
\end{bmatrix}\right)=\begin{bmatrix}-4\\
 & -1\\
 &  & -4
\end{bmatrix}$
\end_inset

,
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Solution*
显然可逆, 因此得到
\begin_inset Formula 
\[
\begin{aligned}B & =\left(A^{*}-2E\right)^{-1}(-8E)A^{-1}=-8\left(A^{*}-2E\right)^{-1}A^{-1}\\
 & =(-8)\begin{bmatrix}-1/4\\
 & -1\\
 &  & -1/4
\end{bmatrix}\begin{bmatrix}1\\
 & -1/2\\
 &  & 1
\end{bmatrix}=\begin{bmatrix}2\\
 & -4\\
 &  & 2
\end{bmatrix}.
\end{aligned}
\]

\end_inset


\end_layout

\begin_layout Remark*
当对角矩阵 
\begin_inset Formula $A=\mathrm{diag}\left(a_{1},a_{2},\cdots,a_{n}\right)$
\end_inset

 可逆时, 其逆矩阵 
\begin_inset Formula $A^{-1}=\mathrm{diag}\left(\frac{1}{a_{1}},\frac{1}{a_{2}},\cdots,\frac{1}{a_{n}}\right)$
\end_inset

.
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Remark*
在矩阵运算中, 提取因式中 
\begin_inset Formula $A^{*}-2E$
\end_inset

 不能忘记后面一项中的单位阵 
\begin_inset Formula $E$
\end_inset

, 尽管有时默认 
\begin_inset Formula $A^{*}-2$
\end_inset

 意味着 
\begin_inset Formula $A^{*}-2E$
\end_inset

, 但不能与 
\begin_inset Formula $A^{*}-\begin{bmatrix}2 & 2 & 2\\
2 & 2 & 2\\
2 & 2 & 2
\end{bmatrix}$
\end_inset

 混淆.
\end_layout

\end_deeper
\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Frame
\begin_inset Argument 4
status open

\begin_layout Plain Layout
矩阵方程
\end_layout

\end_inset


\end_layout

\begin_deeper
\begin_layout Example
设三阶矩阵 
\begin_inset Formula $A,B$
\end_inset

 满足关系: 
\begin_inset Formula $A^{-1}BA=6A+BA$
\end_inset

, 且
\end_layout

\begin_layout Example
\begin_inset Formula 
\[
A=\begin{bmatrix}1/2 & 0 & 0\\
0 & 1/4 & 0\\
0 & 0 & 1/7
\end{bmatrix},
\]

\end_inset


\end_layout

\begin_layout Example
求 
\begin_inset Formula $B$
\end_inset

.
\end_layout

\begin_layout Solution*
由 
\begin_inset Formula $A^{-1}BA-BA=6A\Longrightarrow\left(A^{-1}-E\right)BA=6A\Longrightarrow\left(A^{-1}-E\right)B=6E$
\end_inset

, 所以
\begin_inset Formula 
\begin{align*}
B & =6\left(A^{-1}-E\right)^{-1}=6\left(\begin{bmatrix}2 & 0 & 0\\
0 & 4 & 0\\
0 & 0 & 7
\end{bmatrix}-\begin{bmatrix}1 & 0 & 0\\
0 & 1 & 0\\
0 & 0 & 1
\end{bmatrix}\right)^{-1}=6\begin{bmatrix}1 & 0 & 0\\
0 & 3 & 0\\
0 & 0 & 6
\end{bmatrix}^{-1}\\
 & =6\begin{bmatrix}1 & 0 & 0\\
0 & 1/3 & 0\\
0 & 0 & 1/6
\end{bmatrix}=\begin{bmatrix}6 & 0 & 0\\
0 & 2 & 0\\
0 & 0 & 1
\end{bmatrix}.
\end{align*}

\end_inset


\end_layout

\end_deeper
\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Frame
\begin_inset Argument 4
status open

\begin_layout Plain Layout
矩阵的幂
\end_layout

\end_inset


\end_layout

\begin_deeper
\begin_layout Example
设 
\begin_inset Formula $P=\begin{bmatrix}1 & 2\\
1 & 4
\end{bmatrix}$
\end_inset

, 
\begin_inset Formula $\Lambda=\begin{bmatrix}1 & 0\\
0 & 2
\end{bmatrix}$
\end_inset

, 
\begin_inset Formula $AP=P\Lambda$
\end_inset

, 求 
\begin_inset Formula $A^{n}$
\end_inset

.
\end_layout

\begin_layout Solution*
\begin_inset Formula $|P|=2$
\end_inset

, 
\begin_inset Formula $P^{-1}=\frac{1}{2}\begin{bmatrix}4 & -2\\
-1 & 1
\end{bmatrix}$
\end_inset

.
\begin_inset Formula 
\[
A=P\Lambda P^{-1},\ A^{2}=P\Lambda P^{-1}P\Lambda P^{-1}=P\Lambda^{2}P^{-1},\ \cdots,\!A^{n}=P\Lambda^{n}P^{-1},
\]

\end_inset

而 
\begin_inset Formula $\Lambda^{2}=\begin{bmatrix}1 & 0\\
0 & 2
\end{bmatrix}\begin{bmatrix}1 & 0\\
0 & 2
\end{bmatrix}=\begin{bmatrix}1 & 0\\
0 & 2^{2}
\end{bmatrix}$
\end_inset

, 
\begin_inset Formula $\cdots$
\end_inset

, 
\begin_inset Formula $\Lambda^{n}=\begin{bmatrix}1 & 0\\
0 & 2^{n}
\end{bmatrix}$
\end_inset

, 故
\begin_inset Formula 
\begin{align*}
A^{n} & =\begin{bmatrix}1 & 2\\
1 & 4
\end{bmatrix}\begin{bmatrix}1 & 0\\
0 & 2^{n}
\end{bmatrix}\frac{1}{2}\begin{bmatrix}4 & -2\\
-1 & 1
\end{bmatrix}=\frac{1}{2}\begin{bmatrix}1 & 2^{n+1}\\
1 & 2^{n+2}
\end{bmatrix}\begin{bmatrix}4 & -2\\
-1 & 1
\end{bmatrix}\\
 & =\frac{1}{2}\begin{bmatrix}4-2^{n+1} & 2^{n+1}-2\\
4-2^{n+2} & 2^{n+2}-2
\end{bmatrix}=\begin{bmatrix}2-2^{n} & 2^{n}-1\\
2-2^{n+1} & 2^{n+1}-1
\end{bmatrix}.
\end{align*}

\end_inset


\end_layout

\end_deeper
\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Frame
\begin_inset Argument 4
status open

\begin_layout Plain Layout
矩阵的逆
\end_layout

\end_inset


\end_layout

\begin_deeper
\begin_layout Example
设方阵 
\begin_inset Formula $A$
\end_inset

 满足方程 
\begin_inset Formula $aA^{2}+bA+cE=O$
\end_inset

, 证明 
\begin_inset Formula $A$
\end_inset

 为可逆矩阵, 并求 
\begin_inset Formula $A^{-1}$
\end_inset

, (
\begin_inset Formula $a,b,c$
\end_inset

 为常数, 
\begin_inset Formula $c\neq0)$
\end_inset

.
\end_layout

\begin_layout Proof
由
\begin_inset Formula 
\[
aA^{2}+bA+cE=O\Longrightarrow aA^{2}+bA=-cE\text{, }
\]

\end_inset

又由于
\begin_inset Formula $c\neq0$
\end_inset

, 所以 
\begin_inset Formula $-\frac{a}{c}A^{2}-\frac{b}{c}A=E\Longrightarrow\left(-\frac{a}{c}A-\frac{b}{c}E\right)A=E$
\end_inset

, 由推论 
\begin_inset CommandInset ref
LatexCommand ref
reference "cor:3.2-1"
plural "false"
caps "false"
noprefix "false"

\end_inset

 知, 
\begin_inset Formula $A$
\end_inset

 可逆, 且 
\begin_inset Formula $A^{-1}=-\frac{a}{c}A-\frac{b}{c}E$
\end_inset

.
\end_layout

\end_deeper
\begin_layout Subsection
作业
\end_layout

\begin_layout Frame
\begin_inset Argument 3
status open

\begin_layout Plain Layout
allowframebreaks
\end_layout

\end_inset


\begin_inset Argument 4
status open

\begin_layout Plain Layout
作业
\end_layout

\end_inset


\end_layout

\begin_deeper
\begin_layout Problem
求方阵 
\begin_inset Formula $A=\begin{bmatrix}1 & 2 & 3\\
2 & 2 & 1\\
3 & 4 & 3
\end{bmatrix}$
\end_inset

 的逆矩阵.
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Problem
设 
\begin_inset Formula $A,B,C$
\end_inset

 是同阶矩阵, 且 
\begin_inset Formula $A$
\end_inset

 可逆, 下列结论如果正确, 试证明之, 如果不正确, 试举反例说明之.
\end_layout

\begin_layout Problem
(1) 若 
\begin_inset Formula $AB=O$
\end_inset

, 则 
\begin_inset Formula $B=O$
\end_inset

;
\end_layout

\begin_layout Problem
(2) 若 
\begin_inset Formula $BC=O$
\end_inset

, 则 
\begin_inset Formula $B=O$
\end_inset

.
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Problem
设 
\begin_inset Formula $A,B$
\end_inset

 是 
\begin_inset Formula $n$
\end_inset

 阶方阵, 满足 
\begin_inset Formula $A,B$
\end_inset

 和 
\begin_inset Formula $AB-E$
\end_inset

 都可逆, 证明: 
\end_layout

\begin_layout Problem
(1).
 
\begin_inset Formula $A-B^{-1}$
\end_inset

 可逆, 并求其逆阵; 
\end_layout

\begin_layout Problem
(2).
 
\begin_inset Formula $\left(A-B^{-1}\right)^{-1}-A^{-1}$
\end_inset

 也可逆, 并求其逆阵.
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Problem
设 
\begin_inset Formula $n$
\end_inset

 阶方阵 
\begin_inset Formula $A$
\end_inset

 适合 
\begin_inset Formula 
\[
a_{0}A^{m}+a_{1}A^{m-1}+\cdots+a_{m-1}A+a_{m}E_{n}=0,
\]

\end_inset

其中 
\begin_inset Formula $m\ge1$
\end_inset

, 
\begin_inset Formula $a_{0}a_{m}\neq0$
\end_inset

.
 证明: 方阵 
\begin_inset Formula $A$
\end_inset

 可逆, 并求 
\begin_inset Formula $A^{-1}$
\end_inset

.
 
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Problem
设 
\begin_inset Formula $A$
\end_inset

 为 
\begin_inset Formula $n$
\end_inset

 阶方阵, 如果存在正整数 
\begin_inset Formula $k$
\end_inset

, 使得 
\begin_inset Formula $A^{k}=O$
\end_inset

, 则称 
\begin_inset Formula $A$
\end_inset

 是幂零阵.
 此时, 使得 
\begin_inset Formula $A^{k}=O$
\end_inset

 成立的最小正整数称为方阵 
\begin_inset Formula $A$
\end_inset

 的幂零指数.
 
\end_layout

\begin_layout Problem
设 
\begin_inset Formula $A$
\end_inset

 为幂零矩阵, 且幂零指数是 
\begin_inset Formula $k$
\end_inset

, 证明: 
\begin_inset Formula $E-A$
\end_inset

 可逆, 并求 
\begin_inset Formula $(E-A)^{-1}$
\end_inset

.
 
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Problem
设 
\begin_inset Formula $A$
\end_inset

 是 
\begin_inset Formula $n$
\end_inset

 阶方阵.
 
\end_layout

\begin_layout Problem
(1).
 如果 
\begin_inset Formula $A$
\end_inset

 满足 
\begin_inset Formula $A^{3}+3A^{2}+3A=O$
\end_inset

, 证明: 
\begin_inset Formula $A+2E$
\end_inset

 可逆, 并求 
\begin_inset Formula $(A+2E)^{-1}$
\end_inset

; 
\end_layout

\begin_layout Problem
(2).
 如果 
\begin_inset Formula $A$
\end_inset

 满足 
\begin_inset Formula $A^{3}+E=O$
\end_inset

, 证明: 
\begin_inset Formula $A^{2}+E$
\end_inset

 可逆, 并求 
\begin_inset Formula $\left(A^{2}+E\right)^{-1}$
\end_inset

.
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Problem
求解矩阵方程 
\begin_inset Formula $\begin{bmatrix}1 & -5\\
-1 & 4
\end{bmatrix}X=\begin{bmatrix}3 & 2\\
1 & 4
\end{bmatrix}$
\end_inset

.
\end_layout

\end_deeper
\begin_layout Frame

\end_layout

\end_body
\end_document