习题

$1$.在下述映射中，哪些是线性映射：

$$\begin{align}
a) & [x_1 ,x_2 ,\cdots ,x_n ]\mapsto [x_n ,\cdots ,x_2 ,x_1 ]; \\
b) & [x_1 ,x_2 ,\cdots ,x_n ]\mapsto [x_1 ,x_2^2 ,\cdots ,x_n^n ]; \\
c) & [x_1 ,x_2 ,\cdots ,x_n ]\mapsto [x_1 ,x_1 +x_2 ,\cdots ,x_1 +x_2 +\cdots +x_n ].
\end{align}$$

解：任取$x=[x_1 ,x_2 ,\cdots ,x_n ] ,y=[y_1 ,y_2 ,\cdots ,y_n ] ,\lambda \in \mathbb{R}$，

$a)\;$.根据定义，

$$\varphi (x+y) =\varphi ([x_1 +y_1 ,x_2 +y_2 ,\cdots ,x_n +y_n ])=[x_n +y_n ,\cdots ,x_2 +y_2 ,x_1 +y_1 ],$$

$$\varphi (x) +\varphi (y) =[x_n ,\cdots ,x_2 ,x_1 ] +[y_n ,\cdots ,y_2 ,y_1 ] =[x_n +y_n ,\cdots ,x_2 +y_2 ,x_1 +y_1 ],$$

$$\varphi (\lambda x)=[\lambda x_n \cdots ,\lambda x_2 ,\lambda x_1 ]=\lambda \varphi (x),$$

有$\varphi (x+y)=\varphi (x) +\varphi (y)$且$\varphi (\lambda x)=\lambda \varphi (x)$，满足线性映射的定义.

$b)\;$.因为

$$\begin{align}
\varphi (\lambda x) & =\varphi ([\lambda x_1 ,\lambda x_2 ,\cdots ,\lambda x_n ]) \\
& = [\lambda x_1 ,\lambda ^2 x_2^2 ,\cdots ,\lambda ^n x_n^n ] \\
& =\lambda [x_1 ,\lambda x_2^2 ,\cdots ,\lambda ^{n-1} x_n^n ] \\
& \neq \lambda \varphi (x)
\end{align}$$

所以，不满足线性映射的定义.

$c)\;$.根据定义，

$$\begin{align}
\varphi (x+y) & =\varphi ([x_1 +y_1 ,x_2 +y_2 ,\cdots ,x_n +y_n ]) \\
& = [x_1 +y_1 ,x_1 +y_1 +x_2 +y_2 ,\cdots ,x_1 +y_1 +x_2 +y_2 +\cdots +x_n +y_n ] \\
& = [x_1 ,x_1 +x_2 ,\cdots ,x_1 +x_2 +\cdots +x_n ] +[y_1 ,y_1 +y_2 ,\cdots ,y_1 +y_2 +\cdots +y_n ] \\
& =\varphi (x) +\varphi (y),
\end{align}$$

$$\begin{align}
\varphi (\lambda x) & =\varphi ([\lambda x_1 ,\lambda x_2 ,\cdots ,\lambda x_n ]) \\
& = [\lambda x_1 ,\lambda x_1 +\lambda x_2 ,\cdots ,\lambda x_1 +\lambda x_2 +\lambda x_n ] \\
& =\lambda [x_1 ,x_1 +x_2 ,\cdots ,x_1 +x_2 +x_n ] \\
& =\lambda \varphi (x),
\end{align}$$

有$\varphi (x+y)=\varphi (x) +\varphi (y)$且$\varphi (\lambda x)=\lambda \varphi (x)$，满足线性映射的定义.

$2$.证明

$$\begin{pmatrix} 1 & a & c \\ 0 & 1 & b \\ 0 & 0 & 1 \end{pmatrix} ^m =\begin{pmatrix} 1 & ma & \dfrac{m(m-1)}{2}ab+mc \\ 0 & 1 & mb \\ 0 & 0 & 1 \end{pmatrix} .$$

求矩阵

$$\begin{pmatrix} 1 & a & c \\ 0 & 1 & b \\ 0 & 0 & 1 \end{pmatrix}$$

的逆矩阵.

证明：$(1)\;$.设

$$A=\begin{pmatrix} 1 & a & c \\ 0 & 1 & b \\ 0 & 0 & 1 \end{pmatrix} $$

则

$$A=\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} +\begin{pmatrix} 0 & a & c \\ 0 & 0 & b \\ 0 & 0 & 0 \end{pmatrix} =I+B,$$

其中$$B=\begin{pmatrix} 0 & a & c \\ 0 & 0 & b \\ 0 & 0 & 0 \end{pmatrix} $$

显然

其中$$B^3=\begin{pmatrix} 0 & a & c \\ 0 & 0 & b \\ 0 & 0 & 0 \end{pmatrix} \begin{pmatrix} 0 & a & c \\ 0 & 0 & b \\ 0 & 0 & 0 \end{pmatrix} \begin{pmatrix} 0 & a & c \\ 0 & 0 & b \\ 0 & 0 & 0 \end{pmatrix} =0 $$

因为$I$与$B$可交换，所以由二项式定理得

$$\begin{align}
A^m & =(I+B)^m \\
& = I^m +{m \choose 1} I^{m-1} B+{m \choose 2}I^{m-2} B^2\\
& = I+mIB+\dfrac{m(m-1)}{2} IB^2 \\
& = I+mB+\dfrac{m(m-1)}{2} B^2 \\
& = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}+\begin{pmatrix} 0 & ma & mc \\ 0 & 0 & mb \\ 0 & 0 & 0 \end{pmatrix} +\begin{pmatrix} 0 & 0 & \dfrac{m(m-1)}{2} ab \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix} \\
& = \begin{pmatrix} 1 & ma & \dfrac{m(m-1)}{2}ab+mc \\ 0 & 1 & mb \\ 0 & 0 & 1 \end{pmatrix}
\end{align}$$

$(2)\;$.把$m=-1$代入$(1)$的结果，有

$$\begin{pmatrix} 1 & a & c \\ 0 & 1 & b \\ 0 & 0 & 1 \end{pmatrix} ^1 =\begin{pmatrix} 1 & -a & ab-c \\ 0 & 1 & -b \\ 0 & 0 & 1 \end{pmatrix} .$$

$3$.验证$\begin{pmatrix} 0 & -1 \\ 1 & -1 \end{pmatrix} ^3 =E$.

解：

$$\begin{align}
\begin{pmatrix} 0 & -1 \\ 1 & -1 \end{pmatrix} ^3 & =\begin{pmatrix} 0 & -1 \\ 1 & -1 \end{pmatrix} \begin{pmatrix} 0 & -1 \\ 1 & -1 \end{pmatrix} \begin{pmatrix} 0 & -1 \\ 1 & -1 \end{pmatrix}\\
& = \begin{pmatrix} -1 & 1 \\ -1 & 0 \end{pmatrix} \begin{pmatrix} 0 & -1 \\ 1 & -1 \end{pmatrix}\\
& = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \\
& = E,
\end{align}$$

得证.

$4$.马尔可夫(或随机)矩阵在应用中十分重要：

$$P=(p_{ij}),p_{ij} \geq 0,\displaystyle \sum_{j=1}^n p_{ij} =1 ,i=1,2,\cdots ,n.$$

由马尔可夫矩阵确定的线性变换$\varphi _P$通常作用于概率行向量：

$$X=(x_1 ,\cdots ,x_n ),x_i \geq 0,\displaystyle \sum_{i=1}^n x_i =1.$$

从下述论断可见，这些来自于自然科学问题的定义是协调的论断，即便对$n=2$，也需要证明.

$a)$矩阵$P\in M_n (\mathbb{R} )$是马尔可夫的，当且仅当对任意概率向量$X$，$XP$仍然是概率向量(此处$XP=\varphi _P (X)$).

$b)$如果$P$是正的马尔可夫矩阵(即$\forall i,j,\; p_{ij} > 0$)，那么任意概率向量$X$对应到正的概率向量$XP$(所有的分量严格大于$0$).

$c)$如果$P$和$Q$都是马尔可夫矩阵，那么矩阵$PQ$也是马尔可夫矩阵.特别地，马尔可夫矩阵的任意次方幂$P^k$是马尔可夫矩阵.

证明：$(a)\;(1)$.若矩阵$P\in M_n (\mathbb{R} )$是马尔可夫的，则对于任意的概率向量$X$来说，有

$$\begin{align}
XP & = ( x_1 ,\cdots ,x_n ) \begin{pmatrix} p_{11} & p_{12} & \cdots & p_{1n} \\ p_{21} & p_{22} & \cdots & p_{2n} \\ \vdots & \vdots & & \vdots \\ p_{n1} & p_{n2} & \cdots & p_{nn} \end{pmatrix} \\
& = (x_1 p_{11} +x_2 p_{21} +\cdots +x_n p_{n1} ,\cdots ,x_1 p_{1n} +x_2 p_{2n} +\cdots +x_n p_{nn} )\\
\end{align}$$

把$XP$的所有分量全部加起来，得

$$\begin{align}
& (x_1 p_{11} +x_2 p_{21} +\cdots +x_n p_{n1} )+\cdots +(x_1 p_{1n} +x_2 p_{2n} +\cdots +x_n p_{nn} ) \\
= & x_1 (p_{11} +p_{12} +\cdots +p_{1n} )+\cdots +x_n (p_{n1} +p_{n2} +\cdots +p_{nn} ) \\
& (\because \displaystyle \sum_{j=1}^n p_{ij} =1 ,i=1,2,\cdots ,n ) \\
= & x_1 \cdot 1 +\cdots +x_n \cdot 1 \\
= & 1
\end{align}$$

故$XP$为概率向量.

$(2)$.若对于任意的概率向量$X$来说，$XP$为概率向量.由$(1)$中对$XP$的展开式可知，对应于$x_1 ,\cdots ,x_n$的系数$p_{11} +p_{12} +\cdots +p_{1n} ,\cdots ,p_{n1} +p_{n2} +\cdots +p_{nn}$应该分别为$1,\cdots ,1$.根据定义，得矩阵$P$是马尔可夫矩阵.

$(b)\;$.如果$P$是正的马尔可夫矩阵(即$\forall i,j,\; p_{ij} > 0$)，则对于任意的概率向量$X$来说，有

$$\begin{align}
XP & = ( x_1 ,\cdots ,x_n ) \begin{pmatrix} p_{11} & p_{12} & \cdots & p_{1n} \\ p_{21} & p_{22} & \cdots & p_{2n} \\ \vdots & \vdots & & \vdots \\ p_{n1} & p_{n2} & \cdots & p_{nn} \end{pmatrix} \\
& = (x_1 p_{11} +x_2 p_{21} +\cdots +x_n p_{n1} ,\cdots ,x_1 p_{1n} +x_2 p_{2n} +\cdots +x_n p_{nn} )\\
\end{align}$$

把$XP$的所有分量全部加起来，得

$$\begin{align}
& (x_1 p_{11} +x_2 p_{21} +\cdots +x_n p_{n1} )+\cdots +(x_1 p_{1n} +x_2 p_{2n} +\cdots +x_n p_{nn} ) \\
= & x_1 (p_{11} +p_{12} +\cdots +p_{1n} )+\cdots +x_n (p_{n1} +p_{n2} +\cdots +p_{nn} ) \\
& (\because \displaystyle \sum_{j=1}^n p_{ij} =1 ,i=1,2,\cdots ,n ) \\
= & x_1 \cdot 1 +\cdots +x_n \cdot 1 \\
= & 1
\end{align}$$

由于$\forall i,j,\; p_{ij} > 0$，且$x_i \geq 0,\displaystyle \sum_{i=1}^n x_i =1$，故$XP$的所有分量$x_1 p_{1i} +\cdots +x_n p_{ni}$中，必有某个$x_k \neq 0$，且对应于$p_{ki} > 0$，使得分量$x_1 p_{1i} +\cdots +x_n p_{ni} > 0$，故所有的分量必然全大于零.

因此，任意概率向量$X$对应到正的概率向量$XP$(所有的分量严格大于$0$).

$(c)\;(1)$.如果$P$和$Q$都是马尔可夫矩阵，设

$$P=(p_{ij}),p_{ij} \geq 0,\displaystyle \sum_{j=1}^n p_{ij} =1 ,i=1,2,\cdots ,n,$$

$$Q=(q_{ij}),q_{ij} \geq 0,\displaystyle \sum_{j=1}^n q_{ij} =1 ,i=1,2,\cdots ,n.$$

记$PQ=C=(c_{ij} )$，考察$C$之第$i$行元素之和，则

$$\displaystyle \sum_{j=1}^n c_{ij} =\sum_{j=1}^n \sum_{k=1}^n a_{ik} b_{kj} =\sum_{k=1}^n a_{ik} \left( \sum_{j=1}^n b_{kj} \right) =\sum_{k=1}^n a_{ik} \cdot 1=1 (i=1,2,\cdots ,n)$$

简言之，根据定义，矩阵$PQ$也是马尔可夫矩阵.

$(2)\;$.由$(1)$并用数学归纳法可证.

明显，当$k=2$时，由$(1)$可知论断成立.

设当$k=m$时也成立，下面证明当$k=m+1$时论断也成立.

$$P^{m+1}=P^m \cdot P ,$$

因为$P^m$已经假设是马尔可夫矩阵，$P$本身已经是马尔可夫矩阵，根据$(1)$的结论，两个马尔可夫矩阵的乘积必然也是一个马尔可夫矩阵，故当$k=m+1$时，$P^{m+1}$也是一个马尔可夫矩阵.

由数学归纳法知，马尔可夫矩阵的任意次方幂$P^k$是马尔可夫矩阵.

$5$.若

$$H=\begin{pmatrix} 1 & 1 & 1 & 1 \\ 1 & -1 & 1 & -1 \\ 1 & 1 & -1 & -1 \\ 1 & -1 & -1 & 1 \end{pmatrix} ,$$

求$\sideset{^t}{}H\cdot H$.

解：$$\begin{align}
\sideset{^t}{}H\cdot H & =\begin{pmatrix} 1 & 1 & 1 & 1 \\ 1 & -1 & 1 & -1 \\ 1 & 1 & -1 & -1 \\ 1 & -1 & -1 & 1 \end{pmatrix} \cdot \begin{pmatrix} 1 & 1 & 1 & 1 \\ 1 & -1 & 1 & -1 \\ 1 & 1 & -1 & -1 \\ 1 & -1 & -1 & 1 \end{pmatrix} \\
& = \begin{pmatrix} 4 & 0 & 0 & 0 \\ 0 & 4 & 0 & 0 \\ 0 & 0 & 4 & 0 \\ 0 & 0 & 0 & 4 \end{pmatrix}
\end{align}$$

$6$.由$S_n$中的$n$阶循环(见第$1$章$\S 8$)确定的置换矩阵(行单位阵$E_n$)为

$$P=\begin{pmatrix} 0 & 0 & \cdots & 0 & 1 \\ 1 & 0 & \cdots & 0 & 0 \\ \cdots & \cdots & \cdots & \cdots & \cdots \\ 0 & 0 & \cdots & 0 & 0 \\ 0 & 0 & \cdots & 1 & 0\end{pmatrix} ,$$

验证$P^n =E$.

解：通过计算，矩阵$P$每乘一个$P$，其结果是矩阵$P$里$1$的位置相对往左下角移了一位：

$$P^2=\begin{pmatrix} 0 & 0 & \cdots & 0 & 1 & 0 \\ 0 & 0 & \cdots & 0 & 0 & 1 \\ 1 & 0 & \cdots & 0 & 0 & 0 \\ 0 & 1 & \cdots & 0 & 0 & 0 \\ \cdots & \cdots & \cdots & \cdots & \cdots & \cdots \\ 0 & 0 & \cdots & 0 & 0 & 0 \\ 0 & 0 & \cdots & 1 & 0 & 0 \end{pmatrix} ,$$

$$P^3=\begin{pmatrix} 0 & 0 & \cdots & 1 & 0 & 0 \\ 0 & 0 & \cdots & 0 & 1 & 0 \\ 0 & 0 & \cdots & 0 & 0 & 1 \\ 1 & 0 & \cdots & 0 & 0 & 0 \\ \cdots & \cdots & \cdots & \cdots & \cdots & \cdots \\ 0 & 0 & \cdots & 0 & 0 & 0 \\ 0 & 0 & \cdots & 0 & 0 & 0 \end{pmatrix} ,$$

$$\cdots \cdots \cdots $$

$$P^n =E.$$

$7$.对于任意两个$m\times n$矩阵$A$和$B$，证明

$$\text{rank}(A+B) \leq \text{rank}A +\text{rank}B.$$

证明：令$A,B$的列向量组分别为

$$\alpha _1 ,\alpha _2 ,\cdots ,\alpha _n ;\quad \beta _1 ,\beta _2 ,\cdots ,\beta _n .$$

则$A+B$的列向量组为

$$\alpha _1 +\beta _1 ,\alpha _2 +\beta _2 ,\cdots ,\alpha _n +\beta _n . $$

设$\alpha _{i_1 } ,\alpha _{i_2 } ,\cdots ,\alpha _{i_r }$是向量组$\alpha _1 ,\alpha _2 ,\cdots ,\alpha _n$的一个极大线性无关组；$\beta _{j_1 } ,\beta _{j_2 } ,\cdots ,\beta _{j_t }$是向量组$\beta _1 ,\beta _2 ,\cdots ,\beta _n$的一个极大线性无关组.则$\alpha _1 +\beta _1 ,\alpha _2 +\beta _2 ,\cdots ,\alpha _n +\beta _n$可以由向量组$\alpha _{i_1 } ,\alpha _{i_2 } ,\cdots ,\alpha _{i_r } ,\beta _{j_1 } ,\beta _{j_2 } ,\cdots ,\beta _{j_t }$线性表出.因此

$$\begin{align}
& \text{rank} \lbrace \alpha _1 +\beta _1 ,\alpha _2 +\beta _2 ,\cdots ,\alpha _n +\beta _n \rbrace \\
\leq & \text{rank} \lbrace \alpha _{i_1 } ,\alpha _{i_2 } ,\cdots ,\alpha _{i_r } ,\beta _{j_1 } ,\beta _{j_2 } ,\cdots ,\beta _{j_t } \rbrace \\
\leq & r+t.
\end{align}$$

于是

$$\text{rank}(A+B) \leq \text{rank}A +\text{rank}B.$$

$8$.对于任意的$m\times s$矩阵$A$和$s\times n$矩阵$B$，证明

$$\text{rank}A +\text{rank}B -s\leq \text{rank}AB.$$

证明： 对于分块阵$\begin{pmatrix} E_s & B \\ A & 0_{m\times n } \end{pmatrix}$来说，有

$$\begin{pmatrix} E_s & 0_{s\times n} \\ -A & E_{m\times n}\end{pmatrix} \begin{pmatrix} E_s & B \\ A & 0_{m\times n } \end{pmatrix} \begin{pmatrix} E_s & -B \\ 0_{m\times s} & E_{m\times n} \end{pmatrix} =\begin{pmatrix} E_s & 0_{s\times n} \\ 0_{m\times s} & -AB \end{pmatrix} $$

因此，有

$$\text{rank} \begin{pmatrix} E_s & B \\ A & 0_{m\times n } \end{pmatrix} =\text{rank} \begin{pmatrix} E_s & 0_{s\times n} \\ 0_{m\times s} & -AB \end{pmatrix} =\text{rank} E_s +\text{rank} (-AB) =s+\text{rank} AB .$$

设$\text{rank} A=r ,\text{rank} B=t$，则$A$有一个$r$级子矩阵$A_1 $，使得$\mid A_1 \mid \neq 0$；$B$有一个$t$级子矩阵$B_1 $，使得$\mid B_1 \mid \neq 0$.从而$\begin{pmatrix} E_s & B \\ A & 0 \end{pmatrix}$有一个$r+t$阶子式：

$$\begin{vmatrix} E_1 & B_1 \\ A_1 & 0 \end{vmatrix} =\mid A_1 \mid \mid B_1 \mid \neq 0.$$

于是，$\text{rank} \begin{pmatrix} E_s & B \\ A & 0 \end{pmatrix} \geq \text{rank} A+\text{rank} B$

综上所述，有$s+\text{rank} AB \geq \text{rank} A+\text{rank} B$，即对于任意的$m\times s$矩阵$A$和$s\times n$矩阵$B$，

$$\text{rank}A +\text{rank}B -s\leq \text{rank}AB.$$

$9$.设$A,B,C$是$n$阶方阵，若$ABC=0$，则

$$\text{rank}A +\text{rank}B +\text{rank}C \leq 2n.$$

证明：因为$ABC=0$，$C$的各个列向量都是齐次方程组$ABX=0$的解，故能由它的基础解系线性表出，于是$\text{rank} C \leq $基础解系的秩$n-\text{rank} AB$.即有$\text{rank} AB\leq n-\text{rank} C$.

利用$8$题结论，有

$$\text{rank}A +\text{rank}B -n\leq \text{rank}AB ,$$

因此，

$$\text{rank}A +\text{rank}B -n\leq \text{rank}AB \leq n-\text{rank} C,$$

即

$$\text{rank}A +\text{rank}B +\text{rank}C \leq 2n.$$

$10$.求矩阵

$$A=\begin{pmatrix} x_1 y_1 & x_1 y_2 & \cdots & x_1 y_n \\ x_2 y_1 & x_2 y_2 & \cdots & x_2 y_n \\ \cdots & \cdots & \cdots & \cdots \\ x_n y_1 & x_n y_2 & \cdots & x_n y_n \end{pmatrix} $$

的秩.

提示：$A=[x_1 ,\cdots ,x_n ] (y_1 ,\cdots ,y_n )$.

证明：设$[x_1 ,\cdots ,x_n ] =B,(y_1 ,\cdots ,y_n )=C$.则

由定理$3$：不等式$\text{rank} AB \leq \min{\lbrace \text{rank} A,\text{rank} B\rbrace }$成立，得到

$$\text{rank} A =\text{rank} BC\leq \min{\lbrace \text{rank} B,\text{rank} C\rbrace }$$

因为行向量和列向量本身秩都为$1$，故$\text{rank} A\leq 1$.

$11$.若$A=(a_{ij})$是非退化对称矩阵(即$a_{ij} =a_{ji}$)，证明$A^{-1}$也是对称矩阵.

证明：设$A$是非退化对称矩阵，则$\sideset{^t}{}A =A$.因为逆的转置等于转置的逆，故

$$\sideset{^t}{}(A^{-1}) =(\sideset{^t}{}A )^{-1} =A^{-1}.$$

根据对称的定义，知$A^{-1}$是对称矩阵.

$12$.设

$$A=\begin{pmatrix} 5 & 4 & 3 & 2 & 1 \\ 4 & 8 & 6 & 4 & 2 \\ 3 & 6 & 9 & 6 & 3 \\ 2 & 4 & 6 & 8 & 4 \\ 1 & 2 & 3 & 4 & 5 \end{pmatrix} ,F=\begin{pmatrix} 2 & 3 & 2 & 1 \\ 3 & 6 & 4 & 2 \\ 4 & 8 & 6 & 3 \\ 2 & 4 & 3 & 2 \end{pmatrix} ,$$

求$A^{-1}$和$F^{-1}$.

解：$(1)\;$

$$\begin{align}
(A\mid E)= & \begin{pmatrix} \begin {array}{ccccc|ccccc} 5 & 4 & 3 & 2 & 1 & 1 & 0 & 0 & 0 & 0 \\ 4 & 8 & 6 & 4 & 2 & 0 & 1 & 0 & 0 & 0 \\ 3 & 6 & 9 & 6 & 3 & 0 & 0 & 1 & 0 & 0 \\ 2 & 4 & 6 & 8 & 4 & 0 & 0 & 0 & 1 & 0 \\ 1 & 2 & 3 & 4 & 5 & 0 & 0 & 0 & 0 & 1 \end{array} \end{pmatrix} \\
\overset{F_{1,5} }{\longrightarrow } & \begin{pmatrix} \begin {array}{ccccc|ccccc} 1 & 2 & 3 & 4 & 5 & 0 & 0 & 0 & 0 & 1 \\ 4 & 8 & 6 & 4 & 2 & 0 & 1 & 0 & 0 & 0 \\ 3 & 6 & 9 & 6 & 3 & 0 & 0 & 1 & 0 & 0 \\ 2 & 4 & 6 & 8 & 4 & 0 & 0 & 0 & 1 & 0 \\ 5 & 4 & 3 & 2 & 1 & 1 & 0 & 0 & 0 & 0 \end{array} \end{pmatrix} \\
\overset{F_{2,1} (-4)}{\longrightarrow } & \begin{pmatrix} \begin {array}{ccccc|ccccc} 1 & 2 & 3 & 4 & 5 & 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & -6 & -12 & -18 & 0 & 1 & 0 & 0 & -4 \\ 3 & 6 & 9 & 6 & 3 & 0 & 0 & 1 & 0 & 0 \\ 2 & 4 & 6 & 8 & 4 & 0 & 0 & 0 & 1 & 0 \\ 5 & 4 & 3 & 2 & 1 & 1 & 0 & 0 & 0 & 0 \end{array} \end{pmatrix} \\
\overset{F_{3,1} (-3)}{\longrightarrow } & \begin{pmatrix} \begin {array}{ccccc|ccccc} 1 & 2 & 3 & 4 & 5 & 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & -6 & -12 & -18 & 0 & 1 & 0 & 0 & -4 \\ 0 & 0 & 0 & -6 & -12 & 0 & 0 & 1 & 0 & -3 \\ 2 & 4 & 6 & 8 & 4 & 0 & 0 & 0 & 1 & 0 \\ 5 & 4 & 3 & 2 & 1 & 1 & 0 & 0 & 0 & 0 \end{array} \end{pmatrix} \\
\overset{F_{4,1} (-2)}{\longrightarrow } & \begin{pmatrix} \begin {array}{ccccc|ccccc} 1 & 2 & 3 & 4 & 5 & 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & -6 & -12 & -18 & 0 & 1 & 0 & 0 & -4 \\ 0 & 0 & 0 & -6 & -12 & 0 & 0 & 1 & 0 & -3 \\ 0 & 0 & 0 & 0 & -6 & 0 & 0 & 0 & 1 & -2 \\ 5 & 4 & 3 & 2 & 1 & 1 & 0 & 0 & 0 & 0 \end{array} \end{pmatrix} \\
\overset{F_{5,1} (-5)}{\longrightarrow } & \begin{pmatrix} \begin {array}{ccccc|ccccc} 1 & 2 & 3 & 4 & 5 & 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & -6 & -12 & -18 & 0 & 1 & 0 & 0 & -4 \\ 0 & 0 & 0 & -6 & -12 & 0 & 0 & 1 & 0 & -3 \\ 0 & 0 & 0 & 0 & -6 & 0 & 0 & 0 & 1 & -2 \\ 0 & -6 & -12 & -18 & -24 & 1 & 0 & 0 & 0 & -5 \end{array} \end{pmatrix} \\
\overset{F_{2,5} }{\longrightarrow } & \begin{pmatrix} \begin {array}{ccccc|ccccc} 1 & 2 & 3 & 4 & 5 & 0 & 0 & 0 & 0 & 1 \\ 0 & -6 & -12 & -18 & -24 & 1 & 0 & 0 & 0 & -5 \\ 0 & 0 & 0 & -6 & -12 & 0 & 0 & 1 & 0 & -3 \\ 0 & 0 & 0 & 0 & -6 & 0 & 0 & 0 & 1 & -2 \\ 0 & 0 & -6 & -12 & -18 & 0 & 1 & 0 & 0 & -4 \end{array} \end{pmatrix} \\
\overset{F_2 (-\dfrac{1}{6} ) }{\longrightarrow } & \begin{pmatrix} \begin {array}{ccccc|ccccc} 1 & 2 & 3 & 4 & 5 & 0 & 0 & 0 & 0 & 1 \\ 0 & 1 & 2 & 3 & 4 & -\dfrac{1}{6} & 0 & 0 & 0 & \dfrac{5}{6} \\ 0 & 0 & 0 & -6 & -12 & 0 & 0 & 1 & 0 & -3 \\ 0 & 0 & 0 & 0 & -6 & 0 & 0 & 0 & 1 & -2 \\ 0 & 0 & -6 & -12 & -18 & 0 & 1 & 0 & 0 & -4 \end{array} \end{pmatrix} \\
\overset{F_{3,5} }{\longrightarrow } & \begin{pmatrix} \begin {array}{ccccc|ccccc} 1 & 2 & 3 & 4 & 5 & 0 & 0 & 0 & 0 & 1 \\ 0 & 1 & 2 & 3 & 4 & -\dfrac{1}{6} & 0 & 0 & 0 & \dfrac{5}{6} \\ 0 & 0 & -6 & -12 & -18 & 0 & 1 & 0 & 0 & -4 \\ 0 & 0 & 0 & 0 & -6 & 0 & 0 & 0 & 1 & -2 \\ 0 & 0 & 0 & -6 & -12 & 0 & 0 & 1 & 0 & -3 \end{array} \end{pmatrix} \\
\overset{F_{4,5} }{\longrightarrow } & \begin{pmatrix} \begin {array}{ccccc|ccccc} 1 & 2 & 3 & 4 & 5 & 0 & 0 & 0 & 0 & 1 \\ 0 & 1 & 2 & 3 & 4 & -\dfrac{1}{6} & 0 & 0 & 0 & \dfrac{5}{6} \\ 0 & 0 & -6 & -12 & -18 & 0 & 1 & 0 & 0 & -4 \\ 0 & 0 & 0 & -6 & -12 & 0 & 0 & 1 & 0 & -3 \\ 0 & 0 & 0 & 0 & -6 & 0 & 0 & 0 & 1 & -2 \end{array} \end{pmatrix} \\
\overset{F_3 (-\dfrac16 ) }{\longrightarrow } & \begin{pmatrix} \begin {array}{ccccc|ccccc} 1 & 2 & 3 & 4 & 5 & 0 & 0 & 0 & 0 & 1 \\ 0 & 1 & 2 & 3 & 4 & -\dfrac{1}{6} & 0 & 0 & 0 & \dfrac{5}{6} \\ 0 & 0 & 1 & 2 & 3 & 0 & -\dfrac16 & 0 & 0 & \dfrac46 \\ 0 & 0 & 0 & -6 & -12 & 0 & 0 & 1 & 0 & -3 \\ 0 & 0 & 0 & 0 & -6 & 0 & 0 & 0 & 1 & -2 \end{array} \end{pmatrix} \\
\overset{F_4 (-\dfrac16 ) }{\longrightarrow } & \begin{pmatrix} \begin {array}{ccccc|ccccc} 1 & 2 & 3 & 4 & 5 & 0 & 0 & 0 & 0 & 1 \\ 0 & 1 & 2 & 3 & 4 & -\dfrac{1}{6} & 0 & 0 & 0 & \dfrac{5}{6} \\ 0 & 0 & 1 & 2 & 3 & 0 & -\dfrac16 & 0 & 0 & \dfrac46 \\ 0 & 0 & 0 & 1 & 2 & 0 & 0 & -\dfrac16 & 0 & \dfrac36 \\ 0 & 0 & 0 & 0 & -6 & 0 & 0 & 0 & 1 & -2 \end{array} \end{pmatrix} \\
\overset{F_5 (-\dfrac16 ) }{\longrightarrow } & \begin{pmatrix} \begin {array}{ccccc|ccccc} 1 & 2 & 3 & 4 & 5 & 0 & 0 & 0 & 0 & 1 \\ 0 & 1 & 2 & 3 & 4 & -\dfrac{1}{6} & 0 & 0 & 0 & \dfrac{5}{6} \\ 0 & 0 & 1 & 2 & 3 & 0 & -\dfrac16 & 0 & 0 & \dfrac46 \\ 0 & 0 & 0 & 1 & 2 & 0 & 0 & -\dfrac16 & 0 & \dfrac36 \\ 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & -\dfrac16 & \dfrac26 \end{array} \end{pmatrix} \\
\overset{F_{4,5} (-2) }{\longrightarrow } & \begin{pmatrix} \begin {array}{ccccc|ccccc} 1 & 2 & 3 & 4 & 5 & 0 & 0 & 0 & 0 & 1 \\ 0 & 1 & 2 & 3 & 4 & -\dfrac{1}{6} & 0 & 0 & 0 & \dfrac{5}{6} \\ 0 & 0 & 1 & 2 & 3 & 0 & -\dfrac16 & 0 & 0 & \dfrac46 \\ 0 & 0 & 0 & 1 & 0 & 0 & 0 & -\dfrac16 & \dfrac26 & -\dfrac16 \\ 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & -\dfrac16 & \dfrac26 \end{array} \end{pmatrix} \\
\overset{F_{3,5} (-3) }{\longrightarrow } & \begin{pmatrix} \begin {array}{ccccc|ccccc} 1 & 2 & 3 & 4 & 5 & 0 & 0 & 0 & 0 & 1 \\ 0 & 1 & 2 & 3 & 4 & -\dfrac{1}{6} & 0 & 0 & 0 & \dfrac{5}{6} \\ 0 & 0 & 1 & 2 & 0 & 0 & -\dfrac16 & 0 & \dfrac36 & -\dfrac26 \\ 0 & 0 & 0 & 1 & 0 & 0 & 0 & -\dfrac16 & \dfrac26 & -\dfrac16 \\ 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & -\dfrac16 & \dfrac26 \end{array} \end{pmatrix} \\
\overset{F_{3,4} (-2) }{\longrightarrow } & \begin{pmatrix} \begin {array}{ccccc|ccccc} 1 & 2 & 3 & 4 & 5 & 0 & 0 & 0 & 0 & 1 \\ 0 & 1 & 2 & 3 & 4 & -\dfrac{1}{6} & 0 & 0 & 0 & \dfrac{5}{6} \\ 0 & 0 & 1 & 0 & 0 & 0 & -\dfrac16 & \dfrac26 & -\dfrac16 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 & 0 & -\dfrac16 & \dfrac26 & -\dfrac16 \\ 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & -\dfrac16 & \dfrac26 \end{array} \end{pmatrix} \\
\overset{F_{2,5} (-4) }{\longrightarrow } & \begin{pmatrix} \begin {array}{ccccc|ccccc} 1 & 2 & 3 & 4 & 5 & 0 & 0 & 0 & 0 & 1 \\ 0 & 1 & 2 & 3 & 0 & -\dfrac{1}{6} & 0 & 0 & \dfrac46 & -\dfrac{3}{6} \\ 0 & 0 & 1 & 0 & 0 & 0 & -\dfrac16 & \dfrac26 & -\dfrac16 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 & 0 & -\dfrac16 & \dfrac26 & -\dfrac16 \\ 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & -\dfrac16 & \dfrac26 \end{array} \end{pmatrix} \\
\overset{F_{2,4} (-3) }{\longrightarrow } & \begin{pmatrix} \begin {array}{ccccc|ccccc} 1 & 2 & 3 & 4 & 5 & 0 & 0 & 0 & 0 & 1 \\ 0 & 1 & 2 & 0 & 0 & -\dfrac{1}{6} & 0 & \dfrac36 & -\dfrac26 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 & -\dfrac16 & \dfrac26 & -\dfrac16 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 & 0 & -\dfrac16 & \dfrac26 & -\dfrac16 \\ 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & -\dfrac16 & \dfrac26 \end{array} \end{pmatrix} \\
\overset{F_{2,3} (-2) }{\longrightarrow } & \begin{pmatrix} \begin {array}{ccccc|ccccc} 1 & 2 & 3 & 4 & 5 & 0 & 0 & 0 & 0 & 1 \\ 0 & 1 & 0 & 0 & 0 & -\dfrac{1}{6} & \dfrac26 & -\dfrac16 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 & -\dfrac16 & \dfrac26 & -\dfrac16 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 & 0 & -\dfrac16 & \dfrac26 & -\dfrac16 \\ 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & -\dfrac16 & \dfrac26 \end{array} \end{pmatrix} \\
\overset{F_{1,5} (-5) }{\longrightarrow } & \begin{pmatrix} \begin {array}{ccccc|ccccc} 1 & 2 & 3 & 4 & 0 & 0 & 0 & 0 & \dfrac56 & -\dfrac46 \\ 0 & 1 & 0 & 0 & 0 & -\dfrac{1}{6} & \dfrac26 & -\dfrac16 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 & -\dfrac16 & \dfrac26 & -\dfrac16 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 & 0 & -\dfrac16 & \dfrac26 & -\dfrac16 \\ 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & -\dfrac16 & \dfrac26 \end{array} \end{pmatrix} \\
\overset{F_{1,4} (-4) }{\longrightarrow } & \begin{pmatrix} \begin {array}{ccccc|ccccc} 1 & 2 & 3 & 0 & 0 & 0 & 0 & \dfrac46 & -\dfrac36 & 0 \\ 0 & 1 & 0 & 0 & 0 & -\dfrac{1}{6} & \dfrac26 & -\dfrac16 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 & -\dfrac16 & \dfrac26 & -\dfrac16 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 & 0 & -\dfrac16 & \dfrac26 & -\dfrac16 \\ 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & -\dfrac16 & \dfrac26 \end{array} \end{pmatrix} \\
\overset{F_{1,3} (-3) }{\longrightarrow } & \begin{pmatrix} \begin {array}{ccccc|ccccc} 1 & 2 & 0 & 0 & 0 & 0 & \dfrac36 & -\dfrac26 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & -\dfrac{1}{6} & \dfrac26 & -\dfrac16 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 & -\dfrac16 & \dfrac26 & -\dfrac16 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 & 0 & -\dfrac16 & \dfrac26 & -\dfrac16 \\ 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & -\dfrac16 & \dfrac26 \end{array} \end{pmatrix} \\
\overset{F_{1,2} (-2) }{\longrightarrow } & \begin{pmatrix} \begin {array}{ccccc|ccccc} 1 & 0 & 0 & 0 & 0 & \dfrac26 & -\dfrac16 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & -\dfrac{1}{6} & \dfrac26 & -\dfrac16 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 & -\dfrac16 & \dfrac26 & -\dfrac16 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 & 0 & -\dfrac16 & \dfrac26 & -\dfrac16 \\ 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & -\dfrac16 & \dfrac26 \end{array} \end{pmatrix} ,\\
\end{align}$$

因此，矩阵$A$的逆为$\begin{pmatrix} \dfrac13 & -\dfrac16 & 0 & 0 & 0 \\ -\dfrac{1}{6} & \dfrac13 & -\dfrac16 & 0 & 0 \\ 0 & -\dfrac16 & \dfrac13 & -\dfrac16 & 0 \\ 0 & 0 & -\dfrac16 & \dfrac13 & -\dfrac16 \\ 0 & 0 & 0 & -\dfrac16 & \dfrac13 \end{pmatrix}$.

$(2)\;$

$$\begin{align}
(F\mid E)= & \begin{pmatrix} \begin {array}{cccc|cccc} 2 & 3 & 2 & 1 & 1 & 0 & 0 & 0 \\ 3 & 6 & 4 & 2 & 0 & 1 & 0 & 0 \\ 4 & 8 & 6 & 3 & 0 & 0 & 1 & 0 \\ 2 & 4 & 3 & 2 & 0 & 0 & 0 & 1 \end{array} \end{pmatrix} \\
\overset{F_{2,1} (-1) }{\longrightarrow } & \begin{pmatrix} \begin {array}{cccc|cccc} 2 & 3 & 2 & 1 & 1 & 0 & 0 & 0 \\ 1 & 3 & 2 & 1 & -1 & 1 & 0 & 0 \\ 4 & 8 & 6 & 3 & 0 & 0 & 1 & 0 \\ 2 & 4 & 3 & 2 & 0 & 0 & 0 & 1 \end{array} \end{pmatrix} \\
\overset{F_{3,4} (-2)}{\longrightarrow } & \begin{pmatrix} \begin {array}{cccc|cccc} 2 & 3 & 2 & 1 & 1 & 0 & 0 & 0 \\ 1 & 3 & 2 & 1 & -1 & 1 & 0 & 0 \\ 0 & 0 & 0 & -1 & 0 & 0 & 1 & -2 \\ 2 & 4 & 3 & 2 & 0 & 0 & 0 & 1 \end{array} \end{pmatrix} \\
\overset{F_{4,1} (-1)}{\longrightarrow } & \begin{pmatrix} \begin {array}{cccc|cccc} 2 & 3 & 2 & 1 & 1 & 0 & 0 & 0 \\ 1 & 3 & 2 & 1 & -1 & 1 & 0 & 0 \\ 0 & 0 & 0 & -1 & 0 & 0 & 1 & -2 \\ 0 & 1 & 1 & 1 & -1 & 0 & 0 & 1 \end{array} \end{pmatrix} \\
\overset{F_{1,2} }{\longrightarrow } & \begin{pmatrix} \begin {array}{cccc|cccc} 1 & 3 & 2 & 1 & -1 & 1 & 0 & 0 \\ 2 & 3 & 2 & 1 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & -1 & 0 & 0 & 1 & -2 \\ 0 & 1 & 1 & 1 & -1 & 0 & 0 & 1 \end{array} \end{pmatrix} \\
\overset{F_{2,1} (-2)}{\longrightarrow } & \begin{pmatrix} \begin {array}{cccc|cccc} 1 & 3 & 2 & 1 & -1 & 1 & 0 & 0 \\ 0 & -3 & -2 & -1 & 3 & -2 & 0 & 0 \\ 0 & 0 & 0 & -1 & 0 & 0 & 1 & -2 \\ 0 & 1 & 1 & 1 & -1 & 0 & 0 & 1 \end{array} \end{pmatrix} \\
\overset{F_{1,2} (1)}{\longrightarrow } & \begin{pmatrix} \begin {array}{cccc|cccc} 1 & 0 & 0 & 0 & 2 & -1 & 0 & 0 \\ 0 & -3 & -2 & -1 & 3 & -2 & 0 & 0 \\ 0 & 0 & 0 & -1 & 0 & 0 & 1 & -2 \\ 0 & 1 & 1 & 1 & -1 & 0 & 0 & 1 \end{array} \end{pmatrix} \\
\overset{F_{2,4} (3) }{\longrightarrow } & \begin{pmatrix} \begin {array}{cccc|cccc} 1 & 0 & 0 & 0 & 2 & -1 & 0 & 0 \\ 0 & 0 & 1 & 2 & 0 & -2 & 0 & 3 \\ 0 & 0 & 0 & -1 & 0 & 0 & 1 & -2 \\ 0 & 1 & 1 & 1 & -1 & 0 & 0 & 1 \end{array} \end{pmatrix} \\
\overset{F_{2,4} }{\longrightarrow } & \begin{pmatrix} \begin {array}{cccc|cccc} 1 & 0 & 0 & 0 & 2 & -1 & 0 & 0 \\ 0 & 1 & 1 & 1 & -1 & 0 & 0 & 1 \\ 0 & 0 & 0 & -1 & 0 & 0 & 1 & -2 \\ 0 & 0 & 1 & 2 & 0 & -2 & 0 & 3 \end{array} \end{pmatrix} \\
\overset{F_{3,4} }{\longrightarrow } & \begin{pmatrix} \begin {array}{cccc|cccc} 1 & 0 & 0 & 0 & 2 & -1 & 0 & 0 \\ 0 & 1 & 1 & 1 & -1 & 0 & 0 & 1 \\ 0 & 0 & 1 & 2 & 0 & -2 & 0 & 3 \\ 0 & 0 & 0 & -1 & 0 & 0 & 1 & -2 \end{array} \end{pmatrix} \\
\overset{F_4 (-1) }{\longrightarrow } & \begin{pmatrix} \begin {array}{cccc|cccc} 1 & 0 & 0 & 0 & 2 & -1 & 0 & 0 \\ 0 & 1 & 1 & 1 & -1 & 0 & 0 & 1 \\ 0 & 0 & 1 & 2 & 0 & -2 & 0 & 3 \\ 0 & 0 & 0 & 1 & 0 & 0 & -1 & 2 \end{array} \end{pmatrix} \\
\overset{F_{2,3} (-1) }{\longrightarrow } & \begin{pmatrix} \begin {array}{cccc|cccc} 1 & 0 & 0 & 0 & 2 & -1 & 0 & 0 \\ 0 & 1 & 0 & -1 & -1 & 2 & 0 & -2 \\ 0 & 0 & 1 & 2 & 0 & -2 & 0 & 3 \\ 0 & 0 & 0 & 1 & 0 & 0 & -1 & 2 \end{array} \end{pmatrix} \\
\overset{F_{2,4} (1) }{\longrightarrow } & \begin{pmatrix} \begin {array}{cccc|cccc} 1 & 0 & 0 & 0 & 2 & -1 & 0 & 0 \\ 0 & 1 & 0 & 0 & -1 & 2 & -1 & 0 \\ 0 & 0 & 1 & 2 & 0 & -2 & 0 & 3 \\ 0 & 0 & 0 & 1 & 0 & 0 & -1 & 2 \end{array} \end{pmatrix} \\
\overset{F_{3,4} (-2) }{\longrightarrow } & \begin{pmatrix} \begin {array}{cccc|cccc} 1 & 0 & 0 & 0 & 2 & -1 & 0 & 0 \\ 0 & 1 & 0 & 0 & -1 & 2 & -1 & 0 \\ 0 & 0 & 1 & 0 & 0 & -2 & 2 & -1 \\ 0 & 0 & 0 & 1 & 0 & 0 & -1 & 2 \end{array} \end{pmatrix} \\
\end{align}$$

因此，矩阵$F$的逆为$\begin{pmatrix} 2 & -1 & 0 & 0 \\ -1 & 2 & -1 & 0 \\ 0 & -2 & 2 & -1 \\ 0 & 0 & -1 & 2 \end{pmatrix}$.

$13$.验证

$$A=\begin{pmatrix} a & b \\ c & d \end{pmatrix} ,ad-bc\neq 0\Rightarrow A^{-1}=\dfrac{1}{ad-bc}\begin{pmatrix} d & -b \\ -c & a \end{pmatrix} .$$

特别地，

$$ad-bc=1\Rightarrow A^{-1}=\begin{pmatrix} d & -b \\ -c & a \end{pmatrix} .$$

如果$ad-bc=0$，$A^{-1}$存在吗？

证明：$(1)\;$设$ad-bc\neq 0$，对任意矩阵$A=\begin{pmatrix} a & b \\ c & d \end{pmatrix}$，同时它的逆矩阵为$N=\begin{pmatrix} u & v \\ s & t \end{pmatrix}$.

根据逆矩阵的定义，有

$$AN=\begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} u & v \\ s & t \end{pmatrix} =\begin{pmatrix} au+bs & av+bt \\ cu+ds & cv+dt \end{pmatrix} =\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} .$$

于是，实数$u,v,s,t$必须满足

$$\begin{cases} au+bs=1 \\ cu+ds=0 \\ av+bt=0 \\ cv+dt=1 \end{cases} $$

即$\begin{cases} au+bs=1 \\ cu+ds=0 \end{cases} $且$\begin{cases} av+bt=0 \\ cv+dt=1 \end{cases} $

由于$ad-bc\neq 0$，故解得

$$\begin{cases} u=\dfrac{d}{ad-bc} \\ s=\dfrac{-c}{ad-bc} \end{cases} \quad 且\quad \begin{cases} v=\dfrac{-b}{ad-bc} \\ t=\dfrac{a}{ad-bc} \end{cases}.$$

经验证，有$AN=NA=I$.因此，矩阵$A$的逆矩阵为

$$A^{-1}=\dfrac{1}{ad-bc}\begin{pmatrix} d & -b \\ -c & a \end{pmatrix} .$$

$(2)\;$由$(1)$知，当$ad-bc=1$时，有

$$A^{-1}=\begin{pmatrix} d & -b \\ -c & a \end{pmatrix} .$$

$(3)\;$由$(1)$知，当$ad-bc=0$时，方程组无解，此时矩阵$A$不存在逆矩阵.

$14$.证明任意矩阵

$$A=\begin{pmatrix} a & b \\ c & d \end{pmatrix} $$

满足关系式

$$A^2 =(a+d)A-(ad-bc)E \label{23} \tag{23}$$

(换言之，$A$是二次方程$x^2-(a+d)x+(ad-bc)=0$的一个“根”).

证明：对任意的矩阵$A=\begin{pmatrix} a & b \\ c & d \end{pmatrix}$来说，有

$$A^2=\begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} a & b \\ c & d \end{pmatrix} =\begin{pmatrix} a^2+bc & ab+bd \\ ac+cd & bc+d^2 \end{pmatrix},$$

$$(a+d)A-(ad-bc)E=(a+d)\begin{pmatrix} a & b \\ c & d \end{pmatrix} -(ad-bc)\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}=\begin{pmatrix} a^2+bc & ab+bd \\ ac+cd & bc+d^2 \end{pmatrix},$$

故对任意的矩阵$A=\begin{pmatrix} a & b \\ c & d \end{pmatrix}$来说，

满足关系式

$$A^2 =(a+d)A-(ad-bc)E .$$

$15$.如果$ad-bc\neq 0$，运用关系$\eqref{23}$求逆矩阵$A^{-1}$.

解：由关系$\eqref{23}$得

$$A^2 =(a+d)A-(ad-bc)E ,$$

即

$$(ad-bc)E=(a+d)A-A^2 =A[(a+d)E-A]=[(a+d)E-A]A.$$

因为$ad-bc\neq 0$，故有

$$E=A\dfrac{1}{ad-bc} [(a+d)E-A]=\dfrac{1}{ad-bc} [(a+d)E-A]A.$$

根据逆矩阵的定义，知矩阵$A$的逆矩阵为

$$A^{-1}=\dfrac{1}{ad-bc} [(a+d)E-A] =\dfrac{1}{ad-bc}\begin{pmatrix} d & -b \\ -c & a \end{pmatrix} .$$

$16$.证明若$\begin{pmatrix} a & b \\ c & d \end{pmatrix} ^m =0$，则$\begin{pmatrix} a & b \\ c & d \end{pmatrix} ^2 =0$.

证明：证法$1$.设矩阵$A=\begin{pmatrix} a & b \\ c & d \end{pmatrix}$.

因为$A^m =0$，所以$\mid A^m \mid =\mid A\mid ^m =0$，得到行列式$\mid A\mid =0$.又矩阵$A$是$2\times 2$阶的，故其对应的秩只能为$0$或为$1$.

如果矩阵$A$的秩为$0$，那么显然$A^2 =0$.

如果矩阵$A$的秩为$1$，则矩阵$A$的两行成比例，可设为

$$A=\begin{pmatrix} a & b \\ c & d \end{pmatrix} =\begin{pmatrix} a & b \\ ra & rb \end{pmatrix} =\begin{pmatrix} 1 \\ r \end{pmatrix} (a,b) ,$$

则

$$\begin{align}
A^2 & =A\cdot A \\
& = \begin{pmatrix} 1 \\ r \end{pmatrix} (a,b) \begin{pmatrix} 1 \\ r \end{pmatrix} (a,b) \\
& = \begin{pmatrix} 1 \\ r \end{pmatrix} (a+rb) (a,b) \\
& = (a+rb)\begin{pmatrix} 1 \\ r \end{pmatrix} (a,b) \\
& = (a+rb)A,
\end{align}$$

$$\cdots \cdots \cdots $$

$$\begin{align}
A^m & =A^{m-1} \cdot A \\
& = (a+rb)^{m-2} \begin{pmatrix} 1 \\ r \end{pmatrix} (a,b) \begin{pmatrix} 1 \\ r \end{pmatrix} (a,b) \\
& = (a+rb)^{m-2} \begin{pmatrix} 1 \\ r \end{pmatrix} (a+rb) (a,b) \\
& = (a+rb)^{m-1} \begin{pmatrix} 1 \\ r \end{pmatrix} (a,b) \\
& = (a+rb)^{m-1}A .
\end{align}$$

因为$A^m =0$，故$(a+rb)^{m-1} =0$，即$a+rb=0$，所以$A^2=(a+rb)A =0$.

证法$2$.设$\mu $是矩阵$A$的特征值，$x$为$\mu $所对应的特征向量，由

$$\mu ^m x=A^m x =0$$

及$x\neq 0$可知$\mu =0$，这就是说，矩阵$A$的特征值全为$0$，可见矩阵$A$的特征多项式为$\phi _A (\lambda )=\lambda ^2 $，从而有

$$A^2 =\phi _A (A) =0.$$

$17$.阐明下述论断：设$m\times s$矩阵$X$被水平线和竖直线划分为块(或长方块)，

$$X=\begin{pmatrix} X_{11} & X_{12} & \cdots & X_{1k} \\ X_{21} & X_{22} & \cdots & X_{2k} \\ \cdots & \cdots & \cdots & \cdots \\ X_{l1} & X_{l2} & \cdots & X_{lk} \\ \end{pmatrix} $$

这里$X_{i1} ,\cdots ,X_{ik}$都是$m_i$行矩阵$(m_1 +\cdots +m_l =m)$，而$X_{1j} ,\cdots ,X_{lj}$，都是$s_j$列矩阵$(s_1 +\cdots +s_k =s)$.

如果

$$Y=\begin{pmatrix} Y_{11} & Y_{12} & \cdots & Y_{1r} \\ Y_{21} & Y_{22} & \cdots & Y_{2r} \\ \cdots & \cdots & \cdots & \cdots \\ Y_{k1} & Y_{k2} & \cdots & Y_{kr} \\ \end{pmatrix} $$

是一个$s\times n$矩阵，它的块$Y_{ij}$的阶是$s_i \times n_j$($n_1 +\cdots +n_r =n$)，则乘积$Z=XY$是有意义的，并且矩阵$Z=(z_{ij})$可以分块计算，它的块$Z_{ij}$可按公式$(7)$形式地写出：

$$Z_{ij} =X_{i1} Y_{1j} +X_{i2} Y_{2j} +\cdots +X_{ik} Y_{kj} .$$

由于矩阵$X_{i\nu } ,Y_{\nu j}$的阶所满足的条件，乘积$X_{i\nu } Y_{\nu j}$也是有意义的.矩阵的分块法即便在最简单的情况下也会带来方便，例如

$$\begin{pmatrix} E & A \\ 0 & E \end{pmatrix} \begin{pmatrix} A & 0 \\ -E & B \end{pmatrix} =\begin{pmatrix} 0 & AB \\ -E & B \end{pmatrix} ，$$

此处$A,B,E,0\in M_n(\mathbb{R} )$($E$是单位矩阵，$0$是零矩阵).

证明：矩阵$Z$的行数为$m_1 +m_2 +\cdots +m_l =m$；$Z$的列数为$n_1 +n_2 +\cdots +n_r =n$.于是$XY$与$Z$都是$m\times n$矩阵.

$$(XY)(i;j)=\displaystyle \sum_{w=1}^n x_{iw} y_{wj} $$

设$i=m_1 +\cdots +m_{p-1} +f$，其中$0 < f \leq m_p$；设$j=n_1 +\cdots +n_{q-1} +g$，其中$0 < g \leq n_q$.这意思是$X$的第$i$行属于$X$的第$p$个行组，$Y$的第$j$列属于$Y$的第$q$个列组.于是，

$$\begin{align}
& \displaystyle Z(i;j) \\
= & Z_{pq} (f;g) \\
= & (\sum_{h=1}^k X_{ph} Y_{hq})(f;g) \\
= & \sum_{h=1}^k (X_{ph} Y_{hq})(f;g) \\
= & \sum_{h=1}^k \sum_{a=1}^{s_h } X_{ph} (f;a) Y_{hq}(a;g) \\
= & \sum_{a=1}^{s_1 } X_{p1} (f;a) Y_{1q}(a;g) +\cdots +\sum_{a=1}^{s_k } X_{pk} (f;a) Y_{kq}(a;g) \\
= & \sum_{w=1}^{s_1 } x_{iw} y_{wj} +\cdots +\sum_{w=s_1 +\cdots +s_{k-1} +1}^{s} x_{iw} y_{wj} \\
= & \sum_{w=1}^s x_{iw} y_{wj} \\
= & (XY)(i;j)
\end{align}$$

因此$XY=Z$.

$18$.令矩阵

$$X=(x_{ij}) \in M_n (\mathbb{R} ) ,T=(t_{ij}) \in M_n (\mathbb{R} ).$$

证明$T$左乘$X$得到行$X_{(1)} ,\cdots ,X_{(n)}$的线性组合，而右乘得到列$X^{(1)} ,\cdots ,X^{(n)}$的线性组合，特别注意到如果

$$T=\begin{pmatrix} 1 & t_{12} & t_{13} & \cdots & t_{1n} \\ & 1 & t_{23} & \cdots & t_{2n} \\ \cdots & \cdots & \cdots & \cdots & \cdots \\ 0 & 0 & 0 & \cdots & 1 \end{pmatrix} $$

是上三角矩阵，则

$$TX=\begin{pmatrix} X_{(1)} + t_{12} & X_{(2)} & +\cdots + t_{1n} & X_{(n)} \\ & X_{(2)} & +\cdots + t_{2n} & X_{(n)} \\ & & & \cdots \\ & & & X_{(n)} \end{pmatrix} $$

是从$X$经过一系列(Ⅱ)型初等行变换得到的矩阵.

证明：$(1)\;$如果把矩阵$X$看成是$n\times 1$矩阵的，则

$$X=\begin{pmatrix} X_{(1)} \\ X_{(2)} \\ \vdots \\ X_{(n)} \end{pmatrix} ,$$

于是$T$左乘$X$得到

$$\begin{align}
& TX \\
= & \begin{pmatrix} t_{11} & t_{12} & \cdots & t_{1n} \\ t_{21} & t_{22} & \cdots & t_{2n} \\ \vdots & \vdots & & \vdots \\ t_{n1} & t_{n2} & \cdots & t_{nn} \end{pmatrix} \begin{pmatrix} X_{(1)} \\ X_{(2)} \\ \vdots \\ X_{(n)} \end{pmatrix} \\
= & t_{11} X_{(1)} +\cdots +t_{1n} X_{(n)} +\cdots +t_{n1} X_{(1)} +\cdots +t_{nn} X_{(n)} \\
= & (t_{11} +\cdots +t_{n1} )X_{(1)} +\cdots +(t_{1n} +\cdots +t_{nn} )X_{(n)}
\end{align}$$

即$T$左乘$X$得到行$X_{(1)} ,\cdots ,X_{(n)}$的线性组合.

同样的，如果把矩阵$X$看成是$1\times n$矩阵的，则

$$X=(X^{(1)} ,X^{(2)} ,\cdots ,X^{(n)}) ,$$

于是$T$右乘$X$得到

$$\begin{align}
& XT \\
= & (X^{(1)} ,X^{(2)} ,\cdots ,X^{(n)}) \begin{pmatrix} t_{11} & t_{12} & \cdots & t_{1n} \\ t_{21} & t_{22} & \cdots & t_{2n} \\ \vdots & \vdots & & \vdots \\ t_{n1} & t_{n2} & \cdots & t_{nn} \end{pmatrix} \\
= & t_{11} X^{(1)} +\cdots +t_{n1} X^{(n)} +\cdots +t_{1n} X^{(1)} +\cdots +t_{nn} X^{(n)} \\
= & (t_{11} +\cdots +t_{1n} )X^{(1)} +\cdots +(t_{n1} +\cdots +t_{nn} )X^{(n)}
\end{align}$$

即$T$右乘得到列$X^{(1)} ,\cdots ,X^{(n)}$的线性组合.

$(2)\;$如果

$$T=\begin{pmatrix} 1 & t_{12} & t_{13} & \cdots & t_{1n} \\ & 1 & t_{23} & \cdots & t_{2n} \\ \cdots & \cdots & \cdots & \cdots & \cdots \\ 0 & 0 & 0 & \cdots & 1 \end{pmatrix} $$

是上三角矩阵，并把矩阵$X$看成是$n\times 1$矩阵的，即

$$X=\begin{pmatrix} X_{(1)} \\ X_{(2)} \\ \vdots \\ X_{(n)} \end{pmatrix} .$$

利用(Ⅱ)型初等行变换，对矩阵$X$进行变换.

首先，第$i=2,\cdots ,n$行分别乘以$t_{1i}$加到第$1$行，得

$$\begin{pmatrix} X_{(1)} +t_{12} X_{(2)} +\cdots + & t_{1n} X_{(n)} \\ & X_{(2)} \\ & \vdots \\ & X_{(n)} \end{pmatrix} .$$

然后，第$i=3,\cdots ,n$行分别乘以$t_{2i}$加到第$2$行，得

$$\begin{pmatrix} X_{(1)} + t_{12} & X_{(2)} & +\cdots + t_{1n} & X_{(n)} \\ & X_{(2)} & +\cdots + t_{2n} & X_{(n)} \\ & & & X_{(3)} \\ & & & \cdots \\ & & & X_{(n)} \end{pmatrix} .$$

重复上面的步骤，第$j$次（$j=1,\cdots ,n-1$），把第$i=j+1 ,\cdots ,n $行分别乘以$t_{ji}$加到第$j$行，得

$$\begin{pmatrix} X_{(1)} + t_{12} & X_{(2)} & +\cdots + t_{1n} & X_{(n)} \\ & X_{(2)} & +\cdots + t_{2n} & X_{(n)} \\ & & & \cdots \\ & X_{(j)} + & t_{j,j+1} X_{(j+1)} +\cdots + & t_{j,n} X_{(n)} \\ & & & \cdots \\ & & & X_{(n)} \end{pmatrix} .$$

当第$n-1$次时，把第$n$行乘以$t_{n-1,n}$加到第$n-1$行，得

$$\begin{pmatrix} X_{(1)} + t_{12} & X_{(2)} & +\cdots + t_{1n} & X_{(n)} \\ & X_{(2)} & +\cdots + t_{2n} & X_{(n)} \\ & & & \cdots \\ & & & X_{(n)} \end{pmatrix} ,$$

根据上述对矩阵$X$的一系列(Ⅱ)型初等行变换，可知，矩阵$TX$是从$X$经过一系列(Ⅱ)型初等行变换得到的矩阵.