两道$n$元非齐次线性方程组

青青子衿 · 发表于 2016-8-12 16:33

本帖最后由青青子衿于 2016-10-14 21:49 编辑 \[
\begin{equation}
\left\{
\begin{split}
x_1&+&x_2&+&x_3&+&\cdots&+&x_{n - 2}&+&x_{n - 1}&&&=&\,1\\
x_1&+&x_2&+&x_3&+&\cdots&+&x_{n - 2}&&&+&x_n&=&\,2\\
x_1&+&x_2&+&x_3&+&\cdots&&&+&x_{n - 1}&+&x_n&=&\,3\\
\vdots\,\,\,&&\,\,\vdots&&\,\,\vdots&&
&&\,\,\,\,\vdots&&\,\,\,\,\vdots&&\,\,\,\vdots&&\,\vdots\\
x_1&+&x_2&&&+&\cdots&+&x_{n - 2}&+&x_{n - 1}&+&x_n&=&\,n - 2\\
x_1&&&+&x_3&+&\cdots&+&x_{n - 2}&+&x_{n - 1}&+&x_n&=&\,n - 1\\
&&x_2&+&x_3&+&\cdots&+&x_{n - 2}&+&x_{n - 1}&+&x_n&=&\,n
\end{split}
\right.
\end{equation}
\]

\[
\begin{equation}
\left\{
\begin{split}
&&x_2&+&x_3&+&\cdots&+&x_{n - 2}&+&x_{n - 1}&+&x_{n}&=&\,1\\
x_1&&&+&x_3&+&\cdots&+&x_{n - 2}&+&x_{n - 1}&+&x_n&=&\,2\\
x_1&+&x_2&&&+&\cdots&+&x_{n - 2}&+&x_{n - 1}&+&x_n&=&\,3\\
\vdots\,\,\,&&\,\,\vdots&&\,\,\vdots&&
&&\,\,\,\,\vdots&&\,\,\,\,\vdots&&\,\,\,\vdots&&\,\vdots\\
x_1&+&x_2&+&x_3&+&\cdots&&&+&x_{n - 1}&+&x_n&=&\,n - 2\\
x_1&+&x_2&+&x_3&+&\cdots&+&x_{n - 2}&&&+&x_n&=&\,n - 1\\
x_1&+&x_2&+&x_3&+&\cdots&+&x_{n - 2}&+&x_{n - 1}&&&=&\,n
\end{split}
\right.
\end{equation}
\]

青青子衿 · 发表于 2016-8-12 16:49

本帖最后由青青子衿于 2018-6-16 17:47 编辑 回复 1# 青青子衿
$\Huge{原理一：利用克拉默法则解线性方程组}$

\begin{equation*}
\left\{
\begin{split}
x_1&+&x_2&+&x_3&+&\cdots&+&x_{n - 2}&+&x_{n - 1}&&&=&\,1\\
x_1&+&x_2&+&x_3&+&\cdots&+&x_{n - 2}&&&+&x_n&=&\,2\\
x_1&+&x_2&+&x_3&+&\cdots&&&+&x_{n - 1}&+&x_n&=&\,3\\
\vdots\,\,\,&&\,\,\vdots&&\,\,\vdots&&
&&\,\,\,\,\vdots&&\,\,\,\,\vdots&&\,\,\,\vdots&&\,\vdots\\
x_1&+&x_2&&&+&\cdots&+&x_{n - 2}&+&x_{n - 1}&+&x_n&=&\,n - 2\\
x_1&&&+&x_3&+&\cdots&+&x_{n - 2}&+&x_{n - 1}&+&x_n&=&\,n - 1\\
&&x_2&+&x_3&+&\cdots&+&x_{n - 2}&+&x_{n - 1}&+&x_n&=&\,n
\end{split}
\right.
\end{equation*}
青青子衿发表于 2016-8-12 16:33

第一步：先求系数行列式
方程组$\left(a\right)$的系数行列式为$D_a$
\[D_a=\left|\begin{array}{ccccccc}
1&1&1& \cdots &1&1&0\\
1&1&1& \cdots &1&0&1\\
1&1&1& \cdots &0&1&1\\
\vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\
1&1&0& \cdots &1&1&1\\
1&0&1& \cdots &1&1&1\\
0&1&1& \cdots &1&1&1
\end{array}\right|\]
把行列式$D_a$的第$2$，$3$，$\cdots$，$n$列都加到第$1$列上去，得：
\[D_a=\left|\begin{array}{ccccccc}
n-1&1&1& \cdots &1&1&0\\
n-1&1&1& \cdots &1&0&1\\
n-1&1&1& \cdots &0&1&1\\
\vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\
n-1&1&0& \cdots &1&1&1\\
n-1&0&1& \cdots &1&1&1\\
n-1&1&1& \cdots &1&1&1
\end{array}\right|\]
把行列式$D_a$的第$1$，$2$，$\cdots$，$n-1$行都减去第$n$行，得：
\[D_a=\left|\begin{array}{ccccccc}
0&0&0& \cdots &0&0&-1\\
0&0&0& \cdots &0&-1&0\\
0&0&0& \cdots &-1&0&0\\
\vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\
0&0&-1& \cdots &0&0&0\\
0&-1&0& \cdots &0&0&0\\
n-1&1&1& \cdots &1&1&1
\end{array}\right|=\left(-1\right)^{\frac{n(n-1)}{2}}\left(-1\right)^{n-1}(n-1)\]
第二步：
下面计算行列式${D_a}_{\overset{\phantom,}{i}}$
\[{D_a}_{\overset{\phantom,}{i}}=\begin{array}{ccccccccc}
\begin{split}
\begin{array}{ccccccccc}
\,
\end{array}
\\ \,
\left|
\begin{array}{ccccccccc}
1&1&1& \cdots &1& \cdots &1&1&0\\
1&1&1& \cdots &2& \cdots &1&0&1\\
1&1&1& \cdots &3& \cdots &0&1&1\\
\vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\
1&1&1& \cdots &\color{blue}{n-(i-1)}& \cdots &1&1&1\\
\vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\
1&1&0& \cdots &n-2&\cdots &1&1&1\\
1&0&1& \cdots &n-1& \cdots &1&1&1\\
0&1&1& \cdots &n& \cdots &1&1&1
\end{array}
\right|
\color{red}{倒数}第i行
\\
\begin{array}{ccccccccc}
&第i列&&&&&&&&&&&&
\end{array}
\end{split}
\end{array}\]
将行列式${D_a}_{\overset{\phantom,}{i}}$的第$2$至第$n$行减去第$1$行，得：
\[{D_a}_{\overset{\phantom,}{i}}=\begin{array}{ccccccccc}
\begin{split}
\begin{array}{ccccccccc}
\,
\end{array}
\\ \,
\left|
\begin{array}{ccccccccc}
1&1&1& \cdots &1& \cdots &1&1&0\\
0&0&0& \cdots &1& \cdots &0&-1&1\\
0&0&0& \cdots &2& \cdots &-1&0&1\\
\vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\
0&0&0& \cdots &\color{blue}{n-i}&\cdots &0&0&1\\
\vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\
0&0&-1& \cdots &n-3& \cdots &0&0&1\\
0&-1&0& \cdots &n-2& \cdots &0&0&1\\
-1&0&0& \cdots &n-1& \cdots &0&0&1
\end{array}
\right|
\color{red}{倒数}第i行
\\
\begin{array}{ccccccccc}
&第i列&&&&&&&&&&&&
\end{array}
\end{split}
\end{array}\]
将行列式${D_a}_{\overset{\phantom,}{i}}$的第$1$，$2$，$\cdots$，$i-1$，$i+1$，$\cdots$，$n-1$列分别乘以$n-1$，$n-2$，$\cdots$，$n-i+1$，$n-i-1$，$\cdots$，$1$后加到第$i$列，得：
\[{D_a}_{\overset{\phantom,}{i}}=\begin{array}{ccccccccc}
\begin{split}
\begin{array}{ccccccccc}
\,
\end{array}
\\ \,
\left|
\begin{array}{ccccccccc}
1&1&1& \cdots &
1+1+2+ \cdots+\left(n - i - 1\right)+\left(n-i+1\right) + \cdots+\left(n-1\right)
& \cdots &1&1&0\\
0&0&0& \cdots &0& \cdots &0&-1&1\\
0&0&0& \cdots &0& \cdots &-1&0&1\\
\vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\
0&0&0& \cdots &\color{blue}{n-i}&\cdots &0&0&1\\
\vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\
0&0&-1& \cdots &0& \cdots &0&0&1\\
0&-1&0& \cdots &0& \cdots &0&0&1\\
-1&0&0& \cdots &0& \cdots &0&0&1
\end{array}
\right|
\color{red}{倒数}第i行
\\
\begin{array}{ccccccccc}
&第i列&&&&&&&&&&&&&&&&&&&&&&&&
\end{array}
\end{split}
\end{array}\]
将行列式${D_a}_{\overset{\phantom,}{i}}$的第$1$，$2$，$\cdots$，$i-1$，$i+1$，$\cdots$，$n-1$列加到第$n$列，第$i$列乘以$\frac{1}{i-n}$加到第$n$列，得：
\[{D_a}_{\overset{\phantom,}{i}}=\begin{array}{ccccccccc}
\begin{split}
\begin{array}{ccccccccc}
\,
\end{array}
\\ \,
\left|
\begin{array}{ccccccccc}
1&1&1& \cdots &1+\frac{n(n-1)}{2}-(n-i)& \cdots &1&1&
(n-2)+\frac{\frac{n(n-1)}{2}-n+i+1}{i-n}\\
0&0&0& \cdots &0& \cdots &0&-1&0\\
0&0&0& \cdots &0& \cdots &-1&0&0\\
\vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\
0&0&0& \cdots &\color{blue}{n-i}&\cdots&0&0&0\\
\vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\
0&0&-1& \cdots &0& \cdots &0&0&0\\
0&-1&0& \cdots &0& \cdots &0&0&0\\
-1&0&0& \cdots &0& \cdots &0&0&0
\end{array}
\right|
\color{red}{倒数}第i行
\\
\begin{array}{ccccccccc}
&第i列&&&&&&&&&&&&&&&&&&&&&&&&
\end{array}
\end{split}
\end{array}\]
化简行列式${D_a}_{\overset{\phantom,}{i}}$，得：
\[{D_a}_{\overset{\phantom,}{i}}=\left(-1\right)^{\frac{n(n-1)}{2}}\left(-1\right)^{n-2}\left(n-i\right)\left[(n-2)+\frac{\frac{n(n-1)}{2}-n+i+1}{i-n}\right]\\
=\left(-1\right)^{\frac{n(n-1)}{2}}\left(-1\right)^{n-2}\left[n^2-in-n+i -\frac{n\left(n-1\right)}{2}-1\right]\\
=\left(-1\right)^{\frac{n(n-1)}{2}}\left(-1\right)^{n-2}\left[\frac{n\left(n-1\right)}{2}-1-i(n-1)\right]\\
=\left(-1\right)^{\frac{n(n-1)}{2}}\left(-1\right)^{n-1}\left[1-\frac{n\left(n-1\right)}{2}+i(n-1)\right]\]
由克拉默法则，得：
\[{x_a}_{\overset{\phantom,}{i}}=\frac{{D_a}_{\overset{\phantom,}{i}}}{D_a}=\frac{\left(-1\right)^{\frac{n(n-1)}{2}}\left(-1\right)^{n-1}\left[1-\frac{n\left(n-1\right)}{2}+i(n-1)\right]}{\left(-1\right)^{\frac{n(n-1)}{2}}\left(-1\right)^{n-1}(n-1)}=\frac{1}{n-1}-\frac{n}{2}+i\]
\[ \begin{equation*} \label{a1}\tag*{[Solution#1 of equations a]}
{x_a}_{\overset{\phantom,}{i}}=\frac{1}{n-1}-\frac{n}{2}+i
\end{equation*}\]
\ref{a1}
\begin{equation}
\left\{
\begin{split}
x_1\quad&\ &\ &\ &\ &\ &\ &=&\,\frac{1}{n-1}-\frac{n}{2}+1\\
\,&x_2\quad&\,&\ &\ &\ &\ &=&\,\frac{1}{n-1}-\frac{n}{2}+2\\
\,&\,&x_3\quad&\ &\ &\ &\ &=&\,\frac{1}{n-1}-\frac{n}{2}+3\\
\,&\,&\,&\ddots\quad&\,&\,&\,&=&\,\qquad\quad\vdots\\
\,&\,&\,&\,&x_{n-2}\quad&\,&\,&=&\,\frac{1}{n-1}-\frac{n}{2}+n-2\\
\,&\,&\,&\,&\,&x_{n-1}\quad&\,&=&\,\frac{1}{n-1}-\frac{n}{2}+n-1\\
\,&\,&\,&\,&\,&\quad&x_n&=&\,\frac{1}{n-1}-\frac{n}{2}+n
\end{split}
\right.
\end{equation}

\begin{equation*}
\left\{
\begin{split}
&&x_2&+&x_3&+&\cdots&+&x_{n - 2}&+&x_{n - 1}&+&x_{n}&=&\,1\\
x_1&&&+&x_3&+&\cdots&+&x_{n - 2}&+&x_{n - 1}&+&x_n&=&\,2\\
x_1&+&x_2&&&+&\cdots&+&x_{n - 2}&+&x_{n - 1}&+&x_n&=&\,3\\
\vdots\,\,\,&&\,\,\vdots&&\,\,\vdots&&
&&\,\,\,\,\vdots&&\,\,\,\,\vdots&&\,\,\,\vdots&&\,\vdots\\
x_1&+&x_2&+&x_3&+&\cdots&&&+&x_{n - 1}&+&x_n&=&\,n - 2\\
x_1&+&x_2&+&x_3&+&\cdots&+&x_{n - 2}&&&+&x_n&=&\,n - 1\\
x_1&+&x_2&+&x_3&+&\cdots&+&x_{n - 2}&+&x_{n - 1}&&&=&\,n
\end{split}
\right.
\end{equation*}
青青子衿发表于 2016-8-12 16:33

第一步：先求系数行列式
方程组$\left(b\right)$的系数行列式为$D_b$
\[D_b=\left|\begin{array}{ccccccc}
0&1&1& \cdots &1&1&1\\
1&0&1& \cdots &1&1&1\\
1&1&0& \cdots &1&1&1\\
\vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\
1&1&1& \cdots &0&1&1\\
1&1&1& \cdots &1&0&1\\
1&1&1& \cdots &1&1&0
\end{array}\right|\]
把行列式$D_b$的第$2$，$3$，$\cdots$，$n$列都加到第$1$列上去，得：
\[D_b=\left|\begin{array}{ccccccc}
n-1&1&1& \cdots &1&1&1\\
n-1&0&1& \cdots &1&1&1\\
n-1&1&0& \cdots &1&1&1\\
\vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\
n-1&1&1& \cdots &0&1&1\\
n-1&1&1& \cdots &1&0&1\\
n-1&1&1& \cdots &1&1&0
\end{array}\right|\]
把行列式$D_b$的第$2$，$3$，$\cdots$，$n$行都减去第$1$行，得：
\[D_b=\left|\begin{array}{ccccccc}
n-1&1&1& \cdots &1&1&1\\
0&-1&0& \cdots &0&0&0\\
0&0&-1& \cdots &0&0&0\\
\vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\
0&0&0& \cdots &-1&0&0\\
0&0&0& \cdots &0&-1&0\\
0&0&0& \cdots &0&0&-1
\end{array}\right|=\left(-1\right)^{n-1}(n-1)\]
第二步：
下面计算行列式${D_b}_{\overset{\phantom,}{i}}$
\[{D_b}_{\overset{\phantom,}{i}}=\begin{array}{ccccccccc}
\begin{split}
\begin{array}{ccccccccc}
&第i列&&&&&&&&&
\end{array}
\\ \,
\left|
\begin{array}{ccccccccc}
0&1&1& \cdots &1& \cdots &1&1&1\\
1&0&1& \cdots &2& \cdots &1&1&1\\
1&1&0& \cdots &3& \cdots &1&1&1\\
\vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\
1&1&1& \cdots &i& \cdots &1&1&1\\
\vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\
1&1&1& \cdots &{n - 2}& \cdots &0&1&1\\
1&1&1& \cdots &{n - 1}& \cdots &1&0&1\\
1&1&1& \cdots &n& \cdots &1&1&0
\end{array}
\right|
第i行
\\
\begin{array}{ccccccccc}
\,
\end{array}
\end{split}
\end{array}\]
将行列式${D_b}_{\overset{\phantom,}{i}}$的第$2$至第$n$行减去第$1$行，得：
\[{D_b}_{\overset{\phantom,}{i}}=\begin{array}{ccccccccc}
\begin{split}
\begin{array}{ccccccccc}
&第i列&&&&&&&&&&&&
\end{array}
\\ \,
\left|
\begin{array}{ccccccccc}
0&1&1& \cdots &1& \cdots &1&1&1\\
1&-1&0& \cdots &1& \cdots &0&0&0\\
1&0&-1& \cdots &2& \cdots &0&0&0\\
\vdots & \vdots &\vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\
1&0&0& \cdots &i-1& \cdots &0&0&0\\
\vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\
1&0&0& \cdots &n-3& \cdots &-1&0&0\\
1&0&0& \cdots &n-2& \cdots &0&-1&0\\
1&0&0& \cdots &n-1& \cdots &0&0&-1
\end{array}
\right|
第i行
\\
\begin{array}{ccccccccc}
\,
\end{array}
\end{split}
\end{array}\]
将行列式${D_b}_{\overset{\phantom,}{i}}$的第$2$，$3$，$\cdots$，$i-1$，$i+1$，$\cdots$，$n$列分别乘以$1$，$2$，$\cdots$，$i-2$，$i$，$\cdots$，$n-1$后加到第$i$列，得：
\[{D_b}_{\overset{\phantom,}{i}}=\begin{array}{ccccccccc}
\begin{split}
\begin{array}{ccccccccc}
&第i列&&&&&&&&&&&&&&&&&&&&
\end{array}
\\ \,
\left|
\begin{array}{ccccccccc}
0&1&1& \cdots &1+1+2+\cdots+(i-2)+i+\cdots+(n-1)& \cdots &1&1&0\\
1&-1&0& \cdots &0& \cdots &0&0&0\\
1&0&-1& \cdots &0& \cdots &0&0&0\\
\vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\
1&0&0& \cdots &i-1&\cdots&0&0&0\\
\vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\
1&0&0& \cdots &0& \cdots &-1&0&0\\
1&0&0& \cdots &0& \cdots &0&-1&0\\
1&0&0& \cdots &0& \cdots &0&0&-1
\end{array}
\right|
第i行
\\
\begin{array}{ccccccccc}
\,
\end{array}
\end{split}
\end{array}\]

将行列式${D_b}_{\overset{\phantom,}{i}}$的第$2$，$3$，$\cdots$，$i-1$，$i+1$，$\cdots$，$n$列加到第$1$列，第$i$列乘以$\frac{1}{1-i}$加到第$1$列，得：
\[{D_b}_{\overset{\phantom,}{i}}=\begin{array}{ccccccccc}
\begin{split}
\begin{array}{ccccccccc}
&第i列&&&&&&&&&&&&&&
\end{array}
\\ \,
\left|
\begin{array}{ccccccccc}
(n-2)+\frac{\frac{n(n-1)}{2}-i+2}{1-i}&1&1& \cdots &\frac{n(n-1)}{2}-i+2& \cdots &1&1&1\\
0&-1&0& \cdots &0& \cdots &0&0&0\\
0&0&-1& \cdots &0& \cdots &0&0&0\\
\vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\
0&0&0& \cdots &i-1&\cdots&0&0&0\\
\vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\
0&0&0& \cdots &0& \cdots &-1&0&0\\
0&0&0& \cdots &0& \cdots &0&-1&0\\
0&0&0& \cdots &0& \cdots &0&0&-1
\end{array}
\right|
第i行
\\
\begin{array}{ccccccccc}
\,
\end{array}
\end{split}
\end{array}\]
化简行列式${D_b}_{\overset{\phantom,}{i}}$，得：
\[{D_b}_{\overset{\phantom,}{i}}=\left(-1\right)^{n-2}\left(i-1\right)\left[(n-2)+\frac{\frac{n(n-1)}{2}-i+2}{1-i}\right]\\=\left(-1\right)^{n-1}\left[\frac{n(n-1)}{2}+n(1-i)+i\right]\\=\left(-1\right)^{n-1}\left[\frac{n(n-1)}{2}+n-i(n-1)\right]\]
由克拉默法则，得：
\[{x_b}_{\overset{\phantom,}{i}}=\frac{{D_b}_{\overset{\phantom,}{i}}}{D_b}=\frac{\left(-1\right)^{n-1}\left[\frac{n(n-1)}{2}+n-i(n-1)\right]}{\left(-1\right)^{n-1}(n-1)}=\frac{n}{2}+\frac{n}{n-1}-i\]
\[ \begin{equation*} \label{b1}\tag*{[Solution#1 of equations b]}
{x_b}_{\overset{\phantom,}{i}}=\frac{n}{2}+\frac{n}{n-1}-i
\end{equation*}\]
\ref{b1}
\begin{equation}
\left\{
\begin{split}
x_1\quad&\ &\ &\ &\ &\ &\ &=&\,\frac{n}{n-1}+\frac{n}{2}-1\\
\,&x_2\quad&\,&\ &\ &\ &\ &=&\,\frac{n}{n-1}+\frac{n}{2}-2\\
\,&\,&x_3\quad&\ &\ &\ &\ &=&\,\frac{n}{n-1}+\frac{n}{2}-3\\
\,&\,&\,&\ddots\quad&\,&\,&\,&=&\,\qquad\quad\vdots\\
\,&\,&\,&\,&x_{n-2}\quad&\,&\,&=&\,\frac{n}{n-1}+\frac{n}{2}-\left(n-2\right)\\
\,&\,&\,&\,&\,&x_{n-1}\quad&\,&=&\,\frac{n}{n-1}+\frac{n}{2}-\left(n-1\right)\\
\,&\,&\,&\,&\,&\quad&x_n&=&\,\frac{n}{n-1}+\frac{n}{2}-n
\end{split}
\right.
\end{equation}

青青子衿 · 发表于 2016-8-12 17:03

本帖最后由青青子衿于 2018-6-16 20:02 编辑 回复 1# 青青子衿
$\Huge{原理二：利用逆矩阵解线性方程组}$

\begin{equation*}
\left\{
\begin{split}
x_1&+&x_2&+&x_3&+&\cdots&+&x_{n - 2}&+&x_{n - 1}&&&=&\,1\\
x_1&+&x_2&+&x_3&+&\cdots&+&x_{n - 2}&&&+&x_n&=&\,2\\
x_1&+&x_2&+&x_3&+&\cdots&&&+&x_{n - 1}&+&x_n&=&\,3\\
\vdots\,\,\,&&\,\,\vdots&&\,\,\vdots&&
&&\,\,\,\,\vdots&&\,\,\,\,\vdots&&\,\,\,\vdots&&\,\vdots\\
x_1&+&x_2&&&+&\cdots&+&x_{n - 2}&+&x_{n - 1}&+&x_n&=&\,n - 2\\
x_1&&&+&x_3&+&\cdots&+&x_{n - 2}&+&x_{n - 1}&+&x_n&=&\,n - 1\\
&&x_2&+&x_3&+&\cdots&+&x_{n - 2}&+&x_{n - 1}&+&x_n&=&\,n
\end{split}
\right.
\end{equation*}
青青子衿发表于 2016-8-12 16:33

第一步：
先求系数矩阵($n$阶方阵)$\boldsymbol{A_\mathit{a}}=\left(a_{ij}\right)$的逆矩阵，其中$a_{i,n-i+1}=0$，$1\le i\le n$，$a_{ij}=1$，$1\le i\ne \left(n-i+1\right)\le n$.
构造$n\times2n$矩阵$\boldsymbol{B_\mathit{a}}$
\[\boldsymbol{B_\mathit{a}}=\left(\begin{array}{ccccccc|cccccc}
1&1&1& \cdots &1&1&0&1&0&0&\cdots&0&0&0\\
1&1&1& \cdots &1&0&1&0&1&0&\cdots&0&0&0\\
1&1&1& \cdots &0&1&1&0&0&1&\cdots&0&0&0\\
\vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots
&\vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\
1&1&0& \cdots &1&1&1&0&0&0&\cdots&1&0&0\\
1&0&1& \cdots &1&1&1&0&0&0&\cdots&0&1&0\\
0&1&1& \cdots &1&1&1&0&0&0&\cdots&0&0&1
\end{array}\right)\]
将矩阵$\boldsymbol{B_\mathit{a}}$的第$2$，$3$，$\cdots$，$n$行都加到第$1$行，得：
\[{{\boldsymbol B}_a}_{\overset{\phantom,}{\tt1}}=\left(\begin{array}{ccccccc|cccccc}
n-1&n-1&n-1& \cdots &n-1&n-1&n-1&1&1&1&\cdots&1&1&1\\
1&1&1& \cdots &1&0&1&0&1&0&\cdots&0&0&0\\
1&1&1& \cdots &0&1&1&0&0&1&\cdots&0&0&0\\
\vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots
&\vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\
1&1&0& \cdots &1&1&1&0&0&0&\cdots&1&0&0\\
1&0&1& \cdots &1&1&1&0&0&0&\cdots&0&1&0\\
0&1&1& \cdots &1&1&1&0&0&0&\cdots&0&0&1
\end{array}\right)\]
矩阵${{\boldsymbol B}_a}_{\overset{\phantom,}{\tt1}}$的第$1$行乘以$\frac{1}{n-1}$，得：
\[{{\boldsymbol B}_a}_{\overset{\phantom,}{\tt2}}=\left(\begin{array}{ccccccc|cccccc}
1&1&1& \cdots &1&1&1&\frac{1}{n-1}&\frac{1}{n-1}&\frac{1}{n-1}&\cdots&\frac{1}{n-1}&\frac{1}{n-1}&\frac{1}{n-1}\\
1&1&1& \cdots &1&0&1&0&1&0&\cdots&0&0&0\\
1&1&1& \cdots &0&1&1&0&0&1&\cdots&0&0&0\\
\vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots
&\vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\
1&1&0& \cdots &1&1&1&0&0&0&\cdots&1&0&0\\
1&0&1& \cdots &1&1&1&0&0&0&\cdots&0&1&0\\
0&1&1& \cdots &1&1&1&0&0&0&\cdots&0&0&1
\end{array}\right)\]
将矩阵${{\boldsymbol B}_a}_{\overset{\phantom,}{\tt2}}$的第$2$，$3$，$\cdots$，$n$行都减去第$1$行，得：
\[{{\boldsymbol B}_a}_{\overset{\phantom,}{\tt3}}=\left(\begin{array}{ccccccc|ccccccc}
1&1&1& \cdots &1&1&1&\frac{1}{n-1}&\frac{1}{n-1}&\frac{1}{n-1}&\cdots&\frac{1}{n-1}&\frac{1}{n-1}&\frac{1}{n-1}\\
0&0&0& \cdots &0&-1&0&-\frac{1}{n-1}&1-\frac{1}{n-1}&-\frac{1}{n-1}&\cdots&-\frac{1}{n-1}&-\frac{1}{n-1}&-\frac{1}{n-1}\\
0&0&0&\cdots &-1&0&0&-\frac{1}{n-1}&-\frac{1}{n-1}&1-\frac{1}{n-1}&\cdots&-\frac{1}{n-1}&-\frac{1}{n-1}&-\frac{1}{n-1}\\
\vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots
&\vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\
0&0&-1& \cdots &0&0&0&-\frac{1}{n-1}&-\frac{1}{n-1}&-\frac{1}{n-1}&\cdots&1-\frac{1}{n-1}&-\frac{1}{n-1}&-\frac{1}{n-1}\\
0&-1&0& \cdots &0&0&0&-\frac{1}{n-1}&-\frac{1}{n-1}&-\frac{1}{n-1}&\cdots&-\frac{1}{n-1}&1-\frac{1}{n-1}&-\frac{1}{n-1}\\
-1&0&0& \cdots &0&0&0&-\frac{1}{n-1}&-\frac{1}{n-1}&-\frac{1}{n-1}&\cdots&-\frac{1}{n-1}&-\frac{1}{n-1}&1-\frac{1}{n-1}
\end{array}\right)\]
将矩阵${{\boldsymbol B}_a}_{\overset{\phantom,}{\tt3}}$的第$2$，$3$，$\cdots$，$n$行都加到第$1$行，得：
\[{{\boldsymbol B}_a}_{\overset{\phantom,}{\tt4}}=\left(\begin{array}{ccccccc|ccccccc}
0&0&0& \cdots &0&0&1&-\frac{n-2}{n-1}&\frac{1}{n-1}&\frac{1}{n-1}&\cdots&\frac{1}{n-1}&\frac{1}{n-1}&\frac{1}{n-1}\\
0&0&0& \cdots &0&-1&0&-\frac{1}{n-1}&\frac{n-2}{n-1}&-\frac{1}{n-1}&\cdots&-\frac{1}{n-1}&-\frac{1}{n-1}&-\frac{1}{n-1}\\
0&0&0&\cdots &-1&0&0&-\frac{1}{n-1}&-\frac{1}{n-1}&\frac{n-2}{n-1}&\cdots&-\frac{1}{n-1}&-\frac{1}{n-1}&-\frac{1}{n-1}\\
\vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots
&\vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\
0&0&-1& \cdots &0&0&0&-\frac{1}{n-1}&-\frac{1}{n-1}&-\frac{1}{n-1}&\cdots&\frac{n-2}{n-1}&-\frac{1}{n-1}&-\frac{1}{n-1}\\
0&-1&0& \cdots &0&0&0&-\frac{1}{n-1}&-\frac{1}{n-1}&-\frac{1}{n-1}&\cdots&-\frac{1}{n-1}&\frac{n-2}{n-1}&-\frac{1}{n-1}\\
-1&0&0& \cdots &0&0&0&-\frac{1}{n-1}&-\frac{1}{n-1}&-\frac{1}{n-1}&\cdots&-\frac{1}{n-1}&-\frac{1}{n-1}&\frac{n-2}{n-1}
\end{array}\right)\]
将矩阵${{\boldsymbol B}_a}_{\overset{\phantom,}{\tt4}}$的第$2$，$3$，$\cdots$，$n$行都乘以$-1$，得：
\[{{\boldsymbol B}_a}_{\overset{\phantom,}{\tt5}}=\left(\begin{array}{ccccccc|ccccccc}
0&0&0& \cdots &0&0&1&-\frac{n-2}{n-1}&\frac{1}{n-1}&\frac{1}{n-1}&\cdots&\frac{1}{n-1}&\frac{1}{n-1}&\frac{1}{n-1}\\
0&0&0& \cdots &0&1&0&\frac{1}{n-1}&-\frac{n-2}{n-1}&\frac{1}{n-1}&\cdots&\frac{1}{n-1}&\frac{1}{n-1}&\frac{1}{n-1}\\
0&0&0&\cdots &1&0&0&\frac{1}{n-1}&\frac{1}{n-1}&-\frac{n-2}{n-1}&\cdots&\frac{1}{n-1}&\frac{1}{n-1}&\frac{1}{n-1}\\
\vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots
&\vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\
0&0&1& \cdots &0&0&0&\frac{1}{n-1}&\frac{1}{n-1}&\frac{1}{n-1}&\cdots&-\frac{n-2}{n-1}&\frac{1}{n-1}&\frac{1}{n-1}\\
0&1&0& \cdots &0&0&0&\frac{1}{n-1}&\frac{1}{n-1}&\frac{1}{n-1}&\cdots&\frac{1}{n-1}&-\frac{n-2}{n-1}&\frac{1}{n-1}\\
1&0&0& \cdots &0&0&0&\frac{1}{n-1}&\frac{1}{n-1}&\frac{1}{n-1}&\cdots&\frac{1}{n-1}&\frac{1}{n-1}&-\frac{n-2}{n-1}
\end{array}\right)\]
将矩阵${{\boldsymbol B}_a}_{\overset{\phantom,}{\tt5}}$的第$i$行与第$n-i+1$行交换其中$1\le i\le\left[\frac{n}{2}\right]$，得：
\[{{\boldsymbol B}_a}_{\overset{\phantom,}{\tt6}}=\left(\begin{array}{ccccccc|ccccccc}
1&0&0& \cdots &0&0&0&\frac{1}{n-1}&\frac{1}{n-1}&\frac{1}{n-1}&\cdots&\frac{1}{n-1}&\frac{1}{n-1}&-\frac{n-2}{n-1}\\
0&1&0& \cdots &0&0&0&\frac{1}{n-1}&\frac{1}{n-1}&\frac{1}{n-1}&\cdots&\frac{1}{n-1}&-\frac{n-2}{n-1}&\frac{1}{n-1}\\
0&0&1&\cdots &0&0&0&\frac{1}{n-1}&\frac{1}{n-1}&\frac{1}{n-1}&\cdots&-\frac{n-2}{n-1}&\frac{1}{n-1}&\frac{1}{n-1}\\
\vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots
&\vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\
0&0&0& \cdots &1&0&0&\frac{1}{n-1}&\frac{1}{n-1}&-\frac{n-2}{n-1}&\cdots&\frac{1}{n-1}&\frac{1}{n-1}&\frac{1}{n-1}\\
0&0&0& \cdots &0&1&0&\frac{1}{n-1}&-\frac{n-2}{n-1}&\frac{1}{n-1}&\cdots&\frac{1}{n-1}&\frac{1}{n-1}&\frac{1}{n-1}\\
0&0&0& \cdots &0&0&1&-\frac{n-2}{n-1}&\frac{1}{n-1}&\frac{1}{n-1}&\cdots&\frac{1}{n-1}&\frac{1}{n-1}&\frac{1}{n-1}
\end{array}\right)\]
由此求得，$\boldsymbol{A_\mathit{a}}^\mathrm{-1}=\left(a_{ij}\right)$，其中$a_{i,n-i+1}=-\frac{n-2}{n-1}$，$1\le i\le n$，$a_{ij}=\frac{1}{n-1}$，$1\le i\ne \left(n-i+1\right)\le n$.
第二步：求出解向量
接下来，只需要求系数矩阵的逆矩阵$\boldsymbol{A_\mathit{a}}^\mathrm{-1}$与常数项为列构造的列矩阵$\boldsymbol{\beta}$之积$\boldsymbol{x_\mathit{a}}=\boldsymbol{A_\mathit{a}}^\mathrm{-1}\cdot\boldsymbol{\beta}$
\[\begin{array}{}
\boldsymbol{x_\mathit{a}}=\left(\begin{array}{ccccccc}
\frac{1}{n-1}&\frac{1}{n-1}&\frac{1}{n-1}&\cdots&\frac{1}{n-1}&\frac{1}{n-1}&-\frac{n-2}{n-1}\\
\frac{1}{n-1}&\frac{1}{n-1}&\frac{1}{n-1}&\cdots&\frac{1}{n-1}&-\frac{n-2}{n-1}&\frac{1}{n-1}\\
\frac{1}{n-1}&\frac{1}{n-1}&\frac{1}{n-1}&\cdots&-\frac{n-2}{n-1}&\frac{1}{n-1}&\frac{1}{n-1}\\
\vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\
\frac{1}{n-1}&\frac{1}{n-1}&-\frac{n-2}{n-1}&\cdots&\frac{1}{n-1}&\frac{1}{n-1}&\frac{1}{n-1}\\
\frac{1}{n-1}&-\frac{n-2}{n-1}&\frac{1}{n-1}&\cdots&\frac{1}{n-1}&\frac{1}{n-1}&\frac{1}{n-1}\\
-\frac{n-2}{n-1}&\frac{1}{n-1}&\frac{1}{n-1}&\cdots&\frac{1}{n-1}&\frac{1}{n-1}&\frac{1}{n-1}
\end{array}\right)
\left(
\begin{array}{c}
1\vphantom{\frac{1}{n-1}}\\
2\vphantom{\frac{1}{n-1}}\\
3\vphantom{\frac{1}{n-1}}\\
\vdots\\
n-2\vphantom{\frac{1}{n-1}}\\
n-1\vphantom{\frac{1}{n-1}}\\
n \vphantom{\frac{1}{n-1}}
\end{array}
\right)\\

\boldsymbol{x_\mathit{a}}=
\left(
\begin{array}{c}
x_{a,1}\vphantom{\frac{1}{n-1}}\\
x_{a,2}\vphantom{\frac{1}{n-1}}\\
x_{a,3}\vphantom{\frac{1}{n-1}}\\
\vdots\\
x_{a,n-2}\vphantom{\frac{1}{n-1}}\\
x_{a,n-1}\vphantom{\frac{1}{n-1}}\\
x_{a,n}\vphantom{\frac{1}{n-1}}
\end{array}\right)
=
\left(
\begin{array}{c}
\frac{1}{n-1}-\frac{n}{2}+\color{green}{1}\\
\frac{1}{n-1}-\frac{n}{2}+\color{green}{2}\\
\frac{1}{n-1}-\frac{n}{2}+\color{green}{3}\\
\vdots\\
\frac{1}{n-1}-\frac{n}{2}+\color{green}{n-2}\\
\frac{1}{n-1}-\frac{n}{2}+\color{green}{n-1}\\
\frac{1}{n-1}-\frac{n}{2}+\color{green}{n}
\end{array}\right)
\end{array}\]
故，可以发现：
\[ \begin{equation*} \label{a2}\tag*{[Solution#2 of equations a]}
{x_a}_{\overset{\phantom,}{i}}=\frac{1}{n-1}-\frac{n}{2}+i
\end{equation*}\]
\ref{a2}

\begin{equation*}
\left\{
\begin{split}
&&x_2&+&x_3&+&\cdots&+&x_{n - 2}&+&x_{n - 1}&+&x_{n}&=&\,1\\
x_1&&&+&x_3&+&\cdots&+&x_{n - 2}&+&x_{n - 1}&+&x_n&=&\,2\\
x_1&+&x_2&&&+&\cdots&+&x_{n - 2}&+&x_{n - 1}&+&x_n&=&\,3\\
\vdots\,\,\,&&\,\,\vdots&&\,\,\vdots&&
&&\,\,\,\,\vdots&&\,\,\,\,\vdots&&\,\,\,\vdots&&\,\vdots\\
x_1&+&x_2&+&x_3&+&\cdots&&&+&x_{n - 1}&+&x_n&=&\,n - 2\\
x_1&+&x_2&+&x_3&+&\cdots&+&x_{n - 2}&&&+&x_n&=&\,n - 1\\
x_1&+&x_2&+&x_3&+&\cdots&+&x_{n - 2}&+&x_{n - 1}&&&=&\,n
\end{split}
\right.
\end{equation*}
青青子衿发表于 2016-8-12 16:33

第一步：
先求系数矩阵($n$阶方阵)$\boldsymbol{A_\mathit{b}}=\left(a_{ij}\right)$的逆矩阵，其中$a_{ii}=0$，$1\le i\le n$，$a_{ij}=1$，$1\le i\ne j\le n$.
$\color{red}{\fbox{原题参见《线性代数》(李炯生，中国科学技术大学出版社)\(P118$ 例$1$}}\)
构造$n\times2n$矩阵$\boldsymbol{B_\mathit{b}}$
\[\boldsymbol{B_\mathit{b}}=\left(\begin{array}{ccccccc|cccccc}
0&1&1& \cdots &1&1&1&1&0&0&\cdots&0&0&0\\
1&0&1& \cdots &1&1&1&0&1&0&\cdots&0&0&0\\
1&1&0& \cdots &1&1&1&0&0&1&\cdots&0&0&0\\
\vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots
&\vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\
1&1&1& \cdots &0&1&1&0&0&0&\cdots&1&0&0\\
1&1&1& \cdots &1&0&1&0&0&0&\cdots&0&1&0\\
1&1&1& \cdots &1&1&0&0&0&0&\cdots&0&0&1
\end{array}\right)\]
将矩阵$\boldsymbol{B_\mathit{b}}$的第$2$，$3$，$\cdots$，$n$行都加到第$1$行，得：
\[{{\boldsymbol B}_b}_{\overset{\phantom,}{\tt1}}=\left(\begin{array}{ccccccc|cccccc}
n-1&n-1&n-1& \cdots &n-1&n-1&n-1&1&1&1&\cdots&1&1&1\\
1&0&1& \cdots &1&1&1&0&1&0&\cdots&0&0&0\\
1&1&0& \cdots &1&1&1&0&0&1&\cdots&0&0&0\\
\vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots
&\vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\
1&1&1& \cdots &0&1&1&0&0&0&\cdots&1&0&0\\
1&1&1& \cdots &1&0&1&0&0&0&\cdots&0&1&0\\
1&1&1& \cdots &1&1&0&0&0&0&\cdots&0&0&1
\end{array}\right)\]
矩阵${{\boldsymbol B}_b}_{\overset{\phantom,}{\tt1}}$的第$1$行乘以$\frac{1}{n-1}$，得：
\[{{\boldsymbol B}_b}_{\overset{\phantom,}{\tt2}}=\left(\begin{array}{ccccccc|cccccc}
1&1&1& \cdots &1&1&1&\frac{1}{n-1}&\frac{1}{n-1}&\frac{1}{n-1}&\cdots&\frac{1}{n-1}&\frac{1}{n-1}&\frac{1}{n-1}\\
1&0&1& \cdots &1&1&1&0&1&0&\cdots&0&0&0\\
1&1&0& \cdots &1&1&1&0&0&1&\cdots&0&0&0\\
\vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots
&\vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\
1&1&1& \cdots &0&1&1&0&0&0&\cdots&1&0&0\\
1&1&1& \cdots &1&0&1&0&0&0&\cdots&0&1&0\\
1&1&1& \cdots &1&1&0&0&0&0&\cdots&0&0&1
\end{array}\right)\]
将矩阵${{\boldsymbol B}_b}_{\overset{\phantom,}{\tt2}}$的第$2$，$3$，$\cdots$，$n$行都减去第$1$行，得：
\[{{\boldsymbol B}_b}_{\overset{\phantom,}{\tt3}}=\left(\begin{array}{ccccccc|ccccccc}
1&1&1& \cdots &1&1&1&\frac{1}{n-1}&\frac{1}{n-1}&\frac{1}{n-1}&\cdots&\frac{1}{n-1}&\frac{1}{n-1}&\frac{1}{n-1}\\
0&-1&0& \cdots &0&0&0&-\frac{1}{n-1}&1-\frac{1}{n-1}&-\frac{1}{n-1}&\cdots&-\frac{1}{n-1}&-\frac{1}{n-1}&-\frac{1}{n-1}\\
0&0&-1&\cdots &0&0&0&-\frac{1}{n-1}&-\frac{1}{n-1}&1-\frac{1}{n-1}&\cdots&-\frac{1}{n-1}&-\frac{1}{n-1}&-\frac{1}{n-1}\\
\vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots
&\vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\
0&0&0& \cdots &-1&0&0&-\frac{1}{n-1}&-\frac{1}{n-1}&-\frac{1}{n-1}&\cdots&1-\frac{1}{n-1}&-\frac{1}{n-1}&-\frac{1}{n-1}\\
0&0&0& \cdots &0&-1&0&-\frac{1}{n-1}&-\frac{1}{n-1}&-\frac{1}{n-1}&\cdots&-\frac{1}{n-1}&1-\frac{1}{n-1}&-\frac{1}{n-1}\\
0&0&0& \cdots &0&0&-1&-\frac{1}{n-1}&-\frac{1}{n-1}&-\frac{1}{n-1}&\cdots&-\frac{1}{n-1}&-\frac{1}{n-1}&1-\frac{1}{n-1}
\end{array}\right)\]
将矩阵${{\boldsymbol B}_b}_{\overset{\phantom,}{\tt3}}$的第$2$，$3$，$\cdots$，$n$行都加到第$1$行，得：
\[{{\boldsymbol B}_b}_{\overset{\phantom,}{\tt4}}=\left(\begin{array}{ccccccc|ccccccc}
1&0&0& \cdots &0&0&0&-\frac{n-2}{n-1}&\frac{1}{n-1}&\frac{1}{n-1}&\cdots&\frac{1}{n-1}&\frac{1}{n-1}&\frac{1}{n-1}\\
0&-1&0& \cdots &0&0&0&-\frac{1}{n-1}&\frac{n-2}{n-1}&-\frac{1}{n-1}&\cdots&-1\frac{1}{n-1}&-\frac{1}{n-1}&-\frac{1}{n-1}\\
0&0&-1&\cdots &0&0&0&-\frac{1}{n-1}&-\frac{1}{n-1}&\frac{n-2}{n-1}&\cdots&-\frac{1}{n-1}&-\frac{1}{n-1}&-\frac{1}{n-1}\\
\vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots
&\vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\
0&0&0& \cdots &-1&0&0&-\frac{1}{n-1}&-\frac{1}{n-1}&-\frac{1}{n-1}&\cdots&\frac{n-2}{n-1}&-\frac{1}{n-1}&-\frac{1}{n-1}\\
0&0&0& \cdots &0&-1&0&-\frac{1}{n-1}&-\frac{1}{n-1}&-\frac{1}{n-1}&\cdots&-\frac{1}{n-1}&\frac{n-2}{n-1}&-\frac{1}{n-1}\\
0&0&0& \cdots &0&0&-1&-\frac{1}{n-1}&-\frac{1}{n-1}&-\frac{1}{n-1}&\cdots&-\frac{1}{n-1}&-\frac{1}{n-1}&\frac{n-2}{n-1}
\end{array}\right)\]
将矩阵${{\boldsymbol B}_b}_{\overset{\phantom,}{\tt4}}$的第$2$，$3$，$\cdots$，$n$行都乘以$-1$，得：
\[{{\boldsymbol B}_b}_{\overset{\phantom,}{\tt5}}=\left(\begin{array}{ccccccc|ccccccc}
1&0&0& \cdots &0&0&0&-\frac{n-2}{n-1}&\frac{1}{n-1}&\frac{1}{n-1}&\cdots&\frac{1}{n-1}&\frac{1}{n-1}&\frac{1}{n-1}\\
0&1&0& \cdots &0&0&0&\frac{1}{n-1}&-\frac{n-2}{n-1}&\frac{1}{n-1}&\cdots&\frac{1}{n-1}&\frac{1}{n-1}&\frac{1}{n-1}\\
0&0&1&\cdots &0&0&0&\frac{1}{n-1}&\frac{1}{n-1}&-\frac{n-2}{n-1}&\cdots&\frac{1}{n-1}&\frac{1}{n-1}&\frac{1}{n-1}\\
\vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots
&\vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\
0&0&0& \cdots &1&0&0&\frac{1}{n-1}&\frac{1}{n-1}&\frac{1}{n-1}&\cdots&-\frac{n-2}{n-1}&\frac{1}{n-1}&\frac{1}{n-1}\\
0&0&0& \cdots &0&1&0&\frac{1}{n-1}&\frac{1}{n-1}&\frac{1}{n-1}&\cdots&\frac{1}{n-1}&-\frac{n-2}{n-1}&\frac{1}{n-1}\\
0&0&0& \cdots &0&0&1&\frac{1}{n-1}&\frac{1}{n-1}&\frac{1}{n-1}&\cdots&\frac{1}{n-1}&\frac{1}{n-1}&-\frac{n-2}{n-1}
\end{array}\right)\]
由此求得，${\boldsymbol{A_\mathit{b}}^\mathrm{-1}}=\left(a_{ij}\right)$，其中$a_{ii}=-\frac{n-2}{n-1}$，$1\le i\le n$，$a_{ij}=\frac{1}{n-1}$，$1\le i\ne j\le n$.
第二步：求出解向量
接下来，只需要求系数矩阵的逆矩阵$\boldsymbol{A_\mathit{b}}^{-1}$与常数项为列构造的列矩阵$\boldsymbol{\beta}$之积$\boldsymbol{x_\mathit{b}}=\boldsymbol{A_\mathit{b}^{-1}}\cdot\boldsymbol{\beta}$
\[
\begin{array}{}
\boldsymbol{x_\mathit{b}}
=\left(\begin{array}{ccccccc}
-\frac{n-2}{n-1}&\frac{1}{n-1}&\frac{1}{n-1}&\cdots&\frac{1}{n-1}&\frac{1}{n-1}&\frac{1}{n-1}\\
\frac{1}{n-1}&-\frac{n-2}{n-1}&\frac{1}{n-1}&\cdots&\frac{1}{n-1}&\frac{1}{n-1}&\frac{1}{n-1}\\
\frac{1}{n-1}&\frac{1}{n-1}&-\frac{n-2}{n-1}&\cdots&\frac{1}{n-1}&\frac{1}{n-1}&\frac{1}{n-1}\\
\vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\
\frac{1}{n-1}&\frac{1}{n-1}&\frac{1}{n-1}&\cdots&-\frac{n-2}{n-1}&\frac{1}{n-1}&\frac{1}{n-1}\\
\frac{1}{n-1}&\frac{1}{n-1}&\frac{1}{n-1}&\cdots&\frac{1}{n-1}&-\frac{n-2}{n-1}&\frac{1}{n-1}\\
\frac{1}{n-1}&\frac{1}{n-1}&\frac{1}{n-1}&\cdots&\frac{1}{n-1}&\frac{1}{n-1}&-\frac{n-2}{n-1}
\end{array}\right)
\left(
\begin{array}{c}
1\vphantom{\frac{1}{n-1}}\\
2\vphantom{\frac{1}{n-1}}\\
3\vphantom{\frac{1}{n-1}}\\
\vdots\\
n-2\vphantom{\frac{1}{n-1}}\\
n-1\vphantom{\frac{1}{n-1}}\\
n \vphantom{\frac{1}{n-1}}
\end{array}\right)
\\
\boldsymbol{x_\mathit{b}}=
\left(
\begin{array}{c}
x_{b,1}\vphantom{\frac{1}{n-1}}\\
x_{b,2}\vphantom{\frac{1}{n-1}}\\
x_{b,3}\vphantom{\frac{1}{n-1}}\\
\vdots\\
x_{b,n-2}\vphantom{\frac{1}{n-1}}\\
x_{b,n-1}\vphantom{\frac{1}{n-1}}\\
x_{b,n} \vphantom{\frac{1}{n-1}}
\end{array}\right)
=
\left(
\begin{array}{c}
\frac{n}{n-1}+\frac{n}{2}-\color{green}{1}\\
\frac{n}{n-1}+\frac{n}{2}-\color{green}{2}\\
\frac{n}{n-1}+\frac{n}{2}-\color{green}{3}\\
\vdots\\
\frac{n}{n-1}+\frac{n}{2}-\left(\color{green}{n-2}\right)\\
\frac{n}{n-1}+\frac{n}{2}-\left(\color{green}{n-1}\right)\\
\frac{n}{n-1}+\frac{n}{2}-\color{green}{n}
\end{array}\right)
\end{array}
\]
故，可以发现：
\[ \begin{equation*} \label{b2}\tag*{[Solution#2 of equations b]}
{x_b}_{\overset{\phantom,}{i}}=\frac{n}{2}+\frac{n}{n-1}-i
\end{equation*}\]
\ref{b2}

青青子衿 · 发表于 2016-8-13 21:28

本帖最后由青青子衿于 2018-6-16 21:14 编辑 回复 1# 青青子衿
$\Huge{原理三：利用高斯-约当(Gauss-Jordan)消元法解线性方程组}$

\begin{equation*}
\left\{
\begin{split}
x_1&+&x_2&+&x_3&+&\cdots&+&x_{n - 2}&+&x_{n - 1}&&&=&\,1\\
x_1&+&x_2&+&x_3&+&\cdots&+&x_{n - 2}&&&+&x_n&=&\,2\\
x_1&+&x_2&+&x_3&+&\cdots&&&+&x_{n - 1}&+&x_n&=&\,3\\
\vdots\,\,\,&&\,\,\vdots&&\,\,\vdots&&
&&\,\,\,\,\vdots&&\,\,\,\,\vdots&&\,\,\,\vdots&&\,\vdots\\
x_1&+&x_2&&&+&\cdots&+&x_{n - 2}&+&x_{n - 1}&+&x_n&=&\,n - 2\\
x_1&&&+&x_3&+&\cdots&+&x_{n - 2}&+&x_{n - 1}&+&x_n&=&\,n - 1\\
&&x_2&+&x_3&+&\cdots&+&x_{n - 2}&+&x_{n - 1}&+&x_n&=&\,n
\end{split}
\right.
\end{equation*}
青青子衿发表于 2016-8-12 16:33

第一步：增广矩阵化成行最简
先构造系数矩阵($n$阶方阵)$\boldsymbol{C_\mathit{a}}=\left(a_{ij}\right)$的增广矩阵$\boldsymbol{C_\mathit{a}}=\left(\boldsymbol{A_\mathit{a}},\boldsymbol{\beta}\right)$，其中$a_{i,n-i+1}=0$，$1\le i\le n$，$a_{ij}=1$，$1\le i\ne \left(n-i+1\right)\le n$.
构造$n\times (n+1)$矩阵$\boldsymbol{C_\mathit{a}}$
\[\boldsymbol{C_\mathit{a}}=\left(\begin{array}{ccccccc|cccccc}
1&1&1& \cdots &1&1&0&1\\
1&1&1& \cdots &1&0&1&2\\
1&1&1& \cdots &0&1&1&3\\
\vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots& \vdots\\
1&1&0& \cdots &1&1&1&n-2\\
1&0&1& \cdots &1&1&1&n-1\\
0&1&1& \cdots &1&1&1&n
\end{array}\right)\]
将矩阵$\boldsymbol{C_\mathit{a}}$的第$2$，$3$，$\cdots$，$n$行都加到第$1$行，得：
\[{{\boldsymbol C}_a}_{\overset{\phantom,}{\tt1}}=\left(\begin{array}{ccccccc|c}
n-1&n-1&n-1& \cdots &n-1&n-1&n-1&\frac{n(n+1)}{2}\\
1&1&1& \cdots &1&0&1&2\\
1&1&1& \cdots &0&1&1&3\\
\vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots& \vdots\\
1&1&0& \cdots &1&1&1&n-2\\
1&0&1& \cdots &1&1&1&n-1\\
0&1&1& \cdots &1&1&1&n
\end{array}\right)\]
矩阵${{\boldsymbol C}_a}_{\overset{\phantom,}{\tt1}}$的第$1$行乘以$\frac{1}{n-1}$，得：
\[{{\boldsymbol C}_a}_{\overset{\phantom,}{\tt2}}=\left(\begin{array}{ccccccc|c}
1&1&1& \cdots &1&1&1&\frac{n(n+1)}{2(n-1)}\\
1&1&1& \cdots &1&0&1&2\\
1&1&1& \cdots &0&1&1&3\\
\vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots &\vdots\\
1&1&0& \cdots &1&1&1&n-2\\
1&0&1& \cdots &1&1&1&n-1\\
0&1&1& \cdots &1&1&1&n
\end{array}\right)\]
将矩阵${{\boldsymbol C}_a}_{\overset{\phantom,}{\tt2}}$的第$2$，$3$，$\cdots$，$n$行都减去第$1$行，得：
\[{{\boldsymbol C}_a}_{\overset{\phantom,}{\tt3}}=\left(\begin{array}{ccccccc|c}
1&1&1& \cdots &1&1&1&\frac{n}{2}+\frac{1}{n-1}+1\\
0&0&0& \cdots &0&-1&0&1-\frac{n}{2}-\frac{1}{n-1}\\
0&0&0&\cdots &-1&0&0&2-\frac{n}{2}-\frac{1}{n-1}\\
\vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots &\vdots\\
0&0&-1& \cdots &0&0&0&n-3-\frac{n}{2}-\frac{1}{n-1}\\
0&-1&0& \cdots &0&0&0&n-2-\frac{n}{2}-\frac{1}{n-1}\\
-1&0&0& \cdots &0&0&0&n-1-\frac{n}{2}-\frac{1}{n-1}
\end{array}\right)\]
将矩阵${{\boldsymbol C}_a}_{\overset{\phantom,}{\tt3}}$的第$2$，$3$，$\cdots$，$n$行都加到第$1$行，得：
\[{{\boldsymbol C}_a}_{\overset{\phantom,}{\tt4}}=\left(\begin{array}{ccccccc|c}
0&0&0& \cdots &0&0&1&\frac{n}{2}+\frac{1}{n-1}\\
0&0&0& \cdots &0&-1&0&1-\frac{n}{2}-\frac{1}{n-1}\\
0&0&0&\cdots &-1&0&0&2-\frac{n}{2}-\frac{1}{n-1}\\
\vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots &\vdots\\
0&0&-1& \cdots &0&0&0&n-3-\frac{n}{2}-\frac{1}{n-1}\\
0&-1&0& \cdots &0&0&0&n-2-\frac{n}{2}-\frac{1}{n-1}\\
-1&0&0& \cdots &0&0&0&n-1-\frac{n}{2}-\frac{1}{n-1}
\end{array}\right)\]
将矩阵${{\boldsymbol C}_a}_{\overset{\phantom,}{\tt4}}$的第$2$，$3$，$\cdots$，$n$行都乘以$-1$，得：
\[{{\boldsymbol C}_a}_{\overset{\phantom,}{\tt5}}=\left(\begin{array}{ccccccc|c}
0&0&0& \cdots &0&0&1&\frac{n}{2}+\frac{1}{n-1}\\
0&0&0& \cdots &0&1&0&\frac{n}{2}+\frac{1}{n-1}-1\\
0&0&0&\cdots &1&0&0&\frac{n}{2}+\frac{1}{n-1}-2\\
\vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots &\vdots\\
0&0&1& \cdots &0&0&0&\frac{n}{2}+\frac{1}{n-1}-(n-3)\\
0&1&0& \cdots &0&0&0&\frac{n}{2}+\frac{1}{n-1}-(n-2)\\
1&0&0& \cdots &0&0&0&\frac{n}{2}+\frac{1}{n-1}-(n-1)
\end{array}\right)\]
将矩阵${{\boldsymbol C}_a}_{\overset{\phantom,}{\tt5}}$的第$i$行与第$n-i+1$行交换其中$1\le i\le\left[\frac{n}{2}\right]$，得：
\[{{\boldsymbol C}_a}_{\overset{\phantom,}{\tt6}}=\left(\begin{array}{ccccccc|c}
1&0&0& \cdots &0&0&0&\frac{1}{n-1}-\frac{n}{2}+1\\
0&1&0& \cdots &0&0&0&\frac{1}{n-1}-\frac{n}{2}+2\\
0&0&1&\cdots &0&0&0&\frac{1}{n-1}-\frac{n}{2}+3\\
\vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots &\vdots\\
0&0&0& \cdots &1&0&0&\frac{1}{n-1}-\frac{n}{2}+(n-2)\\
0&0&0& \cdots &0&1&0&\frac{1}{n-1}-\frac{n}{2}+(n-1)\\
0&0&0& \cdots &0&0&1&\frac{1}{n-1}-\frac{n}{2}+n
\end{array}\right)\]
第二步：写出解向量
因此
\[
\boldsymbol{x_\mathit{a}}=
\left(
\begin{array}{c}
x_{a,1}\vphantom{\frac{1}{n-1}}\\
x_{a,2}\vphantom{\frac{1}{n-1}}\\
x_{a,3}\vphantom{\frac{1}{n-1}}\\
\vdots\\
x_{a,n-2}\vphantom{\frac{1}{n-1}}\\
x_{a,n-1}\vphantom{\frac{1}{n-1}}\\
x_{a,n}\vphantom{\frac{1}{n-1}}
\end{array}\right)
=
\left(
\begin{array}{c}
\frac{1}{n-1}-\frac{n}{2}+\color{green}{1}\\
\frac{1}{n-1}-\frac{n}{2}+\color{green}{2}\\
\frac{1}{n-1}-\frac{n}{2}+\color{green}{3}\\
\vdots\\
\frac{1}{n-1}-\frac{n}{2}+\color{green}{n-2}\\
\frac{1}{n-1}-\frac{n}{2}+\color{green}{n-1}\\
\frac{1}{n-1}-\frac{n}{2}+\color{green}{n}
\end{array}\right)
\]
故，可以发现：
\[ \begin{equation*}\label{a 3}\tag*{[Solution#3 of equations a]}
{x_a}_{\overset{\phantom,}{i}}=\frac{1}{n-1}-\frac{n}{2}+i
\end{equation*}\]
\ref{a 3}

\begin{equation*}
\left\{
\begin{split}
&&x_2&+&x_3&+&\cdots&+&x_{n - 2}&+&x_{n - 1}&+&x_{n}&=&\,1\\
x_1&&&+&x_3&+&\cdots&+&x_{n - 2}&+&x_{n - 1}&+&x_n&=&\,2\\
x_1&+&x_2&&&+&\cdots&+&x_{n - 2}&+&x_{n - 1}&+&x_n&=&\,3\\
\vdots\,\,\,&&\,\,\vdots&&\,\,\vdots&&
&&\,\,\,\,\vdots&&\,\,\,\,\vdots&&\,\,\,\vdots&&\,\vdots\\
x_1&+&x_2&+&x_3&+&\cdots&&&+&x_{n - 1}&+&x_n&=&\,n - 2\\
x_1&+&x_2&+&x_3&+&\cdots&+&x_{n - 2}&&&+&x_n&=&\,n - 1\\
x_1&+&x_2&+&x_3&+&\cdots&+&x_{n - 2}&+&x_{n - 1}&&&=&\,n
\end{split}
\right.
\end{equation*}
青青子衿发表于 2016-8-12 16:33

第一步：增广矩阵化成行最简
先构造系数矩阵($n$阶方阵)$\boldsymbol{A_\mathit{b}}=\left(a_{ij}\right)$的增广矩阵$\boldsymbol{C_\mathit{b}}=\left(\boldsymbol{A_\mathit{b}},\boldsymbol{\beta}\right)$，其中$a_{ii}=0$，$1\le i\le n$，$a_{ij}=1$，$1\le i\ne j\le n$.
构造$n\times(n+1)$矩阵$\boldsymbol{C_\mathit{b}}$
\[\boldsymbol{C_\mathit{b}}=\left(\begin{array}{ccccccc|c}
0&1&1& \cdots &1&1&1&1\\
1&0&1& \cdots &1&1&1&2\\
1&1&0& \cdots &1&1&1&3\\
\vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots &\vdots \\
1&1&1& \cdots &0&1&1&n-2\\
1&1&1& \cdots &1&0&1&n-1\\
1&1&1& \cdots &1&1&0&n
\end{array}\right)\]
将矩阵$\boldsymbol{C_\mathit{b}}$的第$2$，$3$，$\cdots$，$n$行都加到第$1$行，得：
\[{{\boldsymbol C}_b}_{\overset{\phantom,}{\tt1}}=\left(\begin{array}{ccccccc|c}
n-1&n-1&n-1& \cdots &n-1&n-1&n-1&\frac{n(n+1)}{2}\\
1&0&1& \cdots &1&1&1&2\\
1&1&0& \cdots &1&1&1&3\\
\vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots &\vdots\\
1&1&1& \cdots &0&1&1&n-2\\
1&1&1& \cdots &1&0&1&n-1\\
1&1&1& \cdots &1&1&0&n
\end{array}\right)\]
矩阵${{\boldsymbol C}_b}_{\overset{\phantom,}{\tt1}}$的第$1$行乘以$\frac{1}{n-1}$，得：
\[{{\boldsymbol C}_b}_{\overset{\phantom,}{\tt2}}=\left(\begin{array}{ccccccc|c}
1&1&1& \cdots &1&1&1&\frac{n(n+1)}{2(n-1)}\\
1&0&1& \cdots &1&1&1&2\\
1&1&0& \cdots &1&1&1&3\\
\vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots &\vdots\\
1&1&1& \cdots &0&1&1&n-2\\
1&1&1& \cdots &1&0&1&n-1\\
1&1&1& \cdots &1&1&0&n
\end{array}\right)\]
将矩阵${{\boldsymbol C}_b}_{\overset{\phantom,}{\tt2}}$的第$2$，$3$，$\cdots$，$n$行都减去第$1$行，得：
\[{{\boldsymbol C}_b}_{\overset{\phantom,}{\tt3}}=\left(\begin{array}{ccccccc|c}
1&1&1& \cdots &1&1&1&\frac{n}{2}+\frac{1}{n-1}+1\\
0&-1&0& \cdots &0&0&0&1-\frac{n}{2}-\frac{1}{n-1}\\
0&0&-1&\cdots &0&0&0&2-\frac{n}{2}-\frac{1}{n-1}\\
\vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots &\vdots\\
0&0&0& \cdots &-1&0&0&n-3-\frac{n}{2}-\frac{1}{n-1}\\
0&0&0& \cdots &0&-1&0&n-2-\frac{n}{2}-\frac{1}{n-1}\\
0&0&0& \cdots &0&0&-1&n-1-\frac{n}{2}-\frac{1}{n-1}
\end{array}\right)\]
将矩阵${{\boldsymbol C}_b}_{\overset{\phantom,}{\tt3}}$的第$2$，$3$，$\cdots$，$n$行都加到第$1$行，得：
\[{{\boldsymbol C}_b}_{\overset{\phantom,}{\tt4}}=\left(\begin{array}{ccccccc|c}
1&0&0& \cdots &0&0&0&\frac{n}{2}+\frac{1}{n-1}\\
0&-1&0& \cdots &0&0&0&1-\frac{n}{2}-\frac{1}{n-1}\\
0&0&-1&\cdots &0&0&0&2-\frac{n}{2}-\frac{1}{n-1}\\
\vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots &\vdots\\
0&0&0& \cdots &-1&0&0&n-3-\frac{n}{2}-\frac{1}{n-1}\\
0&0&0& \cdots &0&-1&0&n-2-\frac{n}{2}-\frac{1}{n-1}\\
0&0&0& \cdots &0&0&-1&n-1-\frac{n}{2}-\frac{1}{n-1}
\end{array}\right)\]
将矩阵${{\boldsymbol C}_b}_{\overset{\phantom,}{\tt4}}$的第$2$，$3$，$\cdots$，$n$行都乘以$-1$，得：
\[{{\boldsymbol C}_b}_{\overset{\phantom,}{\tt5}}=\left(\begin{array}{ccccccc|c}
1&0&0& \cdots &0&0&1&\frac{n}{2}+\frac{1}{n-1}\\
0&1&0& \cdots &0&0&0&\frac{n}{2}+\frac{1}{n-1}-1\\
0&0&1&\cdots &0&0&0&\frac{n}{2}+\frac{1}{n-1}-2\\
\vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots &\vdots\\
0&0&0& \cdots &1&0&0&\frac{n}{2}+\frac{1}{n-1}-(n-3)\\
0&0&0& \cdots &0&1&0&\frac{n}{2}+\frac{1}{n-1}-(n-2)\\
0&0&0& \cdots &0&0&1&\frac{n}{2}+\frac{1}{n-1}-(n-1)
\end{array}\right)\]
第二步：写出解向量
因此
\[
\boldsymbol{x_\mathit{b}}=
\left(
\begin{array}{c}
x_{b,1}\vphantom{\frac{1}{n-1}}\\
x_{b,2}\vphantom{\frac{1}{n-1}}\\
x_{b,3}\vphantom{\frac{1}{n-1}}\\
\vdots\\
x_{b,n-2}\vphantom{\frac{1}{n-1}}\\
x_{b,n-1}\vphantom{\frac{1}{n-1}}\\
x_{b,n}\vphantom{\frac{1}{n-1}}
\end{array}\right)
=
\left(
\begin{array}{c}
\frac{n}{2}+\frac{n}{n-1}-\color{green}{1}\\
\frac{n}{2}+\frac{n}{n-1}-\color{green}{2}\\
\frac{n}{2}+\frac{n}{n-1}-\color{green}{3}\\
\vdots\\
\frac{n}{2}+\frac{n}{n-1}-\left(\color{green}{n-2}\right)\\
\frac{n}{2}+\frac{n}{n-1}-\left(\color{green}{n-1}\right)\\
\frac{n}{2}+\frac{n}{n-1}-\color{green}{n}
\end{array}\right)
\]
故，可以发现：
\begin{equation*} \label{b 3}\tag*{[Solution#3 of equations b]}
{x_b}_{\overset{\phantom,}{i}}=\frac{n}{2}+\frac{n}{n-1}-i
\end{equation*}
\ref{b 3}

青青子衿 · 发表于 2016-10-25 23:24

本帖最后由青青子衿于 2016-11-5 22:10 编辑 其中$\color{orange}{\boldsymbol{A_\mathit{a}}}$的逆矩阵$\color{orange}{\boldsymbol{A_\mathit{a}}^{-1}}$和$\color{violet}{\boldsymbol{A_\mathit{b}}}$的逆矩阵$\color{violet}{\boldsymbol{A_\mathit{b}}^{-1}}$还可以这样求：
回复 3# 青青子衿

回复青青子衿
$\Huge{原理二：利用逆矩阵解线性方程组}$

由此求得，$\boldsymbol{A_\mathit{a}}^{-1}=\left(a_{ij}\right)$，其中$a_{i,n-i+1}=-\frac{n-2}{n-1}$，$1\le i\le n$，$a_{ij}=\frac{1}{n-1}$，$1\le i\ne \left(n-i+1\right)\le n$.

青青子衿发表于 2016-8-12 17:03

\[\color{orange}{\boldsymbol{A_\mathit{a}}}
=\left( {\begin{array}{{c}}
1&1&1&\cdots&0\\
\vdots&\vdots&\vdots& &\vdots\\
1&1&0&\cdots&1\\
1&0&1&\cdots&1\\
0&1&1&\cdots&1
\end{array}} \right)\]
若用$\boldsymbol{J}$表示元素均为$1$的$n$级矩阵（即$\boldsymbol{J} = \boldsymbol{1_n}{\boldsymbol{1_n}}^\prime $），则$\boldsymbol{A_\mathit{a}}=\boldsymbol{J}-\boldsymbol{S}$（其中矩阵$\boldsymbol{S}$为元素均为$1$的$n$级斜对角(Skew Diagonal)矩阵，且$\boldsymbol{S}^2=\boldsymbol{E}$（即矩阵$\boldsymbol{S}$是对合矩阵），$\boldsymbol{JS}=\boldsymbol{SJ}=\boldsymbol{J}$）
于是，$\color{orange}{\boldsymbol{A_\mathit{a}}^{-1}}=a\boldsymbol{S} + b\boldsymbol{J}$当且仅当下式成立：
\[\begin{array}{rl}
\boldsymbol{E}&= \left({\boldsymbol{J}-\boldsymbol{S}} \right)\left({a\boldsymbol{S} +b\boldsymbol{J}} \right) = a\boldsymbol{J}+b\boldsymbol{J}^2 - a\boldsymbol{E} - b\boldsymbol{J}\\
&=\left({a-b}\right)\boldsymbol{J}+b{\boldsymbol{J}^2}-a\boldsymbol{E} = \left( {a - b} \right)\boldsymbol{J}+ b\boldsymbol{1_n}{\boldsymbol{1_n}}^\prime\boldsymbol{1_n}{\boldsymbol{1_n}}^\prime-a\boldsymbol{E}\\
&=\left({a-b}\right)\boldsymbol{J}+nb\boldsymbol{1_n}{\boldsymbol{1_n}}^\prime- a\boldsymbol{E}=\left({a - b}\right)\boldsymbol{J} + nb\boldsymbol{J} - a\boldsymbol{E}\\
&= \left( {a - b + nb} \right)\boldsymbol{J}- a\boldsymbol{E}
\end{array}\]
解得
\[\left\{ \begin{array}{l}
a=-1\\
b=\frac{1}{{n - 1}}
\end{array} \right.\]
因此
\begin{align*}
\color{orange}{\boldsymbol{A_\mathit{a}}^{-1}}
&= \begin{pmatrix}
\frac{1}{n - 1}&\frac{1}{n - 1}&\frac{1}{n - 1}&\cdots&\frac{1}{n - 1} - 1\\
\vdots&\vdots&\vdots& &\vdots\\
\frac{1}{n - 1}&\frac{1}{n - 1}&\frac{1}{n - 1} - 1&\cdots&\frac{1}{n - 1}\\
\frac{1}{n - 1}&\frac{1}{n - 1} - 1&\frac{1}{n - 1}&\cdots&\frac{1}{n - 1}\\
\frac{1}{n - 1} - 1&\frac{1}{n - 1}&\frac{1}{n - 1}&\cdots&\frac{1}{n - 1}
\end{pmatrix}\\
\\
&= \begin{pmatrix}
\frac{1}{n - 1}&\frac{1}{n - 1}&\frac{1}{n - 1}&\cdots&-\frac{n-2}{n-1}\\
\vdots&\vdots&\vdots& &\vdots\\
\frac{1}{n - 1}&\frac{1}{n - 1}&-\frac{n-2}{n-1}&\cdots&\frac{1}{n - 1}\\
\frac{1}{n - 1}&-\frac{n-2}{n-1}&\frac{1}{n - 1}&\cdots&\frac{1}{n - 1}\\
-\frac{n-2}{n-1}&\frac{1}{n - 1}&\frac{1}{n - 1}&\cdots&\frac{1}{n - 1}
\end{pmatrix}
\end{align*}

回复青青子衿
$\Huge{原理二：利用逆矩阵解线性方程组}$

由此求得，$\boldsymbol{A_\mathit{b}}^{-1}=\left(a_{ij}\right)$，其中$a_{ii}=-\frac{n-2}{n-1}$，$1\le i\le n$，$a_{ij}=\frac{1}{n-1}$，$1\le i\ne j\le n$.

青青子衿发表于 2016-8-12 17:03

$\color{red}{\fbox{原题参见《高等代数(上册)——大学高等代数课程创新教材》(丘维声，清华大学出版社)\(P186$ 例7}}\)
\[
\color{violet}{\boldsymbol{A_\mathit{b}}}=
\left( {\begin{array}{{c}}
0&1&1& \cdots &1\\
1&0&1& \cdots &1\\
1&1&0& \cdots &1\\
\vdots & \vdots & \vdots &{}& \vdots \\
1&1&1& \cdots &0
\end{array}} \right)\]
若用$\boldsymbol{J}$表示元素均为$1$的$n$级矩阵（即$\boldsymbol{J}=\boldsymbol{1_n}{\boldsymbol{1_n}}^\prime$），则$\boldsymbol{A_\mathit{b}}=\boldsymbol{J}-\boldsymbol{E}$（其中$\boldsymbol{E}$为$n$级单位矩阵）
于是，$\color{violet}{\boldsymbol{A_\mathit{b}}^{-1}}=a\boldsymbol{E} + b\boldsymbol{J}$当且仅当下式成立：
\[\begin{array}{rl}
\boldsymbol{E}&=\left( {\boldsymbol{J}-\boldsymbol{E}}\right)\left( {a\boldsymbol{E} + b\boldsymbol{J}}\right) = a\boldsymbol{J}+ b\boldsymbol{J}^2- a\boldsymbol{E}- b\boldsymbol{J}\\
&=\left({a-b}\right)\boldsymbol{J}+ b\boldsymbol{J}^2 - a\boldsymbol{E}= \left({a-b} \right)\boldsymbol{J}+ b\boldsymbol{1_n}{\boldsymbol{1_n}}^\prime\boldsymbol{1_n}{\boldsymbol{1_n}}^\prime- a\boldsymbol{E}\\
&= \left({a - b}\right)\boldsymbol{J}+ nb\boldsymbol{1_n}{\boldsymbol{1_n}}^\prime-a\boldsymbol{E}= \left({a - b}\right)\boldsymbol{J} + nb\boldsymbol{J}- a\boldsymbol{E}\\
&= \left( {a- b+nb} \right)\boldsymbol{J}- a\boldsymbol{E}
\end{array}\]
解得
\[\left\{ \begin{array}{l}
a =  - 1\\
b = \frac{1}{{n - 1}}
\end{array} \right.\]
因此
\begin{align*}
\color{violet}{\boldsymbol{A_\mathit{b}}^{-1}}
&= \begin{pmatrix}
\frac{1}{n - 1} - 1&\frac{1}{n - 1}&\frac{1}{n - 1}& \cdots &\frac{1}{n - 1}\\
\frac{1}{n - 1}&\frac{1}{n - 1} - 1&\frac{1}{n - 1}& \cdots &\frac{1}{n - 1}\\
\frac{1}{n - 1}&\frac{1}{n - 1}&\frac{1}{n - 1} - 1& \cdots &\frac{1}{n - 1}\\
\vdots & \vdots & \vdots &{}& \vdots \\
\frac{1}{n - 1}&\frac{1}{n - 1}&\frac{1}{n - 1}& \cdots &\frac{1}{n - 1} - 1
\end{pmatrix}\\
\\
&= \begin{pmatrix}
-\frac{n-2}{n-1}&\frac{1}{n - 1}&\frac{1}{n - 1}& \cdots &\frac{1}{n - 1}\\
\frac{1}{n - 1}&-\frac{n-2}{n-1}&\frac{1}{n - 1}& \cdots &\frac{1}{n - 1}\\
\frac{1}{n - 1}&\frac{1}{n - 1}&-\frac{n-2}{n-1}& \cdots &\frac{1}{n - 1}\\
\vdots & \vdots & \vdots &{}& \vdots \\
\frac{1}{n - 1}&\frac{1}{n - 1}&\frac{1}{n - 1}& \cdots &-\frac{n-2}{n-1}
\end{pmatrix}
\end{align*}

—————————————————————————————————————————————————————————————————————
注：可能有人会好奇：为什么构造$\color{orange}{\boldsymbol{A_\mathit{a}}^{-1}}=a\boldsymbol{S} + b\boldsymbol{J}$和$\color{violet}{\boldsymbol{A_\mathit{b}}^{-1}}=a\boldsymbol{E} + b\boldsymbol{J}$？
实际上有如下定理可以保证。
$Sherman–Morrison\ \ formula$:en.wikipedia.org/wiki/Sherman%E2%80%93Morrison_formula

矩阵$\boldsymbol{A}$为$n$级$\color{red}{可逆}$矩阵，向量$\boldsymbol{u}$与向量$\boldsymbol{v}$为$n$维列向量，且$1+\boldsymbol{v^T}\boldsymbol{A}^{-1}\boldsymbol{u}\ne0$，则有：
\[\left(\boldsymbol{A}+\boldsymbol{uv^T}\right)^{-1}=\boldsymbol{A}^{-1}-\frac{\boldsymbol{A}^{-1}\boldsymbol{uv^T}\boldsymbol{A}^{-1}}{1+\boldsymbol{v^T}\boldsymbol{A}^{-1}\boldsymbol{u}}\]
此题就是$Sherman–Morrison\ \ formula$的特殊形式：
\begin{align*}
\color{orange}{\boldsymbol{A_\mathit{a}}^{-1}}&=\left(-\boldsymbol{S}+\boldsymbol{1_n}{\boldsymbol{1_n}}^\prime\right)^{-1}=-\boldsymbol{S}-\frac{(-\boldsymbol{S})\boldsymbol{1_n}{\boldsymbol{1_n}}^\prime(-\boldsymbol{S})}{1+{\boldsymbol{1_n}}^\prime(-\boldsymbol{S})\boldsymbol{1_n}}\\
&=-\boldsymbol{S}-\frac{\boldsymbol{1_n}{\boldsymbol{1_n}}^\prime}{1-{\boldsymbol{1_n}}^\prime\boldsymbol{1_n}}=-\boldsymbol{S}-\frac{\boldsymbol{J}}{1-n}\\
&=\frac{\boldsymbol{J}}{n-1}-\boldsymbol{S}
\end{align*}
\begin{align*}
\color{violet}{\boldsymbol{A_\mathit{b}}^{-1}}&=\left(-\boldsymbol{E}+\boldsymbol{1_n}{\boldsymbol{1_n}}^\prime\right)^{-1}=-\boldsymbol{E}-\frac{(-\boldsymbol{E})\boldsymbol{1_n}{\boldsymbol{1_n}}^\prime(-\boldsymbol{E})}{1+{\boldsymbol{1_n}}^\prime(-\boldsymbol{E})\boldsymbol{1_n}}\\
&=-\boldsymbol{E}-\frac{\boldsymbol{1_n}{\boldsymbol{1_n}}^\prime}{1-{\boldsymbol{1_n}}^\prime\boldsymbol{1_n}}=-\boldsymbol{E}-\frac{\boldsymbol{J}}{1-n}\\
&=\frac{\boldsymbol{J}}{n-1}-\boldsymbol{E}
\end{align*}

另外，还有其推广：
$Woodbury\ \ matrix\ \ identity$:en.wikipedia.org/wiki/Woodbury_matrix_identity

矩阵$\boldsymbol{A}$为$n$级$\color{red}{可逆}$矩阵，矩阵$\boldsymbol{U}$为$n\times m$矩阵，矩阵$\boldsymbol{B}$为$m$级$\color{red}{可逆}$矩阵，矩阵$\boldsymbol{V}$为$m\times n$矩阵，且矩阵$\boldsymbol{B}^{-1}+\boldsymbol{V}\boldsymbol{A}^{-1}\boldsymbol{U}$$\color{red}{可逆}$，则有：
\[\left(\boldsymbol{A}+\boldsymbol{UBV}\right)^{-1}=\boldsymbol{A}^{-1}-\boldsymbol{A}^{-1}\boldsymbol{U}\left(\boldsymbol{B}^{-1}+\boldsymbol{V}\boldsymbol{A}^{-1}\boldsymbol{U}\right)^{-1}\boldsymbol{V}\boldsymbol{A}^{-1}\]
$Binomial\ \ inverse\ \ theorem$:en.wikipedia.org/wiki/Binomial_inverse_theorem

矩阵$\boldsymbol{A}$为$n$级$\color{red}{可逆}$矩阵，矩阵$\boldsymbol{U}$为$n\times m$矩阵，矩阵$\boldsymbol{B}$为$m$级矩阵，矩阵$\boldsymbol{V}$为$m\times n$矩阵，且矩阵$\boldsymbol{B}+\boldsymbol{B}\boldsymbol{V}\boldsymbol{A}^{-1}\boldsymbol{U}\boldsymbol{B}$$\color{red}{可逆}$，则有：
\[\left(\boldsymbol{A}+\boldsymbol{UBV}\right)^{-1}=\boldsymbol{A}^{-1}-\boldsymbol{A}^{-1}\boldsymbol{U}\boldsymbol{B}\left(\boldsymbol{B}+\boldsymbol{B}\boldsymbol{V}\boldsymbol{A}^{-1}\boldsymbol{U}\boldsymbol{B}\right)^{-1}\boldsymbol{B}\boldsymbol{V}\boldsymbol{A}^{-1}\]

其中$Sherman–Morrison\ \ formula$和$Woodbury\ \ matrix\ \ identity$都可以由分块矩阵求逆(blockwise matrix inversion)公式得到
$Blockwise\ \ matrix\ \ inversion$:en.wikipedia.org/wiki/Invertible_matrix#Blockwise_inversion

矩阵$\boldsymbol{A}$为$n$级$\color{red}{可逆}$矩阵，矩阵$\boldsymbol{B}$为$n\times m$矩阵，矩阵$\boldsymbol{C}$为$m\times n$矩阵，矩阵$\boldsymbol{D}$为$m$级矩阵，且矩阵$\boldsymbol{D}-\boldsymbol{C}\boldsymbol{A}^{-1}\boldsymbol{B}$$\color{red}{可逆}$，则有：
\begin{align*}
\begin{pmatrix}
\boldsymbol{A}&\boldsymbol{B}\\
\boldsymbol{C}&\boldsymbol{D}
\end{pmatrix}^{-1}
= \begin{pmatrix}
\boldsymbol{A}^{-1}+\boldsymbol{A}^{-1}\boldsymbol{B}\left(\boldsymbol{D}-\boldsymbol{C}\boldsymbol{A}^{-1}\boldsymbol{B}\right)^{-1}\boldsymbol{C}\boldsymbol{A}^{-1}&&-\boldsymbol{A}^{-1}\boldsymbol{B}\left(\boldsymbol{D}-\boldsymbol{C}\boldsymbol{A}^{-1}\boldsymbol{B}\right)^{-1}\\
-\left(\boldsymbol{D}-\boldsymbol{C}\boldsymbol{A}^{-1}\boldsymbol{B}\right)^{-1}\boldsymbol{C}\boldsymbol{A}^{-1}&&\left(\boldsymbol{D}-\boldsymbol{C}\boldsymbol{A}^{-1}\boldsymbol{B}\right)^{-1}\end{pmatrix}
\end{align*}

en.wikibooks.org/wiki/LaTeX/Mathematics#Notes

isee · 发表于 2016-10-27 22:53

这个代码是算是大工程了。

个人喜欢三，高效易操作。

矩阵看着美，不好操作，至少对偶。

战巡 · 发表于 2016-10-28 02:26

其实矩阵一点也不麻烦，只是需要技巧，比如5楼那里的办法就很好
1楼的办法也可以大幅简化，强行求行列式那是个笨办法

这里首先介绍Sylvester的行列式定理：若$X_{n\times n}$为可逆矩阵，$A_{n\times m},B_{m\times n}$为另外两矩阵，有：
\[|X+AB|=|X||I_m+BX^{-1}A|\]

为了避免使用反对角的单位阵这种麻烦的东西，对$x$进行重新排列，令$y_k=x_{n-k}$，有
\[(J_n-I_n)y=(1,2,...,n)'\]
由于$J_n=1_n1_n'$,有
\[D_a=|J_n-I_n|=(-1)^n|I_1-1_n'1_n|=(-1)^{n-1}(n-1)\]
又易证
\[D_{ak}=|A_kB_k-I_n|\]
其中
\[A_k=\begin{pmatrix}1 && 0\\1 && 1\\...&&...\\1 && k-2\\1 && k\\1 && k\\...&&...\\1 && n-1\end{pmatrix}\]
\[B_k=\begin{pmatrix}1_n'\\Id_k'\end{pmatrix}\]
这里$Id_k$定义为一个除第$k$项为$1$外，其他项均为$0$的列向量
故有
\[D_{ak}=(-1)^n|I_2-B_kA_k|=(-1)^n|I_2-\begin{pmatrix}n && \frac{n^2-n}{2}+1\\1 && k\end{pmatrix}|=(-1)^n(k(n-1)-\frac{1}{2}n(n+1))\]

青青子衿 · 发表于 2016-11-6 19:16

本帖最后由青青子衿于 2019-12-12 00:09 编辑 回复 7# 战巡

其实矩阵一点也不麻烦，只是需要技巧，比如5楼那里的办法就很好
1楼的办法也可以大幅简化，强行求行列式那是个$\color{red}{笨办法}$
战巡发表于 2016-10-28 02:26

现在，我也觉得强行求行列式那是个$\color{red}{笨办法}$了。

其中构造的那两个矩阵，真是精妙！
谢谢战版分享！
请允许我整理一下

$\Huge{原理一：利用克拉默法则解线性方程组}$

\[D_a=\left|\begin{array}{ccccccc}
1&1&1& \cdots &1&1&0\\
1&1&1& \cdots &1&0&1\\
1&1&1& \cdots &0&1&1\\
\vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\
1&1&0& \cdots &1&1&1\\
1&0&1& \cdots &1&1&1\\
0&1&1& \cdots &1&1&1
\end{array}\right|\]

\[{D_a}_{\overset{\phantom,}{i}}=\begin{array}{ccccccccc}
\begin{split}
\begin{array}{ccccccccc}
&第i列&&&&&&&&&
\end{array}
\\ \,
\left|
\begin{array}{ccccccccc}
1&1&\cdots&1&1&1&\cdots&1&0\\
1&1&\cdots&1&2&1&\cdots&0&1\\
\vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\vdots &\vdots&\vdots\\
1&1&\cdots&1&i-1&0&\cdots&1&1\\
1&1&\cdots&1&\color{red}{i}&1&\cdots&1&1\\
1&1&\cdots&0&i+1&1&\cdots&1&1\\
\vdots&\vdots&\vdots&\vdots&\vdots &\vdots&\vdots&\vdots&\vdots\\
1&0&\cdots&1&{n-1}&1&\cdots&1&1\\
0&1&\cdots&1&n&1&\cdots&1&1
\end{array}
\right|
第i行
\\
\begin{array}{ccccccccc}
\,
\end{array}
\end{split}
\end{array}\]
青青子衿发表于 2016-8-12 17:03

因为$|\boldsymbol{X}+\boldsymbol{u}\boldsymbol{v}^T|=\left(1+\boldsymbol{v}^T\boldsymbol{X}^{-1}\boldsymbol{u}\right)|\boldsymbol{X}|$，${\boldsymbol{S}_n}^2=\boldsymbol{E}\Rightarrow(-\boldsymbol{S}_n)^{-1}=(-\boldsymbol{S}_n)$，所以有：
\[D_a=|\boldsymbol{A_\mathit{a}}|=|\boldsymbol{J}_n-\boldsymbol{S}_n|=|-\boldsymbol{S}_n+\boldsymbol{1_n}{\boldsymbol{1_n}}^\prime|=\left[1+{\boldsymbol{1_n}}^\prime(-\boldsymbol{S}_n)^{-1}\boldsymbol{1_n}\right]|-\boldsymbol{S}_n|=|-\boldsymbol{S}_n|\left[1-{\boldsymbol{1_n}}^\prime\boldsymbol{S}_n\boldsymbol{1_n}\right]\\
=\left(-1\right)^{\frac{n(n-1)}{2}}\left(-1\right)^{n}(1-n)=\color{gold}{\left(-1\right)^{\frac{n(n-1)}{2}}\left(-1\right)^{n-1}(n-1)}\]

因为$|\boldsymbol{X}+\boldsymbol{U}\boldsymbol{W}\boldsymbol{V}^T|=\left|\boldsymbol{W}^{-1}+\boldsymbol{V}^T\boldsymbol{X}^{-1}\boldsymbol{U}\right||\boldsymbol{W}||\boldsymbol{X}|$，${\boldsymbol{S}_n}^2=\boldsymbol{E}\Rightarrow(-\boldsymbol{S}_n)^{-1}=(-\boldsymbol{S}_n)$，所以有：

\begin{split}
{D_a}_{\overset{\phantom,}{i}}&=|{\boldsymbol{U}}_i{{\boldsymbol{V}}_i}^T-\boldsymbol{S}_n|=|-\boldsymbol{S}_n+{\boldsymbol{U}}_i\boldsymbol{E}_2{{\boldsymbol{V}}_i}^T|=\left|\boldsymbol{E}_2+{{\boldsymbol{V}}_i}^T(-\boldsymbol{S}_n)^{-1}{\boldsymbol{U}}_i\right||-\boldsymbol{S}_n|=|-\boldsymbol{S}_n|\left|\boldsymbol{E}_2-{{\boldsymbol{V}}_i}^T\boldsymbol{S}_n{\boldsymbol{U}}_i\right|\\
&=\left(-1\right)^{\frac{n(n-1)}{2}}\left(-1\right)^{n}\left|\boldsymbol{E}_2-\begin{pmatrix}n && \frac{n^2-n}{2}+1\\1 && n-i\end{pmatrix}\right|=\left(-1\right)^{\frac{n(n-1)}{2}}\left(-1\right)^{n}\left|\begin{pmatrix}n&& \frac{n^2-n}{2}+1\\1 && n-i\end{pmatrix}-\boldsymbol{E}_2\right|\\
&=\left(-1\right)^{\frac{n(n-1)}{2}}\left(-1\right)^{n}\left|\begin{pmatrix}n-1&& \frac{n^2-n}{2}+1\\1 && n-i-1\end{pmatrix}\right|=\left(-1\right)^{\frac{n(n-1)}{2}}\left(-1\right)^{n}\left[(n-i-1)(n-1)-\left(\frac{n^2-n}{2}+1\right)\right]\\
&=\left(-1\right)^{\frac{n(n-1)}{2}}\left(-1\right)^{n}\left[(n-1)^2-i(n-1)-\frac{n(n-1)}{2}-1\right]=\color{blue}{\left(-1\right)^{\frac{n(n-1)}{2}}\left(-1\right)^{n}\left[\frac{n(n-3)}{2}-i(n-1)\right]}\\
&=\left(-1\right)^{\frac{n(n-1)}{2}}\left(-1\right)^{n}\left[\frac{n(n-1)}{2}-1-i(n-1)\right]=\left(-1\right)^{\frac{n(n-1)}{2}}\left(-1\right)^{n}(n-1)\left(\frac{n}{2}-\frac{1}{n-1}-i\right)\\
&=\color{gold}{\left(-1\right)^{\frac{n(n-1)}{2}}\left(-1\right)^{n-1}(n-1)}\left(\frac{1}{n-1}-\frac{n}{2}+i\right)

\end{split}
—————————————————————————————————————————————————————————————————————
\[{\boldsymbol{U}}_i=\begin{pmatrix}1 && 0\\1 && 1\\\vdots&&\vdots\\1 && n-i-2\\\color{red}{1}&&\color{red}{n-i}\\1 && n-i\\
\vdots&&\vdots\\1 && n-2\\1 && n-1\end{pmatrix}\ \ \color{red}{倒数第i行}\\
\\
\begin{gather*}
&\color{red}{第i列}&\\
{{\boldsymbol{V}}_i}^T=\begin{pmatrix}{\boldsymbol{1_n}}^\prime\\
{\boldsymbol{\alpha}_i}^\prime\end{pmatrix}=&
\begin{pmatrix}1&1&\cdots&1&\color{red}{1}&1&\cdots&1&1\\
0&0&\cdots&0&\color{red}{1}&0&\cdots&1&1\end{pmatrix}&\\
\end{gather*}
\]

\[
\begin{array}{llclc}

&&{\boldsymbol{U}}_i{{\boldsymbol{V}}_i}^T&-&\boldsymbol{S}_n&\\
&=\color{red}{倒数第i行}&
\begin{pmatrix}1 && 0\\1 && 1\\\vdots&&\vdots\\1 && n-i-2\\\color{red}{1}&&\color{red}{n-i}\\1 &&n-i\\\vdots&&\vdots\\1 && n-2\\1 && n-1\end{pmatrix}

\begin{matrix}\\
\begin{pmatrix}1&1&\cdots&1&\color{red}{1}&1&\cdots&1&1\\
0&0&\cdots&0&\color{red}{1}&0&\cdots&1&1\end{pmatrix}&\\
\color{red}{第i列}
\end{matrix}
&-&\begin{pmatrix}
0&0&\cdots&0&0&0&\cdots&0&1\\
0&0&\cdots&0&0&0&\cdots&1&0\\
\vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\vdots &\vdots&\vdots\\
0&0&\cdots&0&0&1&\cdots&0&0\\
0&0&\cdots&0&\color{red}{1}&0&\cdots&0&0\\
0&0&\cdots&1&0&0&\cdots&0&0\\
\vdots&\vdots&\vdots&\vdots&\vdots &\vdots&\vdots&\vdots&\vdots\\
0&1&\cdots&0&0&0&\cdots&0&0\\
1&0&\cdots&0&0&0&\cdots&0&0\end{pmatrix}&\color{red}{倒数第i行}
\\
&&\color{red}{第i列}&&\color{red}{第i列}&\\
&=\color{red}{倒数第i行}&\begin{pmatrix}
1&1&\cdots&1&1&1&\cdots&1&1\\
1&1&\cdots&1&2&1&\cdots&1&1\\
\vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\vdots &\vdots&\vdots\\
1&1&\cdots&1&n-i-1&1&\cdots&1&1\\
1&1&\cdots&1&\color{red}{n-i+1}&1&\cdots&1&1\\
1&1&\cdots&1&n-i+1&1&\cdots&1&1\\
\vdots&\vdots&\vdots&\vdots&\vdots &\vdots&\vdots&\vdots&\vdots\\
1&1&\cdots&1&{n-1}&1&\cdots&1&1\\
1&1&\cdots&1&n&1&\cdots&1&1\end{pmatrix}
&-&\begin{pmatrix}
0&0&\cdots&0&0&0&\cdots&0&1\\
0&0&\cdots&0&0&0&\cdots&1&0\\
\vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\vdots &\vdots&\vdots\\
0&0&\cdots&0&0&1&\cdots&0&0\\
0&0&\cdots&0&\color{red}{1}&0&\cdots&0&0\\
0&0&\cdots&1&0&0&\cdots&0&0\\
\vdots&\vdots&\vdots&\vdots&\vdots &\vdots&\vdots&\vdots&\vdots\\
0&1&\cdots&0&0&0&\cdots&0&0\\
1&0&\cdots&0&0&0&\cdots&0&0\end{pmatrix}&\color{red}{倒数第i行}
\\

&&\color{red}{第i列}&&\\
&=\color{red}{倒数第i行}&\begin{pmatrix}
0&1&\cdots&1&1&1&\cdots&1&1\\
1&0&\cdots&1&2&1&\cdots&1&1\\
\vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\vdots &\vdots&\vdots\\
1&1&\cdots&0&n-i-1&1&\cdots&1&1\\
1&1&\cdots&1&\color{red}{n-i}&1&\cdots&1&1\\
1&1&\cdots&1&n-i+1&0&\cdots&1&1\\
\vdots&\vdots&\vdots&\vdots&\vdots &\vdots&\vdots&\vdots&\vdots\\
1&1&\cdots&1&{n-1}&1&\cdots&0&1\\
1&1&\cdots&1&n&1&\cdots&1&0\end{pmatrix}&\\

\end{array}
\]

\[
\begin{split}
{{\boldsymbol{V}}_i}^T{\boldsymbol{S}}_n{\boldsymbol{U}}_i
&=\begin{matrix}
\\
\begin{pmatrix}1&1&\cdots&1&\color{red}{1}&1&\cdots&1&1\\
0&0&\cdots&0&\color{red}{1}&0&\cdots&0&0\end{pmatrix}&\\
\color{red}{第i列}
\end{matrix}
\begin{matrix}
\\
\begin{pmatrix}
0&0&\cdots&0&0&0&\cdots&0&1\\
0&0&\cdots&0&0&0&\cdots&1&0\\
\vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\vdots &\vdots&\vdots\\
0&0&\cdots&0&0&1&\cdots&0&0\\
0&0&\cdots&0&\color{red}{1}&0&\cdots&0&0\\
0&0&\cdots&1&0&0&\cdots&0&0\\
\vdots&\vdots&\vdots&\vdots&\vdots &\vdots&\vdots&\vdots&\vdots\\
0&1&\cdots&0&0&0&\cdots&0&0\\
1&0&\cdots&0&0&0&\cdots&0&0\end{pmatrix}\\
\color{red}{倒数第i列}
\end{matrix}
\begin{pmatrix}1 && 0\\1 && 1\\\vdots&&\vdots\\1 && n-i-2\\
\color{red}{1}&&\color{red}{n-i}\\1 && n-i\\\vdots&&\vdots\\1 && n-2\\1 && n-1\end{pmatrix}\color{red}{倒数第i行}\\
&=\begin{matrix}
\\
\begin{pmatrix}1&1&\cdots&1&\color{red}{1}&1&\cdots&1&1\\
0&0&\cdots&0&\color{red}{1}&0&\cdots&0&0\end{pmatrix}&\\
\color{red}{倒数第i列}
\end{matrix}
\begin{pmatrix}1 && 0\\1 && 1\\\vdots&&\vdots\\1 &&n-i-2\\\color{red}{1}&&\color{red}{n-i}\\1 &&n-i\\\vdots&&\vdots\\1 && n-2\\1 && n-1\end{pmatrix}\color{red}{倒数第i行}\\
&=\begin{pmatrix}n && \frac{n^2-n}{2}+1\\1 && n-i\end{pmatrix}
\end{split}
\]

$\Huge{原理一：利用克拉默法则解线性方程组}$

\[D_b=\left|\begin{array}{ccccccc}
0&1&1& \cdots &1&1&1\\
1&0&1& \cdots &1&1&1\\
1&1&0& \cdots &1&1&1\\
\vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\
1&1&1& \cdots &0&1&1\\
1&1&1& \cdots &1&0&1\\
1&1&1& \cdots &1&1&0
\end{array}\right|\]

\[{D_b}_{\overset{\phantom,}{i}}=\begin{array}{ccccccccc}
\begin{split}
\begin{array}{ccccccccc}
&第i列&&&&&&&&&
\end{array}
\\ \,
\left|
\begin{array}{ccccccccc}
0&1&\cdots&1&1&1&\cdots&1&1\\
1&0&\cdots&1&2&1&\cdots&1&1\\
\vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\vdots &\vdots&\vdots\\
1&1&\cdots&0&i-1&1&\cdots&1&1\\
1&1&\cdots&1&\color{red}{i}&1&\cdots&1&1\\
1&1&\cdots&1&i+1&0&\cdots&1&1\\
\vdots&\vdots&\vdots&\vdots&\vdots &\vdots&\vdots&\vdots&\vdots\\
1&1&\cdots&1&{n-1}&1&\cdots&0&1\\
1&1&\cdots&1&n&1&\cdots&1&0
\end{array}
\right|
第i行
\\
\begin{array}{ccccccccc}
\,
\end{array}
\end{split}
\end{array}\]
青青子衿发表于 2016-8-12 17:03

因为$|\lambda\boldsymbol{E}_n-\boldsymbol{A}\boldsymbol{B}|=\lambda^{n-m}|\lambda\boldsymbol{E}_m-\boldsymbol{B}\boldsymbol{A}|$，所以有：
\[D_b=|\boldsymbol{A}|=|\boldsymbol{J}_n-\boldsymbol{E}_n|=|-\boldsymbol{E}_n+\boldsymbol{1_n}{\boldsymbol{1_n}}^\prime|=(-1)^n|-\boldsymbol{E}_1+{\boldsymbol{1_n}}^\prime\boldsymbol{1_n}|=\color{gold}{(-1)^{n-1}(n-1)}\]
\begin{split}
{D_b}_{\overset{\phantom,}{i}}&=|{\boldsymbol{U}}_i{{\boldsymbol{V}}_i}^T-\boldsymbol{E}_n|=|-\boldsymbol{E}_n+{\boldsymbol{U}}_i{{\boldsymbol{V}}_i}^T|=(-1)^{n-2}|-\boldsymbol{E}_2+{{\boldsymbol{V}}_i}^T{\boldsymbol{U}}_i|\\
&=(-1)^{n-2}\left|-\boldsymbol{E}_2+\begin{pmatrix}n && \frac{n^2-n}{2}+1\\1 && i\end{pmatrix}\right|=(-1)^n\left|\begin{pmatrix}n-1&& \frac{n^2-n}{2}+1\\1 && i-1\end{pmatrix}\right|\\
&=(-1)^{n}\left[(i-1)(n-1)-\left(\frac{n^2-n}{2}+1\right)\right]=\color{blue}{(-1)^{n}\left[i(n-1)-\frac{n(n+1)}{2}\right]}\\
&=(-1)^{n}(n-1)\left(i-1-\frac{n}{2}-\frac{1}{n-1}\right)=(-1)^{n}(n-1)\left(i-\frac{n}{2}-\frac{n}{n-1}\right)\\
&=\color{gold}{(-1)^{n-1}(n-1)}\left(\frac{n}{2}+\frac{n}{n-1}-i\right)
\end{split}
—————————————————————————————————————————————————————————————————————
其中\[{\boldsymbol{U}}_i=\begin{pmatrix}1 && 0\\1 && 1\\\vdots&&\vdots\\1 && i-2\\\color{red}{1}&&\color{red}{i}\\1 && i\\\vdots&&\vdots\\1 && n-2\\1 && n-1\end{pmatrix}\ \ \color{red}{第i行}\\
\\
\begin{gather*}
&\color{red}{第i列}&\\
{{\boldsymbol{V}}_i}^T=\begin{pmatrix}{\boldsymbol{1_n}}^\prime\\
{\boldsymbol{\alpha}_i}^\prime\end{pmatrix}=&
\begin{pmatrix}1&1&\cdots&1&\color{red}{1}&1&\cdots&1&1\\
0&0&\cdots&0&\color{red}{1}&0&\cdots&1&1\end{pmatrix}&\\
\end{gather*}
\]

\[
\begin{array}{llclc}

&&{\boldsymbol{U}}_i{{\boldsymbol{V}}_i}^T&-&\boldsymbol{E}_n\\
&=\color{red}{第i行}&
\begin{pmatrix}1 && 0\\1 && 1\\\vdots&&\vdots\\1 && i-2\\\color{red}{1}&&\color{red}{i}\\1 && i\\\vdots&&\vdots\\1 && n-2\\1 && n-1\end{pmatrix}

\begin{matrix}\\
\begin{pmatrix}1&1&\cdots&1&\color{red}{1}&1&\cdots&1&1\\
0&0&\cdots&0&\color{red}{1}&0&\cdots&1&1\end{pmatrix}&\\
\color{red}{第i列}
\end{matrix}
&-&\begin{pmatrix}
1&0&\cdots&0&0&0&\cdots&0&0\\
0&1&\cdots&0&0&0&\cdots&0&0\\
\vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\vdots &\vdots&\vdots\\
0&0&\cdots&1&0&0&\cdots&0&0\\
0&0&\cdots&0&\color{red}{1}&0&\cdots&0&0\\
0&0&\cdots&0&0&1&\cdots&0&0\\
\vdots&\vdots&\vdots&\vdots&\vdots &\vdots&\vdots&\vdots&\vdots\\
0&0&\cdots&0&0&0&\cdots&1&0\\
0&0&\cdots&0&0&0&\cdots&0&1\end{pmatrix}&\color{red}{第i行}
\\
&&\color{red}{第i列}&&\color{red}{第i列}&\\
&=\color{red}{第i行}&\begin{pmatrix}
1&1&\cdots&1&1&1&\cdots&1&1\\
1&1&\cdots&1&2&1&\cdots&1&1\\
\vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\vdots &\vdots&\vdots\\
1&1&\cdots&1&i-1&1&\cdots&1&1\\
1&1&\cdots&1&\color{red}{i+1}&1&\cdots&1&1\\
1&1&\cdots&1&i+1&1&\cdots&1&1\\
\vdots&\vdots&\vdots&\vdots&\vdots &\vdots&\vdots&\vdots&\vdots\\
1&1&\cdots&1&{n-1}&1&\cdots&1&1\\
1&1&\cdots&1&n&1&\cdots&1&1\end{pmatrix}
&-&\begin{pmatrix}
1&0&\cdots&0&0&0&\cdots&0&0\\
0&1&\cdots&0&0&0&\cdots&0&0\\
\vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\vdots &\vdots&\vdots\\
0&0&\cdots&1&0&0&\cdots&0&0\\
0&0&\cdots&0&\color{red}{1}&0&\cdots&0&0\\
0&0&\cdots&0&0&1&\cdots&0&0\\
\vdots&\vdots&\vdots&\vdots&\vdots &\vdots&\vdots&\vdots&\vdots\\
0&0&\cdots&0&0&0&\cdots&1&0\\
0&0&\cdots&0&0&0&\cdots&0&1\end{pmatrix}&\color{red}{第i行}
\\

&&\color{red}{第i列}&&\\
&=\color{red}{第i行}&\begin{pmatrix}
0&1&\cdots&1&1&1&\cdots&1&1\\
1&0&\cdots&1&2&1&\cdots&1&1\\
\vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\vdots &\vdots&\vdots\\
1&1&\cdots&0&i-1&1&\cdots&1&1\\
1&1&\cdots&1&\color{red}{i}&1&\cdots&1&1\\
1&1&\cdots&1&i+1&0&\cdots&1&1\\
\vdots&\vdots&\vdots&\vdots&\vdots &\vdots&\vdots&\vdots&\vdots\\
1&1&\cdots&1&{n-1}&1&\cdots&0&1\\
1&1&\cdots&1&n&1&\cdots&1&0\end{pmatrix}&\\

\end{array}
\]

\[
\begin{split}
{{\boldsymbol{V}}_i}^T{\boldsymbol{U}}_i
&=\begin{matrix}
\\
\begin{pmatrix}1&1&\cdots&1&\color{red}{1}&1&\cdots&1&1\\
0&0&\cdots&0&\color{red}{1}&0&\cdots&0&0\end{pmatrix}&\\
\color{red}{第i列}
\end{matrix}
\begin{pmatrix}1 && 0\\1 && 1\\\vdots&&\vdots\\1 && i-2\\\color{red}{1}&&\color{red}{i}\\1 && i\\\vdots&&\vdots\\1 && n-2\\1 && n-1\end{pmatrix}\color{red}{第i行}\\
&=\begin{pmatrix}n && \frac{n^2-n}{2}+1\\1 && i\end{pmatrix}
\end{split}
\]

—————————————————————————————————————————————————————————————————————
注：发现战版更喜欢用$\boldsymbol{I}$表示单位矩阵，我更喜欢用$\boldsymbol{E}$表示单位矩阵，用$\boldsymbol{I}$表示单位矩阵是因为单位矩阵的英文是Identity matrix，而用$\boldsymbol{E}$表示单位矩阵就不得而知了？于是，好奇心驱使下翻了翻Mathworld，找到了：
An identity matrix may be denoted $\boldsymbol{1}$,$\boldsymbol{I}$, or$\boldsymbol{E}$(the latter being an abbreviation for the German term "$\color{red}{Einheitsmatrix}$"; Courant and Hilbert 1989, p. 7). Identity matrices are sometimes also known as unit matrices (Akivis and Goldberg 1972, p. 71).
mathworld.wolfram.com/IdentityMatrix.html
另外，在Wiki里看见：$|\boldsymbol{E}_n-\boldsymbol{A}\boldsymbol{B}|=|\boldsymbol{E}_m-\boldsymbol{B}\boldsymbol{A}|$才叫Sylvester行列式恒等式(Sylvester's determinant identity)，也叫Weinstein–Aronszajn identity
可以推广为：$|\lambda\boldsymbol{E}_n-\boldsymbol{A}\boldsymbol{B}|=\lambda^{n-m}|\lambda\boldsymbol{E}_m-\boldsymbol{B}\boldsymbol{A}|$
en.wikipedia.org/wiki/Sylvester%27s_determinant_identity
en.wikipedia.org/wiki/Matrix_determinant_lemma
en.wikipedia.org/wiki/Schur_complement

青青子衿 · 发表于 2024-1-14 21:42

青青子衿发表于 2016-11-6 19:16
回复 7# 战巡

\begin{align*}

\\
\boldsymbol{P}&\,\,\colon\!=\begin{pmatrix}
\boldsymbol{E}_n&-t\boldsymbol{B}_{n\times\,\!m}\\
\boldsymbol{A}_{m\times\,\!n}&\boldsymbol{E}_m\\
\end{pmatrix}\\
\\

\boldsymbol{Q}&\,\,\colon\!=\begin{pmatrix}
\boldsymbol{E}_n&\boldsymbol{O}_{n\times\,\!m}\\
-\boldsymbol{A}_{m\times\,\!n}&\boldsymbol{E}_m\\
\end{pmatrix}\\
\\
\boldsymbol{P}\boldsymbol{Q}&=\begin{pmatrix}
\boldsymbol{E}_n+t\boldsymbol{B}_{n\times\,\!m}\boldsymbol{A}_{m\times\,\!n}&-t\boldsymbol{B}_{n\times\,\!m}\\
\boldsymbol{O}_{m\times\,\!n}&\boldsymbol{E}_m\\
\end{pmatrix}\\
\\
\boldsymbol{Q}\boldsymbol{P}&=\begin{pmatrix}
\boldsymbol{E}_n&-t\boldsymbol{B}_{n\times\,\!m}\\
\boldsymbol{O}_{m\times\,\!n}&\boldsymbol{E}_m+t\boldsymbol{A}_{m\times\,\!n}\boldsymbol{B}_{n\times\,\!m}\\
\end{pmatrix}\\
\\
\\
&\>\>\det\left(\boldsymbol{P}\boldsymbol{Q}\right)=\det\left(\boldsymbol{Q}\boldsymbol{P}\right)\\
\\
&|\boldsymbol{E}_m+t\boldsymbol{A}_{m\times\,\!n}\boldsymbol{B}_{n\times\,\!m}|=|\boldsymbol{E}_n+t\boldsymbol{B}_{n\times\,\!m}\boldsymbol{A}_{m\times\,\!n}|\\
\end{align*}

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两道\(n\)元非齐次线性方程组

原理一：利用克拉默法则解线性方程组

原理二：利用逆矩阵解线性方程组

原理三：利用高斯-约当(Gauss-Jordan)消元法解线性方程组