求下面矩阵 $A_n$ 的秩(rank)
\[
A_n=\begin{pmatrix}
1-\frac{1}{n} & -\frac{1}{n} & \cdots & -\frac{1}{n} & -\frac{1}{n} \\
-\frac{1}{n} & 1-\frac{1}{n} & \cdots & -\frac{1}{n} & -\frac{1}{n} \\
\vdots & & \ddots & \vdots & \vdots \\
-\frac{1}{n} & -\frac{1}{n} & \cdots & 1-\frac{1}{n} & -\frac{1}{n}\\
-\frac{1}{n} & -\frac{1}{n} & \cdots & -\frac{1}{n} &1-\frac{1}{n} \\
\end{pmatrix}_{n\times n}
\]
1
\[
|A_n|=\begin{vmatrix}
1-\frac{1}{n} & -\frac{1}{n} & \cdots & -\frac{1}{n} & -\frac{1}{n} \\
-\frac{1}{n} & 1-\frac{1}{n} & \cdots & -\frac{1}{n} & -\frac{1}{n} \\
\vdots & & \ddots & \vdots & \vdots \\
-\frac{1}{n} & -\frac{1}{n} & \cdots & 1-\frac{1}{n} & -\frac{1}{n}\\
-\frac{1}{n} & -\frac{1}{n} & \cdots & -\frac{1}{n} &1-\frac{1}{n} \\
\end{vmatrix}_{n\times n}\xlongequal[i=1,2,3,\ldots,n-1]{r_n+r_i}
\begin{vmatrix}
1-\frac{1}{n} & -\frac{1}{n} & \cdots & -\frac{1}{n} & -\frac{1}{n} \\
-\frac{1}{n} & 1-\frac{1}{n} & \cdots & -\frac{1}{n} & -\frac{1}{n} \\
\vdots & & \ddots & \vdots & \vdots \\
-\frac{1}{n} & -\frac{1}{n} & \cdots & 1-\frac{1}{n} & -\frac{1}{n}\\
0 & 0 & \cdots & 0 &0 \\
\end{vmatrix}_{n\times n}
\]
因此, $|A_n|=0$.
另一方面
\[
\begin{split}
B_{n-1}=\begin{vmatrix}
1-\frac{1}{n} & -\frac{1}{n} & \cdots & -\frac{1}{n} \\
-\frac{1}{n} & 1-\frac{1}{n} & \cdots & -\frac{1}{n} \\
\vdots & & \ddots & \vdots \\
-\frac{1}{n} & -\frac{1}{n} & \cdots & 1-\frac{1}{n} \\
\end{vmatrix}_{(n-1)\times (n-1)}&\xlongequal[i=2,3,\ldots,n-1]{c_1+c_i}
\begin{vmatrix}
1-\frac{n-1}{n} & -\frac{1}{n} & \cdots & -\frac{1}{n} \\
1-\frac{n-1}{n} & 1-\frac{1}{n} & \cdots & -\frac{1}{n} \\
\vdots & & \ddots & \vdots \\
1-\frac{n-1}{n} & -\frac{1}{n} & \cdots & 1-\frac{1}{n}\\
\end{vmatrix}_{(n-1)\times (n-1)}\\
&\xlongequal[i=2,3,\ldots,n-1]{r_i-r_1}
\begin{vmatrix}
1-\frac{n-1}{n} & -\frac{1}{n} & \cdots & -\frac{1}{n} \\
0 & 1 & \cdots & 0 \\
\vdots & & \ddots & \vdots \\
0 & 0 & \cdots & 1\\
\end{vmatrix}_{(n-1)\times (n-1)}\\
\end{split}
\]
因此, 这个 $n-1$ 阶方阵 $B_{n-1}$ 的行列式为 $1-\frac{n-1}{n}=\frac{1}{n}$. 因此原矩阵的秩为 $n-1$.