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设

$D_n = \left|\begin{array}{ccccc} 2 & a & a & \cdots & a\\ a & 2 & a & \cdots & a\\ a & a & 2 & \cdots & a\\ \vdots & \vdots & \vdots & \ddots & \vdots\\ a & a & a & \cdots & 2 \end{array}\right|$D_n = \left|\begin{array}{ccccc} 2 & a & a & \cdots & a\\ a & 2 & a & \cdots & a\\ a & a & 2 & \cdots & a\\ \vdots & \vdots & \vdots & \ddots & \vdots\\ a & a & a & \cdots & 2 \end{array}\right|

对第 $2, 3, \cdots \text{ }, n$2, 3, \cdots \text{ }, n 行分别作变换

$R_i \rightarrow R_i - R_1 (i = 2, 3, \cdots \text{ }, n),$R_i \rightarrow R_i - R_1 (i = 2, 3, \cdots \text{ }, n),

则

${\color{blue}{D_n = \left|\begin{array}{ccccc} 2 & a & a & \cdots & a\\ a - 2 & 2 - a & 0 & \cdots & 0\\ a - 2 & 0 & 2 - a & \cdots & 0\\ \vdots & \vdots & \vdots & \ddots & \vdots\\ a - 2 & 0 & 0 & \cdots & 2 - a \end{array}\right| .}}${\color{blue}{D_n = \left|\begin{array}{ccccc} 2 & a & a & \cdots & a\\ a - 2 & 2 - a & 0 & \cdots & 0\\ a - 2 & 0 & 2 - a & \cdots & 0\\ \vdots & \vdots & \vdots & \ddots & \vdots\\ a - 2 & 0 & 0 & \cdots & 2 - a \end{array}\right| .}}

再对第一列作变换

$C_1 \rightarrow C_1 + C_2 + \cdots + C_n,$C_1 \rightarrow C_1 + C_2 + \cdots + C_n,

得

${\color{blue}{D_n = \left|\begin{array}{ccccc} 2 + (n - 1) a & a & a & \cdots & a\\ 0 & 2 - a & 0 & \cdots & 0\\ 0 & 0 & 2 - a & \cdots & 0\\ \vdots & \vdots & \vdots & \ddots & \vdots\\ 0 & 0 & 0 & \cdots & 2 - a \end{array}\right| .}}${\color{blue}{D_n = \left|\begin{array}{ccccc} 2 + (n - 1) a & a & a & \cdots & a\\ 0 & 2 - a & 0 & \cdots & 0\\ 0 & 0 & 2 - a & \cdots & 0\\ \vdots & \vdots & \vdots & \ddots & \vdots\\ 0 & 0 & 0 & \cdots & 2 - a \end{array}\right| .}}

因此该行列式化为上三角行列式，故

${\color{red}{D_n = [2 + (n - 1) a] (2 - a)^{n - 1} .}}${\color{red}{D_n = [2 + (n - 1) a] (2 - a)^{n - 1} .}}