例 1.18

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• 例 8.48

例 1.18

计算 $n$ 阶行列式：

$$|A|=\left|\begin{array}{cccc}(a_1+b_1{)}^{-1}& (a_1+b_2{)}^{-1}& \cdots & (a_1+b_n{)}^{-1}\\ (a_2+b_1{)}^{-1}& (a_2+b_2{)}^{-1}& \cdots & (a_2+b_n{)}^{-1}\\ ⋮& ⋮& & ⋮\\ (a_n+b_1{)}^{-1}& (a_n+b_2{)}^{-1}& \cdots & (a_n+b_n{)}^{-1}\end{array}\right|.$$

解答

解 记 $|A|$ 为 $D_n$，我们来求 $D_n$ 与 $D_{n-1}$ 之间的递推公式。注意下面计算中的第一步是将行列式的前 $n-1$ 列每列都减去第 $n$ 列；第三步是将行列式的前 $n-1$ 行每行都减去第 $n$ 行；第二步和第四步都是提取公因式。

$$\begin{array}{rl}{D}_n& =\left|\begin{array}{cccc}\frac{1}{a_1+b_1}& \cdots & \frac{1}{a_1+b_{n-1}}& \frac{1}{a_1+b_n}\\ ⋮& & ⋮& ⋮\\ \frac{1}{a_{n-1}+b_1}& \cdots & \frac{1}{a_{n-1}+b_{n-1}}& \frac{1}{a_{n-1}+b_n}\\ \frac{1}{a_n+b_1}& \cdots & \frac{1}{a_n+b_{n-1}}& \frac{1}{a_n+b_n}\end{array}\right|\\ & =\left|\begin{array}{cccc}\frac{b_n-b_1}{(a_1+b_1)(a_1+b_n)}& \cdots & \frac{b_n-b_{n-1}}{(a_1+b_{n-1})(a_1+b_n)}& \frac{1}{a_1+b_n}\\ ⋮& & ⋮& ⋮\\ \frac{b_n-b_1}{(a_{n-1}+b_1)(a_{n-1}+b_n)}& \cdots & \frac{b_n-b_{n-1}}{(a_{n-1}+b_{n-1})(a_{n-1}+b_n)}& \frac{1}{a_{n-1}+b_n}\\ \frac{b_n-b_1}{(a_n+b_1)(a_n+b_n)}& \cdots & \frac{b_n-b_{n-1}}{(a_n+b_{n-1})(a_n+b_n)}& \frac{1}{a_n+b_n}\end{array}\right|\\ & =\frac{{\prod }_{i=1}^{n-1}(b_n-b_i)}{{\prod }_{j=1}^n(a_j+b_n)}\left|\begin{array}{cccc}\frac{1}{a_1+b_1}& \cdots & \frac{1}{a_1+b_{n-1}}& 1\\ ⋮& & ⋮& ⋮\\ \frac{1}{a_{n-1}+b_1}& \cdots & \frac{1}{a_{n-1}+b_{n-1}}& 1\\ \frac{1}{a_n+b_1}& \cdots & \frac{1}{a_n+b_{n-1}}& 1\end{array}\right|\\ & =\frac{{\prod }_{i=1}^{n-1}(b_n-b_i)}{{\prod }_{j=1}^n(a_j+b_n)}\left|\begin{array}{cccc}\frac{a_n-a_1}{(a_1+b_1)(a_n+b_1)}& \cdots & \frac{a_n-a_1}{(a_1+b_{n-1})(a_n+b_{n-1})}& 0\\ ⋮& & ⋮& ⋮\\ \frac{a_n-a_{n-1}}{(a_{n-1}+b_1)(a_n+b_1)}& \cdots & \frac{a_n-a_{n-1}}{(a_{n-1}+b_{n-1})(a_n+b_{n-1})}& 0\\ \frac{1}{a_n+b_1}& \cdots & \frac{1}{a_n+b_{n-1}}& 1\end{array}\right|\\ & =\frac{{\prod }_{i=1}^{n-1}(a_n-a_i)(b_n-b_i)}{{\prod }_{j=1}^n(a_j+b_n){\prod }_{k=1}^{n-1}(a_n+b_k)}\left|\begin{array}{ccc}\frac{1}{a_1+b_1}& \cdots & \frac{1}{a_1+b_{n-1}}\\ ⋮& & ⋮\\ \frac{1}{a_{n-1}+b_1}& \cdots & \frac{1}{a_{n-1}+b_{n-1}}\end{array}\right|\\ & =\frac{{\prod }_{i=1}^{n-1}(a_n-a_i)(b_n-b_i)}{{\prod }_{j=1}^n(a_j+b_n){\prod }_{k=1}^{n-1}(a_n+b_k)}\cdot D_{n-1}.\end{array}$$

不断递推下去即得

$$|A|=\frac{{\prod }_{1\le i<j\le n}(a_j-a_i)(b_j-b_i)}{{\prod }_{i,j=1}^n(a_i+b_j)}.\square$$