本帖最后由 青青子衿 于 2017-10-15 14:22 编辑
真幸运,抢到了本版第200个主帖
“嵌入型”行列式
\[
D_{n,I}=\left|\begin{array}{ccccccc}
a_1&b_1&b_2& \cdots &b_{n-3}&b_{n-2}&b_{n-1}\\
b_1&a_2&b_2& \cdots &b_{n-3}&b_{n-2}&b_{n-1}\\
b_1&b_2&a_3& \cdots &b_{n-3}&b_{n-2}&b_{n-1}\\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots & \vdots \\
b_1&b_2&b_3& \cdots &a_{n-2}&b_{n-2}&b_{n-1}\\
b_1&b_2&b_3& \cdots &b_{n-2}&a_{n-1}&b_{n-1}\\
b_1&b_2&b_3& \cdots &b_{n-2}&b_{n-1}&a_{n}
\end{array}\right|=\left(\sum_{j=1}^{n-1}b_j\frac{\displaystyle\prod_{i=1}^j(a_i-b_i)}{\displaystyle\prod_{i=1}^j(a_{i+1}-b_i)}+a_1\right)\prod_{i=1}^{n-1}(a_{i+1}-b_i)
\]
“替换型”行列式
\[
D_{n,II}=\left|\begin{array}{ccccccc}
a_1&b_2&b_3& \cdots &b_{n-2}&b_{n-1}&b_{n}\\
b_1&a_2&b_3& \cdots &b_{n-2}&b_{n-1}&b_{n}\\
b_1&b_2&a_3& \cdots &b_{n-2}&b_{n-1}&b_{n}\\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots & \vdots \\
b_1&b_2&b_3& \cdots &a_{n-2}&b_{n-1}&b_{n}\\
b_1&b_2&b_3& \cdots &b_{n-2}&a_{n-1}&b_{n}\\
b_1&b_2&b_3& \cdots &b_{n-2}&b_{n-1}&a_{n}
\end{array}\right|=\left(\sum_{j=1}^n\frac{b_j}{a_j-b_j}+1\right)\prod_{i=1}^{n}(a_i-b_i)
\] |