Problem 1

解下述线性方程组:

\[\left\{ \begin{aligned} (1+a_1)x_1+x_2+\cdots+x_n=b_1\\ x_1+(1+a_2)x_2+\cdots+x_n=b_2\\ \cdots\cdots\cdots\cdots\cdots\cdots\\ x_1+x_2+\cdots+(1+a_n)x_n=b_n\\ \end{aligned} \right. \]

其中 \(a_i\not=0,i=1,2,...,n\) 且 \(\frac1{a_1}+\cdots+\frac1{a_n}\not=-1\)

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观察到 \(a_ix_i=b_i-\sum\limits_{i=1}^nx_i\) 即

\[\begin{aligned} x_i=&\frac{b_i}{a_i}-\frac1{a_i}\sum\limits_{i=1}^nx_i\\ \sum\limits_{i=1}^nx_i=&\sum\limits_{i=1}^n\frac{b_i}{a_i}-(\sum\limits_{i=1}^{n}\frac1{a_i})(\sum\limits_{i=1}^nx_i)\\ \sum\limits_{i=1}^nx_i=&\frac{\sum\limits_{i=1}^n\frac{b_i}{a_i}}{1+\sum\limits_{i=1}^n\frac1{a_i}}\\ x_i=&\frac{b_i-\frac{\sum\limits_{i=1}^n\frac{b_i}{a_i}}{1+\sum\limits_{i=1}^n\frac1{a_i}}}{a_i} \end{aligned} \]