某书补充题选做
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 \(\frac1{a_1}+\cdots+\frac1{a_n}\not=-1\)
观察到 \(a_ix_i=b_i-\sum\limits_{i=1}^nx_i\) 即
\[\begin{aligned}
x_i=&\frac{b_i}{a_i}-\frac1{a_i}\sum\limits_{i=1}^nx_i\\
\sum\limits_{i=1}^nx_i=&\sum\limits_{i=1}^n\frac{b_i}{a_i}-(\sum\limits_{i=1}^{n}\frac1{a_i})(\sum\limits_{i=1}^nx_i)\\
\sum\limits_{i=1}^nx_i=&\frac{\sum\limits_{i=1}^n\frac{b_i}{a_i}}{1+\sum\limits_{i=1}^n\frac1{a_i}}\\
x_i=&\frac{b_i-\frac{\sum\limits_{i=1}^n\frac{b_i}{a_i}}{1+\sum\limits_{i=1}^n\frac1{a_i}}}{a_i}
\end{aligned}
\]
Problem 2
解下述线性方程组:
\[\left\{ \begin{aligned} x_1+2x_2+\cdots+nx_n=b_1\\ nx_1+x_2+\cdots+(n-1)x_n=b_2\\ \cdots\cdots\cdots\cdots\cdots\cdots\\ 2x_1+3x_2+\cdots+x_n=b_n\\ \end{aligned} \right. \]
容易观察到 \(\sum\limits_{i=1}^nx_i=\frac{2\sum\limits_{i=1}^nb_i}{n(n+1)}\)
再容易观察到 \(x_i=b_{i\% n+1}-b_i+\frac{2\sum\limits_{i=1}^nb_i}{n(n+1)}\)
