Problem 1
解下述线性方程组:
[left{ egin{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\ cdotscdotscdotscdotscdotscdots\ x_1+x_2+cdots+(1+a_n)x_n=b_n\ end{aligned} ight. ]
其中 (a_i ot=0,i=1,2,...,n) 且 (frac1{a_1}+cdots+frac1{a_n} ot=-1)
观察到 (a_ix_i=b_i-sumlimits_{i=1}^nx_i) 即
[egin{aligned}
x_i=&frac{b_i}{a_i}-frac1{a_i}sumlimits_{i=1}^nx_i\
sumlimits_{i=1}^nx_i=&sumlimits_{i=1}^nfrac{b_i}{a_i}-(sumlimits_{i=1}^{n}frac1{a_i})(sumlimits_{i=1}^nx_i)\
sumlimits_{i=1}^nx_i=&frac{sumlimits_{i=1}^nfrac{b_i}{a_i}}{1+sumlimits_{i=1}^nfrac1{a_i}}\
x_i=&frac{b_i-frac{sumlimits_{i=1}^nfrac{b_i}{a_i}}{1+sumlimits_{i=1}^nfrac1{a_i}}}{a_i}
end{aligned}
]
Problem 2
解下述线性方程组:
[left{ egin{aligned} x_1+2x_2+cdots+nx_n=b_1\ nx_1+x_2+cdots+(n-1)x_n=b_2\ cdotscdotscdotscdotscdotscdots\ 2x_1+3x_2+cdots+x_n=b_n\ end{aligned} ight. ]
容易观察到 (sumlimits_{i=1}^nx_i=frac{2sumlimits_{i=1}^nb_i}{n(n+1)})
再容易观察到 (x_i=b_{i\% n+1}-b_i+frac{2sumlimits_{i=1}^nb_i}{n(n+1)})