三对角、五对角追赶法求解线性方程组

一.三对角

1.1问题引入

问题引入：
$A=\left\{\begin{array}{ccccccc}6& 1& 3& & & & \\ 8& 6& 1& & & & \\ & ...& ...& ...& & & \\ & & ...& ...& ...& & \\ & & & ...& ...& ...& \\ & & & & 8& 6& 1\\ & & & & & 8& 6\end{array}\right\}，$ $b=\left\{\begin{array}{c}7\\ 15\\ ...\\ 15\\ 14\end{array}\right\}$$，则方程有解$$X$使$AX=b$，取n=10求$X$

1.2三对角算法原理

首先将矩阵进行LU分解，设方程组$AX=F$，即
$\left[\begin{array}{ccccccccc}{b}_1& c_1& & & & & & & \\ & a_2& b_2& c_2& & & & & \\ & & ...& ...& ...& & & & \\ & & & a_i& b_i& c_i& & & \\ & & & & ...& ...& ...& & \\ & & & & & a_{n-1}& b_{n-1}& c_{n-1}& \\ & & & & & & a_n& b_n& \end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\\ ...\\ ...\\ ...\\ ...\\ x_n\end{array}\right]=\left[\begin{array}{c}{f}_1\\ f_2\\ ...\\ ...\\ ...\\ ...\\ f_n\end{array}\right]$
则有则有$A=LU$，即
$L=$