随着现代工业和科学的发展，线性方程组的应用出现在经济管理、工程计算等各个领域，许多的应用会导出一些具有特殊结构的稀疏线性方程组的计算问题^{[1, 2]}.伴随着这些方程组的出现，寻找简便而且准确的求解方法就显得十分重要而且具有现实意义.

在对角线性方程组的解法中，追赶法是一类比较常用的方法，因其计算公式简单，运算量和存储量小，在科学领域中被广泛运用，倍受广大科研技术人员关注.近年来，关于对角型线性方程组的追赶法，已经有了比较细致的研究，主要有求解三对角线性方程组、循环三对角线性方程组、五对角线性方程组、拟五对角线性方程组、对称循环五对角线性方程组和七对角线性方程组的各类追赶法，参见文献[3-9].但目前尚未见文献探讨反五对角与拟反五对角方程组的追赶法，而反五对角与拟反五对角方程组是反对角线性方程组中比较常见的一类，在力学、流体力学、工程学等领域有很重要的应用(例如神舟飞船运行轨道的某些计算问题可归结为此类线性方程组的求解问题).因此，本文将借鉴文献[6, 7]的思想，建立求解反五对角线性方程组和拟反五对角线性方程组的追赶法.

定义1  若方阵$A=(a_{ij})$的元素当$1\leq i \leq n-3$, $1\leq j \leq n-i-2$且$4\leq i \leq n$, $n-i+4\leq j \leq n$时, 均有$a_{ij}=0$, 则称此矩阵为反五对角矩阵.

引理^[9]  对于任意阶数不小于3的反五对角矩阵$A$, 一般记:

$ A=\left(\begin{array}{ccccccc} & & & &e_1 &d_1&c_1 \\ & & & e_2 & d_2 & c_2 & b_2 \\ & &e_3 & d_3 & c_3 & b_3 & a_3 \\ & \vdots & \vdots & \vdots & \vdots & \vdots & \\ e_{n-2} &d_{n-2} & c_{n-2} & b_{n-2} & a_{n-2} & & \\ d_{n-1} &c_{n-1} & b_{n-1} & a_{n-1} & & & \\ c_n &b_n & a_n & & & & \end{array} \right). $

若有$d_i, e_i\neq0$, 且$|c_1|\geq|d_1|+|e_1|$, $|c_2|\geq|b_2|+|d_2|+|e_2|$, $|c_{n-1}|\geq|a_{n-1}|+|b_{n-1}|+|d_{n-1}|$, $|c_n|\geq|b_n|+|a_n|$, $|c_i|\geq|a_i|+|b_i|+|d_i|+|e_i|$$(i=3, \cdots, n-2)$, 其中至少有一个不等号严格成立，则此矩阵的各阶顺序主子式不等于零，存在唯一的分解$A=LU$.若上述条件不变，其中$|c_n|>|b_n|+|a_n|$, 则称$A$为具非零元素链对角占优矩阵.

在实际问题中, 经常遇到如下形式的线性方程组

$ \left( \begin{array}{ccccccc} & & & &e_1 &d_1& c_1 \\ & & & e_2 & d_2 & c_2 & b_2 \\ & & e_3 &d_3 & c_3 & b_3 & a_3 \\ & \vdots &\vdots & \vdots & \vdots & \vdots & \\ e_{n-2} & d_{n-2} & c_{n-2} & b_{n-2} & a_{n-2} & & \\ d_{n-1} & c_{n-1} & b_{n-1} & a_{n-1} & & & \\ c_n & b_n & a_n & & & & \\ \end{array} \right) \left( \begin{array}{c} x_1 \\ x_2 \\ \vdots \\ \vdots \\ \vdots \\ \vdots\\ x_n \\ \end{array} \right) = \left( \begin{array}{c} f_1 \\ f_2 \\ \vdots \\ \vdots \\ \vdots \\ \vdots \\ f_n \end{array} \right). $

(2.1)

这种方程组称为反五对角线性方程组, 简记为$Ax=f$.

假设

$ \left\{\begin{array}{ll} |c_1|\geq|d_1|+|e_1|, ~d_1, e_1\neq0;\\ |c_2|\geq|b_2|+|d_2|+|e_2|, ~b_2, d_2, e_2\neq0;\\ |c_i|\geq|a_i|+|b_i|+|d_i|+|e_i|, ~a_i, b_i, d_i, e_i\neq0~~(i=3, 4, \cdots, n-2);\\ |c_{n-1}|\geq|a_{n-1}|+|b_{n-1}|+|d_{n-1}|, ~a_{n-1}, b_{n-1}, d_{n-1}\neq0;\\ |c_n|\geq|b_n|+|a_n|, ~a_n, b_n\neq0. \end{array} \right. $

此时, $A$为具非零元素链对角占优矩阵，可实现矩阵$LU$分解.分解形式为

$ A=\left( \begin{array}{cccccc} & & & & & l_1\\ & & & & l_2 & m_2 \\ & & & l_3 & m_3 & s_3 \\ & &\vdots & \vdots & \vdots & \\ & \vdots &\vdots &\vdots & & \\ l_n & m_n & s_n & & & \\ \end{array} \right) \left( \begin{array}{cccccc} 1 & & & & & \\ p_2 & 1 & & & & \\ q_3 & p_3 & 1 & & & \\ & \vdots & \vdots & \vdots & & \\ & &\vdots &\vdots &\vdots & \\ & & & q_n & p_n & 1 \\ \end{array} \right)=LU. $

(2.2)

利用矩阵乘法，可得

$ \left\{\begin{array}{ll} p_n=d_1/c_1, l_1=c_1, m_2=b_2, l_2=c_2-b_2d_1/c_1, \\ s_i=a_i, m_i=b_i-a_ip_{n-i+3}, l_i=c_i-a_ip_{n-i+3}-m_ip_{n-i+2}, i=3, 4, \cdots, n, \\ p_{n-i+1}=(d_i-m_iq_{n-i+2})/l_i, i=2, 3, \cdots, n-1, \\ q_{n-i+1}=e_i/l_i, i=1, 2, \cdots, n-2. \end{array} \right. $

(2.3)

可将求解方程组$Ax=f$化为依次求解$\left\{\begin{array}{ll} Ly=f, \\ Ux=y. \\ \end{array} \right.$

算法1  

第一步：解方程$Ly=f$, 即“追”过程, 算法如下:

$ \left\{\begin{array}{ll} y_n=f_1/c_1, \\ y_{n-1}=(f_2c_1-b_2y_nc_1)/(c_1c_2-b_2d_1), \\ y_{n-i+1}=(f_i-m_iy_{n-i+2}-a_iy_{n-i+3})/l_i, i=3, 4, \cdots, n. \end{array} \right. $

第二步：解方程$Ux=y$, 即“赶”过程，算法如下：

$ \left\{\begin{array}{ll} x_1=y_1, \\ x_2=y_2-x_1p_2, \\ x_i=y_i-x_{i-2}q_i-x_{i-1}p_i, i=3, 4, \cdots, n, \end{array} \right. $

其中$l_i, m_i, p_i, q_i$的计算见(2.3) 式.

上述算法1就是求解反五对角线性方程组的追赶法.

算例演示：用追赶法求解反五对角线性方程组:

$ \left\{\begin{array}{ll} -2x_5-2x_6+4x_7=6, \\ -x_4-2x_5+5x_6-2x_7=2, \\ -2x_3-x_4+6x_5-x_6-2x_7=0, \\ -2x_2-x_3+6x_4-x_5-2x_6=0, \\ -x_1-x_2+5x_3-2x_4-x_5=-1, \\ -3x_1+6x_2-x_3-2x_4=-2, \\ 4x_1-2x_2-x_3=-3, \end{array} \right. $

有

$ \begin{eqnarray*}&&A=\left( \begin{array}{ccccccc} & & & &-2 &-2&4 \\ & & & -1 & -2 & 5 & -2 \\ & &-2 & -1 & 6 & -1 & -2 \\ &-2 & -1 & 6 & -1 & -2 & \\ -1 &-1 & 5 & -2 & -1 & & \\ -3 &6 & -1 &-2 & & & \\ 4 &-2 &-1 & & & & \\ \end{array} \right), ~~f=\left( \begin{array}{c} 6 \\ 2 \\ 0 \\ 0 \\ -1 \\ -2 \\ -3 \\ \end{array} \right), \\ && y_7=f_1/c_1=6/4=3/2, \\ &&y_6=(f_2c_1-b_2y_7c_1)/(c_1c_2-b_2d_1)\\ & =&[2\times4-(-2)\times(3/2)\times4]/[4\times5-(-2)\times(-2)]=5/4, \end{eqnarray*} $

$ \begin{eqnarray*}&&q_7=e_1/l_1=-1/2, l_2=c_2-b_2d_1/c_1=5-(-2)\times(-2)/4=4, \\ &&p_6=(d_2-m_2q_7)/l_2=[-2-(-2)\times(-1/2)]/4=-3/4, \\ && m_3=b_3-a_3d_1/c_1=(-1)-(-2)\times(-2/4)=-2, \\ &&l_3=c_3-a_3q_7-m_3p_6=6-(-2)\times(-1/2)-(-2)\times(-3/4)=7/2, \\ &&y_5=(f_3-m_3y_6-a_3y_7)/l_3=[0-(-2))\times(5/4)-(-2)\times(3/2)]/(7/2)=11/7, \\ &&q_6=e_2/l_2=-1/4, p_5=(d_3-m_3q_6)/l_3=[-1-(-2)\times(-1/4)]/(7/2)=-3/7, \\ &&m_4=b_4-a_4p_6=(-1)-(-2)\times(-3/4)=-5/2, \\ &&l_4=c_4-a_4q_6-m_4p_5=6-(-2)\times(-1/4)-(-5/2)\times(-3/7)=31/7, \\ &&y_4=(f_4-m_4y_5-a_4y_6)/l_4=[0-(-5/2)\times(11/7)-(-2)\times(5/4)]/(31/7)=45/31, \\ &&q_5=e_3/l_3=-2/(7/2)=-4/7, \\ && p_4=(d_4-m_4q_5)/l_4=[-1-(-5/2)\times(-4/7)]/(31/7)=-17/31, \\ &&m_5=b_5-a_5p_5=-2-(-1)\times(-3/7)=-17/7, \\ &&l_5=c_5-a_5q_5-m_5p_4=5-(-1)\times(-4/7)-(-17/7)\times(-17/31)=96/31, \\ &&y_3=(f_5-m_5y_4-a_5y_5)/l_5=[-1-(-17/7)\times45/31-(-1)\times11/7]/(96/31)=127/96, \\ &&q_4=e_4/l_4=-2/(31/7)=-14/31, \\ &&p_3=(d_5-m_5q_4)/l_5=[-1-(-17/7)\times(-14/31)]/(96/31)=-65/96, \\ &&m_6=b_6-a_6p_4=-1-(-2)\times(-17/31)=-65/31, \\ &&l_6=c_6-a_6q_4-m_6p_3=6-(-2)\times(-14/31)-(-65/31)\times(-65/96)=353/96, \\ &&y_2=(f_6-m_6y_3-a_6y_4)/l_6=[-2-(-65/31)\times127/96-(-2)\times45/31]/(353/96)=1, \\ &&q_3=e_5/l_5=-1/(96/31)=-31/96, \\ &&p_2=(d_6-m_6q_3)/l_6=[-3-(-65/31)\times(-31/96)]/(353/96)=-1, \\ &&m_7=b_7-a_7p_3=-2-(-1)\times(-65/96)=-257/96, \\ &&l_7=c_7-a_7q_3-m_7p_2=4-(-1)\times(-31/96)-(-257/96)\times(-1)=1, \\ &&y_1=(f_7-m_7y_2-a_7y_3)/l_7=[-3-(-257/96)\times1-(-1)\times127/96]/1=1.\end{eqnarray*} $

利用以上结果, 我们可以得到关于$x_i(i=1, 2, 3, 4, 5, 6, 7)$的解:

$ \begin{eqnarray*}&&x_1=y_1=1, x_2=y_2-x_1p_2=1-1\times(-1)=2, \\ &&x_3=y_3-x_1q_3-x_2p_3=127/96-1\times(-31/96)-2\times(-65/96)=3, \\ &&x_4=y_4-x_2q_4-x_3p_4=45/31-2\times(-14/31)-3\times(-17/31)=4, \\ &&x_5=y_5-x_3q_5-x_4p_5=11/7-3\times(-4/7)-4\times(-3/7)=5, \\ &&x_6=y_6-x_4q_6-x_5p_6=5/4-4\times(-1/4)-5\times(-3/4)=6, \\ &&x_7=y_7-x_5q_7-x_6p_7=3/2-5\times(-1/2)-6\times(-1/2)=7.\end{eqnarray*} $

在实际问题中，由于误差的原因，经常遇到如下形式的线性方程组

$ \left( \begin{array}{ccccccc} b_1 &a_1 & & &e_1 &d_1& c_1 \\ a_2 & & & e_2 & d_2 & c_2 & b_2 \\ & & e_3 &d_3 & c_3 & b_3 & a_3 \\ & \vdots &\vdots & \vdots & \vdots & \vdots & \\ e_{n-2} & d_{n-2} & c_{n-2} & b_{n-2} & a_{n-2} & & \\ d_{n-1} & c_{n-1} & b_{n-1} & a_{n-1} & & &e_{n-1} \\ c_n & b_n & a_n & & & e_n &d_n \\ \end{array} \right) \left( \begin{array}{c} x_1 \\ x_2 \\ \vdots \\ \vdots \\ \vdots \\ \vdots\\ x_n \\ \end{array} \right) = \left( \begin{array}{c} f_1 \\ f_2 \\ \vdots \\ \vdots \\ \vdots \\ \vdots \\ f_n \end{array} \right). $

(3.1)

这种方程组称为拟反五对角线性方程组，简记$Ax=f$.可将系数矩阵$A$分解为3个矩阵的乘积$A=LUD$, 其中

$ L\left( \begin{array}{ccccc} & & & &p_1 \\ & & & p_1 &g_2 \\ & & p_3 &g_3 &a_3 \\ & \vdots &\vdots &\vdots & \\ p_n& g_n &a_n & & \\ \end{array} \right), ~~~~U=\left( \begin{array}{ccccccc} 1 & & & & & & \\ 0 & 1& & & & & \\ 0 & 0& 1 & & & & \\ & 0& q_4 & 1 & & & \\ & & h_5 & q_5 & 1 & & \\ & & & \vdots &\vdots & \vdots & \\ & & & & h_n & q_n & 1 \\ \end{array} \right), ~ D=\left( \begin{array}{cccccc} 1 & & & & s_1 & t_1 \\ t_2 & 1 & & & & s_2 \\ t_3 & s_3 & 1 & & & \\ \vdots & \vdots & &\vdots & & \\ \vdots & \vdots & & &\vdots & \\ t_n & s_n & & & & 1 \\ \end{array} \right). $

矩阵元素的计算公式如下:

$ \left\{\begin{array}{ll} p_1=c_1, q_n=d_1/c_1, h_n=e_1/c_1, \omega_n=a_1/c_1, \sigma_n=b_1/c_1; \\ g_2=b_2, p_2=c_2-b_2d_1/c_1, q_{n-1}=(c_1d_2-b_2e_1)/(c_1c_2-b_2d_1), h_{n-1}=c_1e_2/(c_1c_2-b_2d_1);\\ \omega_{n-1}=-a_1b_2/(c_1c_2-b_2d_1), \sigma_{n-1}=(a_2c_1-b_1b_2)/(c_1c_2-b_2d_1); \\ \alpha_i=a_i, g_i=b_i-a_iq_{n-i+3}, p_i=c_i-a_i h_{n-i+3}-g_iq_{n-i+2};\\ q_{n-i+1}=(d_i-g_i h_{n-i+2})/p_i, h_{n-i+1}=e_i/p_i;\\ \omega_{n-i+1}=(-a_i\omega_{n-i+3}-g_i\omega_{n-i+2})/p_i, \sigma_{n-i+1}=(-a_i\sigma_{n-i+3}-g_i\sigma_{n-i+2})/p_i, i=3, 4, \ldots, n-4;\\ g_{n-3}=b_{n-3}-\alpha_{n-3}q_6, p_{n-3}=c_{n-3}-\alpha_{n-3}h_6-g_{n-3}q_5;\\ q_4=(d_{n-3}-g_{n-3}h_5)/p_{n-3};\\ \omega_4=(e_{n-3}-\alpha_{n-3}\omega_6-g_{n-3}\omega_5)/p_{n-3};\\ \sigma_4=(-\alpha_{n-3}\sigma_6-g_{n-3}\sigma_5)/p_{n-3};\\ g_{n-2}=b_{n-2}-\alpha_{n-2}q_5, p_{n-2}=c_{n-2}-\alpha_{n-2}h_5-g_{n-2}q_4;\\ \omega_3= (d_{n-2}-\alpha_{n-2}\omega_5-g_{n-2}\omega_4)/p_{n-2};\\ \sigma_3=(e_{n-2}-\alpha_{n-2}\sigma_5-g_{n-2}\sigma_4)/p_{n-2};\\ g_{n-1}=b_{n-1}-\alpha_{n-1}q_4, p_{n-1}=c_{n-1}-\alpha_{n-1}\omega_4-g_{n-1}\omega_3;\\ \omega_2=e_{n-1}/p_{n-1}, \sigma_2=(d_{n-1}-\alpha_{n-1}\sigma_4-g_{n-1}\sigma_3)/p_{n-1}; \\ g_n=b_n-\alpha_n\omega_3, p_n=c_n-\alpha_n\sigma_3-g_n\sigma_2;\\ \sigma_1=(d_n-g_n\omega_2)/p_n, \omega_1=e_n/p_n;\\ s_i=\omega_i, t_i=\sigma_i, i=1, 2, 3;\\ s_4=\omega_4-q_4s_3, t_4=\sigma_4-q_4t_3;\\ s_i=\omega_i-h_is_{i-2}-q_is_{i-1}, t_i=\sigma_i-h_it_{i-2}-q_it_{i-1}, i=5, 6, \ldots, n. \end{array}\right. $

可将求解方程组$Ax=f$化为依次求解$\left\{\begin{array}{cl} Lz=f, \\ Uy=z, \\ Dx=y.\\ \end{array} \right.$

算法2  

第一步:解方程$Lz=f$, 即“追”过程, 算法如下：

$ \left\{\begin{array}{ll} z_n=f_1/c_1, \\ z_{n-1}=(c_1f_2-c_1b_2z_n)/(c_1c_2-b_2d_1), \\ z_{n-i+1}=(f_i-g_iz_{n-i+2}-a_iz_{n-i+3})/p_i, i=3, 4, \ldots, n.\\ \end{array} \right. $

第二步:依次求解$Uy=z$, $Dx=y$, 即“赶”的过程, 算法如下：

$ \begin{eqnarray*}&& \left\{\begin{array}{ll} y_1=z_1, y_2=z_2, y_3=z_3, \\ y_4=z_4-q_4y_3, \\ y_i=z_i-h_iy_{i-2}-q_iy_{i-1}, i=5, 6, \ldots, n, \end{array} \right.\\ &&\left\{\begin{array}{ll} y_n=\frac{(y_{n-1}-t_{n-1}y_1+s_{n-1}t_2y_1-s_{n-1}y_2)(s_nt_2s_1-t_ns_1)-(y_n-t_ny_1+s_nt_2y_1-s_ny_2)(1-t_{n-1}s_1-s_{n-1}t_2s_1)}{(s_{n-1}t_2t_1-t_{n-1}t_1-s_{n-1}s_2)(s_nt_2s_1-t_ns_1)-(1-t_nt_1+s_nt_2t_1-s_ns_2)(1-t_{n-1}s_1+s_{n-1}t_2s_1)};\\ x_{n-1}=\frac{(y_{n-1}-t_{n-1}y_1+s_{n-1}t_2y_1-s_{n-1}y_2)(1-t_nt_1+s_nt_2t_1-s_ns_2)}{(1-t_{n-1}s_1+s_{n-1}t_2s_1)(1-t_nt_1+s_nt_2t_1-s_ns_2)-(s_nt_2s_1-t_ns_1)(s_{n-1}t_2t_1-t_{n-1}t_1-s_{n-1}s_2)}\\ ~~~~~~~~~~~~-\frac{(y_n-t_ny_1+s_nt_2y_1-s_ny_2)(s_{n-1}t_2t_1-t_{n-1}t_1-s_{n-1}s_2)}{(1-t_{n-1}s_1+s_{n-1}t_2s_1)(1-t_nt_1+s_nt_2t_1-s_ns_2)-(s_nt_2s_1-t_ns_1)(s_{n-1}t_2t_1-t_{n-1}t_1-s_{n-1}s_2)};\\ x_i=y_i-t_ix_1-s_ix_2, i=3, 4, \ldots, n-2;\\ x_2=y_2-t_2y_1+t_2s_1x_{n-1}+t_2t_1x_n-s_2x_n;\\ x_1=y_1-s_1x_{n-1}-t_1x_n, \end{array} \right.\end{eqnarray*} $

其中$p_i, g_i, h_i, q_i, t_i, s_i$的值的计算见(3.3) 式.

上述算法2就是求解拟反五对角线性方程组的追赶法.

算例演示:用追赶法求解拟反五对角线性方程组:

$ \left\{\begin{array}{cl} x_1+x_2-x_4-x_5+4x_6=18, \\ x_1-x_3-x_4+4x_5-x_6=8, \\ -x_2-x_3+4x_4-x_5-x_6=0, \\ -x_1-x_2+4x_3-x_4-x_5=0, \\ -x_1+4x_2-x_3-x_4+x_6=6, \\ 4x_1-x_2-x_3+x_5+x_6=10, \end{array} \right. $

有

$ \begin{eqnarray*}&&A=\left( \begin{array}{cccccc} 1 & 1 & &-1 & -1 & 4 \\ 1 & & -1 &-1 & 4 & -1 \\ & -1 &-1 & 4 & -1 & -1 \\ -1 & -1 & 4 &-1 &-1 & \\ -1 & 4 & -1 & -1 & & 1 \\ 4 & -1 &-1 & & 1 & 1 \\ \end{array} \right), ~~f=\left( \begin{array}{c} 18\\ 8\\ 0\\ 0\\ 6\\ 10\\ \end{array} \right); \\ &&z_6=f_1/c_1=9/2;z_5=(c_1f_2-c_1b_2z_6)/(c_1c_2-b_2d_1)=50/15=10/3;\\ &&q_6=d_1/c_1=-1/4;q_5=(c_1d_2-b_2e_1)/(c_1c_2-b_2d_1)=-5/15=-1/3;\end{eqnarray*} $

$ \begin{eqnarray*}&&h_6=e_1/c_1=-1/4;p_3=c_3-a_3h_6-g_3q_5=10/3, g_3=b_3-a_3q_6=-5/4;\\ &&z_4=(f_3-g_3z_5-a_3z_6)/p_3=(26/3)/(10/3)=13/5;\\ &&q_4=(d_3-g_3h_5)/p_3=(-4/3)/(10/3)=-2/5, h_5=c_1e_1/(c_1c_2-b_2d_1)=-4/15;\\ &&p_4=c_4-a_4h_5-g_4q_4=16/5, g_4=b_4-a_4q_5=-4/3, \\ &&z_3=(f_4-g_4z_4-a_4z_5)/p_4=(34/5)/(16/5)=17/8;\\ &&\omega_6=a_1/c_1=1/4, \omega_5=-a_1b_2/(c_1c_2-b_2d_1)=1/15, \\ &&\omega_4=(e_3-a_3\omega_6-g_3\omega_5)/p_3=(-2/3)/(10/3)=-1/5, \\ &&\omega_3=(d_4-a_4\omega_5-g_5\omega_4)/p_4=(-6/5)/(16/5)=-3/8;\\ &&g_5=b_5-a_5q_4=-7/5, p_5=c_5-a_5\omega_4-g_5\omega_3=131/40;\\ &&z_2=(f_5-g_5z_3-a_5z_4)/p_5=463/131;\\ &&\sigma_6=b_1/c_1=1/4, \sigma_5=(a_2c_1-b_1b_2)/(c_1c_2-b_2d_1)=1/3, \sigma_4=(-a_3\sigma_6-g_3\sigma_5)/p_3=1/5, \\ &&\sigma_3=(e_4-a_4\sigma_5-g_4\sigma_4)/p_4=(-2/5)/(16/5)=-1/8, \\ &&\sigma_2=(d_5-a_5\sigma_4-g_5\sigma_3)/p_5=(-39/40)/(131/40)=-39/131, \\ &&p_6=c_6-a_6\sigma_3-g_6\sigma_2=454/131, g_6=b_6-a_6\omega_3=-11/8, \\ &&z_1=(f_6-g_6z_2-a_6z_3)/p_6=(2225/131)/(454/131)=2225/454;\\ &&y_1=z_1=2225/454, y_2=z_2=463/131, y_3=z_3=17/8, y_4=z_4-q_4y_3=69/20, \\ &&y_5=z_5-h_5y_3-q_5y_4=101/20, y_6=z_6-h_6y_4-q_6y_5=53/8, \\ &&s_1=\omega_1=e_6/p_6=131/454, s_2=\omega_2=e_5/p_5=40/131, s_3=\omega_3=-3/8, \\ &&s_4=\omega_4-q_4s_3=-7/20, s_5=\omega_5-h_5s_3-q_5s_4=-3/20, s_6=\omega_6-h_6s_4-q_6s_5=1/8;\\ &&t_1=\sigma_1=(d_6-g_6\omega_2)/p_6=(186/131)/(454/131)=186/454, t_2=\sigma_2=-39/131, \\ &&t_3=\sigma_3=-1/8, t_4=\sigma_4-q_4t_3=3/20, \\ &&t_5=\sigma_5-q_5t_3-q_5t_4=7/20, t_6=\sigma_6-h_6t_4-q_6t_5=3/8.\end{eqnarray*} $

利用以上结果, 我们最终可以得到关于$x_i(i=1, 2, 3, 4, 5, 6)$的解:

$ \begin{eqnarray*}&&x_6=6; x_5=5; x_1=y_1-s_1x_5-t_1x_6=2225/454-131/454\times5-186/454\times6=1;\\ &&x_2=y_2-t_2y_1+t_2s_1x_5+t_2t_1x_6-s_2x_6\\ &=& 463/131-(-39/131)\times2225/454+(-39/131)\times131/454\\ &&\times5+(-39/131)\times186/454\times6-40/131\times6=2;\\ &&x_3=y_3-t_3x_1-s_3x_2=17/8-(-1/8)\times1-(-3/8)\times2=3;\\ &&x_4=y_4-t_4x_1-s_4x_2=69/20-3/20\times1-(-7/20)\times2=4.\end{eqnarray*} $

本文通过$LU$分解推导出了反五对角线性方程组的追赶法.针对拟反五对角线性方程组的特点，沿用$LU$分解和追赶法的基本思想，把拟反五对角线性方程组的系数矩阵$A$分解成3个矩阵的乘积$LUD$，与$LU$算法相比，虽然增加了一个矩阵$D$，可是简化了推导过程的复杂度.追赶法求解n阶反五对角线性方程组只需要$O(11n)$的运算量，追赶法求解$n$阶拟反五对角线性方程组的运算量仅为$O(39n)$，运算量比较小.