數值計算——係數矩陣部分對角線爲0時線性方程組求解方法（附程序）

求解線性方程組時，我們經常用的方法是高斯消去法，矩陣三角分解，雅克比迭代，以及迭代方法如共軛梯度等。在使用這些方法求解的過程中，通常需要,但是難免會遇到對角線有一些數爲0的情況。本文求解方法大致求解思路還是採取高斯消去法，在高斯消去法的基礎上多了一個矩陣係數對角線爲0的判斷，如果爲0，則與上一行交換，包括列向量對應的位置。

原理參考我之前寫的文章：https://blog.csdn.net/weixin_41788456/article/details/102485139

程序：

//開發人員：陳帥  開發日期：2019.10.15   郵箱：[email protected] 
#include "pch.h"
#include <iostream>//基本數據流輸入/輸出
#include <iomanip> //參數化輸入/輸出 
#include <vector>//STL動態數組容器
using namespace std;
//************************
//改進的高斯消去法公式
//***********************
vector<double> gaussian_elim_mutate(vector<vector<double>>a, vector<double>b); //高斯消去法求解線性方程組AX=B
vector<double> gaussian_elim_mutate(vector<vector<double>>a, vector<double>b)
{
	int n = size(b);
	vector<double>x;    //定義方程組解
	x.resize(n);
	vector<double>mi_k; //定義消去過程中的中間變量
	mi_k.resize(n);
	double sum;
	double line_exchange = 0;
	double m[3];
	//如果對角線爲0，則交換上一行，直到不爲0爲止
	for (int k = 0; line_exchange != -1; k++)
	{
		for (int i = 0; i < n; i++)
		{
			line_exchange = -1;
			if (a[i][i] == 0)
			{

				line_exchange = i;
				if (i == 0)
				{
					for (int j = 0; j < 3; j++)
					{
						m[j] = a[n - 1][j];
						a[n - 1][j] = a[i][j];
						a[i][j] = m[j];
					}
					m[0] = b[n - 1];
					b[n - 1] = b[i];
					b[i] = m[0];
				}
				else
				{
					for (int j = 0; j < 3; j++)
					{
						m[j] = a[i - 1][j];
						a[i - 1][j] = a[i][j];
						a[i][j] = m[j];
					}
					m[0] = b[i - 1];
					b[i - 1] = b[i];
					b[i] = m[0];
				}
			}
		}
	}
	//再次判斷能否用高斯消去法（驗證）
	for (int i = 0; i < n; i++)
	{
		if (a[i][i] == 0)
		{
			cout << "can't use Gaussian meathod" << endl;
		}
	}
	//n-1步消元
	for (int k = 0; k < n - 1; k++)
	{
		//求出第i次初等行變換系數
		for (int j = k + 1; j < n; j++)
		{
			mi_k[j] = a[j][k] / a[k][k];
		}
		for (int i = k + 1; i < n; i++)
		{
			for (int j = 0; j < n; j++)
			{
				a[i][j] = a[i][j] - mi_k[i] * a[k][j];
			}
			b[i] = b[i] - mi_k[i] * b[k];
		}
	}	//回代過程
	x[n - 1] = b[n - 1] / a[n - 1][n - 1];
	for (int i = n - 2; i >= 0; i--)
	{
		sum = 0;
		for (int j = i + 1; j < n; j++)
		{
			sum = sum + a[i][j] * x[j];
		}
		x[i] = (b[i] - sum) / a[i][i];
	}
	return x;
}
int main()
{
	vector<vector<double>>a;
	a.resize(3, vector<double>(3));
	vector<double>b(3);
	vector<double>x;
	a[0] = { 0,1.1,3.1 }, a[1] = { 2.0,0,0.36 }, a[2] = { 5.0,0.96,0 };
	b = { 6.0,0.02,0.96 };

	x = gaussian_elim_mutate(a, b);
	cout << "解爲：" << endl;
	for (int i = 0; i < 3; i++)
		cout <<"x["<<i<<"]="<< fixed << setprecision(2) << setw(5) << x[i] << endl;
}

運行結果：