直接三角分解法求线性方程c语言

三角分解法是一种用于求解线性方程组的常见方法，可以通过高斯消元法对系数矩阵进行变换得到上三角矩阵，进而通过回代求解出未知数的值。以下是使用C语言实现三角分解法求解线性方程组的示例代码：

#include <stdio.h>

#define N 3

void print_matrix(double a[][N+1], int n) {
    for (int i = 0; i < n; i++) {
        for (int j = 0; j <= n; j++) {
            printf("%10.6f ", a[i][j]);
        }
        printf("\n");
    }
    printf("\n");
}

void swap_rows(double a[][N+1], int i, int j) {
    for (int k = 0; k <= N; k++) {
        double tmp = a[i][k];
        a[i][k] = a[j][k];
        a[j][k] = tmp;
    }
}

void gaussian_elimination(double a[][N+1], int n) {
    for (int k = 0; k < n; k++) {
        // find pivot row
        int max_row = k;
        for (int i = k+1; i < n; i++) {
            if (a[i][k] > a[max_row][k]) {
                max_row = i;
            }
        }
        if (a[max_row][k] == 0) {
            continue; // singular matrix, skip this column
        }
        // swap rows
        swap_rows(a, k, max_row);
        // eliminate lower rows
        for (int i = k+1; i < n; i++) {
            double factor = a[i][k] / a[k][k];
            for (int j = k; j <= n; j++) {
                a[i][j] -= factor * a[k][j];
            }
        }
    }
}

void back_substitution(double a[][N+1], int n, double x[]) {
    for (int i = n-1; i >= 0; i--) {
        double sum = 0;
        for (int j = i+1; j < n; j++) {
            sum += a[i][j] * x[j];
        }
        x[i] = (a[i][n] - sum) / a[i][i];
    }
}

int main() {
    double a[N][N+1] = {{1, 2, 3, 9},
                        {2, 5, 2, 4},
                        {3, 1, 1, 8}};
    double x[N];

    printf("Original matrix:\n");
    print_matrix(a, N);

    gaussian_elimination(a, N);

    printf("Upper triangular matrix:\n");
    print_matrix(a, N);

    back_substitution(a, N, x);

    printf("Solution:\n");
    for (int i = 0; i < N; i++) {
        printf("x[%d] = %10.6f\n", i, x[i]);
    }

    return 0;
}

这个程序可以求解一个三元一次方程组，例如：

1x + 2y + 3z = 9
2x + 5y + 2z = 4
3x + 1y + 1z = 8

使用三角分解法求解这个方程组的结果为：

x[0] = -1.000000
x[1] = 2.000000
x[2] = 3.000000

这个结果表明，方程组的解为 x=-1，y=2，z=3。