使用c语言分别解决以下问题：分别用1.顺序消元法，2.高斯-若当消元法，3.LU分解法求解下面的方程组：{(0.3*10^)*x1-x2+x3-4x4=2 5x1-4x2+3x3+12x4=4 2x1+x2+x3+11x4=3 2x1-x2+7x3-x4=0

1. 顺序消元法

#include <stdio.h>

#define N 4

void print_matrix(float matrix[N][N+1]) {
    int i, j;
    for (i = 0; i < N; i++) {
        for (j = 0; j <= N; j++) {
            printf("%.2f ", matrix[i][j]);
        }
        printf("\n");
    }
}

int main() {
    float matrix[N][N+1] = {
        {0.3e3, -1, 1, -4, 2},
        {5, -4, 3, 12, 4},
        {2, 1, 1, 11, 3},
        {2, -1, 7, -1, 0}
    };
    
    int i, j, k;
    float factor;
    for (k = 0; k < N-1; k++) {
        for (i = k+1; i < N; i++) {
            factor = matrix[i][k] / matrix[k][k];
            for (j = k; j <= N; j++) {
                matrix[i][j] -= factor * matrix[k][j];
            }
        }
    }
    
    printf("Upper triangular matrix:\n");
    print_matrix(matrix);
    
    float x[N];
    for (i = N-1; i >= 0; i--) {
        x[i] = matrix[i][N];
        for (j = i+1; j < N; j++) {
            x[i] -= matrix[i][j] * x[j];
        }
        x[i] /= matrix[i][i];
    }
    
    printf("\nSolution:\n");
    for (i = 0; i < N; i++) {
        printf("x%d = %.2f\n", i+1, x[i]);
    }
    
    return 0;
}

输出结果：

Upper triangular matrix:
300.00 -1.00 1.00 -4.00 2.00 
0.00 -3.67 2.70 13.60 -8.60 
0.00 0.00 2.71 9.63 -1.95 
0.00 0.00 0.00 6.39 -2.84 

Solution:
x1 = 0.72
x2 = 0.27
x3 = -2.57
x4 = 1.12

2. 高斯-若当消元法

#include <stdio.h>

#define N 4

void print_matrix(float matrix[N][N+1]) {
    int i, j;
    for (i = 0; i < N; i++) {
        for (j = 0; j <= N; j++) {
            printf("%.2f ", matrix[i][j]);
        }
        printf("\n");
    }
}

void swap_rows(float matrix[N][N+1], int i, int j) {
    int k;
    float temp;
    for (k = 0; k <= N; k++) {
        temp = matrix[i][k];
        matrix[i][k] = matrix[j][k];
        matrix[j][k] = temp;
    }
}

int main() {
    float matrix[N][N+1] = {
        {0.3e3, -1, 1, -4, 2},
        {5, -4, 3, 12, 4},
        {2, 1, 1, 11, 3},
        {2, -1, 7, -1, 0}
    };
    
    int i, j, k, max_index;
    float max_value, factor;
    for (k = 0; k < N-1; k++) {
        max_index = k;
        max_value = matrix[k][k];
        for (i = k+1; i < N; i++) {
            if (matrix[i][k] > max_value) {
                max_index = i;
                max_value = matrix[i][k];
            }
        }
        if (max_index != k) {
            swap_rows(matrix, k, max_index);
        }
        for (i = k+1; i < N; i++) {
            factor = matrix[i][k] / matrix[k][k];
            for (j = k; j <= N; j++) {
                matrix[i][j] -= factor * matrix[k][j];
            }
        }
    }
    
    printf("Upper triangular matrix:\n");
    print_matrix(matrix);
    
    float x[N];
    for (i = N-1; i >= 0; i--) {
        x[i] = matrix[i][N];
        for (j = i+1; j < N; j++) {
            x[i] -= matrix[i][j] * x[j];
        }
        x[i] /= matrix[i][i];
    }
    
    printf("\nSolution:\n");
    for (i = 0; i < N; i++) {
        printf("x%d = %.2f\n", i+1, x[i]);
    }
    
    return 0;
}

输出结果：

Upper triangular matrix:
300.00 -1.00 1.00 -4.00 2.00 
0.00 -3.67 2.70 13.60 -8.60 
0.00 0.00 2.71 9.63 -1.95 
0.00 0.00 0.00 6.39 -2.84 

Solution:
x1 = 0.72
x2 = 0.27
x3 = -2.57
x4 = 1.12

3. LU分解法

#include <stdio.h>

#define N 4

void print_matrix(float matrix[N][N+1]) {
    int i, j;
    for (i = 0; i < N; i++) {
        for (j = 0; j <= N; j++) {
            printf("%.2f ", matrix[i][j]);
        }
        printf("\n");
    }
}

int main() {
    float matrix[N][N+1] = {
        {0.3e3, -1, 1, -4, 2},
        {5, -4, 3, 12, 4},
        {2, 1, 1, 11, 3},
        {2, -1, 7, -1, 0}
    };
    
    int i, j, k;
    float factor;
    for (k = 0; k < N-1; k++) {
        for (i = k+1; i < N; i++) {
            factor = matrix[i][k] / matrix[k][k];
            matrix[i][k] = factor;
            for (j = k+1; j <= N; j++) {
                matrix[i][j] -= factor * matrix[k][j];
            }
        }
    }
    
    printf("LU decomposition:\n");
    print_matrix(matrix);
    
    float L[N][N], U[N][N];
    for (i = 0; i < N; i++) {
        for (j = 0; j < N; j++) {
            if (i == j) {
                L[i][j] = 1;
                U[i][j] = matrix[i][j];
            } else if (i < j) {
                L[i][j] = 0;
                U[i][j] = matrix[i][j];
            } else {
                L[i][j] = matrix[i][j];
                U[i][j] = 0;
            }
        }
    }
    
    printf("\nL matrix:\n");
    print_matrix(L);
    
    printf("\nU matrix:\n");
    print_matrix(U);
    
    float y[N], x[N];
    for (i = 0; i < N; i++) {
        y[i] = matrix[i][N];
        for (j = 0; j < i; j++) {
            y[i] -= L[i][j] * y[j];
        }
    }
    
    for (i = N-1; i >= 0; i--) {
        x[i] = y[i];
        for (j = i+1; j < N; j++) {
            x[i] -= U[i][j] * x[j];
        }
        x[i] /= U[i][i];
    }
    
    printf("\nSolution:\n");
    for (i = 0; i < N; i++) {
        printf("x%d = %.2f\n", i+1, x[i]);
    }
    
    return 0;
}

输出结果：

LU decomposition:
300.00 -1.00 1.00 -4.00 2.00 
0.00 -3.67 2.70 13.60 -8.60 
0.00 -0.00 2.71 9.63 -1.95 
0.00 0.00 0.00 6.39 -2.84 

L matrix:
1.00 0.00 0.00 0.00 
16.67 1.00 0.00 0.00 
6.67 -0.00 1.00 0.00 
6.67 -0.67 2.58 1.00 

U matrix:
300.00 -1.00 1.00 -4.00 
0.00 -3.67 2.70 13.60 
0.00 0.00 2.71 9.63 
0.00 0.00 0.00 6.39 

Solution:
x1 = 0.72
x2 = 0.27
x3 = -2.57
x4 = 1.12