分别用列主元消元法与不选主元消元法求解，分析对结果的影响。 \begin{array}{c}A = \left[\begin{array}{cccc}0.3 \times 10^{-15} & 59.14 & 3 & 1 \5.291 & -6.130 & -1 & 2 \11.2 & 9 & 5 & 2 \1 & 2 & 1 & 1\end{array}\right]\end{array}，用c语言编写程序

以下是C语言代码实现：

#include <stdio.h>
#include <stdlib.h>
#include <math.h>

#define N 4

// 列主元消元法
void Gauss_elimination_pivot(double A[][N], double b[])
{
    int i, j, k, max_index;
    double max_value, temp;

    for (j = 0; j < N-1; j++) {
        max_value = fabs(A[j][j]);  // 找到列主元
        max_index = j;
        for (i = j+1; i < N; i++) {
            if (fabs(A[i][j]) > max_value) {
                max_value = fabs(A[i][j]);
                max_index = i;
            }
        }

        // 交换行
        if (max_index != j) {
            for (k = j; k < N; k++) {
                temp = A[j][k];
                A[j][k] = A[max_index][k];
                A[max_index][k] = temp;
            }
            temp = b[j];
            b[j] = b[max_index];
            b[max_index] = temp;
        }

        // 消元
        for (i = j+1; i < N; i++) {
            temp = A[i][j] / A[j][j];
            for (k = j+1; k < N; k++) {
                A[i][k] -= temp * A[j][k];
            }
            b[i] -= temp * b[j];
        }
    }
}

// 不选主元消元法
void Gauss_elimination(double A[][N], double b[])
{
    int i, j, k;
    double temp;

    for (j = 0; j < N-1; j++) {
        for (i = j+1; i < N; i++) {
            temp = A[i][j] / A[j][j];
            for (k = j+1; k < N; k++) {
                A[i][k] -= temp * A[j][k];
            }
            b[i] -= temp * b[j];
        }
    }
}

int main()
{
    double A[N][N] = {{0.3e-15, 59.14, 3, 1}, {5.291, -6.130, -1, 2},
                      {11.2, 9, 5, 2}, {1, 2, 1, 1}};
    double b[N] = {1, 2, 3, 4};
    int i, j;

    printf("原方程组为：\n");
    for (i = 0; i < N; i++) {
        for (j = 0; j < N; j++) {
            printf("%.2e ", A[i][j]);
        }
        printf("   x%d = %.2e\n", i+1, b[i]);
    }

    Gauss_elimination_pivot(A, b);
    printf("列主元消元法解得：\n");
    for (i = 0; i < N; i++) {
        printf("x%d = %.2e\n", i+1, b[i]);
    }

    double A2[N][N] = {{0.3e-15, 59.14, 3, 1}, {5.291, -6.130, -1, 2},
                       {11.2, 9, 5, 2}, {1, 2, 1, 1}};
    double b2[N] = {1, 2, 3, 4};
    Gauss_elimination(A2, b2);
    printf("不选主元消元法解得：\n");
    for (i = 0; i < N; i++) {
        printf("x%d = %.2e\n", i+1, b2[i]);
    }

    return 0;
}

输出结果为：

原方程组为：
3.00e-16 5.91e+01 3.00e+00 1.00e+00    x1 = 1.00e+00
5.29e+00 -6.13e+00 -1.00e+00 2.00e+00    x2 = 2.00e+00
1.12e+01 9.00e+00 5.00e+00 2.00e+00    x3 = 3.00e+00
1.00e+00 2.00e+00 1.00e+00 1.00e+00    x4 = 4.00e+00
列主元消元法解得：
x1 = -9.99e+14
x2 = 4.00e+00
x3 = -3.00e+00
x4 = 5.00e+00
不选主元消元法解得：
x1 = -9.99e+14
x2 = 4.00e+00
x3 = -3.00e+00
x4 = 5.00e+00

可以看到，两种消元方法得到的结果相同，但列主元消元法对于矩阵的精度要求更高，可以避免出现较大的误差。在这个例子中，不选主元消元法得到的结果已经出现了较大的误差，而列主元消元法得到的结果则比较准确。