用c语言编程，在dev-c++上将A,B俩个四阶方阵化为最简阶梯型行列式，A中的元素为 -6 -7 7 6 3 4 2 3 -4 -2 0 6 1 7 8 3，B中的元素为 -3 4 -2 5 -4 -4 4 2 -3 6 1 6 1 1 -1 9

以下是实现该问题的 C 代码：

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

#define N 4

void swap(double *a, double *b) {
    double temp = *a;
    *a = *b;
    *b = temp;
}

void print_matrix(double matrix[][N], char name) {
    printf("%c = \n", name);
    for (int i = 0; i < N; i++) {
        for (int j = 0; j < N; j++) {
            printf("%5.1f", matrix[i][j]);
        }
        printf("\n");
    }
    printf("\n");
}

void row_reduce(double matrix[][N]) {
    int i, j, k, m;
    double temp;
    for (i = 0; i < N; i++) {
        for (j = 0; j < N; j++) {
            if (i != j) {
                temp = matrix[j][i] / matrix[i][i];
                for (k = i; k < N; k++) {
                    matrix[j][k] -= temp * matrix[i][k];
                }
            }
        }
    }
    for (i = 0; i < N; i++) {
        temp = matrix[i][i];
        if (temp == 0) {
            printf("Error: Singular matrix!\n");
            exit(0);
        }
        for (m = i; m < N; m++) {
            matrix[i][m] /= temp;
        }
    }
}

int main() {
    double A[N][N] = {{-6, -7, 7, 6}, {3, 4, 2, 3}, {-4, -2, 0, 6}, {1, 7, 8, 3}};
    double B[N][N] = {{-3, 4, -2, 5}, {-4, -4, 4, 2}, {-3, 6, 1, 6}, {1, 1, -1, 9}};
    int i, j, k, m, n;
    double temp;
    
    printf("Original matrices:\n");
    print_matrix(A, 'A');
    print_matrix(B, 'B');
    
    // row reduce matrix A
    for (i = 0; i < N; i++) {
        if (A[i][i] == 0) {
            for (j = i + 1; j < N; j++) {
                if (A[j][i] != 0) {
                    for (k = i; k < N; k++) {
                        swap(&A[i][k], &A[j][k]);
                    }
                    break;
                }
            }
            if (j == N) {
                printf("Error: Singular matrix!\n");
                exit(0);
            }
        }
        for (j = i + 1; j < N; j++) {
            temp = A[j][i] / A[i][i];
            for (k = i; k < N; k++) {
                A[j][k] -= temp * A[i][k];
            }
        }
    }
    
    // row reduce matrix B
    for (i = 0; i < N; i++) {
        if (B[i][i] == 0) {
            for (j = i + 1; j < N; j++) {
                if (B[j][i] != 0) {
                    for (k = i; k < N; k++) {
                        swap(&B[i][k], &B[j][k]);
                    }
                    break;
                }
            }
            if (j == N) {
                printf("Error: Singular matrix!\n");
                exit(0);
            }
        }
        for (j = i + 1; j < N; j++) {
            temp = B[j][i] / B[i][i];
            for (k = i; k < N; k++) {
                B[j][k] -= temp * B[i][k];
            }
        }
    }
    
    // row reduce matrix A and B together
    for (i = 0; i < N; i++) {
        if (A[i][i] == 0) {
            for (j = i + 1; j < N; j++) {
                if (A[j][i] != 0) {
                    for (k = i; k < N; k++) {
                        swap(&A[i][k], &A[j][k]);
                        swap(&B[i][k], &B[j][k]);
                    }
                    break;
                }
            }
            if (j == N) {
                printf("Error: Singular matrix!\n");
                exit(0);
            }
        }
        for (j = i + 1; j < N; j++) {
            temp = A[j][i] / A[i][i];
            for (k = i; k < N; k++) {
                A[j][k] -= temp * A[i][k];
                B[j][k] -= temp * B[i][k];
            }
        }
    }
    
    // normalize matrix A and B
    row_reduce(A);
    row_reduce(B);
    
    printf("Row reduced matrices:\n");
    print_matrix(A, 'A');
    print_matrix(B, 'B');
    
    return 0;
}

输出结果为：

Original matrices:
A =
 -6.0 -7.0  7.0  6.0
  3.0  4.0  2.0  3.0
 -4.0 -2.0  0.0  6.0
  1.0  7.0  8.0  3.0

B =
 -3.0  4.0 -2.0  5.0
 -4.0 -4.0  4.0  2.0
 -3.0  6.0  1.0  6.0
  1.0  1.0 -1.0  9.0

Row reduced matrices:
A =
  1.0  0.0  0.0  0.0
  0.0  1.0  0.0  0.0
  0.0  0.0  1.0  0.0
  0.0  0.0  0.0  1.0

B =
  1.0  0.0  0.0  0.0
  0.0  1.0  0.0  0.0
  0.0  0.0  1.0  0.0
  0.0  0.0  0.0  1.0

可以看到，经过高斯消元的处理，两个矩阵都被化为了最简阶梯型行列式，且行列式值均为 1。