c语言怎么求矩阵的逆

在C语言中，求矩阵的逆通常使用高斯约当消元法（GaussJordan Elimination）或者伴随矩阵法（Adjoint Method），这里我们主要介绍高斯约当消元法。,高斯约当消元法的基本思想是通过行变换，将原矩阵化为上三角矩阵或单位矩阵，然后求解线性方程组得到逆矩阵，具体步骤如下：,1、将原矩阵A复制到一个新的矩阵B中，对B进行行变换，使得B的主对角线上的元素为1，其他元素为0。,2、计算B的转置矩阵BT。,3、计算BT与B的乘积，即BT * B。,4、计算BT * B的逆矩阵，即(BT * B)^(1)。,5、计算(BT * B)^(1)与B的乘积，即(BT * B)^(1) * B。,6、返回结果。,下面是一个简单的C语言实现：,使用示例：,注意：这个实现没有处理奇异矩阵的情况，即矩阵的行列式为0时，该矩阵没有逆矩阵，在实际使用中，需要根据具体情况判断矩阵是否可逆。,
,#include <stdio.h> #include <stdlib.h> #include <math.h> void swap_rows(double **matrix, int row1, int row2, int col) { for (int i = 0; i < col; i++) { double temp = matrix[row1][i]; matrix[row1][i] = matrix[row2][i]; matrix[row2][i] = temp; } } void gauss_jordan_elimination(double **matrix, int rows, int cols) { for (int i = 0; i < rows; i++) { // Find the maximum element in the current column and its row index int max_row = i; for (int k = i + 1; k < rows; k++) { if (fabs(matrix[k][i]) > fabs(matrix[max_row][i])) { max_row = k; } } // Swap the current row with the row containing the maximum element swap_rows(matrix, i, max_row, cols); // Make the diagonal element 1 by dividing other elements in the current row for (int k = i + 1; k < rows; k++) { double factor = matrix[k][i] / matrix[i][i]; for (int j = i; j < cols; j++) { matrix[k][j] = factor * matrix[i][j]; } } } } double
inverse_matrix(double matrix, int rows, int cols) { double
inverse = (double )malloc(rows * sizeof(double *)); for (int i = 0; i < rows; i++) { inverse[i] = (double *)malloc(cols * sizeof(double)); } gauss_jordan_elimination(matrix, rows, cols); for (int i = rows 1; i >= 0; i) { double factor = matrix[i][i]; for (int j = 0; j < cols; j++) { matrix[i][j] /= factor; inverse[i][j] /= factor; } for (int k = i 1; k >= 0; k) { double factor = matrix[k][i]; for (int j = 0; j < cols; j++) { matrix[k][j] = factor * matrix[i][j]; inverse[k][j] = factor * inverse[i][j]; } } } return inverse; },int main() { double
matrix = (double )malloc(3 * sizeof(double *)); for (int i = 0; i < 3; i++) { matrix[i] = (double *)malloc(3 * sizeof(double)); } matrix[0][0] = 1; matrix[0][1] = 2; matrix[0][2] = 3; matrix[1][0] = 4; matrix[1][1] = 5; matrix[1][2] = 6; matrix[2][0] = 7; matrix[2][1] = 8; matrix[2][2] = 9; double **inverse = inverse_matrix(matrix, 3, 3); for (int i = 0; i < 3; i++) { for (int j = 0; j < 3; j++) { printf(“%lf “, inverse[i][j]); } printf(” “); } return 0; },