C++计算逆矩阵

1这篇文章包含了逆运算在内的常见矩阵计算：

https://blog.csdn.net/sinat_36219858/article/details/78164606

2

/*************************************************************************
    > File Name: inv.cpp
    > Author: ims
    > Created Time: 2018/8/1 22:06:12
 ************************************************************************/
#include    
#include   
using namespace std;
double det(int n, double *aa)
{
	if (n == 1)
		return aa[0];
	double *bb = new double[(n - 1)*(n - 1)];//创建n-1阶的代数余子式阵bb    
	int mov = 0;//判断行是否移动   
	double sum = 0.0;//sum为行列式的值  
	for (int arow = 0; arow brow ? 0 : 1; //bb中小于arow的行，同行赋值，等于的错过，大于的加一  
			for (int j = 0; jbi)    //ai行的代数余子式是：小于ai的行，aa与bb阵，同行赋值  
						pi = 0;
					else
						pi = 1;     //大于等于ai的行，取aa阵的ai+1行赋值给阵bb的bi行  
					if (aj>bj)    //ai行的代数余子式是：小于ai的行，aa与bb阵，同行赋值  
						pj = 0;
					else
						pj = 1;     //大于等于ai的行，取aa阵的ai+1行赋值给阵bb的bi行  
 
					bb[bi*(n - 1) + bj] = aa[(bi + pi)*n + bj + pj];
				}
			}
			printf("aa[%d][%d]的余子式\n", ai, aj);
			for (int i = 0; i < n - 1; i++)
			{
				for (int j = 0; j < n - 1; j++)
				{
					printf("%lf     ",  bb[i*(n - 1) + j]);
				}
				printf(" \n");
			}
			if ((ai + aj) % 2 == 0)  q = 1;//因为列数为0，所以行数是偶数时候，代数余子式为-1.  
			else  q = (-1);
			adjoint[ai*n + aj] = q*det(n - 1, bb);
		}
	}
    for(int i = 0; i < n; i++)//adjoint 转置
    {
        for(int j = 0; j < i; j++)
        {
            double tem = adjoint[i*n + j];
            adjoint[i*n + j] =  adjoint[j*n + i];
            adjoint[j*n + i] =  tem;
        }
    }
	printf("伴随阵： \n");
	for (int i = 0; i < n; i++) //打印伴随阵  
	{
		for (int j = 0; j < n; j++)
		{
			cout << adjoint[i*n + j] << "\t";
		}
		cout << endl;
	}
	printf("逆矩阵： \n");
	for (int i = 0; i < n; i++) //打印逆矩阵  
	{
		for (int j = 0; j < n; j++)
		{
			aa[i*n + j] = adjoint[i*n + j] / det_aa;
			cout << aa[i*n + j] << "\t";
		}
		cout << endl;
	}
	delete[]adjoint;
	delete[]bb;
}
int main()
{
	int n = 0; //阶数  
	printf("输入阶数:");
	scanf("%d", &n); /*读入阶数*/
	double *aa = new double[n*n];
	printf("输入矩阵:\n");
	for (int i = 0; i < n*n; i++)
		cin >> aa[i];
	inverse(n, aa);
	delete[]aa;
}