高斯消元法求逆矩阵

仅供自己学习记录,没有处理主对角线有0得情况,但也基本够用了。

const int row = 4;
int res[row][row];
void Gauss(float mat[row][row])
{
	float tmp[4][4 * 2];
	//1.构建增广矩阵
	for (int i = 0;i < 4;i++)
	{
		for (int j = 0;j < row * 2;j++)
		{
			if (j >= row)
			{
				tmp[i][j] = j == (row + i) ? 1 : 0;
			}
			else
			{
				tmp[i][j] = mat[i][j];
			}
		}
	}
	//2.化为行阶梯形式
	//特殊处理第一行

	for (int i = 1;i < row;i++)
	{
		for (int z = i;z < row;z++)
		{
			float k = (float)tmp[z][i - 1]/(float)tmp[i - 1][i - 1] ;
			//cout << k <
			tmp[z][i - 1] = 0;
			for (int j = i;j < row * 2;j++)
			{
				tmp[z][j] -= k*tmp[i-1][j];
			}
		}
	}
	//对角线元素归1
	for (int i = 0;i < row;i++)
	{
		if (abs(tmp[i][i] - 1) >= 0.001f)
		{
			float k = tmp[i][i];
			for (int z = 0;z < row * 2;z++)
			{
				tmp[i][z] /= k;
			}
		}
	}

	//3.化为单位矩阵形式
	for (int i = row - 1;i >= 1;i--)
	{
		for (int z = i;z >= 1;z--)
		{
			float k = tmp[z - 1][i] / tmp[i][i];
			//cout << k << " " << tmp[z - 1][i] << " " << tmp[i][i] <<"("<
			for (int j = i+1;j < row * 2;j++)
			{
				tmp[z-1][j] -= k * tmp[i][j];
			}
		}
	}

	cout << "res:<<<<<<<<<<<<<<<<<<<<<<<<<<" << endl;
	/*for (int i = 0;i < row;i++)
	{
		//cout << tmp[i-1][i-1] << "zzzzzz" << endl;
		for (int j = 0;j < row * 2;j++)
		{
			cout << tmp[i][j] << " ";
		}
		cout << endl;
	}*/
	//4.完成并输出
	for (int i = 0;i < row;i++)
	{
		for (int j = 0;j < row ;j++)
		{
			res[i][j] = tmp[i][row+j];
			cout << res[i][j] << " ";
		}
		cout << endl;
	}
}

测试数据:


1 0 0 1
0 1 0 3
0 0 1 2
0 0 0 1

2 1 1
3 2 1
2 1 2


你可能感兴趣的:(线性代数,算法,c++,编程语言)