C語言求多維矩陣的逆

採用高斯消去法求逆

直接上代碼

void Matrix_inverse(double arc[6][6], int n, double ans[6][6])//計算矩陣的逆
{
	int i, j, k;//列
	double max, tempA, tempB, P;
	int max_num;
	double arcs[6][6];
	memcpy(arcs, arc, 288);
	for (i = 0; i < n; i++)
	{
		ans[i][i] = 1;
	}
	for (i = 0; i < n; i++)//第i列
	{
		max = fabs(arcs[i][i]);
		max_num = i;
		for (j = i + 1; j < n; j++)//選出主元
		{
			if (fabs(arcs[j][i]) > max)
			{
				max = fabs(arcs[j][i]);
				max_num = j;
			}
		}
		/*if (max == 0)
		{
			printf("i can't");
			break;
		}*/
		for (k = 0; k < n; k++)//交換行
		{
			tempA = arcs[i][k];
			arcs[i][k] = arcs[max_num][k];
			arcs[max_num][k] = tempA;
			tempB = ans[i][k];
			ans[i][k] = ans[max_num][k];
			ans[max_num][k] = tempB;
		}
		for (k = i + 1; k < n; k++)
		{
			P = arcs[k][i] / arcs[i][i];
			for (j = 0; j < n; j++)
			{
				arcs[k][j] = arcs[k][j] - arcs[i][j] * P;
				ans[k][j] = ans[k][j] - ans[i][j] * P;
			}
		}
	}
	for (i = 0; i < n; i++)//行
	{
		P = arcs[i][i];
		for (j = i; j < n; j++)
		{
			arcs[i][j] = arcs[i][j] / P;
		}
		for (j = 0; j < n; j++)
		{
			ans[i][j] = ans[i][j] / P;
		}
	}
	for (i = n - 1; i > 0; i--)
	{
		for (j = i - 1; j >= 0; j--)
		{
			for (k = 0; k < n; k++)
			{
				ans[j][k] = ans[j][k] - ans[i][k] * arcs[j][i];
			}
		}
	}
}

我寫的是針對6維矩陣的，有需要的話，把6改成其他數字就好了