算法系列：矩阵的求逆方法详解（第五篇）

	/**
	 * 矩阵求逆
	 * 
	 * @param args
	 *            参数a是个浮点型（double）的二维数组，
	 * @return 返回值是一个浮点型二维数组（矩阵a的逆矩阵）
	 */
	public static double[][] inv(double[][] matrix) {
		int n = matrix.length;
		double[][] matrix1 = new double[n][2 * n];
		double[][] result = new double[n][n];
		for (int i = 0; i < n; i++) {
			for (int j = 0; j < n; j++) {
				matrix1[i][j] = matrix[i][j];
			}
		}
		for (int k = 0; k < n; k++) {
			for (int t = n; t < n * 2; t++) {
				if ((t - k) == n) {
					matrix1[k][t] = 1.0;
				} else {
					matrix1[k][t] = 0;
				}
			}
		}
		// get the change of the array
		for (int k = 0; k < n; k++) {
			if (matrix1[k][k] != 1) {
				double bs = matrix1[k][k];
				matrix1[k][k] = 1;
				for (int p = k; p < n * 2; p++) {
					matrix1[k][p] /= bs;
				}
			}
			for (int q = 0; q < n; q++) {
				if (q != k) {
					double bs = matrix1[q][k];
					for (int p = 0; p < n * 2; p++) {
						matrix1[q][p] -= bs * matrix1[k][p];
					}
				} else {
					continue;
				}
			}
		}
		for (int x = 0; x < n; x++) {

			for (int y = n; y < n * 2; y++) {
				result[x][y - n] = matrix1[x][y];
			}
		}

		return result;
	}