boolean lu(double[] a, int[] pivot, int n); //矩阵LU分解 boolean guass(double[] a, int[] p, double[] b, int n);//Guass求线性代数方程组的解 void qr(double[] a, double[] d, int n);//矩阵的QR分解 boolean householder(double[] a, double[] d, double[] b, int n);//householder求线性代数方程组的解 附三道实验的例题。