作业1-线性方程组求解VIP免费

1分别用矩阵求逆、矩阵除法以及矩阵分解求线性方程的解。2下面是一个线性病态方程组：(1)求方程的解。(2)将方程右边向量元素b3改为[0.5：0.01：0.6]，再求解，并比较b3的变化和解的相对变化。(3)计算系数矩阵A和条件数并分析结论。解：1-1>>A=[2,3,5;3,7,4;1,-7,1];B=[10,3,5]X=A\B.'B=1035X=-1.8060-0.53733.0448>>A=[2,3,5;3,7,4;1,-7,1];B=[10,3,5]'B=1035>>[L,U]=lu(A);X=U\(L\B)X=-1.8060-0.53733.04481-2>>clearall>>A=[6,5,-2,5;9,-1,4,-1;3,4,2,-2;3,-9,0,2]A=65-259-14-1342-23-902>>b=[-4,13,1,-1]'b=-4131-1>>x=A\bx=-5.33331.000018.500012.0000>>[L,U]=lu(A);x=U\(L\b)x=-5.33331.000018.500012.00002-1>>clearall>>A=[1/2,1/3,1/4;1/3,1/4,1/5;1/4,1/5,1/6];>>b=[0.95;0.67;0.52]b=0.95000.67000.5200>>B=sym(b)B=19/2067/10013/25>>x=A\Bx=6/53/53/52-2>>A=[1/2,1/3,1/4;1/3,1/4,1/5;1/4,1/5,1/6];>>b=[0.95;0.67;0.52]B=sym(b)b=0.95000.67000.5200B=19/2067/10013/25>>x=A\Bx=6/53/53/5>>b3=[0.5:0.01:0.6]b1=0.95b2=0.67fori=1:length(b3)b(1)=b1;b(2)=b2;b(3)=b3(i);x0=A\b;x1(i)=x0(1);x2(i)=x0(2);x3(i)=x0(3);endb3=0.50000.51000.52000.53000.54000.55000.56000.57000.58000.59000.6000b1=0.9500b2=0.6700>>plot(b3,x3)0.50.510.520.530.540.550.560.570.580.590.6-20-1001020304050