2 1 -1 1 -3
1 1 1 0 1
-3 -2 0 1 -2
r1-2r2,r3+3r2
0 -1 -3 1 -5
1 1 1 0 1
0 1 3 1 1
r1+r3,r2-r3,r1*(1/2)
0 0 0 1 -2
1 0 -2 -1 0
0 1 3 1 1
r2+r1,r3-r1
0 0 0 1 -2
1 0 -2 0 -2
0 1 3 0 3
交换行
1 0 -2 0 -2
0 1 3 0 3
0 0 0 1 -2
AX=0的基础解系为:a1=(2,-3,1,0,0)',a2=(2,-3,0,2,1)'
所以其解空间的维数是2.
将 a1,a2 用Schmidt正交化过程正交化得
b1=a1=(2,-3,1,0,0)'
b2=a2-(a2,b1)/(b1,b1)b1=(2,-3,-13,28,14)'/14
单位化得
c1=(2,-3,1,0,0)'/√14
c2=(2,-3,-13,28,14)'/√1162
正交规范解为 c1,c2.
O,太麻烦了