d=det（aij）大学行列式.求解

|x|^2=x1^2+x2^2+...+xn^2
D=
1-x1^2/|x|^2 -x1x2/|x|^2 ... -x1xn/|x|^2
-x2x1/|x|^2 1-x2^2/|x|^2 ... -x2xn/|x|^2
....
-xnx1/|x|^2 -xnx2/|x|^2 ... 1-xn^2/|x|^2
加边
1 x1/|x|^2 x2/|x|^2 ... xn/|x|^2
0 1-x1^2/|x|^2 -x1x2/|x|^2 ... -x1xn/|x|^2
0 -x2x1/|x|^2 1-x2^2/|x|^2 ... -x2xn/|x|^2
....
0 -xnx1/|x|^2 -xnx2/|x|^2 ... 1-xn^2/|x|^2
ri + xi-1r1, i=2,3,...,n+1
1 x1/|x|^2 x2/|x|^2 ... xn/|x|^2
x1 1 0 ... 0
x2 0 1 ... 0
...
xn 0 0 ... 1
c1-x1c2-x2c3-...-xncn+1
1-x1^2/|x|^2-...-xn^2/|x|^2 x1/|x|^2 x2/|x|^2 ... xn/|x|^2
0 1 0 ... 0
0 0 1 ... 0
...
0 0 0 ... 1
=
0 x1/|x|^2 x2/|x|^2 ... xn/|x|^2
0 1 0 ... 0
0 0 1 ... 0
...
0 0 0 ... 1
= 0