用配方法将二次型化为标准型并求出相应的可逆变换
f(x1,x2,x3)
=x12-x22+2x32-2x1x2+4x1x3
解： f=(x12-2x1x2+4x1x3)-x22+2x32 =[x12-2x1(X2-2x3)+(X2-2x3)2]-(X2-2x3)2-x22+2x33 =(x1-x2+2x3)2-2x22+4x2x3-2x32 =(x1-x2+2x3)2-2(x2-x3)2 令 {y1=x1-x2+2x3 y2=x2-X3 y3=x3， 即Y= (1 -1 2 0 1 -1 0 0 1)Y 所以经可逆线性变换 X= (1 -1 2 0 1 -1 0 0 1)-1． Y= (1 1 -1 0 1 1 0 0 1)Y 即 {X,1=yl+y2-y3 X2=y2+y3 X3=y3 二次型化为标准型Y12-2y22