x1x2+x1x3-3x2x3二次型化为标准型,

此题用配方法即可.令x1=y1+y2,x2=y1-y2,y3=x3 则f = (y1+y2)(y1-y2)+(y1+y2)y3-3(y1-y2)y3= y1^2-y2^2-2y1y3+4y2y3= (y1-y3)^2-y2^2-y3^3+4y2y3= (y1-y3)^2 - (y2-2y3)^2 + 3y3^2令 z1=y1-y2,z2=y2-2y3,z3=y3,则f = ...首先谢谢您~但是还是想追问一下~应该令Z1=y1-y3吧？不是Z1=Y1-Y2吧~然后C2是怎么求出来的？我求的是C2=1 0 -10 1 20 0 1之后X和Z是怎么变得就没看懂，为什么是C2C1^-1？？？谢谢，麻烦了~第7行笔误 应该是 令 z1=y1-y3, z2=y2-2y3,z3=y3, 则....不好意思, 对应 C2 =1 0 -10 1 -200 1因为1. (x1,x2,x3)' =C1 (y1,y2,y3)' 2. (z1,z2,z3)' = C2(y1,y2,y3)'由2得 C2^-1(z1,z2,z3)' = (y1,y2,y3)'代入1中得 (x1,x2,x3)' =C1 (y1,y2,y3)'= C1C2^-1(z1,z2,z3)' 不好意思了哈