x1x2+x1x3-3x2x3二次型化为标准型,

kzl4192022-10-04 11:39:541条回答

x1x2+x1x3-3x2x3二次型化为标准型,
顺便我想问一下规范型和交换矩阵怎么求?尽快哟~会追分的

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lengxue001 共回答了17个问题 | 采纳率100%: 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 = z1^2-z2^2+3z3^2
变换矩阵分2步:
1.(x1,x2,x3)' =C1 (y1,y2,y3)'
C1 =
1 1 0
1 -1 0
0 0 1
2.(z1,z2,z3)' = C2(y1,y2,y3)'
C2 =
1 0 0
-1 1 0
0 -2 1
所以 (z1,z2,z3)' = C2C1^-1 (x1,x2,x3)' =C1
变换矩阵为C2C^-1
规范型为 w1^2+w2^2-w3^2.; 1年前