高等代数(北大版)第5章习题参考答案[1]

2020-03-26 23:08

第五章 二次型

1.用非退化线性替换化下列二次型为标准形,并利用矩阵验算所得结果。 1)?4x1x2?2x1x3?2x2x3;

2)x2?2x2211x2?2x2?4x2x3?4x3;

3)x2?3x212?2x1x2?2x1x3?6x2x3;

4)8x1x4?2x3x4?2x2x3?8x2x4; 5)x1x2?x1x3?x1x4?x2x3?x2x4?x3x4;

6)x221?2x2?x24?4x1x2?4x1x3?2x1x4?2x2x3?2x2x4?2x3x4;7)x22221?x2?x3?x4?2x1x2?2x2x3?2x3x4。

解 1)已知 f?x1,x2,x3???4x1x2?2x1x3?2x2x3, 先作非退化线性替换

?x1?y1?y2 ??x2?y1?y2 (1)

??x3?y3则

f?x221,x2,x3???4y1?4y2?4y1y3

??4y22221?4y1y3?y3?y3?4y2

???2y321?y3??y3?4y22,

再作非退化线性替换

??y1?112z1?2z3 ??y2?z2 ??y3?z3?则原二次型的标准形为

f?x,x22212,x3???z1?4z2?z3,

最后将(2)代入(1),可得非退化线性替换为

2) (

11?x?z?z?z312?122?11? ?x2?z1?z2?z3 (3)

22??x3?z3??于是相应的替换矩阵为 ?1? T??1?0?1?10?10????20??0?1???0?010?11???22??10????2?1??0???0?101??2?1?, 2?1???且有

??1? T?AT??0?0?0400??0?。 1??222 2)已知f?x1,x2,x3??x1?2x1x2?2x2?4x2x3?4x3,

由配方法可得

2222 f?x1,x2,x3???x1?2x1x2?x2???x2?4x2x3?4x3?

??x1?x2???x2?2x3?,

22于是可令

?y1?x1?x2? ?y2?x2?2x3,

?y?x3?3则原二次型的标准形为

22 f?x1,x2,x3??y1?y2,

且非退化线性替换为

?x1?y1?y2?2y3? ?x2?y2?2y3,

?x?y3?3相应的替换矩阵为 ?1?T? ?0?0??1102???2?, 1??

且有

?1? T?AT???1?2?01?20??1??0??1?1???01220??1??2??0?4???0?1102??1???2???0?1???00100??0?。 0??2 (3)已知f?x1,x2,x3??x12?3x2?2x1x2?2x1x3?6x2x3,

由配方法可得

2222 f?x1,x2,x3???x12?2x1x2?2x1x3?2x2x3?x2?x3???4x2?4x2x3?x3?

??x1?x2?x3???2x2?x3?,

22于是可令

?y1?x1?x2?x3? ?y2?2x2?x3,

?y?x3?3则原二次型的标准形为

22 f?x1,x2,x3??y1?y2,

且非退化线性替换为

13?x?y?y?y312?122?11? ?x2?y2?y3,

22??x3?y3??相应的替换矩阵为 ??1? T??0??0??121203??2?1??, 2?1????121203??2??11?????02??1??0???且有

???11T?AT???2?3???20121?2??0??1??0??1????11???1?3?3??11????3??0??0??0??0?100??0?。 0??(4)已知f?x1,x2,x3,x4??8x1x2?2x3x4?2x2x3?8x2x4,

先作非退化线性替换 ?x1??x2 ??x3?x?4?y1?y4?y2?y3?y4,

2 f?x1,x2,x3,x4??8y1y4?8y4?2y3y4?2y2y3?8y2y4

2?21111?1??1?? ?8?y4?2y4?y1?y2?y3???y1?y2?y3??

2828?2??2?????11?1? ?8?y1?y2?y3??2y2y3

28?2?111?1??? ?8?y1?y2?y3?y4??2?y1?y2?y3??2y2y3,

284?2???222再作非退化线性替换

?y1??y2 ??y3?y?4?z1?z2?z3?z2?z3?z4,

5353?1??? f?x1,x2,x3,x4??8?z1?z2?z3?z4??2?z1?z2?z3?

8844?2???22 ?2z2?2z3,

22再令

??w1??w2 ??w3??w4??z1??z2?z3?12z1?58z2?38z3?z454x2?34x3,

则原二次型的标准形为

2222 f?x1,x2,x3,x4???2w1?2w2?2w3?8w4,

且非退化线性替换为

??x1??x2 ??x3??x4??12w1?54w2?34w3?w4?w2?w3?w2?w3??12w1?w4,

相应的替换矩阵为 ?1??20 T???0?1???2?54110?341?1??0?, ?0?1???10且有

??2??0 T?AT??0??0?020000?200??0?。 ?0?8??(5)已知f?x1,x2,x3,x4??x1x2?x1x3?x1x4?x2x3?x2x4?x3x4, 先作非退化线性替换

?x1??x2 ??x3?x?4?2y1?y2?y2?y3?y4,

2 f?x1,x2,x3,x4??2y1y2?y2?2y1y3?2y2y3?2y1y4?2y2y4?y3y4

??y1?y2?y3?y4?再作非退化线性替换

?z1?z2?? ??z3???z4?y12132??2??y3?y4??y4?y1,

24??2?y1?y2?y3?y4?y3??y412y4,


高等代数(北大版)第5章习题参考答案[1].doc 将本文的Word文档下载到电脑 下载失败或者文档不完整,请联系客服人员解决!

下一篇:中考物理 综合练习(七)(无答案)中考物理 苏教版

相关阅读
本类排行
× 注册会员免费下载(下载后可以自由复制和排版)

马上注册会员

注:下载文档有可能“只有目录或者内容不全”等情况,请下载之前注意辨别,如果您已付费且无法下载或内容有问题,请联系我们协助你处理。
微信: QQ: