?4?20??x1??0???23?2??x???0?, ???2?????0?22????0???x3???其基础解系为?3=(1,2,2)T, 单位化得
??122?p3?3??,,?,
?3?333?得正交阵
T?2?3?1T???3???2?3?232?3131?3??2?,?3?2??3?
?2??.
T?1AT??5????1???19.将下列二次型用矩阵形式表示.
(1) f(x1,x2,x3)?x1?2x2?5x3?2x1x2?6x2x3?2x3x1； (2) f(x1,x2,x3,x4)?x1x2?x2x3?x3x4?x4x1；
(3) f(x1,x2,x3,x4)?6x1?3x1x2?2x1x3?5x1x4?2x2?x2x4. 【解】
22222?111??x1?????(1)f(x1,x2,x3)?(x1,x2,x3)1?23x2; ??????135????x3????0??1?2(2) f(x1,x2,x3,x4)?(x1,x2,x3,x4)??0??1??21201200120121?2???x?1?0????x2?; ???1?x3??2??x4??0??
3?6?2??32(3) f(x1,x2,x3,x4)?(x1,x2,x3,x4)?2??10??51???22
20.写出下列各二次型的矩阵.
?10005?2??x??11?????x2?. 2???x30???x???4?0??(1)f(x1,x2,x3)?x1?4x2?x3?4x1x2?2x1x3?4x2x3; (2) f(x1,x2,x3)?x1?x2?7x3?2x1x2?4x2x3.
222222?1 2 1??x1?????解：（1）由f(x1,x2,x3)?(x1,x2,x3)?2 4 2??x2?
?1 2 1??x????3??1 2 1???所以二次型的矩阵为?2 4 2?
?1 2 1?????1 ?1 0??x1?????（2）由f(x1,x2,x3)?(x1,x2,x3)??1 1 ?2??x2?
? 0 ?2 ?7??x????3???1 ?1 0???所以二次型的矩阵为??1 1 ?2?
? 0 ?2 ?7???
21. 当t为何值时，二次型f(x1,x2,x3)?x1?6x1x2?4x1x3?x2?2x2x3?tx3的秩为2.
222?132??x1?????【解】 f(x1,x2,x3)?(x1,x2,x3)311x2 ??????21t????x3??2??132??131?r2?????r3?2r1?0?8?5?????8?A??311?????? r?3r??21?????21t???0?5t?4??
2?13?2??13??5???5?r3?5r2?01?01?????8? 8???25??0?5t?4??00t?4????8??7?R(A)?2?t?.
8
22.用正交变换把下列二次型化为标准形,并写出所作的变换. (1) f(x1,x2,x3,x4)=2x1x2-2x3x4;
(2) f(x1,x2,x3)?x1?2x2?3x3?4x1x2?4x2x3.
222?0 1 0 0???1 0 0 0?，A的特征多项式 解：(1)f的正交矩阵A???0 0 0 ?1????0 0 ?1 0??? 1 0 0A??E? 1 ?? 0 0 0 0 ?? ?1 0 0 ?1 ??于是A的全部特征值为?1?1(二重),?2??1(二重).
?(?2?1)2?0，
?1?? 0?????1 0??????,???1?1,解（A-E）x=0，得基础解系1 ?0?2??1?????0??? 1??1?? 0??2????1?? 0? 0????????1? 0??1??1???,?2?,??正交化得?1?，单位化得?1???? ?0???1?2?2?2?????????0??1??0?? 1??0?? ?2??????1??0????? 10????,???2??1解（A+E）x=0,得基础解系?3? ? 0?4?1????? 0???1?
?1??0????????1??0?20???????? 10?1??1?,??正交化 ?3???,?4???，单位化得：?3??
? 0??1?2?4?2????????? 0???1?? 0??1?? 0??????2?1?1? 0 ? 0?2?2??1?1? 0 0?2?2? 取正交矩阵T?(?1,?2,?3,?4)??11??0 ? 0 ?22????0 1 0 1???22??令x=Ty，得f?x?Ax?y?(T?AT)y?y1?y2?y3?y3
2222? 1 ?2 0???(2)f的矩阵A???2 2 ?2?
? 0 ?2 3???特征多项式 A??E?0，得特征值?1??1,?2?2,?3?5
?2???当?1??1，解（A+E）x=0，得基础解系?1??2?
?1?????2???当?2?2，解（A-2E）x=0，得基础解系?2?? 1?
? 2????1?? 2???当?3?5，解（A-5E）x=0，得基础解系?3???1?
? 1??????2??2??1???3??3?? 3???????212单位化得?1???,?2?? ?,?3????
?3??3??3???????122??????? ??? ???3??3??3?
取正交矩阵T?(?1,?2,?3)，令x=Ty，得
22f?x?Ax?y?(T?AT)y??y12?2y2?5y3
23. 用配方法把下列二次型化为标准型，并求所作变换.
22(1)f(x1,x2,x3)?2x1x2?4x1x3?x2?8x3;(2)f(x1,x2,x3)?2x1x2?4x1x3.【解】
22(1)f(x1,x2,x3)?2x1x2?4x1x3?x2?8x3;
22 ??(x2?2x1x2?x12)?x12?4x1x3?8x32??(x2?x1)2?(x1?2x3)2?12x3令
?y1??x1?x2?x1?y2?2y3??y?x?2x??2?x2?y1?y2?2y3 13?y?x?x3?y333???110由于102?0 010∴ 上面交换为可逆变换.
得f(x1,x2,x3)??y1?y2?12y3.
222(2)f(x1,x2,x3)?2x1x2?4x1x3.
x1x2?y??12?2,?x1?y1?y2??x1x2 令?x2?y1?y2为可逆线性变换??y??,?x?y?2223?3??y3?x3.22f(x1,x2,x3)?2(y12?y2)?4(y1?y2)y3?2y12?2y2?4y1y3?4y2y3222?2(y12?2y1y3?y3)?2y2?4y2y3?2y3
?2(y1?y3)2?2(y2?y3)2??1?y1?y3?22令??2?y2?y3为可逆线性交换f(?1,?2,?3)?2?1?2?2 ???y33?所作线性交换为