?y1?z1??y2??z1?z2?z3?1z4?2 ?，
?y?z?1z34?32?y?z4?4则原二次型的标准形为
222 f?x1,x2,x3,x4???z1?z2?z3?32z4， 4且非退化线性替换为
1?x?z?z?z?z4123?12?1?x??z?z?z?z4?2123 ?2，
?1x?z?z4?332???x4?z4相应的替换矩阵为
??1???1 T????0??0且有
1?1?1??2?11?1??2?， 1?01??2?001?0??100?0?10?。
3?00??4?00??1??0 T?AT??0??0?222 （6）已知f?x1,x2,x3,x4??x1?2x2?x4?4x1x2?4x1x3?2x1x4
?2x2x3?2x2x4?2x3x4， 由配方法可得
f?x1,x2,x3,x4??x12?2x1?2x2?2x3?x4???2x2?2x3?x4?
222 ??2x2?2x3?x4??2x2?x4?2x2x3?2x2x4?2x3x4
2??
??x1?2x2?2x3?x4?于是可令
231?1?2?2?x2?x3?x4???x3?x4?，
22?2?2?y1?x1?2x2?2x3?x4?31??y2?x2?x3?x4 ?， 22?y3?x3?x4???y4?x4则原二次型的标准形为
f?y1?2y2?且非退化线性替换为
2212y3， 2?x1?y1?2y2?y3?y4?3??x2?y2?y3?y4 ?， 2?x3?y3?y4???x4?y4故替换矩阵为
?1?21?3?01?2 T??1?00?000?且有
?1??1??， ?1?1??001200??0?。 0??0???10??0?2 T?AT??00??00?2222 （7）已知f?x1,x2,x3,x4??x1?x2?x3?x4?2x1x2?2x2x3?2x3x4，
由配方法可得
22 f?x1,x2,x3,x4??x2 ?2x2?x1?x3???x1?x3??2x1x3?2x3x4?x42222 ??x1?x2?x3??2x1x3?x3 ?2x3x4?x4?x322 ??x1?x2?x3???x3?x4??2x1x3?x3?x12?x12
222 ?x1??x1?x2?x3???x3?x4???x1?x3?，
222????于是可令
?y1?x1?y?x?x?x?2123 ?，
?y3?x3?x4??y4?x1?x3则原二次型的标准形为
f?y1?y2?y2?y4， 且非退化线性替换为
2222?x1?y1?x?y?y?224 ?，
x??y?y14?3??x4?y1?y3?y4相应的替换矩阵为
?1??0 T???1??1?且有
0??10?1?，
001??01?1??0??100?。 ?010?00?1??0000?1??0