工程数学--线性代数课后题答案 第五版

?1?52???2??1C??00??

2???0?21???

(2) f(x1? x2? x3)?x12?2x32?2x1x3?2x2x3? 解 f(x1? x2? x3)?x12?2x32?2x1x3?2x2x3 ?(x1?x3)2?x32?2x2x3? ?(x1?x3)2?x22?(x2?x3)2?

???y1?x1?x3?x1?y1?y2?y3令 ?y2?x2? 即?x2?y2?

???y3?x2?x3?x3??y2?y3二次型化为规范形

f?y12?y22?y32?

所用的变换矩阵为

?11?1?C??010?? ?0?11??? (3) f(x1? x2? x3)?2x12?x22?4x32?2x1x2?2x2x3? 解 f(x1? x2? x3)?2x12?x22?4x32?2x1x2?2x2x3?

22?4x3?2x2x3 ?2(x1?1x2)2?1x2222 ?2(x1?1x2)2?1(x2?2x3)2?2x3?

22?y??1?令 ?y2???y3???x?12(x1?x2)?121(x?2x)? 即?x??2322?2x3?x3??1y?1y?1y212223? 2y2?2y321y23二次型化为规范形

f?y12?y22?y32?

所用的变换矩阵为

?1?1?1?1C??022?? 2?001??? 31? 设

f?x12?x22?5x32?2ax1x2?2x1x3?4x2x3

为正定二次型? 求a?

?1a?1? 解 二次型的矩阵为A??a12?? 其主子式为

??125???1a?11a2 a11?1? ?1?a? a12??a(5a?4)?

a1?125 因为f为正主二次型? 所以必有1?a2?0且?a(5a?4)?0? 解之得?4?a?0?

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32? 判别下列二次型的正定性? (1) f??2x12?6x22?4x32?2x1x2?2x1x3?

??211? 解 二次型的矩阵为A??1?60?? 因为

?10?4???a11??2?0? ?21?11?0? |A|??38?0? 1?6所以f为负定?

(2) f?x12?3x22?9x32?19x42?2x1x2?4x1x3?2x1x4?6x2x4?12x3x4? ?1??1 解 二次型的矩阵为A??2?1??130?3209?61??3?? 因为 ?6?19??1?121?1?4?0, ?130?6?0, A?24?0? a11?1?0?

?13209所以f为正定?

33? 证明对称阵A为正定的充分必要条件是? 存在可逆矩阵U? 使A?U TU? 即A与单位阵E合同?

证明 因为对称阵A为正定的? 所以存在正交矩阵P使

PTAP?diag(?1? ?2? ? ? ?? ?n)??? 即A?P?PT?

其中?1? ?2? ? ? ?? ?n均为正数?

令?1?diag(?1, ?2, ? ? ? ,?n)? 则???1?1? A?P?1?1TPT? 再令U??1TPT? 则U可逆? 且A?UTU?