?k?1k??1??????0?kk?10???10???10???? ?????k??k?1??2??0?2??0?2?故命题成立. ii)左边?f(?)?a0E?a1??a2?2???am?m
m??10???10??10???a0??01???a1??0??????am?0?m??
???2??2?2m?a0?a1?1?a2?1????am?10?? ??2m?0a0?a1?2?a2?2???am?2??0??f(?1)?=右边 ???0?f(?2)??(2) i) 利用数学归纳法.当k?2时
A2?P?P?1P?P?1?P?2P?1成立
假设k时成立,则k?1时
Ak?1?Ak?A?P?kP?1P?P?1?P?k?1P?1成立,故命题成立,
kk?1即 A?P?P
ii) 证明 右边?Pf(?)P?1
?P(a0E?a1??a2?2???am?m)P?1
?a0PEP?1?a1P?P?1?a2P?2P?1???amP?mP?1 ?a0E?a1A?a2A2???amAm?f(A)=左边
?
19．设n阶矩阵A的伴随矩阵为A，证明： (1) 若(2) 证明
(1) 用反证法证明．假设由此得A?这与有
A?0,则A??0;
n?1A??A.
A??0则有A?(A?)?1?E
AA?(A?)?1?AE(A?)?1?O?A??O
A??0矛盾，故当A?0时
A??0
(2)
1?由于A?A, 则AA??AE
A?1取行列式得到: 若若
AA??An
A?0 则A??An?1
A?0由(1)知A??0此时命题也成立
故有
A??An?1
26
?120．取A?B??C?D???0?101010AB检验: ?CD?1010?10AB11??0 而
CD11故
AB0??,验证??CD1?02001020?0?10110?10ACBD
002010??4 002011ABA?CDCBD
?34??O??4?3?,求A8及A4
21．设A??20??O?22???34??O??34??20?4?3??解 A?，令A1???4?3?? A2???22?? ?20??????O?22???A1O?则A???OA??
?2?8?AOAO???118?故A???OA????OA8??
?2??2?8888A8?A1A2?A1A2?1016
8
4?A1A4???O??540??O?4?O?05? ??44??20?A2??O?64??22???1
?OA?22．设n阶矩阵A及s阶矩阵B都可逆,求??BO?????1?C1C2??OA?解 将??BO??分块为??CC??
??4??3其中 C1为s?n矩阵, C2为s?s矩阵
C3为n?n矩阵, C4为n?s矩阵
．
27
An?n??C1C2??EnO????? ?E???????O??C3C4??OEs??AC3?En?C3?A?1??1AC?O?C?O(A存在)?44由此得到?
?1?BC1?O?C1?O(B存在)?BC?E?C?B?1s2?2?1?OB?1??OA?故 ??BO?????A?1O??．
?????O则??B?s?s
第三章 矩阵的初等变换与线性方程组
1．把下列矩阵化为行最简形矩阵：
(1)
(3)
?1??2?3??1??3?2??3?02?1??0??031?; (2) ?0?004?3????13?43??2???35?41??1; (4) ??3?23?20????2?34?2?1??2?31??3?43?; 4?7?1??31?320?2?283?374?7???4?. ?0?3??
解
?102?1?r2?(?2)r1?102?1?????1?(1) ?203~?00?13?
?304?3?r3?(?3)r1?00?20?????r2?(?1)?102?1?r3?r2?102?1?????~?001?3?~?001?3? r3?(?2)?0010??0003?????r3?3?102?1?r2?3r3?102?1?????~?001?3?~?0010?
?0001??0001?????r1?(?2)r2?1000???~?0010? r1?r3?0001???
(2)
?02?31?r2?2?(?3)r1?02?31?????~?0013? ?03?43?
?04?7?1?r3?(?2)r1?00?1?3?????28
0r3?r2??~?0r1?3r2?0??1?1??3?3(3)
?2?2??3?3?2003534010?r1?2?010??13?~?001?00000????43??1?r2?3r1??41??0 ~??20r3?2r1?0????2?1?r4?3r1??03111?4?2?2?23??r1?3r22?~?2r3?r2?2??r4?r25??3? 0???10003?43???48?8? ??36?6??510?10??02?3??1?22? ?000?000???1?1r2?(?4)??00~?00r3?(?3)??00r4?(?5)?
?1?1??00?00??00?(4)
?23??12?3?2??2?3??0r2?2r1??1~?0r3?8r1??0r4?7r1??1r2?r3??0~?0??0?1?3?7??0?1?r1?2r2?0?2?4??12~ ?830r3?3r2?0?8???743?r4?2r2??0?7?1111??10?r1?r2?020?2??01~?0014r2?(?1)?00???0014?r4?r3??00020?2??1?103? ?0014?0000??111??0?2?4? ?8912?7811??20?2???1?1?1? ?014?000??
2．在秩是r的矩阵中,有没有等于0的r?1阶子式?有没有等于0的r阶 子式?
解 在秩是r的矩阵中,可能存在等于0的r?1阶子式,也可能存在等 于0的r阶子式.
?1000????0100?例如，???0010?
???0000??0000???R(?)?3同时存在等于0的3阶子式和2阶子式.
3．从矩阵A中划去一行得到矩阵B,问A,B的秩的关系怎样?
29
R(A)?R(B)
设R(B)?r，且B的某个r阶子式Dr?0.矩阵B是由矩阵A划去一行得 到的，所以在A中能找到与Dr相同的r阶子式Dr，由于Dr?Dr?0， 故而R(A)?R(B).
解
4．求作一个秩是4的方阵,它的两个行向量是(1,0,1,0,0),(1,?1,0,0,0) 解 设?1,?2,?3,?4,?5为五维向量,且?1?(1,0,1,0,0), ??1?????2??2?(1,?1,0,0,0),则所求方阵可为A???3?,秩为4,不妨设
????4?????5???3?(0,0,0,x4,0)???4?(0,0,0,0,x5)取x4?x5?1 ???(0,0,0,0,0)?5?10100????1?1000?故满足条件的一个方阵为?00010?
???00001??00000???
5．求下列矩阵的秩,并求一个最高阶非零子式：
(1)
(3)
02??31???1?12?1?; (2) ?13?44????21837????2?307?5??3?2580?. ??10320????1?13?14402?42?6?6?32?1?3?1???1?3?； ?2?13?705?1?8???
?3?解 (1) ?1?1??1?3r2r1?~?0r3?r1?0?2???1?r1r2??1?~?3?14????1??1??r3r2?5?~?0?05????12103?4?124?600?1??2? 4???1??5?秩为2 0??30
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