线性代数课后习题答案全习题详解

1r1?r2??0~?r2?(?1)?0?0r?r?43021?100000?2?r?r?23?1?1?~14?00???1??0?0?0?021?100000?2??03?

14?00??

?010??101??123???????2.设?100?A?010???456?，求A。

?001??001??789????????010??123??101??452?????????解：A=?100??456??010?=?122?

?001??789??001??782?????????

3．试利用矩阵的初等变换，求下列方阵的逆矩阵：

?1?1?321???(1) ?315?; (2)

?323????3?20?1???0221??. ?1?2?3?2??0121????321100???解 （1）?315010??323001????321100??~?0?14?110???002?101???31??0??320?22?~?0?1011?2?? ?002?101?????723?79????2???300??10063222???~?0?1011?2~010?1?12? ????001?101??001?101???22?22?????3??72???632??故逆矩阵为??1?12?

??101??22????3?20?11?02210(2) ??1?2?3?20?01210?

010000100??0?0?1???1?2?3?2?0121~??0011?00?2?1?33

0010010??001?

0?3?4?10?2???1?2?3?2?0121~??0011?0001?0012010??1?2??001??01~0?3?4??00?1?6?10???00?1?0~??0?0?01000010001000010001?10?1210?12?1?2?2??1`0?1?

?136?1?6?10??1?2?4??10?1?

?136?1?6?10???1?0故逆矩阵为???1?2?

1?2?4??10?1?

?136?1?6?10???41?2??1?3?????4．(1) 设A??221?,B??22?,求X使AX?B;

?31?1??3?1??????021????123?(2) 设A??2?13?,B???,求X使XA?B.

2?31????33?4???解

?41?21?3?初等行变换?100102????(1) ?AB???22122?~?010?15?3??

?31?13?1??001124?????2??10???X?A?1B???15?3?

?124????0??2?A???3(2) ?????B??1?2??2?132?31??3??4?3??1????1?0初等列变换?0~??2???4??00??10?01?

?1?1??74????2?1?1??X?BA?1???．

?474??

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?1?10???1?1?，AX=2X+A,求X。 5.设A??0??101???1?1??0???11? 解：由AX=2X+A得：X＝(A?2E)A＝??10?1?10???

6．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.

?1??0例如，???0?0??0

01000001000??0?0?. R(?)?3同时存在等于0的3阶子式和2阶子式. 0??0?7．从矩阵A中划去一行得到矩阵B, 问A,B的秩的关系怎样? 解 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).

8．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),?2?(1,?1,0,0,0),则所求方阵可为

??1?????3?(0,0,0,x4,0)??2??A???3?, 秩为4, 不妨设??4?(0,0,0,0,x5) 取x4?x5?1

??????5?(0,0,0,0,0)4????5? 35

?1??1故满足条件的一个方阵为?0?0??0

0?100010000001000??0?0? 1??0?9．求下列矩阵的秩,并求一个最高阶非零子式：

?3102???(1) ?1?12?1?; (2)

?13?44???

?32?1?3?1???2?131?3??； (3) ?705?1?8????218??2?30?3?25?103?37??7?5?.

80?20???3102?r1?r2??解 (1) ?1?12?1?~?13?44????1?12?1?r2?3r1?1?12?1?????310204?65??~?? ?13?44?r3?r1?04?65?????r3?r2?1?12?1?3104?65秩为2??4． . 二阶子式?~?1?1?0000?r1?r2?32?1?3?2?r2?2r1?13?4?41???(2) ?2?131?3?~?0?7119?5?

?705?1?8??7??r3r1?0?213327?15??1r?3r?0~?0323?4?41?32?7119?5?秩为2. 二阶子式??7．

2?10000??218?2?30(3) ??3?25?103?37?r?2r4?17?5?~80?r?2r2420??r3?3r42?17?r?3r?011??20?3?63?5??~?0?2?420?r?2r1?10?3320???0??0?0?1?10002?10000327??16? 14?0??r1?r21r4?r1??0~?r3?14?0?0r4?16?r4?r30100322?100000??7? 秩为3 1?0?? 36