线性代数课后习题解答第三章习题解答

20 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) 340313021201; (2) 174034301320;

(3)

12433023221453334311; (4)

34732038234202173132.

解 (1) 3403130212011312)3()2(~rrrr020031001201 )2()1(32~rr01003100120123~rr300031001201

33~r100031001201323~rr1000010012013121)2(~rrrr100001000001

(2) 174034301320

1312)2()3(2~rrrr31003100132021233~rrrr000031001002021~r000031005010

(3)

12433023221453334311

141312323~rrrrrr1010500663008840034311)5()3()4(432~rrr22100221002210034311

2423213~rrrrrr00000000002210032011

(4)

34732038234202173132

242321232~rrrrrr1187701298804202111110141312782~rrrrrr41000410002020111110

34221)1(~rrrrr0000041000111102020132~rr00000410003011020201

21 2.设987654321100010101100001010A，求A。

解：A=11100010101987654321100001010=287221254

3．试利用矩阵的初等变换，求下列方阵的逆矩阵：

(1) 323513123; (2)

1210232112201023.

解 （1）100010001323513123101011001200410123~10121121023200010023~

2102121129227100010003~21021211233267100010001~

故逆矩阵为21021211233267

(2)

1000010000100001121023211220102320104301100001001200110012102321~

106124301100001001000110012102321~10612631110`1022111000010000100021~

106126311101042111000010000100001~

故逆矩阵为10612631110104211

22 4．(1) 设132231,113122214BA,求X使BAX;

(2) 设132321,433312120BA,求X使BXA.

解

(1) 132231113122214BA初等行变换~412315210100010001

4123152101BAX

(2)

132321433312120BA初等列变换~474112100010001

4741121BAX．

5.设101110011A，AX=2X+A,求X。

解：由AX=2X+A得：X＝AEA1)2(＝011101110

6．在秩是r的矩阵中,有没有等于0的1r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1r阶子式,也可能存在等于0的r阶子式.

例如，00000000010000100001. 3)(R同时存在等于0的3阶子式和2阶子式.

7．从矩阵A中划去一行得到矩阵B, 问BA,的秩的关系怎样?

解 )(AR)(BR

设rBR)(，且B的某个r阶子式0Dr.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0DDrr，

故而)()(BRAR.

8．求作一个秩是4的方阵,它的两个行向量是)0,0,1,0,1(,)0,0,0,1,1( 23 解 设54321,,,,为五维向量,且)0,0,1,0,1(1,)0,0,0,1,1(2,则所求方阵可为

,54321A 秩为4, 不妨设)0,0,0,0,0(),0,0,0,0()0,,0,0,0(55443xx 取154xx

故满足条件的一个方阵为0000010000010000001100101

9．求下列矩阵的秩,并求一个最高阶非零子式：

(1) 443112112013; (2) 815073131213123； (3)

02301085235703273812.

解 (1) 443112112013rr21~443120131211564056401211~12133rrrr

2000056401211~23秩为rr. 二阶子式41113．

(2) 8150731312231231527332105911701443127~122113rrrrrr

200000591170144313~23秩为rr. 二阶子式71223．

(3)

02301085235703273812434241322~rrrrrr02301024205363071210131223~rrrr0230114000016000071210

344314211614~rrrrrrrr00000100007121002301 秩为3

三阶子式07023855023085570．

24 10.设A、B都是nm矩阵，证明BA~的充分必要条件是)()(BRAR。

证：必要性即定理3，故需证明充分性，设)()(BRAR=r，由矩阵的等价标准型理论知矩阵A、B具有相同的标准型，nmrEF000，于是FA~，FB~，从而由等价关系的对称性和传递性，知BA~。

11.设32321321kkkA，问k为何值时，可使：

(1) 1)(AR； (2) 2)(AR； (3) 3)(AR。

解：对A作初等变换，32321321kkkA～)2)(1(300)1(3)1(20321kkkkk，

于是，由定理3，(1) 当k＝1时，1)(AR； (2) 当k＝-2时，2)(AR； (3) 当21kk且时，3)(AR。

12．求解下列齐次线性方程组:

(1)

;0222,02,02432143214321xxxxxxxxxxxx (2)

;05105,0363,02432143214321xxxxxxxxxxxx

(3)

;0742,0634,0723,05324321432143214321xxxxxxxxxxxxxxxx (4).0327,01613114,02332,075434321432143214321xxxxxxxxxxxxxxxx

解 (1) 对系数矩阵实施行变换:

2122111212113410013100101~, 即得4443424134334xxxxxxxx . 故方程组的解为

1343344321kxxxx.

(2) 对系数矩阵实施行变换:

5110531631121000001001021~ 即得4432242102xxxxxxxx

故方程组的解为

10010012214321kkxxxx

(3) 对系数矩阵实施行变换: