线性代数第二章习题答案

第二章 矩阵及其运算课后习题答案

1．已知线性变换：

,323,53,22321332123211yyyxyyyxyyyx

求从变量321,,xxx到变量321,,yyy的线性变换．

解

由已知：221321323513122yyyxxx

故

3211221323513122xxxyyy321423736947yyy

321332123211423736947xxxyxxxyxxxy

2．已知两个线性变换

,54,232,232133212311yyyxyyyxyyx

,3,2,3323312211zzyzzyzzy

求从321,,zzz到321,,xxx的线性变换．

解 由已知

221321514232102yyyxxx321310102013514232102zzz321161109412316zzz

因此有

3213321232111610941236zzzxzzzxzzzx

3．设111111111A, ,150421321B 求.23BAAABT及

解 AAB2315042132111111111131111111112 0926508503111111111222942017222132

150421321111111111BAT092650850

4．计算以下乘积：

(1)127075321134; (2)1233,2,1;

(3)2,1312; (4)20413121013143110412;

(5)321332313232212131211321),,(xxxaaaaaaaaaxxx; (6)30003200121013013000120010100121.

解 (1)127075321134102775132)2(7111237449635

(2)123321)10()132231(

(3)2131223)1(321)1(122)1(2632142

(4)204131210131431104126520876

(5)321332313232212131211321xxxaaaaaaaaaxxx

333223113323222112313212111xaxaxaxaxaxaxaxaxa321xxx

322331132112233322222111222xxaxxaxxaxaxaxa

(6)

300032001210130130001200101001219000340042102521 5．设3121A, 2101B，问：

(1)BAAB吗?

(2)2222)(BABABA吗?

(3)22))((BABABA吗?

解 (1)3121A, 2101B. 则6443AB 8321BA BAAB

(2) 52225222)(2BA2914148

但222BABA4301128861148327151610

故2222)(BABABA

(3) ))((BABA102052229060

而 22BA4301114837182

故 22))((BABABA

6．举反列说明以下命题是错误的:

（１）假设02A,那么0A;

（２）假设AA2,那么0A或EA;

（３）假设AYAX,且0A,那么YX.

解 (1) 取0010A, 02A,但0A

(2) 取0011A, AA2,但0A且EA

(3) 取0001A, 1111X, 1011Y. AYAX且0A 但YX.

7．设101A,求kAAA,,,32.

解 12011011012A; 1301101120123AAA

利用数学归纳法证明: 101kAk

当1k时,显然成立,假设k时成立,那么1k时

1)1(01101101kkAAAkk

由数学归纳法原理知:101kAk 8．设001001A,求kA.

解 第一观看

0010010010012A222002012,

3232323003033AAA

由此推测

kkkkkkkkkkkA0002)1(121 )2(k

用数学归纳法证明:

当2k时,显然成立.

假设k时成立，那么1k时，

0010010002)1(1211kkkkkkkkkkkkAAA11111100)1(02)1()1(kkkkkkkkkk

由数学归纳法原理知:

kkkkkkkkkkkA0002)1(121

9．设BA,为n阶矩阵，且A为对称矩阵，证明ABBT也是对称矩阵.

证明 已知：AAT

那么 ABBBABABBABBTTTTTTTT)()(

从而 ABBT也是对称矩阵.

10．设BA,都是n阶对称矩阵，证明AB是对称矩阵的充分必要条件是BAAB.

证明 由已知：AAT BBT

充分性：BAABABABTT)(ABABT

即AB是对称矩阵.

必要性：ABABT)(ABABTTABBA.

11．求以下矩阵的逆矩阵： (1)5221； (2)cossinsincos； (3)145243121;

(4)naaa0021)0(21aaan

解 (1) 5221A, 1A.

.1 ),1(2 ),1(2 ,522122111AAAA

122522122111AAAAA. AAA111225

(2) 01A 故1A存在

cossinsincos22122111AAAA

从而 cossinsincos1A

(3) 2A, 故1A存在

024312111AAA

1613322212AAA

21432332313AAA

故 AAA111716213213012

(4)naaaA0021. 由对角矩阵的性质知

naaaA10011211

12．解以下矩阵方程：

(1) 12643152X; (2) 234311111012112X;

(3) 101311022141X; (4)

021102341010100001100001010X.

解 (1) 126431521X1264215380232

(2) 1111012112234311X

0332321012343113132538122

(3)

11110210132141X2101101311421212101036612104111

(4) 11010100001021102341100001010X010100001021102341100001010