线性代数习题解答
张应应 胡佩 2013-3-1
目 录
第一章 第二章 第三章 第四章 第五章 第六章
行列式 .................................................................................................................... 1 矩阵 ...................................................................................................................... 22 向量组的线性相关性 .......................................................................................... 50 线性方程组 .......................................................................................................... 69 矩阵的相似对角化 .............................................................................................. 91 二次型 ................................................................................................................ 114
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附录:习题参考答案 ........................................................................................................... 129
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教材:段正敏,颜军,阴文革:《线性代数》,高等教育出版社,2010。
第一章 行列式
1.填空题:
(1)3421的逆序数为 5 ;
解:该排列的逆序数为t?0?0?2?3?5. (2)517924的逆序数为 7 ;
解:该排列的逆序数为t?0?1?0?0?3?3?7. (3)设有行列式
21D?65?130?12045?432=?(aij), 10078?11132含因子a12a31a45的项为 -1440,0 ;
解:(?1)t(23154)a12a23a31a45a54?(?1)3?5?2?6?8?3??1440
(?1)t(24153)a12a24a31a45a53?(?1)4?5?0?6?8?1?0
所以D含因子a12a31a45的项为-1440和0.
(4)若n阶行列式Dn??(aij)?a,则D??(?aij)?解:
行列式D中每一行可提出一个公因子?1,
nn??1?na;
?D??(?aij)???1??(aij)???1?a.
11(5)设f(x)?111112?2x,则f(x)?0的根为 1,2,-2 ; 244x8?8x3解:f(x)是一个Vandermonde行列式,
?f(x)?(x?1)(x?2)(x?2)(?2?1)(?2?2)(2?1)?0的根为1,2,-2.
(6)设x1,x2,x3是方程x?px?q?0的三个根,则行列式
3x1 x3x2x1x3x3x2? 0 ; x1x2解:根据条件有x3?px?q?(x?x1)(x?x2)(x?x3)?x3?(x1?x2?x3)x2?ax?x1x2x3 比较系数可得:x1?x2?x3?0,x1x2x3??q
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?x13??px1?q?3再根据条件得:?x2??px2?q
?x3??px?q3?3333原行列式=x1?x2?x3?3x1x2x3??p(x1?x2?x3)?3q?3?(?q)?0.
x(7)设有行列式?123x0=0,则x= 1,2 ;
x10x解:?123x0?x2?3x?2?(x?1)(x?2)?0 x10?x?1,2.
a11(8)设f(x)?a12a22xa42a13xa33a43xa24,则多项式f(x)中x3的系数为 0 ; a34a44a21a31x解:按第一列展开f(x)?a11A11?a21A21?a31A31?xA41,
A11,A21,A31中最多只含有x2项,?含有x3的项只可能是xA41
a12a22xa13xa33xa24a34xA41?x(?1)4?1
3?????????x?xaa?aa?aa?x?a13a22a34?a12a24a33??????123413242233?xA41不含x3项,?f(x)中x3的系数为0. 16(9)如果
0016解:
00250034232500342343=0,则x= 2 ; x343122x???(5?12)(6?3x)?0 x65333?x?2.
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0b(10)
000b0000c0000d00c0000da0= -abcd ; 00解:将行列式按第一行展开:
ab000?a?(?1)1?40c0??abcd. 000d0a31a?3b?3c?321??r1?3r3??r2?2r3(11)如果b01=1,则521??A?ATca31解:b1??abc3021114= 1 ; 1a?3b?3c?3512141?1.
01c21a11a12a22a32a132a112a122a222a322a12?2a132a22?2a23= -16 , 2a32?2a33(12)如a21a312a112a122a13a23=2,则2a212a31a33a21?a31a22?a32a23?a33a13a21?3a11a22?3a12a23?3a13a11a12a22a320a11= -4 ,a12a130a21a22a23a110a31a32a33a21a22a2321= -4 ; 23a31a32??1a33解:A?a21a31a23??1?2?3?AT?a12a33a13?2?3?2
2a112a122a12?2a132a22?2a23?2?12?22a32?2a332?2?2?3?23?1?2?2??3??3??8?0?A???16
2a212a222a312a32?????????????????????????????????????????8??1?2?2??1?23
2a112a122a13a21?3a11a22?3a12a23?3a13a21?a31a22?a32?2?1a23?a33?2?3?1?2??3?2?1?2?3?1?2??3?2?2??3??1?3?1?2??3??????????????????????????????????????????????2??1?????????????????????????????????????????????2?1?2?2??3?2??1?2?2??1?2?3???????????????????????????????????????????????2AT??40a11a12a130a21a22a230a31a32a3321??按第一行展开??23ab1?42(?-1)AT??4.
(13)设n阶行列式D=a?0,且D中的每列的元素之和为b,则行列式D中的第二行的代数余子式之和为=??????;
a11解:
a12a22an2a1na11a12ban2a1nbann=ba111an1a121an2a1n1ann
a21an1a2n??每行元素加到第二行??bannan1??按第二行展开??b?A21?A22??A2n??a?0
?b?0,且A21?A22??A21?A22??A2n??A2n?0
a b?Ain?a,i?1,2,b,n.
实际上,由上述证明过程可知任意行代数余子式之和Ai1?Ai2?a110(14)如果
00a22a23a24a32a33a34a42a12a22a32a42a13a23a33a43a140a22a32a42a12a23a33a43a23a24a34= -1 , a44a24a240=1,则a340a44a111a43=????????;
a11a44a22解:令B?a32a23a33a43a24a34,则 a444
a42
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