人大版线性代数课后习题答案

线性代数第五章课后习题及解答

第五章课后习题及解答1. 求下列矩阵的特征值和特征向量：(1) ;1332⎪⎪⎭⎫⎝⎛-- 解：,07313322=--=--=-λλλλλA I2373,237321-=+=λλ ,001336371237121371⎪⎪⎭⎫ ⎝⎛→→⎪⎪⎭⎫⎝⎛=-++- A I λ 所以，0)(1=-x A I λ的基础解系为：.)371,6(T-因此，A 的属于1λ的所有特征向量为：).0()371,6(11≠-k k T,001336371237123712⎪⎪⎭⎫ ⎝⎛→→⎪⎪⎭⎫⎝⎛-=---+ A I λ 所以，0)(2=-x A I λ的基础解系为：.)371,6(T+因此，A 的属于2λ的所有特征向量为：).0()371,6(22≠+k k T(2) ;211102113⎪⎪⎪⎭⎫ ⎝⎛--解：2)2)(1(21112113--==------=-λλλλλλ A I所以，特征值为：11=λ(单根)，22=λ(二重根)⎪⎪⎪⎭⎫ ⎝⎛-→→⎪⎪⎪⎭⎫ ⎝⎛------=-0001100011111121121 A I λ所以，0)(1=-x A I λ的基础解系为：.)1,1,0(T因此，A 的属于1λ的所有特征向量为：).0()1,1,0(11≠k k T⎪⎪⎪⎭⎫ ⎝⎛-→→⎪⎪⎪⎭⎫ ⎝⎛-----=-0001000110111221112 A I λ所以，0)(2=-x A I λ的基础解系为：.)0,1,1(T因此，A 的属于2λ的所有特征向量为：).0()0,1,1(22≠k k T(3) ;311111002⎪⎪⎪⎭⎫ ⎝⎛-解：3)2(31111102-==------=-λλλλλ A I所以，特征值为：21=λ(三重根)⎪⎪⎪⎭⎫⎝⎛-→→⎪⎪⎪⎭⎫ ⎝⎛----=-0000001111111110001 A I λ所以，0)(1=-x A I λ的基础解系为：.)1,0,1(,)0,1,1(TT-因此，A 的属于1λ的所有特征向量为：TTk k )1,0,1()0,1,1(21-+(21,k k 为不全为零的任意常数)。

线性代数课后习题答案全解.pdf

第一章行列式1. 利用对角线法则计算下列三阶行列式: (1)381141102−−−;解 381141102−−−=2×(−4)×3+0×(−1)×(−1)+1×1×8 −0×1×3−2×(−1)×8−1×(−4)×(−1) =−24+8+16−4=−4. (2)b a c a c b cb a ;解 ba c a cb cb a=acb +bac +cba −bbb −aaa −ccc =3abc −a 3−b 3−c 3. (3)222111c b a c b a ;解 222111c b a c b a=bc 2+ca 2+ab 2−ac 2−ba 2−cb =(a −b )(b −c )(c −a ). 2(4)y x y x x y x y yx y x +++.解 yx y x x y x y yx y x +++=x (x +y )y +yx (x +y )+(x +y )yx −y 3−(x +y )3−x =3xy (x +y )−y 3 3−3x 2 y −x 3−y 3−x =−2(x 3 3+y 3 2. 按自然数从小到大为标准次序, 求下列各排列的逆序数:).(1)1 2 3 4; 解逆序数为0 (2)4 1 3 2;解逆序数为4: 41, 43, 42, 32. (3)3 4 2 1;解逆序数为5: 3 2, 3 1, 4 2, 4 1, 2 1. (4)2 4 1 3;解逆序数为3: 2 1, 4 1, 4 3. (5)1 3 ⋅ ⋅ ⋅ (2n −1) 2 4 ⋅ ⋅ ⋅ (2n );解逆序数为2)1(−n n : 3 2 (1个) 5 2, 5 4(2个) 7 2, 7 4, 7 6(3个)⋅ ⋅ ⋅ ⋅ ⋅ ⋅(2n −1)2, (2n −1)4, (2n −1)6, ⋅ ⋅ ⋅, (2n −1)(2n −2) (n −1个)(6)1 3 ⋅ ⋅ ⋅ (2n −1) (2n ) (2n −2) ⋅ ⋅ ⋅ 2. 解逆序数为n (n −1) : 3 2(1个) 5 2, 5 4 (2个) ⋅ ⋅ ⋅ ⋅ ⋅ ⋅(2n −1)2, (2n −1)4, (2n −1)6, ⋅ ⋅ ⋅, (2n −1)(2n −2) (n −1个) 4 2(1个) 6 2, 6 4(2个) ⋅ ⋅ ⋅ ⋅ ⋅ ⋅(2n )2, (2n )4, (2n )6, ⋅ ⋅ ⋅, (2n )(2n −2) (n −1个) 3. 写出四阶行列式中含有因子a 11a 23 解含因子a 的项. 11a 23(−1)的项的一般形式为t a 11a 23a 3r a 4s 其中rs 是2和4构成的排列, 这种排列共有两个, 即24和42. ,所以含因子a 11a 23 (−1)的项分别是t a 11a 23a 32a 44=(−1)1a 11a 23a 32a 44=−a 11a 23a 32a 44 (−1), t a 11a 23a 34a 42=(−1)2a 11a 23a 34a 42=a 11a 23a 34a 42 4. 计算下列各行列式:.(1)71100251020214214; 解 71100251020214214010014231020211021473234−−−−−======c c c c 34)1(143102211014+−×−−−= 143102211014−−=01417172001099323211=−++======c c c c .(2)2605232112131412−; 解 2605232112131412−26053212213041224−−=====c c 041203212213041224−−=====r r 0000003212213041214=−−=====r r . (3)efcf bf de cd bd aeac ab −−−;解 ef cf bf de cd bd ae ac ab −−−ec b e c b ec b adf −−−=abcdef adfbce 4111111111=−−−=.(4)dc b a 100110011001−−−. 解d c b a 100110011001−−−dc b aab ar r 10011001101021−−−++===== d c a ab 101101)1)(1(12−−+−−=+01011123−+−++=====cd c ada ab dc ccdad ab +−+−−=+111)1)(1(23=abcd +ab +cd +ad +1. 5. 证明:(1)1112222b b a a b ab a +=(a −b )3 证明;1112222b b a a b ab a +00122222221213a b a b a a b a ab a c c c c −−−−−−=====ab a b a b a ab 22)1(22213−−−−−=+21))((a b a a b a b +−−==(a −b )3 (2) . y x z x z y zy x b a bz ay by ax bx az by ax bx az bz ay bx az bz ay by ax )(33+=+++++++++;证明bzay by ax bx az by ax bx az bz ay bxaz bz ay by ax +++++++++bz ay by ax x by ax bx az z bxaz bz ay y b bz ay by ax z by ax bx az y bx az bz ay x a +++++++++++++=bz ay y x by ax x z bxaz z y b y by ax z x bx az y z bz ay x a +++++++=22z y x y x z xz y b y x z x z y z y x a 33+=y x z x z y zy x b y x z x z y z y x a 33+=y x z x z y zy x b a )(33+=.(3)0)3()2()1()3()2()1()3()2()1()3()2()1(2222222222222222=++++++++++++d d d d c c c c b b b b a a a a ; 证明 2222222222222222)3()2()1()3()2()1()3()2()1()3()2()1(++++++++++++d d d d c c c c b b b b a a a a (c 4−c 3, c 3−c 2, c 2−c 1 得) 5232125232125232125232122222++++++++++++=d d d d c c c c b b b b a a a a (c 4−c 3, c 3−c 2得)022122212221222122222=++++=d d c c b b a a . (4)444422221111d c b a d c b a d c b a =(a −b )(a −c )(a −d )(b −c )(b −d )(c −d )(a +b +c +d ); 证明 444422221111d c b a d c b a d c b a )()()(0)()()(001111222222222a d d a c c a b b a d d a c c a b b ad a c a b −−−−−−−−−=)()()(111))()((222a d d a c c a b b dc b ad a c a b +++−−−= ))(())((00111))()((a b d b d d a b c b c c bd b c a d a c a b ++−++−−−−−−= )()(11))()()()((a b d d a b c c b d b c a d a c a b ++++−−−−−= =(a −b )(a −c )(a −d )(b −c )(b −d )(c −d )(a +b +c +d ). (5)12211 000 00 1000 01a x a a a a x x xn n n+⋅⋅⋅−⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅−⋅⋅⋅−−− =x n +a 1x n −1+ ⋅ ⋅ ⋅ +a n −1x +a n .证明用数学归纳法证明.当n =2时, 2121221a x a x a x a x D ++=+−=, 命题成立. 假设对于(n −1)阶行列式命题成立, 即 D n −1=x n −1+a 1 x n −2+ ⋅ ⋅ ⋅ +a n −2x +a n −1则D , n 按第一列展开, 有 11100 100 01)1(11−⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅−⋅⋅⋅−−+=+−x x a xD D n n n n =xD n −1+a n =x n +a 1x n −1+ ⋅ ⋅ ⋅ +a n −1x +a n 因此, 对于n 阶行列式命题成立. .6. 设n 阶行列式D =det(a ij ), 把D 上下翻转、或逆时针旋转90°、或依副对角线翻转, 依次得n nn n a a a a D 11111 ⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅=, 11112 n nnn a a a a D ⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅= , 11113 a a a a D n n nn ⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅=,证明D D D n n 2)1(21)1(−−==, D 3 证明因为D =det(a =D .ij ), 所以 nnn n n n nnnn a a a a a a a a a a D 2211111111111 )1( ⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅−=⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅=−⋅⋅⋅=⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅−−=−− )1()1(331122111121nnn n nn n n a a a a a a a a D D n n n n 2)1()1()2( 21)1()1(−−+−+⋅⋅⋅++−=−=.同理可证 nnn n n n a a a a D ⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅−=− )1(11112)1(2D D n n T n n 2)1(2)1()1()1(−−−=−=. D D D D D n n n n n n n n =−=−−=−=−−−−)1(2)1(2)1(22)1(3)1()1()1()1(.7. 计算下列各行列式(D k (1)为k 阶行列式): aa D n 1 1⋅⋅⋅=, 其中对角线上元素都是a , 未写出的元素都是0; 解 aa a a a D n 010 000 00 000 0010 00⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅=(按第n 行展开) )1()1(10 000 00 000 0010 000)1(−×−+⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅−=n n n aa a )1()1(2 )1(−×−⋅⋅⋅⋅−+n n n a a an n n n n a a a+⋅⋅⋅−⋅−=−−+)2)(2(1)1()1(=a n −a n −2=a n −2(a 2−1).(2)xa aa x a a a xD n ⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅= ; 解将第一行乘(−1)分别加到其余各行, 得 ax x a ax x a a x x a aa a x D n −−⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅−−⋅⋅⋅−−⋅⋅⋅=000 0 00 0, 再将各列都加到第一列上, 得ax ax a x aaa a n x D n −⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅−⋅⋅⋅−⋅⋅⋅−+=0000 0 000 00 )1(=[x +(n −1)a ](x −a )n −1 (3). 111 1 )( )1()( )1(1111⋅⋅⋅−⋅⋅⋅⋅⋅⋅⋅⋅⋅−⋅⋅⋅⋅⋅⋅−⋅⋅⋅−−⋅⋅⋅−=−−−+n a a a n a a a n a a a D n n n n nn n ; 解根据第6题结果, 有 nnn n n n n n n n a a a n a a a n a a aD )( )1()( )1( 11 11)1(1112)1(1−⋅⋅⋅−−⋅⋅⋅⋅⋅⋅⋅⋅⋅−⋅⋅⋅⋅⋅⋅−⋅⋅⋅−⋅⋅⋅−=−−−++此行列式为范德蒙德行列式.∏≥>≥++++−−+−−=112)1(1)]1()1[()1(j i n n n n j a i a D∏≥>≥++−−−=112)1()]([)1(j i n n n j i∏≥>≥++⋅⋅⋅+−++−⋅−⋅−=1121)1(2)1()()1()1(j i n n n n n j i∏≥>≥+−=11)(j i n j i .(4)nnnnn d c d c b a b a D ⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅=11112; 解nnnnn d c d c b a b a D ⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅=11112(按第1行展开) nn n n n nd d c d c b a b a a 00011111111−−−−⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅=0)1(1111111112c d c d c b a b a b nn n n n nn −−−−+⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅−+. 再按最后一行展开得递推公式D 2n =a n d n D 2n −2−b n c n D 2n −2, 即D 2n =(a n d n −b n c n )D 2n −2于是 . ∏=−=ni i i i i n D c b d a D 222)(.而 111111112c b d a d c b a D −==,所以 ∏=−=ni i i i i n c b d a D 12)(.(5) D =det(a ij ), 其中a ij 解 a =|i −j |; ij =|i −j |, 043214 01233 10122 21011 3210)det(⋅⋅⋅−−−−⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅−⋅⋅⋅−⋅⋅⋅−⋅⋅⋅−⋅⋅⋅==n n n n n n n n a D ij n 04321 1 11111 11111 11111 1111 2132⋅⋅⋅−−−−⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅−−−−⋅⋅⋅−−−⋅⋅⋅−−⋅⋅⋅−−⋅⋅⋅−=====n n n n r r r r15242321 0 22210 02210 00210 0001 1213−⋅⋅⋅−−−−⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅−−−−⋅⋅⋅−−−⋅⋅⋅−−⋅⋅⋅−+⋅⋅⋅+=====n n n n n c c c c =(−1)n −1(n −1)2n −2 (6).nn a a a D +⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅+⋅⋅⋅+=1 11 1 1111121, 其中a 1a 2 ⋅ ⋅ ⋅ a n≠0.解nn a a a D +⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅+⋅⋅⋅+=1 11 1 1111121 nn n n a a a a a a a a a c c c c +−⋅⋅⋅−⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅−⋅⋅⋅−⋅⋅⋅−⋅⋅⋅−=====−−100001 000 100 0100 0100 0011332212132 1111312112111000011 000 00 11000 01100 001 −−−−−−+−⋅⋅⋅−⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅−⋅⋅⋅−⋅⋅⋅⋅⋅⋅=nn n a a a a a a a a∑=−−−−−−+⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅=n i i n n a a a a a a a a 1111131******** 00010 000 00 10000 01000 001)11)((121∑=+=ni i n a a a a .8. 用克莱姆法则解下列方程组: (1) =+++−=−−−−=+−+=+++01123253224254321432143214321x x x x x x x x x x x x x x x x ;解因为 14211213513241211111−=−−−−=D , 142112105132412211151−=−−−−−−=D , 284112035122412111512−=−−−−−=D , 426110135232422115113−=−−−−=D , 14202132132212151114=−−−−−=D , 所以 111==D D x , 222==D Dx , 333==DD x , 144−==D D x .(2)=+=++=++=++=+150650650651655454343232121x x x x x x x x x x x x x .解因为 665510006510006510065100065==D , 15075100165100065100065000611==D , 114551010651000650000601000152−==D , 703511650000601000051001653==D , 39551601000051000651010654−==D , 2121100005100065100651100655==D , 所以66515071=x , 66511452−=x , 6657033=x , 6653954−=x , 6652124=x .9. 问λ, µ取何值时, 齐次线性方程组 =++=++=++0200321321321x x x x x x x x x µµλ有非零解？解系数行列式为µλµµµλ−==1211111D .令D =0, 得 µ=0或λ=1.于是, 当µ=0或λ=1时该齐次线性方程组有非零解.10. 问λ取何值时, 齐次线性方程组 =−++=+−+=+−−0)1(0)3(2042)1(321321321x x x x x x x x x λλλ有非零解？解系数行列式为λλλλλλλ−−+−−=−−−−=101112431111132421D=(1−λ)3 =(1−λ)+(λ−3)−4(1−λ)−2(1−λ)(−3−λ) 3+2(1−λ)2 令D =0, 得+λ−3. λ=0, λ=2或λ=3.于是, 当λ=0, λ=2或λ=3时, 该齐次线性方程组有非零解.第二章矩阵及其运算1. 已知线性变换:++=++=++=3213321232113235322y y y x y y y x y y y x , 求从变量x 1, x 2, x 3到变量y 1, y 2, y 3 解由已知:的线性变换.= 221321323513122y y y x x x ,故= −3211221323513122x x x y y y−−−−=321423736947y y y ,−+=−+=+−−=321332123211423736947x x x y x x x y x x x y .2. 已知两个线性变换++=++−=+=32133212311542322y y y x y y y x y y x ,+−=+=+−=323312211323z z y z z y z z y , 求从z 1, z 2, z 3到x 1, x 2, x 3 解由已知的线性变换.−= 221321514232102y y y x x x−− −=321310102013514232102z z z−−−−=321161109412316z z z ,所以有 +−−=+−=++−=3213321232111610941236z z z x z z z x z z z x .3. 设 −−=111111111A ,−−=150421321B , 求3AB −2A 及A T 解 B .−−− −− −−=−1111111112150421321111111111323A AB−−−−= −−− −=2294201722213211111111120926508503,−= −− −−=092650850150421321111111111B A T.4. 计算下列乘积: (1)−127075321134;解 −127075321134 ×+×+××+×−+××+×+×=102775132)2(71112374=49635.(2)123)321(;解123)321(=(1×3+2×2+3×1)=(10).(3))21(312−;解 )21(312−×−××−××−×=23)1(321)1(122)1(2−−−=632142. (4)−−−−20413121013143110412 ; 解−−− −20413121013143110412 −−−=6520876. (5)321332313232212131211321)(x x x a a a a a a a a a x x x ;解321332313232212131211321)(x x x a a a a a a a a a x x x=(a 11x 1+a 12x 2+a 13x 3 a 12x 1+a 22x 2+a 23x 3a 13x 1+a 23x 2+a 33x 3321x x x )322331132112233322222111222x x a x x a x x a x a x a x a +++++=.5. 设 =3121A ,=2101B , 问: (1)AB =BA 吗? 解 AB ≠BA . 因为=6443AB ,=8321BA , 所以AB ≠BA .(2)(A +B )2=A 2+2AB +B 2 解 (A +B )吗? 2≠A 2+2AB +B 2 因为.=+5222B A ,=+52225222)(2B A=2914148,但 + +=++43011288611483222B AB A=27151610,所以(A +B )2≠A 2+2AB +B 2 (3)(A +B )(A −B )=A . 2−B 2 解 (A +B )(A −B )≠A 吗? 2−B 2 因为.=+5222B A ,=−1020B A ,==−+906010205222))((B A B A ,而= −=−718243011148322B A ,故(A +B )(A −B )≠A 2−B 2 6. 举反列说明下列命题是错误的:.(1)若A 2 解取=0, 则A =0;=0010A , 则A 2 (2)若A =0, 但A ≠0. 2 解取=A , 则A =0或A =E ;=0011A , 则A 2 (3)若AX =AY , 且A ≠0, 则X =Y .=A , 但A ≠0且A ≠E . 解取=0001A , −=1111X ,=1011Y , 则AX =AY , 且A ≠0, 但X ≠Y .7. 设=101λA , 求A 2, A 3, ⋅ ⋅ ⋅, A k 解 . ==12011011012λλλA , ===1301101120123λλλA A A , ⋅ ⋅ ⋅ ⋅ ⋅ ⋅,=101λk A k . 8. 设=λλλ001001A , 求A k 解首先观察. =λλλλλλ0010010010012A=222002012λλλλλ,=⋅=3232323003033λλλλλλA A A ,=⋅=43423434004064λλλλλλA A A ,=⋅=545345450050105λλλλλλA A A , ⋅ ⋅ ⋅ ⋅ ⋅ ⋅,=k A k k k k k k k k k k λλλλλλ0002)1(121−−−−. 用数学归纳法证明:当k =2时, 显然成立.假设k 时成立，则k +1时，−=⋅=−−−+λλλλλλλλλ0010010002)1(1211k k k k k k k k k k k k A A A+++=+−+−−+11111100)1(02)1()1(k k k k k k k k k k λλλλλλ, 由数学归纳法原理知:−=−−−k k k k k k k k k k k A λλλλλλ0002)1(121. 9. 设A , B 为n 阶矩阵，且A 为对称矩阵，证明B T 证明因为A AB 也是对称矩阵.T (B =A , 所以T AB )T =B T (B T A )T =B T A T B =B T 从而B AB ,T 10. 设A , B 都是n 阶对称矩阵，证明AB 是对称矩阵的充分必要条件是AB =BA .AB 是对称矩阵.证明充分性: 因为A T =A , B T (AB )=B , 且AB =BA , 所以 T =(BA )T =A T B T 即AB 是对称矩阵.=AB ,必要性: 因为A T =A , B T =B , 且(AB )T AB =(AB )=AB , 所以T =B T A T 11. 求下列矩阵的逆矩阵:=BA .(1)5221; 解=5221A . |A |=1, 故A −1 存在. 因为−−= =1225*22122111A A A A A ,故 *||11A A A =−−−=1225. (2)−θθθθcos sin sin cos ; 解−=θθθθcos sin sin cos A . |A |=1≠0, 故A −1 存在. 因为−= =θθθθcos sin sin cos *22122111A A A A A , 所以 *||11A A A =−−=θθθθcos sin sin cos . (3)−−−145243121; 解−−−=145243121A . |A |=2≠0, 故A −1 存在. 因为−−−−−= =214321613024*332313322212312111A A A A A A A A A A , 所以 *||11A A A =−−−−−−=1716213213012. (4)n a a a 0021(a 1a 2⋅ ⋅ ⋅a n ≠0) .解=n a a a A 0021, 由对角矩阵的性质知=−n a a a A 10011211 . 12. 解下列矩阵方程:(1) −=12643152X ; 解 −=−126431521X − −−=12642153 −=80232. (2) −=−−234311*********X ; 解 1111012112234311−−− −=X−−− −=03323210123431131 −−−=32538122. (3) −= − −101311022141X ;解 11110210132141−− − − −=X− −=210110131142121 =21010366121=04111. (4)−−−= 021102341010100001100001010X . 解 11010100001021102341100001010−−−−− =X −−− =010100001021102341100001010 −−−=201431012. 13. 利用逆矩阵解下列线性方程组:(1) =++=++=++3532522132321321321x x x x x x x x x ; 解方程组可表示为= 321153522321321x x x , 故 = = −0013211535223211321x x x ,从而有 ===001321x x x . (2) =−+=−−=−−05231322321321321x x x x x x x x x . 解方程组可表示为=−−−−−012523312111321x x x , 故 =−−−−−= −3050125233121111321x x x , 故有 ===305321x x x . 14. 设A k =O (k 为正整数), 证明(E −A )−1=E +A +A 2+⋅ ⋅ ⋅+A k −1 证明因为A . k =O , 所以E −A k E −A =E . 又因为k =(E −A )(E +A +A 2+⋅ ⋅ ⋅+A k −1所以 (E −A )(E +A +A ),2+⋅ ⋅ ⋅+A k −1由定理2推论知(E −A )可逆, 且)=E ,(E −A )−1=E +A +A 2+⋅ ⋅ ⋅+A k −1.证明一方面, 有E =(E −A )−1 另一方面, 由A (E −A ).k E =(E −A )+(A −A =O , 有2)+A 2−⋅ ⋅ ⋅−A k −1+(A k −1−A k )=(E +A +A 2+⋅ ⋅ ⋅+A k −1故 (E −A ))(E −A ),−1(E −A )=(E +A +A 2+⋅ ⋅ ⋅+A k −1两端同时右乘(E −A ))(E −A ),−1 (E −A ), 就有−1(E −A )=E +A +A 2+⋅ ⋅ ⋅+A k −1.15. 设方阵A 满足A 2−A −2E =O , 证明A 及A +2E 都可逆, 并求A −1及(A +2E )−1 证明由A .2 A −A −2E =O 得2或 −A =2E , 即A (A −E )=2E ,E E A A =−⋅)(21, 由定理2推论知A 可逆, 且)(211E A A −=−. 由A 2 A −A −2E =O 得2或 −A −6E =−4E , 即(A +2E )(A −3E )=−4E ,E A E E A =−⋅+)3(41)2( 由定理2推论知(A +2E )可逆, 且)3(41)2(1A E E A −=+−.证明由A 2−A −2E =O 得A 2 |A −A =2E , 两端同时取行列式得 2即 |A ||A −E |=2,−A |=2,故 |A |≠0,所以A 可逆, 而A +2E =A 2, |A +2E |=|A 2|=|A |2由 A ≠0, 故A +2E 也可逆. 2 ⇒A −A −2E =O ⇒A (A −E )=2E−1A (A −E )=2A −1)(211E A A −=−E ⇒,又由 A 2 ⇒ (A +2E )(A −3E )=−4 E ,−A −2E =O ⇒(A +2E )A −3(A +2E )=−4E所以 (A +2E )−1(A +2E )(A −3E )=−4(A +2 E )−1 ,)3(41)2(1A E E A −=+−.16. 设A 为3阶矩阵, 21||=A , 求|(2A )−1 解因为−5A *|.*||11A A A =−, 所以 |||521||*5)2(|111−−−−=−A A A A A |2521|11−−−=A A=|−2A −1|=(−2)3|A −1|=−8|A |−1 17. 设矩阵A 可逆, 证明其伴随阵A *也可逆, 且(A *)=−8×2=−16.−1=(A −1 证明由)*.*||11A A A =−, 得A *=|A |A −1 |A *|=|A |, 所以当A 可逆时, 有n |A −1|=|A |n −1从而A *也可逆.≠0,因为A *=|A |A −1 (A *), 所以−1=|A |−1又A .*)(||)*(||1111−−−==A A A A A , 所以(A *)−1=|A |−1A =|A |−1|A |(A −1)*=(A −1 18. 设n 阶矩阵A 的伴随矩阵为A *, 证明:)*.(1)若|A |=0, 则|A *|=0;(2)|A *|=|A |n −1 证明.(1)用反证法证明. 假设|A *|≠0, 则有A *(A *)−1 A =A A *(A *)=E , 由此得 −1=|A |E (A *)−1所以A *=O , 这与|A *|≠0矛盾，故当|A |=0时, 有|A *|=0.=O ,(2)由于*||11A A A =−, 则AA *=|A |E , 取行列式得到 |A ||A *|=|A |n 若|A |≠0, 则|A *|=|A |.n −1 若|A |=0, 由(1)知|A *|=0, 此时命题也成立.;因此|A *|=|A |n −1.19. 设−=321011330A , AB =A +2B , 求B . 解由AB =A +2E 可得(A −2E )B =A , 故− −−−=−=−−321011330121011332)2(11A E A B −=011321330. 20. 设 =101020101A , 且AB +E =A 2+B , 求B .解由AB +E =A 2 (A −E )B =A +B 得 2即 (A −E )B =(A −E )(A +E ).−E , 因为01001010100||≠−==−E A , 所以(A −E )可逆, 从而=+=201030102E A B .21. 设A =diag(1, −2, 1), A *BA =2BA −8E , 求B . 解由A *BA =2BA −8E 得 (A *−2E )BA =−8E , B =−8(A *−2E )−1A =−8[A (A *−2E )]−1 =−8(AA *−2A )−1 =−8(|A |E −2A )−1 =−8(−2E −2A )−1 =4(E +A )−1 =4[diag(2, −1, 2)]−1−1)21 ,1 ,21(diag 4−==2diag(1, −2, 1).22. 已知矩阵A 的伴随阵−=8030010100100001*A , 且ABA −1=BA −1+3E , 求B .解由|A *|=|A |3 由ABA =8, 得|A |=2. −1=BA −1 AB =B +3A ,+3E 得 B =3(A −E )−1A =3[A (E −A −1)]−1 A 11*)2(6*)21(3−−−=−=A E A E−=−−=−1030060600600006603001010010000161. 23. 设P −1 −−=1141P AP =Λ, 其中,−=Λ2001, 求A 11 解由P . −1AP =Λ, 得A =P ΛP −1, 所以A 11= A =P Λ11P −1 |P |=3, .−=1141*P ,−−=−1141311P ,而−= −=Λ11111120 012001,故−− −−−=31313431200111411111A −−=68468327322731. 24. 设AP =P Λ, 其中−−=111201111P ,−=Λ511,求ϕ(A )=A 8(5E −6A +A 2 解 ϕ(Λ)=Λ). 8(5E −6Λ+Λ2 =diag(1,1,5)8)[diag(5,5,5)−diag(−6,6,30)+diag(1,1,25)]=diag(1,1,58 ϕ(A )=P ϕ(Λ)P )diag(12,0,0)=12diag(1,0,0).−1 *)(||1P P P Λ=ϕ−−−−−− −−−=1213032220000000011112011112=1111111114.25. 设矩阵A 、B 及A +B 都可逆, 证明A −1+B −1 证明因为也可逆, 并求其逆阵.A −1(A +B )B −1=B −1+A −1=A −1+B −1而A ,−1(A +B )B −1是三个可逆矩阵的乘积, 所以A −1(A +B )B −1可逆, 即A −1+B −1 (A 可逆.−1+B −1)−1=[A −1(A +B )B −1]−1=B (A +B )−1 26. 计算A .−−−30003200121013013000120010100121. 解设 =10211A , =30122A , −=12131B ,−−=30322B ,则 2121B O B E A O E A+=222111B A O B B A A ,而 −= −−+−=+4225303212131021211B B A ,−−= −− =90343032301222B A , 所以 2121B O B E A O E A +=222111B A O B B A A−−−=9000340042102521, 即−−−30003200121013013000120010100121−−−=9000340042102521. 27. 取==−==1001D C B A , 验证|||||||| D C B A D C B A ≠.解 4100120021010*********0021010010110100101==−−=−−=D C B A , 而 01111|||||||| ==D C B A ,故 ||||||||D C B A D C B A ≠. 28. 设 −=22023443O O A , 求|A 8|及A 4解令. −=34431A ,=22022A , 则=21A O O A A ,故 8218=A O O A A=8281A O O A ,1682818281810||||||||||===A A A A A .= =464444241422025005O O A O O A A . 29. 设n 阶矩阵A 及s 阶矩阵B 都可逆, 求 (1)1−O B A O ; 解设 =−43211C C C C O B A O , 则O B A O 4321C C C C = =s n E O O E BC BC AC AC 2143. 由此得====s n EBC OBC O AC E AC 2143⇒ ====−−121413B C O C O C A C ,所以= −−−O A B O O B A O 111. (2)1−B C O A . 解设 =−43211D D D D B C O A , 则 = ++= s nE O O E BD CD BD CD AD AD D D D D B C O A 4231214321.由此得=+=+==s nEBD CD O BD CD O AD E AD 423121⇒ =−===−−−−14113211B D CA B D O D A D ,所以−= −−−−−11111B CA B O A BC O A . 30. 求下列矩阵的逆阵: (1)2500380000120025; 解设 =1225A , =2538B , 则−−= =−−5221122511A ,−−==−−8532253811B .于是 −−−−= = =−−−−850032000052002125003800001200251111B A B A .(2)4121031200210001. 解设 =2101A ,=4103B ,=2112C , 则−= =−−−−−−1111114121031200210001B CA B O A BC O A−−−−−=411212458103161210021210001.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;解−−340313021201(下一步: r 2+(−2)r 1, r 3+(−3)r 1 ~. )−−−020*********(下一步: r 2÷(−1), r 3 ~÷(−2). )−−010*********(下一步: r 3−r 2 ~. )−−300031001201(下一步: r 3 ~÷3. )−−100031001201(下一步: r 2+3r 3 ~. )−100001001201(下一步: r 1+(−2)r 2, r 1+r 3 ~. )100001000001.(2)−−−−174034301320;解−−−−174034301320(下一步: r 2×2+(−3)r 1, r 3+(−2)r 1 ~. )−−−310031001320(下一步: r 3+r 2, r 1+3r 2 ~. )0000310010020(下一步: r 1 ~÷2. )000031005010.(3)−−−−−−−−−12433023221453334311;解−−−−−−−−−12433023221453334311(下一步: r 2−3r 1, r 3−2r 1, r 4−3r 1~. )−−−−−−−−1010500663008840034311(下一步: r 2÷(−4), r 3÷(−3) , r 4~÷(−5). )−−−−−22100221002210034311(下一步: r 1−3r 2, r 3−r 2, r 4−r 2~. )−−−00000000002210032011.(4)−−−−−−34732038234202173132. 解−−−−−−34732038234202173132(下一步: r 1−2r 2, r 3−3r 2, r 4−2r 2~. )−−−−−1187701298804202111110(下一步: r 2+2r 1, r 3−8r 1, r 4−7r 1 ~. )−−41000410002020111110(下一步: r 1↔r 2, r 2×(−1), r 4−r 3~. )−−−−00000410001111020201(下一步: r 2+r 3~. )−−00000410003011020201. 2. 设= 987654321100010101100001010A , 求A .解100001010是初等矩阵E (1, 2), 其逆矩阵就是其本身.100010101是初等矩阵E (1, 2(1)), 其逆矩阵是E (1, 2(−1))−=100010101.− =100010101987654321100001010A= − =287221254100010101987321654.3. 试利用矩阵的初等变换, 求下列方阵的逆矩阵: (1)323513123;解 100010001323513123~−−−101011001200410123~ −−−−1012002110102/102/3023~−−−−2/102/11002110102/922/7003~−−−−2/102/11002110102/33/26/7001故逆矩阵为−−−−21021211233267.(2)−−−−−1210232112201023.解−−−−−10000100001000011210232112201023~−−−−00100301100001001220594012102321~−−−−−−−−20104301100001001200110012102321~ −−−−−−−106124301100001001000110012102321 ~−−−−−−−−−−10612631110`1022111000010000100021 ~−−−−−−−106126311101042111000010000100001故逆矩阵为−−−−−−−10612631110104211. 4. (1)设 −−=113122214A ,−−=132231B , 求X 使AX =B ;解因为−−−−=132231 113122214) ,(B A−−412315210 100010001 ~r ,所以−−==−4123152101B A X .(2)设−−−=433312120A , −=132321B , 求X 使XA =B . 解考虑A T X T =B T . 因为−−−−=134313*********) ,(T T B A−−−411007101042001 ~r ,所以−−−==−417142)(1T T T B A X ,从而−−−==−4741121BA X . 5. 设−−−=101110011A , AX =2X +A , 求X .解原方程化为(A −2E )X =A . 因为−−−−−−−−−=−101101110110011011) ,2(A E A−−−011100101010110001~,所以−−−=−=−011101110)2(1A E A X .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. 例如,=010*********A , R (A )=3.0000是等于0的2阶子式, 010001000是等于0的3阶子式. 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 ).这是因为B 的非零子式必是A 的非零子式, 故A 的秩不会小于B 的秩.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).解用已知向量容易构成一个有4个非零行的5阶下三角矩阵:−0000001000001010001100001, 此矩阵的秩为4, 其第2行和第3行是已知向量.9. 求下列矩阵的秩, 并求一个最高阶非零子式: (1)−−−443112112013;解−−−443112112013(下一步: r 1↔r 2 ~. )−−−443120131211(下一步: r 2−3r 1, r 3−r 1 ~. )−−−−564056401211(下一步: r 3−r 2 ~. )−−−000056401211, 矩阵的2秩为, 41113−=−是一个最高阶非零子式.(2)−−−−−−−815073*********;解−−−−−−−815073*********(下一步: r 1−r 2, r 2−2r 1, r 3−7r 1 ~. )−−−−−−15273321059117014431(下一步: r 3−3r 2~. )−−−−0000059117014431, 矩阵的秩是2, 71223−=−是一个最高阶非零子式.(3)−−−02301085235703273812. 解−−−02301085235703273812(下一步: r 1−2r 4, r 2−2r 4, r 3−3r 4~. )−−−−−−023*********63071210(下一步: r 2+3r 1, r 3+2r 1~. )−0230114000016000071210(下一步: r 2÷16r 4, r 3−16r 2. )~−02301000001000071210 ~−00000100007121002301, 矩阵的秩为3, 070023085570≠=−是一个最高阶非零子式.10. 设A 、B 都是m ×n 矩阵, 证明A ~B 的充分必要条件是R (A )=R (B ).证明根据定理3, 必要性是成立的.充分性. 设R (A )=R (B ), 则A 与B 的标准形是相同的. 设A 与B 的标准形为D , 则有A ~D , D ~B .由等价关系的传递性, 有A ~B .11. 设−−−−=32321321k k k A , 问k 为何值, 可使(1)R (A )=1; (2)R (A )=2; (3)R (A )=3.解 −−−−=32321321k k k A+−−−−−)2)(1(0011011 ~k k k k k r . (1)当k =1时, R (A )=1; (2)当k =−2且k ≠1时, R (A )=2;(3)当k ≠1且k ≠−2时, R (A )=3.12. 求解下列齐次线性方程组: (1) =+++=−++=−++02220202432143214321x x x x x x x x x x x x ;解对系数矩阵A 进行初等行变换, 有 A = −−212211121211~ −−−3/410013100101,于是 ==−==4443424134334x x x x x x x x ,故方程组的解为−= 1343344321k x x x x (k 为任意常数).(2) =−++=−−+=−++05105036302432143214321x x x x x x x x x x x x ;解对系数矩阵A 进行初等行变换, 有 A = −−−−5110531631121~−000001001021,于是 ===+−=4432242102x x x xx x x x ,故方程组的解为+−= 10010*********k k x x x x (k 1, k 2 (3)为任意常数).=−+−=+−+=−++=+−+07420634072305324321432143214321x x x x x x x x x x x x x x x x ;解对系数矩阵A 进行初等行变换, 有 A =−−−−−7421631472135132~1000010000100001,于是 ====0004321x x x x ,故方程组的解为 ====00004321x x x x .(4) =++−=+−+=−+−=+−+03270161311402332075434321432143214321x x x x x x x x x x x x x x x x .解对系数矩阵A 进行初等行变换, 有 A =−−−−−3127161311423327543~−−000000001720171910171317301,于是 ==−=−=4433432431172017191713173x x x x x x x xx x ,故方程组的解为−−+= 1017201713011719173214321k k x x x x (k 1, k 2为任意常数).13. 求解下列非齐次线性方程组: (1) =+=+−=−+83111021322421321321x x x x x x x x ;解对增广矩阵B 进行初等行变换, 有。

《线性代数》课后习题答案

《线性代数》课后习题答案第一章行列式习题1.11. 证明：（1）首先证明)3(Q 是数域。

因为)3(Q Q ?，所以)3(Q 中至少含有两个复数。

任给两个复数)3(3,32211Q b a b a ∈++，我们有3)()3()3)(3(3)()()3()3(3)()()3()3(21212121221121212211212122 11b a a b b b a a b a b a b b a a b a b a b b a a b a b a +++=++-+-=+-++++=+++。

因为Q 是数域，所以有理数的和、差、积仍然为有理数，所以)3(3)()3()3)(3()3(3)()()3()3()3(3)()()3()3(2121212122112121221 121212211Q b a a b b b a a b a b a Q b b a a b a b a Q b b a a b a b a ∈+++=++∈-+-=+-+∈+++=+++。

如果0322≠+b a ，则必有22,b a 不同时为零，从而0322≠-b a 。

又因为有理数的和、差、积、商仍为有理数，所以)3(33)(3)3()3)(3()3)(3(332222212122222121222222112211Q b a b a a b b a b b a a b a b a b a b a b a b a ∈--+--=-+-+=++。

综上所述，我们有)3(Q 是数域。

（2）类似可证明)(p Q 是数域，这儿p 是一个素数。

（3）下面证明：若q p ,为互异素数，则)()(q Q p Q ?。

（反证法）如果)()(q Qp Q ?，则q b a p Q b a +=?∈?,，从而有q ab qb a p p 2)()(222++==。

由于上式左端是有理数，而q 是无理数，所以必有02=q ab 。

所以有0=a 或0=b 。

线性代数课后习题答案全解.pdf

第一章行列式1. 利用对角线法则计算下列三阶行列式: (1)381141102−−−;解 381141102−−−=2×(−4)×3+0×(−1)×(−1)+1×1×8 −0×1×3−2×(−1)×8−1×(−4)×(−1) =−24+8+16−4=−4. (2)b a c a c b cb a ;解 ba c a cb cb a=acb +bac +cba −bbb −aaa −ccc =3abc −a 3−b 3−c 3. (3)222111c b a c b a ;解 222111c b a c b a=bc 2+ca 2+ab 2−ac 2−ba 2−cb =(a −b )(b −c )(c −a ). 2(4)y x y x x y x y yx y x +++.解 yx y x x y x y yx y x +++=x (x +y )y +yx (x +y )+(x +y )yx −y 3−(x +y )3−x =3xy (x +y )−y 3 3−3x 2 y −x 3−y 3−x =−2(x 3 3+y 3 2. 按自然数从小到大为标准次序, 求下列各排列的逆序数:).(1)1 2 3 4; 解逆序数为0 (2)4 1 3 2;解逆序数为4: 41, 43, 42, 32. (3)3 4 2 1;解逆序数为5: 3 2, 3 1, 4 2, 4 1, 2 1. (4)2 4 1 3;解逆序数为3: 2 1, 4 1, 4 3. (5)1 3 ⋅ ⋅ ⋅ (2n −1) 2 4 ⋅ ⋅ ⋅ (2n );解逆序数为2)1(−n n : 3 2 (1个) 5 2, 5 4(2个) 7 2, 7 4, 7 6(3个)⋅ ⋅ ⋅ ⋅ ⋅ ⋅(2n −1)2, (2n −1)4, (2n −1)6, ⋅ ⋅ ⋅, (2n −1)(2n −2) (n −1个)(6)1 3 ⋅ ⋅ ⋅ (2n −1) (2n ) (2n −2) ⋅ ⋅ ⋅ 2. 解逆序数为n (n −1) : 3 2(1个) 5 2, 5 4 (2个) ⋅ ⋅ ⋅ ⋅ ⋅ ⋅(2n −1)2, (2n −1)4, (2n −1)6, ⋅ ⋅ ⋅, (2n −1)(2n −2) (n −1个) 4 2(1个) 6 2, 6 4(2个) ⋅ ⋅ ⋅ ⋅ ⋅ ⋅(2n )2, (2n )4, (2n )6, ⋅ ⋅ ⋅, (2n )(2n −2) (n −1个) 3. 写出四阶行列式中含有因子a 11a 23 解含因子a 的项. 11a 23(−1)的项的一般形式为t a 11a 23a 3r a 4s 其中rs 是2和4构成的排列, 这种排列共有两个, 即24和42. ,所以含因子a 11a 23 (−1)的项分别是t a 11a 23a 32a 44=(−1)1a 11a 23a 32a 44=−a 11a 23a 32a 44 (−1), t a 11a 23a 34a 42=(−1)2a 11a 23a 34a 42=a 11a 23a 34a 42 4. 计算下列各行列式:.(1)71100251020214214; 解 71100251020214214010014231020211021473234−−−−−======c c c c 34)1(143102211014+−×−−−= 143102211014−−=01417172001099323211=−++======c c c c .(2)2605232112131412−; 解 2605232112131412−26053212213041224−−=====c c 041203212213041224−−=====r r 0000003212213041214=−−=====r r . (3)efcf bf de cd bd aeac ab −−−;解 ef cf bf de cd bd ae ac ab −−−ec b e c b ec b adf −−−=abcdef adfbce 4111111111=−−−=.(4)dc b a 100110011001−−−. 解d c b a 100110011001−−−dc b aab ar r 10011001101021−−−++===== d c a ab 101101)1)(1(12−−+−−=+01011123−+−++=====cd c ada ab dc ccdad ab +−+−−=+111)1)(1(23=abcd +ab +cd +ad +1. 5. 证明:(1)1112222b b a a b ab a +=(a −b )3 证明;1112222b b a a b ab a +00122222221213a b a b a a b a ab a c c c c −−−−−−=====ab a b a b a ab 22)1(22213−−−−−=+21))((a b a a b a b +−−==(a −b )3 (2) . y x z x z y zy x b a bz ay by ax bx az by ax bx az bz ay bx az bz ay by ax )(33+=+++++++++;证明bzay by ax bx az by ax bx az bz ay bxaz bz ay by ax +++++++++bz ay by ax x by ax bx az z bxaz bz ay y b bz ay by ax z by ax bx az y bx az bz ay x a +++++++++++++=bz ay y x by ax x z bxaz z y b y by ax z x bx az y z bz ay x a +++++++=22z y x y x z xz y b y x z x z y z y x a 33+=y x z x z y zy x b y x z x z y z y x a 33+=y x z x z y zy x b a )(33+=.(3)0)3()2()1()3()2()1()3()2()1()3()2()1(2222222222222222=++++++++++++d d d d c c c c b b b b a a a a ; 证明 2222222222222222)3()2()1()3()2()1()3()2()1()3()2()1(++++++++++++d d d d c c c c b b b b a a a a (c 4−c 3, c 3−c 2, c 2−c 1 得) 5232125232125232125232122222++++++++++++=d d d d c c c c b b b b a a a a (c 4−c 3, c 3−c 2得)022122212221222122222=++++=d d c c b b a a . (4)444422221111d c b a d c b a d c b a =(a −b )(a −c )(a −d )(b −c )(b −d )(c −d )(a +b +c +d ); 证明 444422221111d c b a d c b a d c b a )()()(0)()()(001111222222222a d d a c c a b b a d d a c c a b b ad a c a b −−−−−−−−−=)()()(111))()((222a d d a c c a b b dc b ad a c a b +++−−−= ))(())((00111))()((a b d b d d a b c b c c bd b c a d a c a b ++−++−−−−−−= )()(11))()()()((a b d d a b c c b d b c a d a c a b ++++−−−−−= =(a −b )(a −c )(a −d )(b −c )(b −d )(c −d )(a +b +c +d ). (5)12211 000 00 1000 01a x a a a a x x xn n n+⋅⋅⋅−⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅−⋅⋅⋅−−− =x n +a 1x n −1+ ⋅ ⋅ ⋅ +a n −1x +a n .证明用数学归纳法证明.当n =2时, 2121221a x a x a x a x D ++=+−=, 命题成立. 假设对于(n −1)阶行列式命题成立, 即 D n −1=x n −1+a 1 x n −2+ ⋅ ⋅ ⋅ +a n −2x +a n −1则D , n 按第一列展开, 有 11100 100 01)1(11−⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅−⋅⋅⋅−−+=+−x x a xD D n n n n =xD n −1+a n =x n +a 1x n −1+ ⋅ ⋅ ⋅ +a n −1x +a n 因此, 对于n 阶行列式命题成立. .6. 设n 阶行列式D =det(a ij ), 把D 上下翻转、或逆时针旋转90°、或依副对角线翻转, 依次得n nn n a a a a D 11111 ⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅=, 11112 n nnn a a a a D ⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅= , 11113 a a a a D n n nn ⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅=,证明D D D n n 2)1(21)1(−−==, D 3 证明因为D =det(a =D .ij ), 所以 nnn n n n nnnn a a a a a a a a a a D 2211111111111 )1( ⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅−=⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅=−⋅⋅⋅=⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅−−=−− )1()1(331122111121nnn n nn n n a a a a a a a a D D n n n n 2)1()1()2( 21)1()1(−−+−+⋅⋅⋅++−=−=.同理可证 nnn n n n a a a a D ⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅−=− )1(11112)1(2D D n n T n n 2)1(2)1()1()1(−−−=−=. D D D D D n n n n n n n n =−=−−=−=−−−−)1(2)1(2)1(22)1(3)1()1()1()1(.7. 计算下列各行列式(D k (1)为k 阶行列式): aa D n 1 1⋅⋅⋅=, 其中对角线上元素都是a , 未写出的元素都是0; 解 aa a a a D n 010 000 00 000 0010 00⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅=(按第n 行展开) )1()1(10 000 00 000 0010 000)1(−×−+⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅−=n n n aa a )1()1(2 )1(−×−⋅⋅⋅⋅−+n n n a a an n n n n a a a+⋅⋅⋅−⋅−=−−+)2)(2(1)1()1(=a n −a n −2=a n −2(a 2−1).(2)xa aa x a a a xD n ⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅= ; 解将第一行乘(−1)分别加到其余各行, 得 ax x a ax x a a x x a aa a x D n −−⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅−−⋅⋅⋅−−⋅⋅⋅=000 0 00 0, 再将各列都加到第一列上, 得ax ax a x aaa a n x D n −⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅−⋅⋅⋅−⋅⋅⋅−+=0000 0 000 00 )1(=[x +(n −1)a ](x −a )n −1 (3). 111 1 )( )1()( )1(1111⋅⋅⋅−⋅⋅⋅⋅⋅⋅⋅⋅⋅−⋅⋅⋅⋅⋅⋅−⋅⋅⋅−−⋅⋅⋅−=−−−+n a a a n a a a n a a a D n n n n nn n ; 解根据第6题结果, 有 nnn n n n n n n n a a a n a a a n a a aD )( )1()( )1( 11 11)1(1112)1(1−⋅⋅⋅−−⋅⋅⋅⋅⋅⋅⋅⋅⋅−⋅⋅⋅⋅⋅⋅−⋅⋅⋅−⋅⋅⋅−=−−−++此行列式为范德蒙德行列式.∏≥>≥++++−−+−−=112)1(1)]1()1[()1(j i n n n n j a i a D∏≥>≥++−−−=112)1()]([)1(j i n n n j i∏≥>≥++⋅⋅⋅+−++−⋅−⋅−=1121)1(2)1()()1()1(j i n n n n n j i∏≥>≥+−=11)(j i n j i .(4)nnnnn d c d c b a b a D ⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅=11112; 解nnnnn d c d c b a b a D ⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅=11112(按第1行展开) nn n n n nd d c d c b a b a a 00011111111−−−−⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅=0)1(1111111112c d c d c b a b a b nn n n n nn −−−−+⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅−+. 再按最后一行展开得递推公式D 2n =a n d n D 2n −2−b n c n D 2n −2, 即D 2n =(a n d n −b n c n )D 2n −2于是 . ∏=−=ni i i i i n D c b d a D 222)(.而 111111112c b d a d c b a D −==,所以 ∏=−=ni i i i i n c b d a D 12)(.(5) D =det(a ij ), 其中a ij 解 a =|i −j |; ij =|i −j |, 043214 01233 10122 21011 3210)det(⋅⋅⋅−−−−⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅−⋅⋅⋅−⋅⋅⋅−⋅⋅⋅−⋅⋅⋅==n n n n n n n n a D ij n 04321 1 11111 11111 11111 1111 2132⋅⋅⋅−−−−⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅−−−−⋅⋅⋅−−−⋅⋅⋅−−⋅⋅⋅−−⋅⋅⋅−=====n n n n r r r r15242321 0 22210 02210 00210 0001 1213−⋅⋅⋅−−−−⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅−−−−⋅⋅⋅−−−⋅⋅⋅−−⋅⋅⋅−+⋅⋅⋅+=====n n n n n c c c c =(−1)n −1(n −1)2n −2 (6).nn a a a D +⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅+⋅⋅⋅+=1 11 1 1111121, 其中a 1a 2 ⋅ ⋅ ⋅ a n≠0.解nn a a a D +⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅+⋅⋅⋅+=1 11 1 1111121 nn n n a a a a a a a a a c c c c +−⋅⋅⋅−⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅−⋅⋅⋅−⋅⋅⋅−⋅⋅⋅−=====−−100001 000 100 0100 0100 0011332212132 1111312112111000011 000 00 11000 01100 001 −−−−−−+−⋅⋅⋅−⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅−⋅⋅⋅−⋅⋅⋅⋅⋅⋅=nn n a a a a a a a a∑=−−−−−−+⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅=n i i n n a a a a a a a a 1111131******** 00010 000 00 10000 01000 001)11)((121∑=+=ni i n a a a a .8. 用克莱姆法则解下列方程组: (1) =+++−=−−−−=+−+=+++01123253224254321432143214321x x x x x x x x x x x x x x x x ;解因为 14211213513241211111−=−−−−=D , 142112105132412211151−=−−−−−−=D , 284112035122412111512−=−−−−−=D , 426110135232422115113−=−−−−=D , 14202132132212151114=−−−−−=D , 所以 111==D D x , 222==D Dx , 333==DD x , 144−==D D x .(2)=+=++=++=++=+150650650651655454343232121x x x x x x x x x x x x x .解因为 665510006510006510065100065==D , 15075100165100065100065000611==D , 114551010651000650000601000152−==D , 703511650000601000051001653==D , 39551601000051000651010654−==D , 2121100005100065100651100655==D , 所以66515071=x , 66511452−=x , 6657033=x , 6653954−=x , 6652124=x .9. 问λ, µ取何值时, 齐次线性方程组 =++=++=++0200321321321x x x x x x x x x µµλ有非零解？解系数行列式为µλµµµλ−==1211111D .令D =0, 得 µ=0或λ=1.于是, 当µ=0或λ=1时该齐次线性方程组有非零解.10. 问λ取何值时, 齐次线性方程组 =−++=+−+=+−−0)1(0)3(2042)1(321321321x x x x x x x x x λλλ有非零解？解系数行列式为λλλλλλλ−−+−−=−−−−=101112431111132421D=(1−λ)3 =(1−λ)+(λ−3)−4(1−λ)−2(1−λ)(−3−λ) 3+2(1−λ)2 令D =0, 得+λ−3. λ=0, λ=2或λ=3.于是, 当λ=0, λ=2或λ=3时, 该齐次线性方程组有非零解.第二章矩阵及其运算1. 已知线性变换:++=++=++=3213321232113235322y y y x y y y x y y y x , 求从变量x 1, x 2, x 3到变量y 1, y 2, y 3 解由已知:的线性变换.= 221321323513122y y y x x x ,故= −3211221323513122x x x y y y−−−−=321423736947y y y ,−+=−+=+−−=321332123211423736947x x x y x x x y x x x y .2. 已知两个线性变换++=++−=+=32133212311542322y y y x y y y x y y x ,+−=+=+−=323312211323z z y z z y z z y , 求从z 1, z 2, z 3到x 1, x 2, x 3 解由已知的线性变换.−= 221321514232102y y y x x x−− −=321310102013514232102z z z−−−−=321161109412316z z z ,所以有 +−−=+−=++−=3213321232111610941236z z z x z z z x z z z x .3. 设 −−=111111111A ,−−=150421321B , 求3AB −2A 及A T 解 B .−−− −− −−=−1111111112150421321111111111323A AB−−−−= −−− −=2294201722213211111111120926508503,−= −− −−=092650850150421321111111111B A T.4. 计算下列乘积: (1)−127075321134;解 −127075321134 ×+×+××+×−+××+×+×=102775132)2(71112374=49635.(2)123)321(;解123)321(=(1×3+2×2+3×1)=(10).(3))21(312−;解 )21(312−×−××−××−×=23)1(321)1(122)1(2−−−=632142. (4)−−−−20413121013143110412 ; 解−−− −20413121013143110412 −−−=6520876. (5)321332313232212131211321)(x x x a a a a a a a a a x x x ;解321332313232212131211321)(x x x a a a a a a a a a x x x=(a 11x 1+a 12x 2+a 13x 3 a 12x 1+a 22x 2+a 23x 3a 13x 1+a 23x 2+a 33x 3321x x x )322331132112233322222111222x x a x x a x x a x a x a x a +++++=.5. 设 =3121A ,=2101B , 问: (1)AB =BA 吗? 解 AB ≠BA . 因为=6443AB ,=8321BA , 所以AB ≠BA .(2)(A +B )2=A 2+2AB +B 2 解 (A +B )吗? 2≠A 2+2AB +B 2 因为.=+5222B A ,=+52225222)(2B A=2914148,但 + +=++43011288611483222B AB A=27151610,所以(A +B )2≠A 2+2AB +B 2 (3)(A +B )(A −B )=A . 2−B 2 解 (A +B )(A −B )≠A 吗? 2−B 2 因为.=+5222B A ,=−1020B A ,==−+906010205222))((B A B A ,而= −=−718243011148322B A ,故(A +B )(A −B )≠A 2−B 2 6. 举反列说明下列命题是错误的:.(1)若A 2 解取=0, 则A =0;=0010A , 则A 2 (2)若A =0, 但A ≠0. 2 解取=A , 则A =0或A =E ;=0011A , 则A 2 (3)若AX =AY , 且A ≠0, 则X =Y .=A , 但A ≠0且A ≠E . 解取=0001A , −=1111X ,=1011Y , 则AX =AY , 且A ≠0, 但X ≠Y .7. 设=101λA , 求A 2, A 3, ⋅ ⋅ ⋅, A k 解 . ==12011011012λλλA , ===1301101120123λλλA A A , ⋅ ⋅ ⋅ ⋅ ⋅ ⋅,=101λk A k . 8. 设=λλλ001001A , 求A k 解首先观察. =λλλλλλ0010010010012A=222002012λλλλλ,=⋅=3232323003033λλλλλλA A A ,=⋅=43423434004064λλλλλλA A A ,=⋅=545345450050105λλλλλλA A A , ⋅ ⋅ ⋅ ⋅ ⋅ ⋅,=k A k k k k k k k k k k λλλλλλ0002)1(121−−−−. 用数学归纳法证明:当k =2时, 显然成立.假设k 时成立，则k +1时，−=⋅=−−−+λλλλλλλλλ0010010002)1(1211k k k k k k k k k k k k A A A+++=+−+−−+11111100)1(02)1()1(k k k k k k k k k k λλλλλλ, 由数学归纳法原理知:−=−−−k k k k k k k k k k k A λλλλλλ0002)1(121. 9. 设A , B 为n 阶矩阵，且A 为对称矩阵，证明B T 证明因为A AB 也是对称矩阵.T (B =A , 所以T AB )T =B T (B T A )T =B T A T B =B T 从而B AB ,T 10. 设A , B 都是n 阶对称矩阵，证明AB 是对称矩阵的充分必要条件是AB =BA .AB 是对称矩阵.证明充分性: 因为A T =A , B T (AB )=B , 且AB =BA , 所以 T =(BA )T =A T B T 即AB 是对称矩阵.=AB ,必要性: 因为A T =A , B T =B , 且(AB )T AB =(AB )=AB , 所以T =B T A T 11. 求下列矩阵的逆矩阵:=BA .(1)5221; 解=5221A . |A |=1, 故A −1 存在. 因为−−= =1225*22122111A A A A A ,故 *||11A A A =−−−=1225. (2)−θθθθcos sin sin cos ; 解−=θθθθcos sin sin cos A . |A |=1≠0, 故A −1 存在. 因为−= =θθθθcos sin sin cos *22122111A A A A A , 所以 *||11A A A =−−=θθθθcos sin sin cos . (3)−−−145243121; 解−−−=145243121A . |A |=2≠0, 故A −1 存在. 因为−−−−−= =214321613024*332313322212312111A A A A A A A A A A , 所以 *||11A A A =−−−−−−=1716213213012. (4)n a a a 0021(a 1a 2⋅ ⋅ ⋅a n ≠0) .解=n a a a A 0021, 由对角矩阵的性质知=−n a a a A 10011211 . 12. 解下列矩阵方程:(1) −=12643152X ; 解 −=−126431521X − −−=12642153 −=80232. (2) −=−−234311*********X ; 解 1111012112234311−−− −=X−−− −=03323210123431131 −−−=32538122. (3) −= − −101311022141X ;解 11110210132141−− − − −=X− −=210110131142121 =21010366121=04111. (4)−−−= 021102341010100001100001010X . 解 11010100001021102341100001010−−−−− =X −−− =010100001021102341100001010 −−−=201431012. 13. 利用逆矩阵解下列线性方程组:(1) =++=++=++3532522132321321321x x x x x x x x x ; 解方程组可表示为= 321153522321321x x x , 故 = = −0013211535223211321x x x ,从而有 ===001321x x x . (2) =−+=−−=−−05231322321321321x x x x x x x x x . 解方程组可表示为=−−−−−012523312111321x x x , 故 =−−−−−= −3050125233121111321x x x , 故有 ===305321x x x . 14. 设A k =O (k 为正整数), 证明(E −A )−1=E +A +A 2+⋅ ⋅ ⋅+A k −1 证明因为A . k =O , 所以E −A k E −A =E . 又因为k =(E −A )(E +A +A 2+⋅ ⋅ ⋅+A k −1所以 (E −A )(E +A +A ),2+⋅ ⋅ ⋅+A k −1由定理2推论知(E −A )可逆, 且)=E ,(E −A )−1=E +A +A 2+⋅ ⋅ ⋅+A k −1.证明一方面, 有E =(E −A )−1 另一方面, 由A (E −A ).k E =(E −A )+(A −A =O , 有2)+A 2−⋅ ⋅ ⋅−A k −1+(A k −1−A k )=(E +A +A 2+⋅ ⋅ ⋅+A k −1故 (E −A ))(E −A ),−1(E −A )=(E +A +A 2+⋅ ⋅ ⋅+A k −1两端同时右乘(E −A ))(E −A ),−1 (E −A ), 就有−1(E −A )=E +A +A 2+⋅ ⋅ ⋅+A k −1.15. 设方阵A 满足A 2−A −2E =O , 证明A 及A +2E 都可逆, 并求A −1及(A +2E )−1 证明由A .2 A −A −2E =O 得2或 −A =2E , 即A (A −E )=2E ,E E A A =−⋅)(21, 由定理2推论知A 可逆, 且)(211E A A −=−. 由A 2 A −A −2E =O 得2或 −A −6E =−4E , 即(A +2E )(A −3E )=−4E ,E A E E A =−⋅+)3(41)2( 由定理2推论知(A +2E )可逆, 且)3(41)2(1A E E A −=+−.证明由A 2−A −2E =O 得A 2 |A −A =2E , 两端同时取行列式得 2即 |A ||A −E |=2,−A |=2,故 |A |≠0,所以A 可逆, 而A +2E =A 2, |A +2E |=|A 2|=|A |2由 A ≠0, 故A +2E 也可逆. 2 ⇒A −A −2E =O ⇒A (A −E )=2E−1A (A −E )=2A −1)(211E A A −=−E ⇒,又由 A 2 ⇒ (A +2E )(A −3E )=−4 E ,−A −2E =O ⇒(A +2E )A −3(A +2E )=−4E所以 (A +2E )−1(A +2E )(A −3E )=−4(A +2 E )−1 ,)3(41)2(1A E E A −=+−.16. 设A 为3阶矩阵, 21||=A , 求|(2A )−1 解因为−5A *|.*||11A A A =−, 所以 |||521||*5)2(|111−−−−=−A A A A A |2521|11−−−=A A=|−2A −1|=(−2)3|A −1|=−8|A |−1 17. 设矩阵A 可逆, 证明其伴随阵A *也可逆, 且(A *)=−8×2=−16.−1=(A −1 证明由)*.*||11A A A =−, 得A *=|A |A −1 |A *|=|A |, 所以当A 可逆时, 有n |A −1|=|A |n −1从而A *也可逆.≠0,因为A *=|A |A −1 (A *), 所以−1=|A |−1又A .*)(||)*(||1111−−−==A A A A A , 所以(A *)−1=|A |−1A =|A |−1|A |(A −1)*=(A −1 18. 设n 阶矩阵A 的伴随矩阵为A *, 证明:)*.(1)若|A |=0, 则|A *|=0;(2)|A *|=|A |n −1 证明.(1)用反证法证明. 假设|A *|≠0, 则有A *(A *)−1 A =A A *(A *)=E , 由此得 −1=|A |E (A *)−1所以A *=O , 这与|A *|≠0矛盾，故当|A |=0时, 有|A *|=0.=O ,(2)由于*||11A A A =−, 则AA *=|A |E , 取行列式得到 |A ||A *|=|A |n 若|A |≠0, 则|A *|=|A |.n −1 若|A |=0, 由(1)知|A *|=0, 此时命题也成立.;因此|A *|=|A |n −1.19. 设−=321011330A , AB =A +2B , 求B . 解由AB =A +2E 可得(A −2E )B =A , 故− −−−=−=−−321011330121011332)2(11A E A B −=011321330. 20. 设 =101020101A , 且AB +E =A 2+B , 求B .解由AB +E =A 2 (A −E )B =A +B 得 2即 (A −E )B =(A −E )(A +E ).−E , 因为01001010100||≠−==−E A , 所以(A −E )可逆, 从而=+=201030102E A B .21. 设A =diag(1, −2, 1), A *BA =2BA −8E , 求B . 解由A *BA =2BA −8E 得 (A *−2E )BA =−8E , B =−8(A *−2E )−1A =−8[A (A *−2E )]−1 =−8(AA *−2A )−1 =−8(|A |E −2A )−1 =−8(−2E −2A )−1 =4(E +A )−1 =4[diag(2, −1, 2)]−1−1)21 ,1 ,21(diag 4−==2diag(1, −2, 1).22. 已知矩阵A 的伴随阵−=8030010100100001*A , 且ABA −1=BA −1+3E , 求B .解由|A *|=|A |3 由ABA =8, 得|A |=2. −1=BA −1 AB =B +3A ,+3E 得 B =3(A −E )−1A =3[A (E −A −1)]−1 A 11*)2(6*)21(3−−−=−=A E A E−=−−=−1030060600600006603001010010000161. 23. 设P −1 −−=1141P AP =Λ, 其中,−=Λ2001, 求A 11 解由P . −1AP =Λ, 得A =P ΛP −1, 所以A 11= A =P Λ11P −1 |P |=3, .−=1141*P ,−−=−1141311P ,而−= −=Λ11111120 012001,故−− −−−=31313431200111411111A −−=68468327322731. 24. 设AP =P Λ, 其中−−=111201111P ,−=Λ511,求ϕ(A )=A 8(5E −6A +A 2 解 ϕ(Λ)=Λ). 8(5E −6Λ+Λ2 =diag(1,1,5)8)[diag(5,5,5)−diag(−6,6,30)+diag(1,1,25)]=diag(1,1,58 ϕ(A )=P ϕ(Λ)P )diag(12,0,0)=12diag(1,0,0).−1 *)(||1P P P Λ=ϕ−−−−−− −−−=1213032220000000011112011112=1111111114.25. 设矩阵A 、B 及A +B 都可逆, 证明A −1+B −1 证明因为也可逆, 并求其逆阵.A −1(A +B )B −1=B −1+A −1=A −1+B −1而A ,−1(A +B )B −1是三个可逆矩阵的乘积, 所以A −1(A +B )B −1可逆, 即A −1+B −1 (A 可逆.−1+B −1)−1=[A −1(A +B )B −1]−1=B (A +B )−1 26. 计算A .−−−30003200121013013000120010100121. 解设 =10211A , =30122A , −=12131B ,−−=30322B ,则 2121B O B E A O E A+=222111B A O B B A A ,而 −= −−+−=+4225303212131021211B B A ,−−= −− =90343032301222B A , 所以 2121B O B E A O E A +=222111B A O B B A A−−−=9000340042102521, 即−−−30003200121013013000120010100121−−−=9000340042102521. 27. 取==−==1001D C B A , 验证|||||||| D C B A D C B A ≠.解 4100120021010*********0021010010110100101==−−=−−=D C B A , 而 01111|||||||| ==D C B A ,故 ||||||||D C B A D C B A ≠. 28. 设 −=22023443O O A , 求|A 8|及A 4解令. −=34431A ,=22022A , 则=21A O O A A ,故 8218=A O O A A=8281A O O A ,1682818281810||||||||||===A A A A A .= =464444241422025005O O A O O A A . 29. 设n 阶矩阵A 及s 阶矩阵B 都可逆, 求 (1)1−O B A O ; 解设 =−43211C C C C O B A O , 则O B A O 4321C C C C = =s n E O O E BC BC AC AC 2143. 由此得====s n EBC OBC O AC E AC 2143⇒ ====−−121413B C O C O C A C ,所以= −−−O A B O O B A O 111. (2)1−B C O A . 解设 =−43211D D D D B C O A , 则 = ++= s nE O O E BD CD BD CD AD AD D D D D B C O A 4231214321.由此得=+=+==s nEBD CD O BD CD O AD E AD 423121⇒ =−===−−−−14113211B D CA B D O D A D ,所以−= −−−−−11111B CA B O A BC O A . 30. 求下列矩阵的逆阵: (1)2500380000120025; 解设 =1225A , =2538B , 则−−= =−−5221122511A ,−−==−−8532253811B .于是 −−−−= = =−−−−850032000052002125003800001200251111B A B A .(2)4121031200210001. 解设 =2101A ,=4103B ,=2112C , 则−= =−−−−−−1111114121031200210001B CA B O A BC O A−−−−−=411212458103161210021210001.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;解−−340313021201(下一步: r 2+(−2)r 1, r 3+(−3)r 1 ~. )−−−020*********(下一步: r 2÷(−1), r 3 ~÷(−2). )−−010*********(下一步: r 3−r 2 ~. )−−300031001201(下一步: r 3 ~÷3. )−−100031001201(下一步: r 2+3r 3 ~. )−100001001201(下一步: r 1+(−2)r 2, r 1+r 3 ~. )100001000001.(2)−−−−174034301320;解−−−−174034301320(下一步: r 2×2+(−3)r 1, r 3+(−2)r 1 ~. )−−−310031001320(下一步: r 3+r 2, r 1+3r 2 ~. )0000310010020(下一步: r 1 ~÷2. )000031005010.(3)−−−−−−−−−12433023221453334311;解−−−−−−−−−12433023221453334311(下一步: r 2−3r 1, r 3−2r 1, r 4−3r 1~. )−−−−−−−−1010500663008840034311(下一步: r 2÷(−4), r 3÷(−3) , r 4~÷(−5). )−−−−−22100221002210034311(下一步: r 1−3r 2, r 3−r 2, r 4−r 2~. )−−−00000000002210032011.(4)−−−−−−34732038234202173132. 解−−−−−−34732038234202173132(下一步: r 1−2r 2, r 3−3r 2, r 4−2r 2~. )−−−−−1187701298804202111110(下一步: r 2+2r 1, r 3−8r 1, r 4−7r 1 ~. )−−41000410002020111110(下一步: r 1↔r 2, r 2×(−1), r 4−r 3~. )−−−−00000410001111020201(下一步: r 2+r 3~. )−−00000410003011020201. 2. 设= 987654321100010101100001010A , 求A .解100001010是初等矩阵E (1, 2), 其逆矩阵就是其本身.100010101是初等矩阵E (1, 2(1)), 其逆矩阵是E (1, 2(−1))−=100010101.− =100010101987654321100001010A= − =287221254100010101987321654.3. 试利用矩阵的初等变换, 求下列方阵的逆矩阵: (1)323513123;解 100010001323513123~−−−101011001200410123~ −−−−1012002110102/102/3023~−−−−2/102/11002110102/922/7003~−−−−2/102/11002110102/33/26/7001故逆矩阵为−−−−21021211233267.(2)−−−−−1210232112201023.解−−−−−10000100001000011210232112201023~−−−−00100301100001001220594012102321~−−−−−−−−20104301100001001200110012102321~ −−−−−−−106124301100001001000110012102321 ~−−−−−−−−−−10612631110`1022111000010000100021 ~−−−−−−−106126311101042111000010000100001故逆矩阵为−−−−−−−10612631110104211. 4. (1)设 −−=113122214A ,−−=132231B , 求X 使AX =B ;解因为−−−−=132231 113122214) ,(B A−−412315210 100010001 ~r ,所以−−==−4123152101B A X .(2)设−−−=433312120A , −=132321B , 求X 使XA =B . 解考虑A T X T =B T . 因为−−−−=134313*********) ,(T T B A−−−411007101042001 ~r ,所以−−−==−417142)(1T T T B A X ,从而−−−==−4741121BA X . 5. 设−−−=101110011A , AX =2X +A , 求X .解原方程化为(A −2E )X =A . 因为−−−−−−−−−=−101101110110011011) ,2(A E A−−−011100101010110001~,所以−−−=−=−011101110)2(1A E A X .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. 例如,=010*********A , R (A )=3.0000是等于0的2阶子式, 010001000是等于0的3阶子式. 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 ).这是因为B 的非零子式必是A 的非零子式, 故A 的秩不会小于B 的秩.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).解用已知向量容易构成一个有4个非零行的5阶下三角矩阵:−0000001000001010001100001, 此矩阵的秩为4, 其第2行和第3行是已知向量.9. 求下列矩阵的秩, 并求一个最高阶非零子式: (1)−−−443112112013;解−−−443112112013(下一步: r 1↔r 2 ~. )−−−443120131211(下一步: r 2−3r 1, r 3−r 1 ~. )−−−−564056401211(下一步: r 3−r 2 ~. )−−−000056401211, 矩阵的2秩为, 41113−=−是一个最高阶非零子式.(2)−−−−−−−815073*********;解−−−−−−−815073*********(下一步: r 1−r 2, r 2−2r 1, r 3−7r 1 ~. )−−−−−−15273321059117014431(下一步: r 3−3r 2~. )−−−−0000059117014431, 矩阵的秩是2, 71223−=−是一个最高阶非零子式.(3)−−−02301085235703273812. 解−−−02301085235703273812(下一步: r 1−2r 4, r 2−2r 4, r 3−3r 4~. )−−−−−−023*********63071210(下一步: r 2+3r 1, r 3+2r 1~. )−0230114000016000071210(下一步: r 2÷16r 4, r 3−16r 2. )~−02301000001000071210 ~−00000100007121002301, 矩阵的秩为3, 070023085570≠=−是一个最高阶非零子式.10. 设A 、B 都是m ×n 矩阵, 证明A ~B 的充分必要条件是R (A )=R (B ).证明根据定理3, 必要性是成立的.充分性. 设R (A )=R (B ), 则A 与B 的标准形是相同的. 设A 与B 的标准形为D , 则有A ~D , D ~B .由等价关系的传递性, 有A ~B .11. 设−−−−=32321321k k k A , 问k 为何值, 可使(1)R (A )=1; (2)R (A )=2; (3)R (A )=3.解 −−−−=32321321k k k A+−−−−−)2)(1(0011011 ~k k k k k r . (1)当k =1时, R (A )=1; (2)当k =−2且k ≠1时, R (A )=2;(3)当k ≠1且k ≠−2时, R (A )=3.12. 求解下列齐次线性方程组: (1) =+++=−++=−++02220202432143214321x x x x x x x x x x x x ;解对系数矩阵A 进行初等行变换, 有 A = −−212211121211~ −−−3/410013100101,于是 ==−==4443424134334x x x x x x x x ,故方程组的解为−= 1343344321k x x x x (k 为任意常数).(2) =−++=−−+=−++05105036302432143214321x x x x x x x x x x x x ;解对系数矩阵A 进行初等行变换, 有 A = −−−−5110531631121~−000001001021,于是 ===+−=4432242102x x x xx x x x ,故方程组的解为+−= 10010*********k k x x x x (k 1, k 2 (3)为任意常数).=−+−=+−+=−++=+−+07420634072305324321432143214321x x x x x x x x x x x x x x x x ;解对系数矩阵A 进行初等行变换, 有 A =−−−−−7421631472135132~1000010000100001,于是 ====0004321x x x x ,故方程组的解为 ====00004321x x x x .(4) =++−=+−+=−+−=+−+03270161311402332075434321432143214321x x x x x x x x x x x x x x x x .解对系数矩阵A 进行初等行变换, 有 A =−−−−−3127161311423327543~−−000000001720171910171317301,于是 ==−=−=4433432431172017191713173x x x x x x x xx x ,故方程组的解为−−+= 1017201713011719173214321k k x x x x (k 1, k 2为任意常数).13. 求解下列非齐次线性方程组: (1) =+=+−=−+83111021322421321321x x x x x x x x ;解对增广矩阵B 进行初等行变换, 有。

线性代数课后习题答案全解.pdf

第一章行列式1. 利用对角线法则计算下列三阶行列式: (1)381141102−−−;解 381141102−−−=2×(−4)×3+0×(−1)×(−1)+1×1×8 −0×1×3−2×(−1)×8−1×(−4)×(−1) =−24+8+16−4=−4. (2)b a c a c b cb a ;解 ba c a cb cb a=acb +bac +cba −bbb −aaa −ccc =3abc −a 3−b 3−c 3. (3)222111c b a c b a ;解 222111c b a c b a=bc 2+ca 2+ab 2−ac 2−ba 2−cb =(a −b )(b −c )(c −a ). 2(4)y x y x x y x y yx y x +++.解 yx y x x y x y yx y x +++=x (x +y )y +yx (x +y )+(x +y )yx −y 3−(x +y )3−x =3xy (x +y )−y 3 3−3x 2 y −x 3−y 3−x =−2(x 3 3+y 3 2. 按自然数从小到大为标准次序, 求下列各排列的逆序数:).(1)1 2 3 4; 解逆序数为0 (2)4 1 3 2;解逆序数为4: 41, 43, 42, 32. (3)3 4 2 1;解逆序数为5: 3 2, 3 1, 4 2, 4 1, 2 1. (4)2 4 1 3;解逆序数为3: 2 1, 4 1, 4 3. (5)1 3 ⋅ ⋅ ⋅ (2n −1) 2 4 ⋅ ⋅ ⋅ (2n );解逆序数为2)1(−n n : 3 2 (1个) 5 2, 5 4(2个) 7 2, 7 4, 7 6(3个)⋅ ⋅ ⋅ ⋅ ⋅ ⋅(2n −1)2, (2n −1)4, (2n −1)6, ⋅ ⋅ ⋅, (2n −1)(2n −2) (n −1个)(6)1 3 ⋅ ⋅ ⋅ (2n −1) (2n ) (2n −2) ⋅ ⋅ ⋅ 2. 解逆序数为n (n −1) : 3 2(1个) 5 2, 5 4 (2个) ⋅ ⋅ ⋅ ⋅ ⋅ ⋅(2n −1)2, (2n −1)4, (2n −1)6, ⋅ ⋅ ⋅, (2n −1)(2n −2) (n −1个) 4 2(1个) 6 2, 6 4(2个) ⋅ ⋅ ⋅ ⋅ ⋅ ⋅(2n )2, (2n )4, (2n )6, ⋅ ⋅ ⋅, (2n )(2n −2) (n −1个) 3. 写出四阶行列式中含有因子a 11a 23 解含因子a 的项. 11a 23(−1)的项的一般形式为t a 11a 23a 3r a 4s 其中rs 是2和4构成的排列, 这种排列共有两个, 即24和42. ,所以含因子a 11a 23 (−1)的项分别是t a 11a 23a 32a 44=(−1)1a 11a 23a 32a 44=−a 11a 23a 32a 44 (−1), t a 11a 23a 34a 42=(−1)2a 11a 23a 34a 42=a 11a 23a 34a 42 4. 计算下列各行列式:.(1)71100251020214214; 解 71100251020214214010014231020211021473234−−−−−======c c c c 34)1(143102211014+−×−−−= 143102211014−−=01417172001099323211=−++======c c c c .(2)2605232112131412−; 解 2605232112131412−26053212213041224−−=====c c 041203212213041224−−=====r r 0000003212213041214=−−=====r r . (3)efcf bf de cd bd aeac ab −−−;解 ef cf bf de cd bd ae ac ab −−−ec b e c b ec b adf −−−=abcdef adfbce 4111111111=−−−=.(4)dc b a 100110011001−−−. 解d c b a 100110011001−−−dc b aab ar r 10011001101021−−−++===== d c a ab 101101)1)(1(12−−+−−=+01011123−+−++=====cd c ada ab dc ccdad ab +−+−−=+111)1)(1(23=abcd +ab +cd +ad +1. 5. 证明:(1)1112222b b a a b ab a +=(a −b )3 证明;1112222b b a a b ab a +00122222221213a b a b a a b a ab a c c c c −−−−−−=====ab a b a b a ab 22)1(22213−−−−−=+21))((a b a a b a b +−−==(a −b )3 (2) . y x z x z y zy x b a bz ay by ax bx az by ax bx az bz ay bx az bz ay by ax )(33+=+++++++++;证明bzay by ax bx az by ax bx az bz ay bxaz bz ay by ax +++++++++bz ay by ax x by ax bx az z bxaz bz ay y b bz ay by ax z by ax bx az y bx az bz ay x a +++++++++++++=bz ay y x by ax x z bxaz z y b y by ax z x bx az y z bz ay x a +++++++=22z y x y x z xz y b y x z x z y z y x a 33+=y x z x z y zy x b y x z x z y z y x a 33+=y x z x z y zy x b a )(33+=.(3)0)3()2()1()3()2()1()3()2()1()3()2()1(2222222222222222=++++++++++++d d d d c c c c b b b b a a a a ; 证明 2222222222222222)3()2()1()3()2()1()3()2()1()3()2()1(++++++++++++d d d d c c c c b b b b a a a a (c 4−c 3, c 3−c 2, c 2−c 1 得) 5232125232125232125232122222++++++++++++=d d d d c c c c b b b b a a a a (c 4−c 3, c 3−c 2得)022122212221222122222=++++=d d c c b b a a . (4)444422221111d c b a d c b a d c b a =(a −b )(a −c )(a −d )(b −c )(b −d )(c −d )(a +b +c +d ); 证明 444422221111d c b a d c b a d c b a )()()(0)()()(001111222222222a d d a c c a b b a d d a c c a b b ad a c a b −−−−−−−−−=)()()(111))()((222a d d a c c a b b dc b ad a c a b +++−−−= ))(())((00111))()((a b d b d d a b c b c c bd b c a d a c a b ++−++−−−−−−= )()(11))()()()((a b d d a b c c b d b c a d a c a b ++++−−−−−= =(a −b )(a −c )(a −d )(b −c )(b −d )(c −d )(a +b +c +d ). (5)12211 000 00 1000 01a x a a a a x x xn n n+⋅⋅⋅−⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅−⋅⋅⋅−−− =x n +a 1x n −1+ ⋅ ⋅ ⋅ +a n −1x +a n .证明用数学归纳法证明.当n =2时, 2121221a x a x a x a x D ++=+−=, 命题成立. 假设对于(n −1)阶行列式命题成立, 即 D n −1=x n −1+a 1 x n −2+ ⋅ ⋅ ⋅ +a n −2x +a n −1则D , n 按第一列展开, 有 11100 100 01)1(11−⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅−⋅⋅⋅−−+=+−x x a xD D n n n n =xD n −1+a n =x n +a 1x n −1+ ⋅ ⋅ ⋅ +a n −1x +a n 因此, 对于n 阶行列式命题成立. .6. 设n 阶行列式D =det(a ij ), 把D 上下翻转、或逆时针旋转90°、或依副对角线翻转, 依次得n nn n a a a a D 11111 ⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅=, 11112 n nnn a a a a D ⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅= , 11113 a a a a D n n nn ⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅=,证明D D D n n 2)1(21)1(−−==, D 3 证明因为D =det(a =D .ij ), 所以 nnn n n n nnnn a a a a a a a a a a D 2211111111111 )1( ⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅−=⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅=−⋅⋅⋅=⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅−−=−− )1()1(331122111121nnn n nn n n a a a a a a a a D D n n n n 2)1()1()2( 21)1()1(−−+−+⋅⋅⋅++−=−=.同理可证 nnn n n n a a a a D ⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅−=− )1(11112)1(2D D n n T n n 2)1(2)1()1()1(−−−=−=. D D D D D n n n n n n n n =−=−−=−=−−−−)1(2)1(2)1(22)1(3)1()1()1()1(.7. 计算下列各行列式(D k (1)为k 阶行列式): aa D n 1 1⋅⋅⋅=, 其中对角线上元素都是a , 未写出的元素都是0; 解 aa a a a D n 010 000 00 000 0010 00⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅=(按第n 行展开) )1()1(10 000 00 000 0010 000)1(−×−+⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅−=n n n aa a )1()1(2 )1(−×−⋅⋅⋅⋅−+n n n a a an n n n n a a a+⋅⋅⋅−⋅−=−−+)2)(2(1)1()1(=a n −a n −2=a n −2(a 2−1).(2)xa aa x a a a xD n ⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅= ; 解将第一行乘(−1)分别加到其余各行, 得 ax x a ax x a a x x a aa a x D n −−⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅−−⋅⋅⋅−−⋅⋅⋅=000 0 00 0, 再将各列都加到第一列上, 得ax ax a x aaa a n x D n −⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅−⋅⋅⋅−⋅⋅⋅−+=0000 0 000 00 )1(=[x +(n −1)a ](x −a )n −1 (3). 111 1 )( )1()( )1(1111⋅⋅⋅−⋅⋅⋅⋅⋅⋅⋅⋅⋅−⋅⋅⋅⋅⋅⋅−⋅⋅⋅−−⋅⋅⋅−=−−−+n a a a n a a a n a a a D n n n n nn n ; 解根据第6题结果, 有 nnn n n n n n n n a a a n a a a n a a aD )( )1()( )1( 11 11)1(1112)1(1−⋅⋅⋅−−⋅⋅⋅⋅⋅⋅⋅⋅⋅−⋅⋅⋅⋅⋅⋅−⋅⋅⋅−⋅⋅⋅−=−−−++此行列式为范德蒙德行列式.∏≥>≥++++−−+−−=112)1(1)]1()1[()1(j i n n n n j a i a D∏≥>≥++−−−=112)1()]([)1(j i n n n j i∏≥>≥++⋅⋅⋅+−++−⋅−⋅−=1121)1(2)1()()1()1(j i n n n n n j i∏≥>≥+−=11)(j i n j i .(4)nnnnn d c d c b a b a D ⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅=11112; 解nnnnn d c d c b a b a D ⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅=11112(按第1行展开) nn n n n nd d c d c b a b a a 00011111111−−−−⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅=0)1(1111111112c d c d c b a b a b nn n n n nn −−−−+⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅−+. 再按最后一行展开得递推公式D 2n =a n d n D 2n −2−b n c n D 2n −2, 即D 2n =(a n d n −b n c n )D 2n −2于是 . ∏=−=ni i i i i n D c b d a D 222)(.而 111111112c b d a d c b a D −==,所以 ∏=−=ni i i i i n c b d a D 12)(.(5) D =det(a ij ), 其中a ij 解 a =|i −j |; ij =|i −j |, 043214 01233 10122 21011 3210)det(⋅⋅⋅−−−−⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅−⋅⋅⋅−⋅⋅⋅−⋅⋅⋅−⋅⋅⋅==n n n n n n n n a D ij n 04321 1 11111 11111 11111 1111 2132⋅⋅⋅−−−−⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅−−−−⋅⋅⋅−−−⋅⋅⋅−−⋅⋅⋅−−⋅⋅⋅−=====n n n n r r r r15242321 0 22210 02210 00210 0001 1213−⋅⋅⋅−−−−⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅−−−−⋅⋅⋅−−−⋅⋅⋅−−⋅⋅⋅−+⋅⋅⋅+=====n n n n n c c c c =(−1)n −1(n −1)2n −2 (6).nn a a a D +⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅+⋅⋅⋅+=1 11 1 1111121, 其中a 1a 2 ⋅ ⋅ ⋅ a n≠0.解nn a a a D +⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅+⋅⋅⋅+=1 11 1 1111121 nn n n a a a a a a a a a c c c c +−⋅⋅⋅−⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅−⋅⋅⋅−⋅⋅⋅−⋅⋅⋅−=====−−100001 000 100 0100 0100 0011332212132 1111312112111000011 000 00 11000 01100 001 −−−−−−+−⋅⋅⋅−⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅−⋅⋅⋅−⋅⋅⋅⋅⋅⋅=nn n a a a a a a a a∑=−−−−−−+⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅=n i i n n a a a a a a a a 1111131******** 00010 000 00 10000 01000 001)11)((121∑=+=ni i n a a a a .8. 用克莱姆法则解下列方程组: (1) =+++−=−−−−=+−+=+++01123253224254321432143214321x x x x x x x x x x x x x x x x ;解因为 14211213513241211111−=−−−−=D , 142112105132412211151−=−−−−−−=D , 284112035122412111512−=−−−−−=D , 426110135232422115113−=−−−−=D , 14202132132212151114=−−−−−=D , 所以 111==D D x , 222==D Dx , 333==DD x , 144−==D D x .(2)=+=++=++=++=+150650650651655454343232121x x x x x x x x x x x x x .解因为 665510006510006510065100065==D , 15075100165100065100065000611==D , 114551010651000650000601000152−==D , 703511650000601000051001653==D , 39551601000051000651010654−==D , 2121100005100065100651100655==D , 所以66515071=x , 66511452−=x , 6657033=x , 6653954−=x , 6652124=x .9. 问λ, µ取何值时, 齐次线性方程组 =++=++=++0200321321321x x x x x x x x x µµλ有非零解？解系数行列式为µλµµµλ−==1211111D .令D =0, 得 µ=0或λ=1.于是, 当µ=0或λ=1时该齐次线性方程组有非零解.10. 问λ取何值时, 齐次线性方程组 =−++=+−+=+−−0)1(0)3(2042)1(321321321x x x x x x x x x λλλ有非零解？解系数行列式为λλλλλλλ−−+−−=−−−−=101112431111132421D=(1−λ)3 =(1−λ)+(λ−3)−4(1−λ)−2(1−λ)(−3−λ) 3+2(1−λ)2 令D =0, 得+λ−3. λ=0, λ=2或λ=3.于是, 当λ=0, λ=2或λ=3时, 该齐次线性方程组有非零解.第二章矩阵及其运算1. 已知线性变换:++=++=++=3213321232113235322y y y x y y y x y y y x , 求从变量x 1, x 2, x 3到变量y 1, y 2, y 3 解由已知:的线性变换.= 221321323513122y y y x x x ,故= −3211221323513122x x x y y y−−−−=321423736947y y y ,−+=−+=+−−=321332123211423736947x x x y x x x y x x x y .2. 已知两个线性变换++=++−=+=32133212311542322y y y x y y y x y y x ,+−=+=+−=323312211323z z y z z y z z y , 求从z 1, z 2, z 3到x 1, x 2, x 3 解由已知的线性变换.−= 221321514232102y y y x x x−− −=321310102013514232102z z z−−−−=321161109412316z z z ,所以有 +−−=+−=++−=3213321232111610941236z z z x z z z x z z z x .3. 设 −−=111111111A ,−−=150421321B , 求3AB −2A 及A T 解 B .−−− −− −−=−1111111112150421321111111111323A AB−−−−= −−− −=2294201722213211111111120926508503,−= −− −−=092650850150421321111111111B A T.4. 计算下列乘积: (1)−127075321134;解 −127075321134 ×+×+××+×−+××+×+×=102775132)2(71112374=49635.(2)123)321(;解123)321(=(1×3+2×2+3×1)=(10).(3))21(312−;解 )21(312−×−××−××−×=23)1(321)1(122)1(2−−−=632142. (4)−−−−20413121013143110412 ; 解−−− −20413121013143110412 −−−=6520876. (5)321332313232212131211321)(x x x a a a a a a a a a x x x ;解321332313232212131211321)(x x x a a a a a a a a a x x x=(a 11x 1+a 12x 2+a 13x 3 a 12x 1+a 22x 2+a 23x 3a 13x 1+a 23x 2+a 33x 3321x x x )322331132112233322222111222x x a x x a x x a x a x a x a +++++=.5. 设 =3121A ,=2101B , 问: (1)AB =BA 吗? 解 AB ≠BA . 因为=6443AB ,=8321BA , 所以AB ≠BA .(2)(A +B )2=A 2+2AB +B 2 解 (A +B )吗? 2≠A 2+2AB +B 2 因为.=+5222B A ,=+52225222)(2B A=2914148,但 + +=++43011288611483222B AB A=27151610,所以(A +B )2≠A 2+2AB +B 2 (3)(A +B )(A −B )=A . 2−B 2 解 (A +B )(A −B )≠A 吗? 2−B 2 因为.=+5222B A ,=−1020B A ,==−+906010205222))((B A B A ,而= −=−718243011148322B A ,故(A +B )(A −B )≠A 2−B 2 6. 举反列说明下列命题是错误的:.(1)若A 2 解取=0, 则A =0;=0010A , 则A 2 (2)若A =0, 但A ≠0. 2 解取=A , 则A =0或A =E ;=0011A , 则A 2 (3)若AX =AY , 且A ≠0, 则X =Y .=A , 但A ≠0且A ≠E . 解取=0001A , −=1111X ,=1011Y , 则AX =AY , 且A ≠0, 但X ≠Y .7. 设=101λA , 求A 2, A 3, ⋅ ⋅ ⋅, A k 解 . ==12011011012λλλA , ===1301101120123λλλA A A , ⋅ ⋅ ⋅ ⋅ ⋅ ⋅,=101λk A k . 8. 设=λλλ001001A , 求A k 解首先观察. =λλλλλλ0010010010012A=222002012λλλλλ,=⋅=3232323003033λλλλλλA A A ,=⋅=43423434004064λλλλλλA A A ,=⋅=545345450050105λλλλλλA A A , ⋅ ⋅ ⋅ ⋅ ⋅ ⋅,=k A k k k k k k k k k k λλλλλλ0002)1(121−−−−. 用数学归纳法证明:当k =2时, 显然成立.假设k 时成立，则k +1时，−=⋅=−−−+λλλλλλλλλ0010010002)1(1211k k k k k k k k k k k k A A A+++=+−+−−+11111100)1(02)1()1(k k k k k k k k k k λλλλλλ, 由数学归纳法原理知:−=−−−k k k k k k k k k k k A λλλλλλ0002)1(121. 9. 设A , B 为n 阶矩阵，且A 为对称矩阵，证明B T 证明因为A AB 也是对称矩阵.T (B =A , 所以T AB )T =B T (B T A )T =B T A T B =B T 从而B AB ,T 10. 设A , B 都是n 阶对称矩阵，证明AB 是对称矩阵的充分必要条件是AB =BA .AB 是对称矩阵.证明充分性: 因为A T =A , B T (AB )=B , 且AB =BA , 所以 T =(BA )T =A T B T 即AB 是对称矩阵.=AB ,必要性: 因为A T =A , B T =B , 且(AB )T AB =(AB )=AB , 所以T =B T A T 11. 求下列矩阵的逆矩阵:=BA .(1)5221; 解=5221A . |A |=1, 故A −1 存在. 因为−−= =1225*22122111A A A A A ,故 *||11A A A =−−−=1225. (2)−θθθθcos sin sin cos ; 解−=θθθθcos sin sin cos A . |A |=1≠0, 故A −1 存在. 因为−= =θθθθcos sin sin cos *22122111A A A A A , 所以 *||11A A A =−−=θθθθcos sin sin cos . (3)−−−145243121; 解−−−=145243121A . |A |=2≠0, 故A −1 存在. 因为−−−−−= =214321613024*332313322212312111A A A A A A A A A A , 所以 *||11A A A =−−−−−−=1716213213012. (4)n a a a 0021(a 1a 2⋅ ⋅ ⋅a n ≠0) .解=n a a a A 0021, 由对角矩阵的性质知=−n a a a A 10011211 . 12. 解下列矩阵方程:(1) −=12643152X ; 解 −=−126431521X − −−=12642153 −=80232. (2) −=−−234311*********X ; 解 1111012112234311−−− −=X−−− −=03323210123431131 −−−=32538122. (3) −= − −101311022141X ;解 11110210132141−− − − −=X− −=210110131142121 =21010366121=04111. (4)−−−= 021102341010100001100001010X . 解 11010100001021102341100001010−−−−− =X −−− =010100001021102341100001010 −−−=201431012. 13. 利用逆矩阵解下列线性方程组:(1) =++=++=++3532522132321321321x x x x x x x x x ; 解方程组可表示为= 321153522321321x x x , 故 = = −0013211535223211321x x x ,从而有 ===001321x x x . (2) =−+=−−=−−05231322321321321x x x x x x x x x . 解方程组可表示为=−−−−−012523312111321x x x , 故 =−−−−−= −3050125233121111321x x x , 故有 ===305321x x x . 14. 设A k =O (k 为正整数), 证明(E −A )−1=E +A +A 2+⋅ ⋅ ⋅+A k −1 证明因为A . k =O , 所以E −A k E −A =E . 又因为k =(E −A )(E +A +A 2+⋅ ⋅ ⋅+A k −1所以 (E −A )(E +A +A ),2+⋅ ⋅ ⋅+A k −1由定理2推论知(E −A )可逆, 且)=E ,(E −A )−1=E +A +A 2+⋅ ⋅ ⋅+A k −1.证明一方面, 有E =(E −A )−1 另一方面, 由A (E −A ).k E =(E −A )+(A −A =O , 有2)+A 2−⋅ ⋅ ⋅−A k −1+(A k −1−A k )=(E +A +A 2+⋅ ⋅ ⋅+A k −1故 (E −A ))(E −A ),−1(E −A )=(E +A +A 2+⋅ ⋅ ⋅+A k −1两端同时右乘(E −A ))(E −A ),−1 (E −A ), 就有−1(E −A )=E +A +A 2+⋅ ⋅ ⋅+A k −1.15. 设方阵A 满足A 2−A −2E =O , 证明A 及A +2E 都可逆, 并求A −1及(A +2E )−1 证明由A .2 A −A −2E =O 得2或 −A =2E , 即A (A −E )=2E ,E E A A =−⋅)(21, 由定理2推论知A 可逆, 且)(211E A A −=−. 由A 2 A −A −2E =O 得2或 −A −6E =−4E , 即(A +2E )(A −3E )=−4E ,E A E E A =−⋅+)3(41)2( 由定理2推论知(A +2E )可逆, 且)3(41)2(1A E E A −=+−.证明由A 2−A −2E =O 得A 2 |A −A =2E , 两端同时取行列式得 2即 |A ||A −E |=2,−A |=2,故 |A |≠0,所以A 可逆, 而A +2E =A 2, |A +2E |=|A 2|=|A |2由 A ≠0, 故A +2E 也可逆. 2 ⇒A −A −2E =O ⇒A (A −E )=2E−1A (A −E )=2A −1)(211E A A −=−E ⇒,又由 A 2 ⇒ (A +2E )(A −3E )=−4 E ,−A −2E =O ⇒(A +2E )A −3(A +2E )=−4E所以 (A +2E )−1(A +2E )(A −3E )=−4(A +2 E )−1 ,)3(41)2(1A E E A −=+−.16. 设A 为3阶矩阵, 21||=A , 求|(2A )−1 解因为−5A *|.*||11A A A =−, 所以 |||521||*5)2(|111−−−−=−A A A A A |2521|11−−−=A A=|−2A −1|=(−2)3|A −1|=−8|A |−1 17. 设矩阵A 可逆, 证明其伴随阵A *也可逆, 且(A *)=−8×2=−16.−1=(A −1 证明由)*.*||11A A A =−, 得A *=|A |A −1 |A *|=|A |, 所以当A 可逆时, 有n |A −1|=|A |n −1从而A *也可逆.≠0,因为A *=|A |A −1 (A *), 所以−1=|A |−1又A .*)(||)*(||1111−−−==A A A A A , 所以(A *)−1=|A |−1A =|A |−1|A |(A −1)*=(A −1 18. 设n 阶矩阵A 的伴随矩阵为A *, 证明:)*.(1)若|A |=0, 则|A *|=0;(2)|A *|=|A |n −1 证明.(1)用反证法证明. 假设|A *|≠0, 则有A *(A *)−1 A =A A *(A *)=E , 由此得 −1=|A |E (A *)−1所以A *=O , 这与|A *|≠0矛盾，故当|A |=0时, 有|A *|=0.=O ,(2)由于*||11A A A =−, 则AA *=|A |E , 取行列式得到 |A ||A *|=|A |n 若|A |≠0, 则|A *|=|A |.n −1 若|A |=0, 由(1)知|A *|=0, 此时命题也成立.;因此|A *|=|A |n −1.19. 设−=321011330A , AB =A +2B , 求B . 解由AB =A +2E 可得(A −2E )B =A , 故− −−−=−=−−321011330121011332)2(11A E A B −=011321330. 20. 设 =101020101A , 且AB +E =A 2+B , 求B .解由AB +E =A 2 (A −E )B =A +B 得 2即 (A −E )B =(A −E )(A +E ).−E , 因为01001010100||≠−==−E A , 所以(A −E )可逆, 从而=+=201030102E A B .21. 设A =diag(1, −2, 1), A *BA =2BA −8E , 求B . 解由A *BA =2BA −8E 得 (A *−2E )BA =−8E , B =−8(A *−2E )−1A =−8[A (A *−2E )]−1 =−8(AA *−2A )−1 =−8(|A |E −2A )−1 =−8(−2E −2A )−1 =4(E +A )−1 =4[diag(2, −1, 2)]−1−1)21 ,1 ,21(diag 4−==2diag(1, −2, 1).22. 已知矩阵A 的伴随阵−=8030010100100001*A , 且ABA −1=BA −1+3E , 求B .解由|A *|=|A |3 由ABA =8, 得|A |=2. −1=BA −1 AB =B +3A ,+3E 得 B =3(A −E )−1A =3[A (E −A −1)]−1 A 11*)2(6*)21(3−−−=−=A E A E−=−−=−1030060600600006603001010010000161. 23. 设P −1 −−=1141P AP =Λ, 其中,−=Λ2001, 求A 11 解由P . −1AP =Λ, 得A =P ΛP −1, 所以A 11= A =P Λ11P −1 |P |=3, .−=1141*P ,−−=−1141311P ,而−= −=Λ11111120 012001,故−− −−−=31313431200111411111A −−=68468327322731. 24. 设AP =P Λ, 其中−−=111201111P ,−=Λ511,求ϕ(A )=A 8(5E −6A +A 2 解 ϕ(Λ)=Λ). 8(5E −6Λ+Λ2 =diag(1,1,5)8)[diag(5,5,5)−diag(−6,6,30)+diag(1,1,25)]=diag(1,1,58 ϕ(A )=P ϕ(Λ)P )diag(12,0,0)=12diag(1,0,0).−1 *)(||1P P P Λ=ϕ−−−−−− −−−=1213032220000000011112011112=1111111114.25. 设矩阵A 、B 及A +B 都可逆, 证明A −1+B −1 证明因为也可逆, 并求其逆阵.A −1(A +B )B −1=B −1+A −1=A −1+B −1而A ,−1(A +B )B −1是三个可逆矩阵的乘积, 所以A −1(A +B )B −1可逆, 即A −1+B −1 (A 可逆.−1+B −1)−1=[A −1(A +B )B −1]−1=B (A +B )−1 26. 计算A .−−−30003200121013013000120010100121. 解设 =10211A , =30122A , −=12131B ,−−=30322B ,则 2121B O B E A O E A+=222111B A O B B A A ,而 −= −−+−=+4225303212131021211B B A ,−−= −− =90343032301222B A , 所以 2121B O B E A O E A +=222111B A O B B A A−−−=9000340042102521, 即−−−30003200121013013000120010100121−−−=9000340042102521. 27. 取==−==1001D C B A , 验证|||||||| D C B A D C B A ≠.解 4100120021010*********0021010010110100101==−−=−−=D C B A , 而 01111|||||||| ==D C B A ,故 ||||||||D C B A D C B A ≠. 28. 设 −=22023443O O A , 求|A 8|及A 4解令. −=34431A ,=22022A , 则=21A O O A A ,故 8218=A O O A A=8281A O O A ,1682818281810||||||||||===A A A A A .= =464444241422025005O O A O O A A . 29. 设n 阶矩阵A 及s 阶矩阵B 都可逆, 求 (1)1−O B A O ; 解设 =−43211C C C C O B A O , 则O B A O 4321C C C C = =s n E O O E BC BC AC AC 2143. 由此得====s n EBC OBC O AC E AC 2143⇒ ====−−121413B C O C O C A C ,所以= −−−O A B O O B A O 111. (2)1−B C O A . 解设 =−43211D D D D B C O A , 则 = ++= s nE O O E BD CD BD CD AD AD D D D D B C O A 4231214321.由此得=+=+==s nEBD CD O BD CD O AD E AD 423121⇒ =−===−−−−14113211B D CA B D O D A D ,所以−= −−−−−11111B CA B O A BC O A . 30. 求下列矩阵的逆阵: (1)2500380000120025; 解设 =1225A , =2538B , 则−−= =−−5221122511A ,−−==−−8532253811B .于是 −−−−= = =−−−−850032000052002125003800001200251111B A B A .(2)4121031200210001. 解设 =2101A ,=4103B ,=2112C , 则−= =−−−−−−1111114121031200210001B CA B O A BC O A−−−−−=411212458103161210021210001.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;解−−340313021201(下一步: r 2+(−2)r 1, r 3+(−3)r 1 ~. )−−−020*********(下一步: r 2÷(−1), r 3 ~÷(−2). )−−010*********(下一步: r 3−r 2 ~. )−−300031001201(下一步: r 3 ~÷3. )−−100031001201(下一步: r 2+3r 3 ~. )−100001001201(下一步: r 1+(−2)r 2, r 1+r 3 ~. )100001000001.(2)−−−−174034301320;解−−−−174034301320(下一步: r 2×2+(−3)r 1, r 3+(−2)r 1 ~. )−−−310031001320(下一步: r 3+r 2, r 1+3r 2 ~. )0000310010020(下一步: r 1 ~÷2. )000031005010.(3)−−−−−−−−−12433023221453334311;解−−−−−−−−−12433023221453334311(下一步: r 2−3r 1, r 3−2r 1, r 4−3r 1~. )−−−−−−−−1010500663008840034311(下一步: r 2÷(−4), r 3÷(−3) , r 4~÷(−5). )−−−−−22100221002210034311(下一步: r 1−3r 2, r 3−r 2, r 4−r 2~. )−−−00000000002210032011.(4)−−−−−−34732038234202173132. 解−−−−−−34732038234202173132(下一步: r 1−2r 2, r 3−3r 2, r 4−2r 2~. )−−−−−1187701298804202111110(下一步: r 2+2r 1, r 3−8r 1, r 4−7r 1 ~. )−−41000410002020111110(下一步: r 1↔r 2, r 2×(−1), r 4−r 3~. )−−−−00000410001111020201(下一步: r 2+r 3~. )−−00000410003011020201. 2. 设= 987654321100010101100001010A , 求A .解100001010是初等矩阵E (1, 2), 其逆矩阵就是其本身.100010101是初等矩阵E (1, 2(1)), 其逆矩阵是E (1, 2(−1))−=100010101.− =100010101987654321100001010A= − =287221254100010101987321654.3. 试利用矩阵的初等变换, 求下列方阵的逆矩阵: (1)323513123;解 100010001323513123~−−−101011001200410123~ −−−−1012002110102/102/3023~−−−−2/102/11002110102/922/7003~−−−−2/102/11002110102/33/26/7001故逆矩阵为−−−−21021211233267.(2)−−−−−1210232112201023.解−−−−−10000100001000011210232112201023~−−−−00100301100001001220594012102321~−−−−−−−−20104301100001001200110012102321~ −−−−−−−106124301100001001000110012102321 ~−−−−−−−−−−10612631110`1022111000010000100021 ~−−−−−−−106126311101042111000010000100001故逆矩阵为−−−−−−−10612631110104211. 4. (1)设 −−=113122214A ,−−=132231B , 求X 使AX =B ;解因为−−−−=132231 113122214) ,(B A−−412315210 100010001 ~r ,所以−−==−4123152101B A X .(2)设−−−=433312120A , −=132321B , 求X 使XA =B . 解考虑A T X T =B T . 因为−−−−=134313*********) ,(T T B A−−−411007101042001 ~r ,所以−−−==−417142)(1T T T B A X ,从而−−−==−4741121BA X . 5. 设−−−=101110011A , AX =2X +A , 求X .解原方程化为(A −2E )X =A . 因为−−−−−−−−−=−101101110110011011) ,2(A E A−−−011100101010110001~,所以−−−=−=−011101110)2(1A E A X .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. 例如,=010*********A , R (A )=3.0000是等于0的2阶子式, 010001000是等于0的3阶子式. 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 ).这是因为B 的非零子式必是A 的非零子式, 故A 的秩不会小于B 的秩.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).解用已知向量容易构成一个有4个非零行的5阶下三角矩阵:−0000001000001010001100001, 此矩阵的秩为4, 其第2行和第3行是已知向量.9. 求下列矩阵的秩, 并求一个最高阶非零子式: (1)−−−443112112013;解−−−443112112013(下一步: r 1↔r 2 ~. )−−−443120131211(下一步: r 2−3r 1, r 3−r 1 ~. )−−−−564056401211(下一步: r 3−r 2 ~. )−−−000056401211, 矩阵的2秩为, 41113−=−是一个最高阶非零子式.(2)−−−−−−−815073*********;解−−−−−−−815073*********(下一步: r 1−r 2, r 2−2r 1, r 3−7r 1 ~. )−−−−−−15273321059117014431(下一步: r 3−3r 2~. )−−−−0000059117014431, 矩阵的秩是2, 71223−=−是一个最高阶非零子式.(3)−−−02301085235703273812. 解−−−02301085235703273812(下一步: r 1−2r 4, r 2−2r 4, r 3−3r 4~. )−−−−−−023*********63071210(下一步: r 2+3r 1, r 3+2r 1~. )−0230114000016000071210(下一步: r 2÷16r 4, r 3−16r 2. )~−02301000001000071210 ~−00000100007121002301, 矩阵的秩为3, 070023085570≠=−是一个最高阶非零子式.10. 设A 、B 都是m ×n 矩阵, 证明A ~B 的充分必要条件是R (A )=R (B ).证明根据定理3, 必要性是成立的.充分性. 设R (A )=R (B ), 则A 与B 的标准形是相同的. 设A 与B 的标准形为D , 则有A ~D , D ~B .由等价关系的传递性, 有A ~B .11. 设−−−−=32321321k k k A , 问k 为何值, 可使(1)R (A )=1; (2)R (A )=2; (3)R (A )=3.解 −−−−=32321321k k k A+−−−−−)2)(1(0011011 ~k k k k k r . (1)当k =1时, R (A )=1; (2)当k =−2且k ≠1时, R (A )=2;(3)当k ≠1且k ≠−2时, R (A )=3.12. 求解下列齐次线性方程组: (1) =+++=−++=−++02220202432143214321x x x x x x x x x x x x ;解对系数矩阵A 进行初等行变换, 有 A = −−212211121211~ −−−3/410013100101,于是 ==−==4443424134334x x x x x x x x ,故方程组的解为−= 1343344321k x x x x (k 为任意常数).(2) =−++=−−+=−++05105036302432143214321x x x x x x x x x x x x ;解对系数矩阵A 进行初等行变换, 有 A = −−−−5110531631121~−000001001021,于是 ===+−=4432242102x x x xx x x x ,故方程组的解为+−= 10010*********k k x x x x (k 1, k 2 (3)为任意常数).=−+−=+−+=−++=+−+07420634072305324321432143214321x x x x x x x x x x x x x x x x ;解对系数矩阵A 进行初等行变换, 有 A =−−−−−7421631472135132~1000010000100001,于是 ====0004321x x x x ,故方程组的解为 ====00004321x x x x .(4) =++−=+−+=−+−=+−+03270161311402332075434321432143214321x x x x x x x x x x x x x x x x .解对系数矩阵A 进行初等行变换, 有 A =−−−−−3127161311423327543~−−000000001720171910171317301,于是 ==−=−=4433432431172017191713173x x x x x x x xx x ,故方程组的解为−−+= 1017201713011719173214321k k x x x x (k 1, k 2为任意常数).13. 求解下列非齐次线性方程组: (1) =+=+−=−+83111021322421321321x x x x x x x x ;解对增广矩阵B 进行初等行变换, 有。

线性代数_中国人民大学中国大学mooc课后章节答案期末考试题库2023年

线性代数_中国人民大学中国大学mooc课后章节答案期末考试题库2023年1.下列结论中不正确的是【图片】参考答案:若n阶实矩阵A的列向量组两两正交, 则A必为正交矩阵2.下列矩阵中为正交矩阵的有参考答案:_3.若向量组【图片】线性无关, 则在每个向量的相同位置去掉若干分量后仍会线性无关.参考答案:错误4.设A为n阶实矩阵对称，满足【图片】=0,且【图片】,则【图片】（）参考答案:5.【图片】中的任意一个向量与自身的内积必大于零参考答案:错误6.任何实二次型都可经过可逆线性替换化为规范形，且规范形唯一。

参考答案:正确7.若m与n不相等且均大于1,则以下哪些矩阵可以视为向量？参考答案:1行n列的矩阵_m行1列的矩阵8.设【图片】是【图片】矩阵，【图片】的秩，【图片】，则下列说法中正确的是参考答案:当时，将按列分块后, 的所有列所组成的向量组必线性相关.9.已知矩阵【图片】，且【图片】，则【图片】参考答案:10.本周视频中【百鸡问题】对应的线性方程组理论上一定有无穷多解，但是只有有限个解符合实际含义。

参考答案:正确11.当用初等变换法求解n元线性方程组时, 若方程组的增广矩阵经初等行变换化成的阶梯形矩阵的最后一列不含主元，则该方程组可能存在以下什么情况？参考答案:有无穷多个解_有唯一解12.任意一个向量组都有极大线性无关组参考答案:错误13.对于【图片】中的任意两个向量【图片】,都必有【图片】参考答案:正确14.齐次线性方程组的基础解系所含解向量的个数等于参考答案:该方程组系数矩阵的列数减去系数矩阵的秩_该方程组中自由未知量的个数15.当线性方程组的系数矩阵为以下什么方阵时,对该方程组一定可以运用克拉默法则？参考答案:可逆矩阵_初等矩阵16.设矩阵【图片】的秩为【图片】，则下列叙述正确的是参考答案:至少有一个阶子式不等于 017.下列下列叙述正确的是（）参考答案:单位矩阵既是对角矩阵又是对称矩阵18.下列论断中是矩阵【图片】正定的充分必要条件的是（）.参考答案:正定19.一个二次型可能有多个标准型。

线性代数课后习题答案全习题详解

线性代数课后习题答案全习题详解(总92页)--本页仅作为文档封面，使用时请直接删除即可----内页可以根据需求调整合适字体及大小--第一章行列式1.利用对角线法则计算下列三阶行列式：（1）381141102---；（2）b a c a c b c b a ; （3）222111c b a c b a ；（4）y x y x x y x yyx y x +++. 解（1）=---381141102811)1()1(03)4(2⨯⨯+-⨯-⨯+⨯-⨯)1()4(18)1(2310-⨯-⨯-⨯-⨯-⨯⨯- =416824-++-=4-（2）=ba c a cb cb a ccc aaa bbb cba bac acb ---++3333c b a abc ---=（3）=222111c b a c b a 222222cb ba ac ab ca bc ---++))()((a c c b b a ---=（4）yx y x x y x y yx y x +++yx y x y x yx y y x x )()()(+++++=333)(x y x y -+-- 33322333)(3x y x x y y x y y x xy ------+= )(233y x +-=2.按自然数从小到大为标准次序，求下列各排列的逆序数：（1）1 2 3 4；（2）4 1 3 2；（3）3 4 2 1；（4）2 4 1 3；（5）1 3 … )12(-n 2 4 … )2(n ；（6）1 3 … )12(-n )2(n )22(-n … 2. 解（1）逆序数为0（2）逆序数为4：4 1，4 3，4 2，3 2（3）逆序数为5：3 2，3 1，4 2，4 1,2 1 （4）逆序数为3：2 1，4 1，4 3（5）逆序数为2)1(-n n ：3 2 1个 5 2，54 2个 7 2，7 4，7 6 3个 ……………… … )12(-n 2，)12(-n 4，)12(-n 6，…，)12(-n )22(-n )1(-n 个（6）逆序数为)1(-n n3 2 1个 5 2，54 2个 ……………… … )12(-n 2，)12(-n 4，)12(-n 6，…，)12(-n )22(-n )1(-n 个4 2 1个 6 2，6 4 2个 ……………… … )2(n 2，)2(n 4，)2(n 6，…，)2(n )22(-n )1(-n 个3.写出四阶行列式中含有因子2311a a 的项.解由定义知，四阶行列式的一般项为43214321)1(p p p p t a a a a -，其中t 为4321p p p p 的逆序数．由于3,121==p p 已固定，4321p p p p 只能形如13□□，即1324或1342.对应的t 分别为10100=+++或22000=+++∴44322311a a a a -和42342311a a a a 为所求.4.计算下列各行列式：（1）⎥⎥⎥⎥⎦⎥⎢⎢⎢⎢⎣⎢71100251020214214；（2）⎥⎥⎥⎥⎦⎥⎢⎢⎢⎢⎣⎢-265232112131412；（3）⎥⎥⎦⎥⎢⎢⎣⎢---ef cf bf de cd bd ae acab ；（4）⎥⎥⎥⎥⎦⎥⎢⎢⎢⎢⎣⎢---d c b a 100110011001 解(1)7110025*******21434327c c c c --0100142310202110214---=34)1(143102211014+-⨯---=143102211014-- 321132c c c c ++1417172001099-=0(2)265232112131412-24c c -2605032122130412-24r r -0412032122130412- 14r r -0000032122130412-=0(3)ef cf bf de cd bd ae ac ab ---=e c b e c b e c b adf ---=111111111---adfbce =abcdef 4(4)d c b a 100110011001---21ar r +dc b a ab 100110011010---+=12)1)(1(+--dc a ab 10111--+ 23dc c +010111-+-+cd c ada ab =23)1)(1(+--cdadab +-+111=1++++ad cd ab abcd5.证明: (1)1112222b b a a b ab a +=3)(b a -;(2)bz ay by ax bx az by ax bx az bz ay bx az bz ay by ax +++++++++=yx z x z y zy x b a )(33+;(3)0)3()2()1()3()2()1()3()2()1()3()2()1(2222222222222222=++++++++++++d d d d c c c c b b b b a a a a ;(4)444422221111d c b a d c b a d c b a ))()()()((d b c b d a c a b a -----=))((d c b a d c +++-⋅;(5)1221100000100001a x a a a a x x x n n n +-----n n n n a x a x a x ++++=--111 .证明(1)00122222221312a b a b a a b a ab a c c c c ------=左边a b a b a b a ab 22)1(22213-----=+21))((a b a a b a b +--= 右边=-=3)(b a(2)bz ay by ax z by ax bx az y bx az bz ay x a ++++++分开按第一列左边bzay by ax x by ax bx az z bxaz bz ay y b +++++++ ++++++002y by ax z x bx az y z bz ay x a 分别再分bz ay y x by ax x z bx az z y b +++zy x y x z xz y b y x z x z y z y x a 33+分别再分右边=-+=233)1(yx z x z y zy x b y x z x z y z y x a(3) 2222222222222222)3()2()12()3()2()12()3()2()12()3()2()12(++++++++++++++++=d d d d d c c c c c b b b b b a a a a a 左边9644129644129644129644122222141312++++++++++++---d d d d c c c c b b b b a a a a c c c c c c 964496449644964422222++++++++d d d d c c c c b b b b a a a a 分成二项按第二列964419644196441964412222+++++++++d d d c c c b b b a a a 949494949464222224232423d d c c b b a a c c c c c c c c ----第二项第一项06416416416412222=+ddd c c c bb b a a a (4) 444444422222220001ad a c a b a ad a c a b a a d a c a b a ---------=左边=)()()(222222222222222a d d a c c a b b a d a c a b ad a c a b --------- =)()()(111))()((222a d d a c c a b b a d a c ab a d ac a b ++++++--- =⨯---))()((ad a c a b )()()()()(00122222a b b a d d a b b a c c a b b bd b c a b +-++-++--+ =⨯-----))()()()((b d b c a d a c a b )()()()(112222b d a b bd d b c a b bc c ++++++++=))()()()((d b c b d a c a b a -----))((d c b a d c +++-(5) 用数学归纳法证明.,1,2212122命题成立时当a x a x a x a x D n ++=+-==假设对于)1(-n 阶行列式命题成立，即 ,122111-----++++=n n n n n a x a x a x D:1列展开按第则n D1110010001)1(11----+=+-x xa xD D n n n n 右边=+=-n n a xD 1 所以，对于n 阶行列式命题成立.6.设n 阶行列式)det(ij a D =,把D 上下翻转、或逆时针旋转 90、或依副对角线翻转，依次得n nn n a a a a D 11111 =， 11112n nn n a a a a D = ，11113a a a a D n nnn =，证明D D D D D n n =-==-32)1(21,)1(.证明 )det(ij a D =nnnn nn n nn n a a a a a a a a a a D 2211111111111)1(--==∴ =--=--nnn n nnn n a a a a a a a a 331122111121)1()1( nnn n n n a a a a 111121)1()1()1(---=--D D n n n n 2)1()1()2(21)1()1(--+-+++-=-= 同理可证nnn n n n a a a a D 11112)1(2)1(--=D D n n Tn n 2)1(2)1()1()1(---=-= D D D D D n n n n n n n n =-=--=-=----)1(2)1(2)1(22)1(3)1()1()1()1(7.计算下列各行列式（阶行列式为k D k ）：(1)aaD n 11 =,其中对角线上元素都是a ，未写出的元素都是0；(2)xaaax aa a x D n=; (3) 1111)()1()()1(1111n a a a n a a a n a a a D n n n nn n n ------=---+; 提示：利用范德蒙德行列式的结果． (4) nnn nn d c d c b a b a D000011112=; (5)j i a a D ij ij n -==其中),det(;(6)nn a a a D +++=11111111121 ,021≠n a a a 其中.解(1) aa a a a D n 010000000000001000=按最后一行展开)1()1(1000000000010000)1(-⨯-+-n n n aa a)1)(1(2)1(--⋅-+n n n a a a (再按第一行展开)n n n nn a a a+-⋅-=--+)2)(2(1)1()1(2--=n n a a )1(22-=-a a n(2)将第一行乘)1(-分别加到其余各行，得ax x a ax x a a x x a aa a x D n ------=0000000 再将各列都加到第一列上，得ax ax a x aaa a n x D n ----+=000000000)1( )(])1([1a x a n x n --+=- (3) 从第1+n 行开始，第1+n 行经过n 次相邻对换，换到第1行，第n 行经)1(-n 次对换换到第2行…，经2)1(1)1(+=++-+n n n n 次行交换，得nn n n n n n n n n a a a n a a a n a a aD )()1()()1(1111)1(1112)1(1-------=---++此行列式为范德蒙德行列式∏≥>≥++++--+--=112)1(1)]1()1[()1(j i n n n n j a i a D∏∏≥>≥+++-++≥>≥++-•-•-=---=111)1(2)1(112)1()][()1()1()]([)1(j i n n n n n j i n n n j i j i∏≥>≥+-=11)(j i n j i(4) n nnnn d c d c b a b a D 011112=nn n n n nd d c d c b a b a a 0000000011111111----展开按第一行0000)1(1111111112c d c d c b a b a b nn n n n nn ----+-+2222 ---n n n n n n D c b D d a 都按最后一行展开由此得递推公式：222)(--=n n n n n n D c b d a D 即 ∏=-=ni i i i i n D c b d a D 222)(而 111111112c b d a d c b a D -==得 ∏=-=ni i i i i n c b d a D 12)((5)j i a ij -=0432********0122210113210)det( --------==n n n n n n n n a D ij n ,3221r r r r --0432111111111111111111111 --------------n n n n,,141312c c c c c c +++152423210222102210002100001---------------n n n n n =212)1()1(----n n n(6)nn a a D a +++=11111111121 ,,433221c c c c c c ---n n n n a a a a a a a a a a +-------10000100010000100010001000011433221展开（由下往上）按最后一列))(1(121-+n n a a a a nn n a a a a a a a a a --------00000000000000000000000000022433221 nn n a a a a a a a a ----+--000000000000000001133221 ++ nn n a a a a a a a a -------000000000000000001143322n n n n n n a a a a a a a a a a a a 322321121))(1(++++=---)11)((121∑=+=ni in a a a a8.用克莱姆法则解下列方程组：⎪⎪⎩⎪⎪⎨⎧=+++-=----=+-+=+++;01123,2532,242,5)1(4321432143214321x x x x x x x x x x x x x x x x ⎪⎪⎪⎩⎪⎪⎪⎨⎧=+=++=++=++=+.15,065,065,065,165)2(5454343232121x x x x x x x x x x x x x解 (1)11213513241211111----=D 8120735032101111------=145008130032101111---=1421420005410032101111-=---= 112105132412211151------=D 11210513290501115----=1121023313090509151------=2331309050112109151------=1202300461000112109151-----=14200038100112109151----=142-= 112035122412111512-----=D 811507312032701151-------=3139011230023101151-=2842840001910023101151-=----=426110135232422115113-=----=D ; 14202132132212151114=-----=D1,3,2,144332211-========∴DDx D D x D D x D D x (2) 510006510006510065100065=D 展开按最后一行61000510065100655-'D D D ''-'=65 D D D ''-'''-''=6)65(5D D '''-''=3019D D ''''-'''=1146566551141965=⨯-⨯=(,11的余子式中为行列式a D D ',11的余子式中为a D D ''''类推D D ''''''',) 5100165100065100650000611=D 展开按第一列6510065100650006+'D 46+'=D 460319+''''-'''=D 1507= 51165100065000601000152=D 展开按第二列5100651006500061-6510065000610005-365510651065⨯-=1145108065-=--= 51100650000601000051001653=D 展开按第三列5100650006100051650061000510065+6100510656510650061+= 703114619=⨯+= 51000601000051000651010654=D 展开按第四列61000510065100655000610005100651--51065106565--=395-= 110051000651000651100655=D 展开按最后一列D '+10005100651006512122111=+= 665212;665395;665703;6651145;665150744321=-==-==∴x x x x x ． 9.齐次线性方程组取何值时问,,μλ⎪⎩⎪⎨⎧=++=++=++0200321321321x x x x x x x x x μμλ有非零解解 μλμμμλ-==12111113D ，齐次线性方程组有非零解，则03=D即 0=-μλμ 得 10==λμ或不难验证，当,10时或==λμ该齐次线性方程组确有非零解.10.齐次线性方程组取何值时问,λ⎪⎩⎪⎨⎧=-++=+-+=+--0)1(0)3(2042)1(321321321x x x x x x x x x λλλ 有非零解解λλλ----=111132421D λλλλ--+--=101112431)3)(1(2)1(4)3()1(3λλλλλ-------+-=3)1(2)1(23-+-+-=λλλ齐次线性方程组有非零解，则0=D 得 32,0===λλλ或不难验证，当32,0===λλλ或时，该齐次线性方程组确有非零解.第二章矩阵及其运算1 已知线性变换⎪⎩⎪⎨⎧++=++=++=3213321232113235322y y y x y y y x y y y x求从变量x 1 x 2 x 3到变量y 1 y 2 y 3的线性变换解由已知⎪⎪⎭⎫⎝⎛⎪⎪⎭⎫ ⎝⎛=⎪⎪⎭⎫ ⎝⎛221321323513122y y y x x x故 ⎪⎪⎭⎫ ⎝⎛⎪⎪⎭⎫ ⎝⎛=⎪⎪⎭⎫ ⎝⎛-3211221323513122x x x y y y ⎪⎪⎭⎫ ⎝⎛⎪⎪⎭⎫ ⎝⎛----=321423736947y y y⎪⎩⎪⎨⎧-+=-+=+--=321332123211423736947x x x y x x x y x x x y2 已知两个线性变换⎪⎩⎪⎨⎧++=++-=+=32133212311542322y y y x y y y x y y x ⎪⎩⎪⎨⎧+-=+=+-=323312211323z z y z z y z z y求从z 1 z 2 z 3到x 1 x 2 x 3的线性变换解由已知⎪⎪⎭⎫ ⎝⎛⎪⎪⎭⎫ ⎝⎛-=⎪⎪⎭⎫ ⎝⎛221321514232102y y y x x x ⎪⎪⎭⎫⎝⎛⎪⎪⎭⎫ ⎝⎛--⎪⎪⎭⎫ ⎝⎛-=321310102013514232102z z z⎪⎪⎭⎫⎝⎛⎪⎪⎭⎫ ⎝⎛----=321161109412316z z z所以有⎪⎩⎪⎨⎧+--=+-=++-=3213321232111610941236z z z x z z z x z z z x3 设⎪⎪⎭⎫ ⎝⎛--=111111111A ⎪⎪⎭⎫⎝⎛--=150421321B 求3AB 2A 及A TB解 ⎪⎪⎭⎫⎝⎛---⎪⎪⎭⎫ ⎝⎛--⎪⎪⎭⎫ ⎝⎛--=-1111111112150421321111111111323A AB⎪⎪⎭⎫⎝⎛----=⎪⎪⎭⎫ ⎝⎛---⎪⎪⎭⎫ ⎝⎛-=2294201722213211111111120926508503⎪⎪⎭⎫ ⎝⎛-=⎪⎪⎭⎫ ⎝⎛--⎪⎪⎭⎫ ⎝⎛--=092650850150421321111111111B A T4 计算下列乘积(1)⎪⎪⎭⎫⎝⎛⎪⎪⎭⎫ ⎝⎛-127075321134解 ⎪⎪⎭⎫ ⎝⎛⎪⎪⎭⎫ ⎝⎛-127075321134⎪⎪⎭⎫ ⎝⎛⨯+⨯+⨯⨯+⨯-+⨯⨯+⨯+⨯=102775132)2(71112374⎪⎪⎭⎫ ⎝⎛=49635(2)⎪⎪⎭⎫⎝⎛123)321(解 ⎪⎪⎭⎫⎝⎛123)321((132231)(10)(3))21(312-⎪⎪⎭⎫⎝⎛解 )21(312-⎪⎪⎭⎫⎝⎛⎪⎪⎭⎫ ⎝⎛⨯-⨯⨯-⨯⨯-⨯=23)1(321)1(122)1(2⎪⎪⎭⎫ ⎝⎛---=632142(4)⎪⎪⎪⎭⎫ ⎝⎛---⎪⎭⎫ ⎝⎛-20413121013143110412解 ⎪⎪⎪⎭⎫⎝⎛---⎪⎭⎫ ⎝⎛-20413121013143110412⎪⎭⎫⎝⎛---=6520876(5)⎪⎪⎭⎫⎝⎛⎪⎪⎭⎫ ⎝⎛321332313232212131211321)(x x x a a a a a a a a a x x x解⎪⎪⎭⎫ ⎝⎛⎪⎪⎭⎫ ⎝⎛321332313232212131211321)(x x x a a a a a a a a a x x x(a 11x 1a 12x 2a 13x 3 a 12x 1a 22x 2a 23x 3 a 13x 1a 23x 2a 33x 3)⎪⎪⎭⎫ ⎝⎛321x x x322331132112233322222111222x x a x x a x x a x a x a x a +++++=5 设⎪⎭⎫ ⎝⎛=3121A ⎪⎭⎫⎝⎛=2101B 问(1)AB BA 吗解 AB BA 因为⎪⎭⎫ ⎝⎛=6443AB ⎪⎭⎫⎝⎛=8321BA 所以AB BA(2)(A B)2A 22AB B 2吗解 (A B)2A 22AB B 2 因为⎪⎭⎫ ⎝⎛=+5222B A⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛=+52225222)(2B A ⎪⎭⎫ ⎝⎛=2914148但⎪⎭⎫ ⎝⎛+⎪⎭⎫ ⎝⎛+⎪⎭⎫ ⎝⎛=++43011288611483222B AB A ⎪⎭⎫⎝⎛=27151610所以(A B)2A 22AB B 2 (3)(A B)(A B)A 2B 2吗解 (A B)(A B)A 2B 2因为⎪⎭⎫ ⎝⎛=+5222B A⎪⎭⎫ ⎝⎛=-1020B A⎪⎭⎫ ⎝⎛=⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛=-+906010205222))((B A B A而 ⎪⎭⎫ ⎝⎛=⎪⎭⎫ ⎝⎛-⎪⎭⎫ ⎝⎛=-718243011148322B A故(A B)(A B)A 2B 26 举反列说明下列命题是错误的 (1)若A 20 则A 0 解取⎪⎭⎫ ⎝⎛=0010A 则A 20 但A 0(2)若A 2A 则A 0或A E 解取⎪⎭⎫ ⎝⎛=0011A 则A 2A 但A 0且A E(3)若AX AY 且A 0 则X Y 解取⎪⎭⎫ ⎝⎛=0001A⎪⎭⎫ ⎝⎛-=1111X ⎪⎭⎫ ⎝⎛=1011Y则AX AY 且A 0 但X Y7 设⎪⎭⎫ ⎝⎛=101λA 求A 2A 3Ak解 ⎪⎭⎫ ⎝⎛=⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛=12011011012λλλA⎪⎭⎫ ⎝⎛=⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛==1301101120123λλλA A A⎪⎭⎫ ⎝⎛=101λk A k8设⎪⎪⎭⎫ ⎝⎛=λλλ001001A 求A k解首先观察⎪⎪⎭⎫ ⎝⎛⎪⎪⎭⎫ ⎝⎛=λλλλλλ0010010010012A ⎪⎪⎭⎫ ⎝⎛=222002012λλλλλ⎪⎪⎭⎫ ⎝⎛=⋅=3232323003033λλλλλλA A A ⎪⎪⎭⎫ ⎝⎛=⋅=43423434004064λλλλλλA A A ⎪⎪⎭⎫ ⎝⎛=⋅=545345450050105λλλλλλA A A⎝⎛=kA k k kk k k k k k k λλλλλλ0002)1(121----⎪⎪⎪⎭⎫用数学归纳法证明当k 2时显然成立假设k 时成立，则k 1时，⎪⎪⎭⎫ ⎝⎛⎪⎪⎪⎪⎭⎫ ⎝⎛-=⋅=---+λλλλλλλλλ0010010002)1(1211k k k k k k k k k k k k A A A⎪⎪⎪⎪⎭⎫ ⎝⎛+++=+-+--+11111100)1(02)1()1(k k k k k k k k k k λλλλλλ 由数学归纳法原理知⎪⎪⎪⎪⎭⎫ ⎝⎛-=---k k k k k k k k k k k A λλλλλλ0002)1(1219 设A B 为n 阶矩阵，且A 为对称矩阵，证明B T AB 也是对称矩阵证明因为A T A 所以(B T AB)T B T (B T A)T B T A T B B T AB 从而B T AB 是对称矩阵10 设A B 都是n 阶对称矩阵，证明AB 是对称矩阵的充分必要条件是AB BA 证明充分性因为A T A B T B 且AB BA 所以(AB)T (BA)T A T B T AB 即AB 是对称矩阵必要性因为A T A B T B 且(AB)T AB 所以 AB (AB)T B T A T BA 11 求下列矩阵的逆矩阵 (1)⎪⎭⎫⎝⎛5221解⎪⎭⎫ ⎝⎛=5221A |A|1 故A 1存在因为 ⎪⎭⎫ ⎝⎛--=⎪⎭⎫ ⎝⎛=1225*22122111A A A A A故 *||11A A A =-⎪⎭⎫ ⎝⎛--=1225 (2)⎪⎭⎫ ⎝⎛-θθθθcos sin sin cos 解 ⎪⎭⎫ ⎝⎛-=θθθθcos sin sin cos A |A|10 故A 1存在因为⎪⎭⎫ ⎝⎛-=⎪⎭⎫ ⎝⎛=θθθθcos sin sin cos *22122111A A A A A所以*||11A A A =-⎪⎭⎫ ⎝⎛-=θθθθcos sin sin cos(3)⎪⎪⎭⎫⎝⎛---145243121解 ⎪⎪⎭⎫⎝⎛---=145243121A |A|20 故A 1存在因为⎪⎪⎭⎫ ⎝⎛-----=⎪⎪⎭⎫ ⎝⎛=214321613024*332313322212312111A A A A A A A A A A所以 *||11A A A =-⎪⎪⎪⎭⎫⎝⎛-----=1716213213012(4)⎪⎪⎪⎭⎫ ⎝⎛n a a a 0021(a 1a 2a n0)解 ⎪⎪⎪⎭⎫ ⎝⎛=n a a a A 0021由对角矩阵的性质知 ⎪⎪⎪⎪⎪⎪⎭⎫ ⎝⎛=-n a a a A 10011211 12 解下列矩阵方程 (1)⎪⎭⎫⎝⎛-=⎪⎭⎫⎝⎛12643152X解⎪⎭⎫ ⎝⎛-⎪⎭⎫ ⎝⎛=-126431521X ⎪⎭⎫ ⎝⎛-⎪⎭⎫ ⎝⎛--=12642153⎪⎭⎫ ⎝⎛-=80232(2)⎪⎭⎫ ⎝⎛-=⎪⎪⎭⎫ ⎝⎛--234311111012112X 解 1111012112234311-⎪⎪⎭⎫ ⎝⎛--⎪⎭⎫ ⎝⎛-=X⎪⎪⎭⎫ ⎝⎛---⎪⎭⎫ ⎝⎛-=03323210123431131 ⎪⎪⎭⎫⎝⎛---=32538122 (3)⎪⎭⎫ ⎝⎛-=⎪⎭⎫ ⎝⎛-⎪⎭⎫ ⎝⎛-101311022141X解 11110210132141--⎪⎭⎫ ⎝⎛-⎪⎭⎫ ⎝⎛-⎪⎭⎫ ⎝⎛-=X ⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-⎪⎭⎫ ⎝⎛-=210110131142121⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛=21010366121⎪⎪⎭⎫ ⎝⎛=04111 (4)⎪⎪⎭⎫ ⎝⎛---=⎪⎪⎭⎫ ⎝⎛⎪⎪⎭⎫ ⎝⎛021102341010100001100001010X解 11010100001021102341100001010--⎪⎪⎭⎫ ⎝⎛⎪⎪⎭⎫ ⎝⎛---⎪⎪⎭⎫ ⎝⎛=X⎪⎪⎭⎫ ⎝⎛⎪⎪⎭⎫ ⎝⎛---⎪⎪⎭⎫ ⎝⎛=010100001021102341100001010⎪⎪⎭⎫ ⎝⎛---=20143101213 利用逆矩阵解下列线性方程组(1)⎪⎩⎪⎨⎧=++=++=++3532522132321321321x x x x x x x x x解方程组可表示为⎪⎪⎭⎫⎝⎛=⎪⎪⎭⎫ ⎝⎛⎪⎪⎭⎫ ⎝⎛321153522321321x x x故 ⎪⎪⎭⎫ ⎝⎛=⎪⎪⎭⎫ ⎝⎛⎪⎪⎭⎫ ⎝⎛=⎪⎪⎭⎫ ⎝⎛-0013211535223211321x x x从而有 ⎪⎩⎪⎨⎧===001321x x x(2)⎪⎩⎪⎨⎧=-+=--=--05231322321321321x x x x x x x x x解方程组可表示为⎪⎪⎭⎫⎝⎛=⎪⎪⎭⎫ ⎝⎛⎪⎪⎭⎫ ⎝⎛-----012523312111321x x x故 ⎪⎪⎭⎫ ⎝⎛=⎪⎪⎭⎫ ⎝⎛⎪⎪⎭⎫ ⎝⎛-----=⎪⎪⎭⎫ ⎝⎛-3050125233121111321x x x故有 ⎪⎩⎪⎨⎧===35321x x x14 设A k O (k 为正整数) 证明(E A)1E A A 2 A k1证明因为A k O 所以E A k E 又因为E A k (E A)(E A A 2A k 1)所以 (E A)(E A A 2 A k 1)E 由定理2推论知(E A)可逆且(E A)1E A A 2A k1证明一方面有E (E A)1(E A) 另一方面由A k O 有 E (E A)(A A 2)A 2A k1(A k1A k )(E A A 2 A k 1)(E A)故 (E A)1(E A)(E A A 2 A k 1)(E A) 两端同时右乘(E A)1就有(E A)1(E A)E A A 2A k115 设方阵A 满足A 2A 2E O 证明A 及A 2E 都可逆并求A 1及(A 2E)1证明由A 2A 2E O 得A 2A 2E 即A(A E)2E 或E E A A =-⋅)(21由定理2推论知A 可逆且)(211E A A -=-由A 2A 2E O 得 A 2A 6E 4E 即(A 2E)(A 3E)4E或 E A E E A =-⋅+)3(41)2( 由定理2推论知(A 2E)可逆且)3(41)2(1A E E A -=+-证明由A 2A 2E O 得A 2A 2E 两端同时取行列式得 |A 2A|2 即 |A||A E|2 故 |A|0所以A 可逆而A 2E A 2 |A 2E||A 2||A|20 故A 2E 也可逆由 A 2A 2E O A(A E)2E A 1A(A E)2A 1E)(211E A A -=-又由 A 2A 2E O (A 2E)A 3(A 2E)4E(A 2E)(A 3E)4 E所以 (A 2E)1(A 2E)(A 3E)4(A 2 E)1)3(41)2(1A E E A -=+- 16 设A 为3阶矩阵 21||=A 求|(2A)15A*|解因为*||11A A A =- 所以 |||521||*5)2(|111----=-A A A A A |2521|11---=A A|2A 1|(2)3|A 1|8|A|1821617 设矩阵A 可逆证明其伴随阵A*也可逆且(A*)1(A 1)*证明由*||11A A A =- 得A*|A|A 1所以当A 可逆时有|A*||A|n |A 1||A|n 1从而A*也可逆因为A*|A|A 1所以(A*)1|A|1A又*)(||)*(||1111---==A A A A A 所以(A*)1|A|1A |A|1|A|(A 1)*(A 1)*18 设n 阶矩阵A 的伴随矩阵为A* 证明 (1)若|A|0 则|A*|0 (2)|A*||A|n 1证明(1)用反证法证明假设|A*|0 则有A*(A*)1E 由此得A A A*(A*)1|A|E(A*)1O所以A*O 这与|A*|0矛盾，故当|A|0时有|A*|0 (2)由于*||11A A A =- 则AA*|A|E 取行列式得到|A||A*||A|n 若|A|0 则|A*||A|n 1若|A|0 由(1)知|A*|0 此时命题也成立因此|A*||A|n119设⎪⎪⎭⎫ ⎝⎛-=321011330A AB A 2B 求B解由AB A 2E 可得(A 2E)B A 故⎪⎪⎭⎫ ⎝⎛-⎪⎪⎭⎫ ⎝⎛---=-=--321011330121011332)2(11A E AB ⎪⎪⎭⎫⎝⎛-=01132133020 设⎪⎪⎭⎫⎝⎛=101020101A 且AB E A 2B 求B解由AB E A 2B 得 (A E)B A 2E 即 (A E)B (A E)(A E)因为01001010100||≠-==-E A 所以(A E)可逆从而⎪⎪⎭⎫⎝⎛=+=201030102E A B21 设A diag(1 2 1) A*BA 2BA 8E 求B解由A*BA 2BA 8E 得(A*2E)BA 8EB 8(A*2E)1A 1 8[A(A*2E)]1 8(AA*2A)1 8(|A|E 2A)18(2E 2A)14(E A)14[diag(2 1 2)]1)21 ,1 ,21(diag 4-=2diag(1 2 1)22已知矩阵A 的伴随阵⎪⎪⎪⎭⎫⎝⎛-=8030010100100001*A且ABA 1BA13E 求B 解由|A*||A|38 得|A|2 由ABA1BA13E 得AB B 3AB 3(A E)1A 3[A(E A 1)]1A11*)2(6*)21(3---=-=A E A E⎪⎪⎪⎭⎫⎝⎛-=⎪⎪⎪⎭⎫⎝⎛--=-103006060060000660300101001000016123 设P 1AP 其中⎪⎭⎫ ⎝⎛--=1141P ⎪⎭⎫⎝⎛-=Λ2001求A 11解由P 1AP得A P P 1所以A 11 A=P 11P 1.|P|3 ⎪⎭⎫⎝⎛-=1141*P ⎪⎭⎫ ⎝⎛--=-1141311P而 ⎪⎭⎫⎝⎛-=⎪⎭⎫⎝⎛-=Λ11111120 012001故 ⎪⎪⎪⎭⎫⎝⎛--⎪⎭⎫ ⎝⎛-⎪⎭⎫ ⎝⎛--=31313431200111411111A ⎪⎭⎫ ⎝⎛--=6846832732273124 设AP P 其中⎪⎪⎭⎫⎝⎛--=111201111P ⎪⎪⎭⎫ ⎝⎛-=Λ511 求(A)A 8(5E 6A A 2) 解 ()8(5E 62)diag(1158)[diag(555)diag(6630)diag(1125)]diag(1158)diag(1200)12diag(100) (A)P ()P 1*)(||1P P P Λ=ϕ⎪⎪⎭⎫ ⎝⎛------⎪⎪⎭⎫ ⎝⎛⎪⎪⎭⎫ ⎝⎛---=1213032220000000011112011112⎪⎪⎭⎫⎝⎛=111111111425 设矩阵A 、B 及A B 都可逆证明A 1B 1也可逆并求其逆阵证明因为 A 1(A B)B 1B1A1A1B1而A 1(A B)B 1是三个可逆矩阵的乘积所以A 1(A B)B 1可逆即A1B 1可逆(A1B 1)1[A 1(A B)B 1]1B(A B)1A26 计算⎪⎪⎪⎭⎫⎝⎛---⎪⎪⎪⎭⎫⎝⎛30003200121013013000120010100121解设⎪⎭⎫ ⎝⎛=10211A ⎪⎭⎫ ⎝⎛=30122A ⎪⎭⎫ ⎝⎛-=12131B ⎪⎭⎫ ⎝⎛--=30322B则 ⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛2121B O B E A O E A ⎪⎭⎫ ⎝⎛+=222111B A O B B A A而 ⎪⎭⎫ ⎝⎛-=⎪⎭⎫ ⎝⎛--+⎪⎭⎫ ⎝⎛-⎪⎭⎫ ⎝⎛=+4225303212131021211B B A ⎪⎭⎫ ⎝⎛--=⎪⎭⎫ ⎝⎛--⎪⎭⎫ ⎝⎛=90343032301222B A 所以 ⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛2121B O B E A O E A ⎪⎭⎫ ⎝⎛+=222111B A O B B A A ⎪⎪⎪⎭⎫⎝⎛---=9000340042102521 即 ⎪⎪⎪⎭⎫ ⎝⎛---⎪⎪⎪⎭⎫ ⎝⎛30003200121013013000120010100121⎪⎪⎪⎭⎫ ⎝⎛---=9000340042102521 27 取⎪⎭⎫ ⎝⎛==-==1001D C B A 验证|||||||| D C B A D C B A ≠解41001200210100101002000021010010110100101==--=--=D C B A而01111|||||||| ==D C B A故 |||||||| D C B A D C B A ≠28 设⎪⎪⎪⎭⎫ ⎝⎛-=22023443O O A 求|A 8|及A 4解令⎪⎭⎫ ⎝⎛-=34431A ⎪⎭⎫ ⎝⎛=22022A 则 ⎪⎭⎫ ⎝⎛=21A O O A A故 8218⎪⎭⎫ ⎝⎛=A O O A A ⎪⎭⎫ ⎝⎛=8281A O O A 1682818281810||||||||||===A A A A A⎪⎪⎪⎭⎫⎝⎛=⎪⎭⎫ ⎝⎛=464444241422025005O O A O O A A29 设n 阶矩阵A 及s 阶矩阵B 都可逆求 (1)1-⎪⎭⎫⎝⎛O B A O解设⎪⎭⎫ ⎝⎛=⎪⎭⎫⎝⎛-43211C C C C O B A O 则 ⎪⎭⎫ ⎝⎛O B A O ⎪⎭⎫ ⎝⎛4321C C C C ⎪⎭⎫ ⎝⎛=⎪⎭⎫ ⎝⎛=s n E O O E BC BC AC AC 2143由此得 ⎪⎩⎪⎨⎧====s n E BC O BC O AC E AC 2143⎪⎩⎪⎨⎧====--121413B C O C O C A C所以 ⎪⎭⎫ ⎝⎛=⎪⎭⎫ ⎝⎛---O A B O O B A O 111(2)1-⎪⎭⎫ ⎝⎛B C O A解设⎪⎭⎫ ⎝⎛=⎪⎭⎫⎝⎛-43211D D D D B C O A 则 ⎪⎭⎫ ⎝⎛=⎪⎭⎫ ⎝⎛++=⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛s n E O O E BD CD BD CD AD AD D D D D B C O A 4231214321由此得 ⎪⎩⎪⎨⎧=+=+==s nE BD CD O BD CD OAD E AD 423121⎪⎩⎪⎨⎧=-===----14113211B D CA B D O D A D所以 ⎪⎭⎫ ⎝⎛-=⎪⎭⎫ ⎝⎛-----11111B CA B O A BC O A30 求下列矩阵的逆阵(1)⎪⎪⎪⎭⎫⎝⎛2500380000120025 解设⎪⎭⎫ ⎝⎛=1225A ⎪⎭⎫ ⎝⎛=2538B 则⎪⎭⎫⎝⎛--=⎪⎭⎫ ⎝⎛=--5221122511A ⎪⎭⎫⎝⎛--=⎪⎭⎫⎝⎛=--8532253811B于是 ⎪⎪⎪⎭⎫ ⎝⎛----=⎪⎭⎫ ⎝⎛=⎪⎭⎫ ⎝⎛=⎪⎪⎪⎭⎫ ⎝⎛----850032000052002125003800001200251111B A B A(2)⎪⎪⎪⎭⎫ ⎝⎛4121031200210001解设⎪⎭⎫ ⎝⎛=2101A ⎪⎭⎫ ⎝⎛=4103B ⎪⎭⎫ ⎝⎛=2112C 则⎪⎭⎫ ⎝⎛-=⎪⎭⎫ ⎝⎛=⎪⎪⎪⎭⎫⎝⎛------1111114121031200210001B CA B O A BC O A⎪⎪⎪⎪⎪⎪⎭⎫⎝⎛-----=411212458103161210021210001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) ⎪⎪⎪⎭⎫ ⎝⎛--340313*********2)3()2(~r r r r -+-+⎪⎪⎪⎭⎫ ⎝⎛---020********* )2()1(32~-÷-÷r r ⎪⎪⎪⎭⎫ ⎝⎛--01003100120123~r r -⎪⎪⎪⎭⎫⎝⎛--300031001201 33~÷r ⎪⎪⎪⎭⎫ ⎝⎛--100031001201323~r r +⎪⎪⎪⎭⎫ ⎝⎛-1000010012013121)2(~r r r r +-+⎪⎪⎪⎭⎫⎝⎛100001000001(2) ⎪⎪⎪⎭⎫ ⎝⎛----1740343013201312)2()3(2~r r r r -+-+⨯⎪⎪⎪⎭⎫ ⎝⎛---31003100132021233~r r r r ++⎪⎪⎪⎭⎫ ⎝⎛000031001002021~÷r ⎪⎪⎪⎭⎫⎝⎛000031005010 (3) ⎪⎪⎪⎪⎭⎫⎝⎛---------12433023221453334311141312323~rr r r rr ---⎪⎪⎪⎪⎭⎫ ⎝⎛--------1010500663008840034311)5()3()4(432~-÷-÷-÷r r r ⎪⎪⎪⎪⎭⎫⎝⎛-----22100221002210034311 2423213~r r r r r r ---⎪⎪⎪⎪⎭⎫⎝⎛---000000000022********(4) ⎪⎪⎪⎪⎭⎫⎝⎛------34732038234202173132 242321232~rr r r rr ---⎪⎪⎪⎪⎭⎫ ⎝⎛-----1187701298804202111110141312782~rr r r r r --+⎪⎪⎪⎪⎭⎫⎝⎛--410004100020201111134221)1(~r r r r r --⨯↔⎪⎪⎪⎪⎭⎫⎝⎛----0000041000111102020132~rr +⎪⎪⎪⎪⎭⎫⎝⎛--000004100030110202012.设⎪⎪⎪⎭⎫ ⎝⎛=⎪⎪⎪⎭⎫ ⎝⎛⎪⎪⎪⎭⎫ ⎝⎛987654321100010101100001010A ，求A 。

线性代数课后练习参考答案(初稿)

线性代数课后习题参考答案(初稿)习题一1. 用行列式定义计算下列各题（1）4245322635-=-⨯-⨯=-（2）12130111110101(1)(1)21011110++=-+-=（3）1312001002020030(1)3002(1)243000040040004++=-=⨯-=- （4）11121310000230234645(1)4562(1)3(1)4045681089891078910+++=-=⨯-+⨯-= 2. 利用行列式的性质计算下列各题（1）214121413121506201232123250625062-== （2）28512851105131025319061906512511310805120512121100107609712--------==---=----=----------（3）111111111abac aebcebdcdde adf b c e adfbce bfcfef b c e ----=-=----111024020adfbce adfbce -== （4）3300011()()01000a b b b a b b b ab a b a b a a b a a b a a b a a b b a a b b b b a b a b a -==--=-------- （5）x a a aa x aa a ax a a a ax =(1)(1)(1)(1)x n a a aax n a x a ax n a a x a x n a a ax+-+-+-+- =[(1)]x n a+-1111a aa x a a a x a a ax=[(1)]x na +-1001001001x ax a x a---[(1)]x n a =+-1()n x a --（6）22222222222222222222(1)(2)(3)212325(1)(2)(3)2123250(1)(2)(3)212325(1)(2)(3)212325a a a a a a a ab b b b b b b bc c c c c c c cd d d d d d d d ++++++++++++==++++++++++++（7）12311000011231110001223110200(1)!1232110020123111001n n n n n n n n n n n n n nn -+-+-==--+----+-(8)0121111110001012111112002131111122012301230123241n n n n n n n n n n n n n --------==-----------------12(1)2(1)n n n --=--3. 证明下列各题（1）111111111111111122222222222222223333333333333333a b b c c a a b c c a b b c c a a b b c c a a b c c a b b c c a a b b c c a a b c c a b b c c a ++++++++++=++++++++++++111111111111112222222222222233333333333333a b c c b c c a a b c b c a a b c c b c c a a b c b c a a b c c b c c a a b c b c a ++=+++=+++ 1112223332a b c a b c a b c = （2）00()()()()00x y z x z yx y z y z x z x y x y z y z x z y x =-+++-+-+-（证明略）（3）11111111111111111110111111111110111111111110111x x x x x y y y y y y+---=++++--- 21000111111111001111110111001111110111000x x x x y xy x y y y y y y y-⎛-⎫- ⎪=++=++++ ⎪ ⎪---⎝⎭- 222222210011001100y xy x y x xy xy x y x y y y ⎛⎫+ ⎪=+-=-+= ⎪- ⎪-⎝⎭（4）设01211000100010n n n a a x D a x a x----=-，则按最后一行展开，可得01113210001101(1)0011n n n n n a a x xD a x a x x a x+-------=-+--211122122()n n n n n n n n a xD a x a xD a xa x D --------=+=++=++. 332123223321123210n n n n n n n n n n n a xa a x a xx D a xa a x a x a x a x -----------==+++++=++++++4. 解法参考例 1.11.5. 问齐次线性方程组123123123(1)2402(3)0(1)0x x x x x x x x x λλλ--+=⎧⎪+-+=⎨⎪++-=⎩ 有非零解时，必须满足什么条件？解：齐次线性方程组有非零解，当且仅当1242310111λλλ---=-. 又124111111231231012111112403(1)(3)λλλλλλλλλλλλ-----=--=--------+-(2)(3)0,λλλ=---=解得，0,λ=或2λ=，或3λ=.所以，当0,λ=或2λ=，或3λ=，齐次线性方程组有非零解.习题二 1. 1654127,2211210712A B A B -⎛⎫⎛⎫+=-=⎪ ⎪---⎝⎭⎝⎭2. 解：由A X B +=，得020133.221X B A -⎛⎫⎪=-=-- ⎪ ⎪--⎝⎭ 3. 解：213220583221720,0564292290T AB A A B -⎛⎫⎛⎫ ⎪ ⎪-=--=- ⎪ ⎪ ⎪ ⎪-⎝⎭⎝⎭ 4. 解：（1）()31,2,32132231101⎛⎫ ⎪=⨯+⨯+⨯= ⎪ ⎪⎝⎭ （2）()22411,212336-⎛⎫⎛⎫ ⎪ ⎪-=- ⎪ ⎪ ⎪ ⎪-⎝⎭⎝⎭，（3）12110162134021311491231042217--⎛⎫⎛⎫⎛⎫ ⎪⎪ ⎪=⎪⎪ ⎪ ⎪⎪ ⎪-⎝⎭⎝⎭⎝⎭（4） 131********78113413120510402⎛⎫⎪--⎛⎫⎛⎫⎪= ⎪ ⎪ ⎪---⎝⎭⎝⎭⎪⎝⎭5. 解：（1）错误，令1101,,0111A B ⎛⎫⎛⎫== ⎪ ⎪⎝⎭⎝⎭则有AB BA ≠；（2）错误，令1101,,0111A B ⎛⎫⎛⎫== ⎪ ⎪⎝⎭⎝⎭则有222()2.A B A AB B +≠++(3) 错误，令1101,,0111A B ⎛⎫⎛⎫== ⎪ ⎪⎝⎭⎝⎭则可得22()().A B A B A B +-≠- (4) 错误，设00,10A ⎛⎫= ⎪⎝⎭则有20A =，但0.A ≠（5）错误，设10,00A ⎛⎫= ⎪⎝⎭则有2A A =，但.A I ≠6．解：2221010(),0101AB A B -⎛⎫⎛⎫== ⎪ ⎪-⎝⎭⎝⎭7．证明：因为A 为对称矩阵，所以T A A =. 故(),T T T T T B AB B A B B AB ==因此，T B AB 是对称矩阵.8. 证明：因为(),(),T T T T T T A A A A AA AA == 所以,T T A A AA 是对称矩阵.9. 解：由32,A X B -=得43/211(3)15/2127/211/25/2X B A -⎛⎫ ⎪=--=- ⎪ ⎪⎝⎭. 10. 2cos 2sin 2,sin 2cos 2A θθθθ-⎛⎫=⎪⎝⎭cos sin sin cos n n n A n n θθθθ-⎛⎫= ⎪⎝⎭对n 作数学归纳法. 当2n =时,22222cos 2sin 2cos sin 2cos sin sin 2cos 22cos sin cos sin A θθθθθθθθθθθθ-⎛⎫--⎛⎫==⎪ ⎪-⎝⎭⎝⎭, 结论成立. 假设, 当n k =时, 结论成立, 即cos sin sin cos k k k A k k θθθθ-⎛⎫=⎪⎝⎭. 下证1n k =+结论成也立. 由归纳假设可得,1k A+=cos sin cos sin sin cos sin cos k k k A A k k θθθθθθθθ--⎛⎫⎛⎫=⎪⎪⎝⎭⎝⎭cos cos sin sin cos sin sin cos cos sin sin cos cos cos sin sin k k k k k k k k θθθθθθθθθθθθθθθθ---⎛⎫=⎪+-⎝⎭cos(1)sin(1)sin(1)cos(1)k k k k θθθθ+-+⎛⎫=⎪++⎝⎭因此，由归纳法可得cos sin sin cos n n n A n n θθθθ-⎛⎫=⎪⎝⎭. 11. （1）解：由初等行变换可得，11103111031110311007221240012200122001043314500244000390001311118002150000000000A -------⎛⎫⎛⎫⎛⎫⎛⎫⎪⎪ ⎪⎪----⎪ ⎪ ⎪ ⎪=→→→⎪ ⎪ ⎪ ⎪------ ⎪⎪ ⎪⎪-⎝⎭⎝⎭⎝⎭⎝⎭（2）解：由初等行变换可得，111111107125016016234000000⎛⎫⎛⎫⎛⎫ ⎪ ⎪ ⎪-→-→- ⎪ ⎪ ⎪ ⎪ ⎪ ⎪-⎝⎭⎝⎭⎝⎭12. 解法见第38页例2.14. 13. (1)解：22222311111111111011111110111λλλλλλλλλλλλλλλλλλλ⎛⎫⎛⎫⎛⎫ ⎪ ⎪ ⎪→→---⎪ ⎪ ⎪ ⎪ ⎪ ⎪---⎝⎭⎝⎭⎝⎭2221101100(1)(2)(1)(1)λλλλλλλλλλ⎛⎫ ⎪→--- ⎪ ⎪-+-+⎝⎭，当2λ=-时，方程组无解，当1λ=时，方程组的增广矩阵为111100000000⎛⎫⎪ ⎪ ⎪⎝⎭因此方程组的解为12111010001k k --⎛⎫⎛⎫⎛⎫⎪ ⎪ ⎪++ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭， 12,k k 为任意常数，当1λ≠，且2λ≠-时，方程组有唯一解，221211(1)(1),,222x x x λλλλλλλ+++=-=-+=-+++（2）解：322111************213221λλλλλλλλλλλλ---⎛⎫⎛⎫ ⎪⎪--→-- ⎪ ⎪ ⎪ ⎪---⎝⎭⎝⎭ 112111210111011101(2)(1)2(1)00(1)(3)1λλλλλλλλλλλλλλλ--⎛⎫⎛⎫⎪ ⎪→-+--→--- ⎪ ⎪ ⎪ ⎪-------⎝⎭⎝⎭当1λ=时，方程组无解，方程组的增广矩阵为111100000000⎛⎫⎪ ⎪ ⎪⎝⎭因此方程组的解为12111010001k k --⎛⎫⎛⎫⎛⎫⎪ ⎪ ⎪++ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭， 12,k k 为任意常数，当3λ=时，方程组无解，当3λ≠且1λ≠时，方程组有唯一解，123411,,.33x x x λλλ-=-==-- 14. 解：通过初等变换，可得A 的标准型矩阵为，17100010101002800105100015⎛⎫- ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪-⎝⎭15. 解析：通过初等行变换可将矩阵()A I 化为()()A I I B →，则1A B -= 例如（1）通过初等行变换，121012101052250101210121-⎛⎫⎛⎫⎛⎫→→ ⎪ ⎪ ⎪--⎝⎭⎝⎭⎝⎭，故 112522521--⎛⎫⎛⎫= ⎪ ⎪-⎝⎭⎝⎭相类似的方法可求的其余矩阵的逆矩阵，答案见教材第177页. 16. 解：原线性方程组可写成123123122103430x x x ⎛⎫⎛⎫⎛⎫⎪⎪ ⎪= ⎪⎪ ⎪ ⎪⎪ ⎪⎝⎭⎝⎭⎝⎭，因此，11231123132210234301x x x -⎛⎫⎛⎫⎛⎫⎛⎫ ⎪⎪ ⎪ ⎪ ⎪==- ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭⎝⎭17.（1）由原矩阵方程可得121122111321182431511133X --⎛⎫-⎛⎫-⎛⎫ ⎪ ⎪== ⎪ ⎪ ⎪-- ⎪⎝⎭ ⎪-⎝⎭⎝⎭，（2）由原矩阵方程可得1111143120112011104X --⎛⎫⎛⎫⎛⎫⎛⎫ ⎪== ⎪ ⎪⎪ ⎪---⎝⎭⎝⎭⎝⎭⎝⎭（3）由原矩阵方程可得11010143100210100201001134001120010102X ----⎛⎫⎛⎫⎛⎫⎛⎫⎪ ⎪⎪ ⎪=-=- ⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪--⎝⎭⎝⎭⎝⎭⎝⎭18证明：因为21()()k k I A I A A A I A I +-++++=-=，所以12()()k I A I A A A --=++++19．解：由220A A I --=，得()2A I AI -=，3(2)4A IA I I -+=-，因此，1(),2A I A --=13(2)4A IA I --+=- 20. 证明：由220A AB B ++=，且B 可逆得，22[()],()A A B B E B A A B E ---+=-+=，因此，,A A B +可逆，且1212(),().A A B B A B B ----=-++=-21. 令11123,01121001B C ⎛⎫⎛⎫ ⎪== ⎪ ⎪⎝⎭ ⎪⎝⎭，则111311044,0111100122B C --⎛⎫-⎛⎫- ⎪ ⎪==-⎪ ⎪ ⎪ ⎪- ⎪⎝⎭⎝⎭，因此1111130004411000002200001100001101B B A A A ----⎛⎫- ⎪ ⎪ ⎪-⎛⎫⎛⎫⎪=== ⎪⎪ ⎪⎝⎭⎝⎭- ⎪ ⎪- ⎪⎝⎭. 22. 证明：若,B C 可逆，则有11000B C I CB --⎛⎫⎛⎫= ⎪ ⎪⎝⎭⎝⎭，所以A 可逆，且1110.0C A B---⎛⎫= ⎪⎝⎭ 反之，若A 可逆, 设其逆为XY Z V ⎛⎫⎪⎝⎭，则， 000B X Y I o C Z V I ⎛⎫⎛⎫⎛⎫= ⎪⎪ ⎪⎝⎭⎝⎭⎝⎭，因此，,BZ I CY I ==，因此,B C 可逆.23. 证明：用反证法. 假设A 是奇异矩阵，则由2A A =，得211A A AA --=，即A E =，这与已知条件矛盾，所以A 是非奇异矩阵.习题三 1. (3,8,7)T β=2. 解: 设11223344,x x x x βαααα=+++ 即12341111121111,1111111111x x x x ⎛⎫⎛⎫⎛⎫⎛⎫⎛⎫ ⎪ ⎪ ⎪ ⎪ ⎪-- ⎪ ⎪ ⎪ ⎪ ⎪=+++ ⎪ ⎪ ⎪ ⎪ ⎪-- ⎪ ⎪ ⎪ ⎪ ⎪--⎝⎭⎝⎭⎝⎭⎝⎭⎝⎭ 解得, 12345111,,,4444x x x x ===-=-, 因此12345111.4444βαααα=+--3. 解: 由3(),αβαβ-=+ 得117(1,,2,)222T αα=-=---. 4. 类似第2题的解法，可得1234243.βαααα=+-+ 5. (1) 解: 设1122330,x x x ααα++= 即1231111260133x x x ⎛⎫⎛⎫⎛⎫⎪ ⎪ ⎪++= ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭，上面方程组只有零解，所以123,,ααα线性无关. (2) 因为111111111141406120612117024000A ⎛⎫⎛⎫⎛⎫ ⎪ ⎪ ⎪=-→-→- ⎪ ⎪ ⎪ ⎪ ⎪ ⎪-⎝⎭⎝⎭⎝⎭, 所以秩(A)=2, 故123,,ααα线性相关. 6. 用反证法容易证明结论成立.7. 证明: (1) 设11220,m m x x x βββ+++= 则有11220,m m x x x ααα+++= 又因为12,,,m ααα线性无关, 所以120,m x x x ==== 因此12,,,,mβββ线性无关.(2) 若12,,,,m βββ线性相关, 则存在不全为零的数12,,,,m x x x 使得11220,m m x x x βββ+++= 因此11220,m m x x x ααα+++= 故而12,,,m ααα线性相关.8. 证明: ()⇒设112223331()()()0,k k k αααααα+++++= 整理得,131122233()()()0k k k k k k ααα+++++=,因为123,,ααα线性无关, 所以131223000k k k k k k +=⎧⎪+=⎨⎪+=⎩ 又因为1011100011≠, 所以上面方程组只有零解, 故122331,,αααααα+++线性无关.()⇐ 设1122330,k k k ααα++= 整理得,123121232312331111()()()()()()0,222k k k k k k k k k αααααα+-++-++++-++= 又因为122331,,αααααα+++线性无关，所以123123123(000k k k k k k k k k +-=⎧⎪-++=⎨⎪-+=⎩ 解得上面方程组只有零解，因此，123,,ααα线性无关. 证明： 9.（⇒）设1mi i i k αα==∑，和10.mi i i l α==∑ 则，111()mmmi i i i i i i i i i k l k l αααα====+=+∑∑∑，又α的表达式唯一，因此,i i i k l k += 即0,i l = 故，12,,,m ααα 线性无关.（⇐）设11m m i i i i i i k l ααα====∑∑，则1()0mi i i i k l α=-=∑，因为12,,,m ααα 线性无关，所以，,i i k l =故α的表达式唯一.10. 证明：因为12,,,m ααα 线性相关，则存在不全为零的数12,,,m k k k 使得，10.mi i i k α==∑若有某个0i k =，不妨设10k =，则有20,mi i i k α==∑ 又任一1m -向量都线性无关，因此230m k k k ====，这与12,,,m k k k 不全为零矛盾，因此12,,,m k k k 全不为零，命题得证. 11. 答案见教材178页. 12. 解: (1) 因为13213213221307107132076005A c c c ⎛⎫⎛⎫⎛⎫ ⎪ ⎪ ⎪=-→--→-- ⎪ ⎪ ⎪ ⎪ ⎪ ⎪--+-+⎝⎭⎝⎭⎝⎭所以，当50,c -+≠ 即5c ≠时，123,,ααα线性无关.(2 ) 当5c =时，123,,ααα线性相关，且312111.77ααα=+ 13. 解：（1）因为234411231123112311232344050100501032613261050100000102110210120000A ------⎛⎫⎛⎫⎛⎫⎛⎫⎪ ⎪ ⎪⎪-- ⎪ ⎪ ⎪ ⎪=→→→⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪⎪------⎝⎭⎝⎭⎝⎭⎝⎭因此，向量组1234,,,αααα的秩为2， 12,αα是一个极大线性无关组，且314122,2.ααααα==-+用类似的方法可求（2），（3），答案见教材.14. (1) 因为120131(,)1224αα⎛⎫ ⎪- ⎪= ⎪ ⎪⎝⎭，有一个二阶子式01331=--，所以秩（12,αα）=2，即12,αα线性无关.（2）容易计算124,,ααα线性无关. 15. 答案见教材.16. (1)任取()()12121,,,,,,,,,n n x x x y y y V k R ∈∈则有11220n n x y x y x y ++++++=，120n kx kx kx +++=所以()()()121211221,,,,,,,,,n n n n x x x y y y x y x y x y V +=+++∈，12121(,,,)(,,,)n n k x x x kx kx kx V =∈，因此，1V 是线性空间.(2) 任取()()12122,,,,,,,n n x x x y y y V ∈,则有11222n n x y x y x y ++++++=，因此， ()()()121211222,,,,,,,,,.n n n n x x x y y y x y x y x y V +=+++∉ 因此，2V 不是线性空间. 17. 证明：因为111111101101211110011==-=--，所以123,,ααα线性无关，即秩（123,,ααα）=3，故123,,ααα生成的子空间就是R .18. 因为 12311160,032-=-≠ 所以，秩（123,,ααα）=3，故123,,ααα是R 的一组基.令1112233k k k βααα=++，即123(5,0,7)(1,1,0)(2,1,3)(3,1,2).k k k =-++ 因此123123232350327k k k k k k k k ++=⎧⎪-++=⎨⎪+=⎩，解得，1232,3,1,k k k ===- 所以112323βααα=+-.19. 方法见例3.17. 20. 见教材答案21. 证明：因为A 是正交阵，所以21,1T A A A -==.又*,A A A E = 即*1A A A -=.因此，2**()T A A A E E ==，故*A 是正交阵. 习题四 1. 解（1）1251251251320170171490214000378017000⎛⎫⎛⎫⎛⎫⎪ ⎪⎪---⎪ ⎪ ⎪→→⎪ ⎪ ⎪-- ⎪ ⎪⎪-⎝⎭⎝⎭⎝⎭, 所以，原方程组与下面方程组同解，1232325070x x x x x ++=⎧⎨-=⎩选取3x 作为自由未知量，解得基础解系为1971-⎛⎫ ⎪ ⎪ ⎪⎝⎭，因此，方程组的解为1971k -⎛⎫ ⎪ ⎪ ⎪⎝⎭（2）313411311131159815980467113131340000------⎛⎫⎛⎫⎛⎫⎪ ⎪ ⎪--→--→-- ⎪ ⎪ ⎪ ⎪ ⎪ ⎪----⎝⎭⎝⎭⎝⎭，选取选取34,x x 作为自由未知量，解得基础解系为3/23/43/27/4,1001-⎛⎫⎛⎫⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎝⎭⎝⎭故方程组的同解为123/23/43/27/41001k k -⎛⎫⎛⎫⎪ ⎪ ⎪ ⎪+ ⎪ ⎪ ⎪ ⎪⎝⎭⎝⎭（3）见教材答案（4）见教材答案2. （1）对增广矩阵做行初等变换得1121011210(,)211210*********/200031/2A b --⎛⎫⎛⎫ ⎪ ⎪=--→ ⎪ ⎪ ⎪ ⎪----⎝⎭⎝⎭解得特解为5/6101/6⎛⎫⎪ ⎪ ⎪ ⎪-⎝⎭，对应的齐次线性方程组的基础解系为3510-⎛⎫ ⎪- ⎪ ⎪ ⎪⎝⎭，因此方程组的同解为5/6101/6⎛⎫ ⎪ ⎪ ⎪ ⎪-⎝⎭+3510k -⎛⎫ ⎪- ⎪ ⎪ ⎪⎝⎭（2）答案见教材 3. （略）4. 证明：令i e 为n 阶单位矩阵的第i 列，即(0,0,,1,0,,0)Ti ie =, 则有0,1,2,,i Ae i n ==，因此12(,,,)0,n A e e e AI == 故0A =。

线性代数课后习题3部分答案

1 1 r3 r4 1 r3 r1r1 0 r2 2 0 0 0 0
1 1 0 0
1 1 1 0
向量组的秩为3等于向量的个数，所以1 , 2 , 3线性无关。
• (2)
1 1 1 r r 1 1 1 1 3 1 r2 r1 (1 , 2 , 3 ) 1 3 1 0 2 0 0 1 3 k 0 2 k-1 0
1 r2 ( 2) r3 3 r2 0 r4 5 r2 0 0
2 1 r 1 1 3 r34 (r32) 0 r2 r3 0 0 2 4 0 2 4 0 3 0
2 1 0 1 0 1 2 0 0 0 3 0
1 0 1 0 2 0 1 1 0 2 0 0 0 1 1r2
r ( A) 3 5,
x3 , x5为自由未知量
(x3 ,x5 )T 分别取为(1,0)T ,(0,1)T , 得到两个解：
5 (1)解：将向量按列构成矩阵：
1 1 (1 , 2 , 3 , ) 1 2 2 1 3 0 2 1 2 4 r32 rr11 0 2 2 6 r r4 2 r1 0 3 1 5 0 1 3 1 3 7 0 5 3 11 3 0
1.(3)解：
1 2 3 4 4 1 2 3 4 4 r3 r1 0 1 1 1 3 0 1 1 1 3 A （A，b)= 0 5 3 5 3 1 3 0 1 1 0 7 3 1 3 0 7 3 1 3
+2 =(1,0,3,-2) +(-6,2,0,4) =(-5,2,3,2)

线性代数(含全部课后题详细答案)1第一章一元多项式习题及解答.docx

A 组1.判别Q (厉)二｛0 +勿亦|0,处0｝是否为数域？解是.2.设/(x) = x3 4-x2 4-x+l, g(兀)=兀2+3兀+ 2,求 /(兀)+ g(x)，/(x)-g(x), f(x)g(x). 解/(x) + g (x) = x3 4- 2x2 + 4x + 3 ,/(兀)-g(x)"-2x-l,f(x)g(x) = x5 +4x4 +6兀'+6兀$ +5x + 2 .3.设/(%) = (5x-4),993(4x2 -2x-l),994 (8x3 -1 lx+2)'995,求 /(%)的展开式中各项系数的和.解由于/(兀)的各项系数的和等于/⑴，所以/(I) = (5-4严3(4-2- 1尸94(8-11 + 2)1995 =-1.4.求g(兀)除以/(兀)的商q(x)与余式心).(1)f (x) —— 3%2— x — 1, g(兀)=3F - 2兀+1 ；(2)/(x) = x4 -2x4-5, g(x) = x2 -x + 2 .解（1）用多项式除法得到x 73x~ — 2x +13_93X + 3—x —x-i3 37 ° 14 7-- 无_+ —x --3 9 926 2-- X ---9 9所以'恥）十岭心）W（2）用多项式除法得到x4— 2x + 5兀4 —”丫" + 2 兀2— 2x~ — 2 兀+5 jy?—兀~ + 2 兀-x2-4x4-5-兀? + X - 2—5x + 7所以，q(x) = x2 +x-l, r(x) = -5x + 7 .5.设是两个不相等的常数，证明多项式/(兀)除以(x-a)(x-b)所得余式为af(b)_bg)a-b a-h证明依题意可设/(x) = (x - a)(x - b)q(x) + cx+d,则”(a) = ca + d,[f(b) = cb + d.解得F=(/a) --,\d = (af(b)-bf(a))/(a-b).故所得余式为a-b a-b6.问m,p,q适合什么条件时，/(兀)能被g(x)整除？(1) /(x) = x3 + px + q , g(x) = x2 + nvc-1；(2) f(x) = x4 + px2 +q , g(兀)=x2 + mx+l.解(1)由整除的定义知，要求余式r(x) = 0 .所以先做多项式除法,3x2 + mx -1x-in“+ “X + q3 2x + mx^ - x-mx1 +(〃 + l)x + g2 2一 mx_ — m^x + m°(# +1 + 加〜)兀 + (g —m)要求厂(x) = (/? + l +加2)兀+ (§ —加)=0 ,所以(“ + 1 +加2) = 0, q-m = 0.即p = -l-m2, q - m时, 可以整除.（2）方法同上.先做多项式除法，所得余式为厂（兀）=加（2 — ”一nr ）兀+ （1 + @ —卩一加〜），所以 m （2-p-/772） = 0, 1 + ^ - p - m 2= 0 ,即 m = 0, p = q + \ 或“二 2— 加［q = l 时，可以整除.7. 求/（兀）与gCr ）的最大公因式:(1) f (x) — x 4 + — 3%2 — 4x — 1, g (x)=兀彳 + — x — 1 ； (2) f(x) = x 4— 4x 3+ 1, g(x) = x 3— 3x 2+1 ；(3) /(x) = x 4 -10x 2 +1, g(x) = x 4 -4A /2X 3 +6X 2 +4A /2X +1 .解（1）用辗转相除法得到用等式写出來，就是所以(/(x),g(x)) = x + l ・（2）同样地,<8 4 / 3 3= -X + — — -X-—(3 344-2x 2-3x-l1 1 --- X 4——2 -- 4 X 3+ X 2- X - 1 x 4 + x 3- 3x 2- 4x- 11 2 3 , -2x 2 — 3兀—12 21 2 3 1 -- X ----- X ---—2兀~ — 2兀2 4 433-- X ----X -144一丄 184—X H - 3 3 0心宀丄兀2 24 3 2牙+牙-X - Xf(x) = xg(x)^(-2x 2-3x-l),g(x) =所以(/⑴,g (兀)) = 1.⑶ 同样用辗转相除法，可得(/(x),g(x)) = F —2血兀一1.8.求 w(x),仄兀)使 w(x) f\x) + v(x)g(ji) = (/(x), g(%))：(1) f (x) = %4 4- 2x^ — %2 — 4x — 2, (x) = %4 + x — x~ — 2x — 2 : (2) /(x) = 4x 4-2x 3-16x 2+5x4-9, g(x) = 2兀3-x 2-5x+4:(3) /(x) = x A-x 3-4x 2 +4x + l, g (兀)=x 2 -x-l.解(1)利用辗转相除法，可以得到/(x) = g (A ：) + (x 3-2x)'g (兀)=(x+l)(x 3 - 2x) + (x 2 -2),x — 2兀=x(^x~ — 2).因而，(/(x),g(x)) = x 2-2,并且(/(兀),g (兀))=/ 一 2 = g (兀)_ (兀+1)(疋 _ 2兀) =g (兀)一(X +1) (f(x) -g (兀))=(一兀一 1)/(兀)+ (兀+2)g(x),所以 u(x) = -x-\, v(x) = x + 21 10 -- X H --- 3 9x 3 - 3x 2x-13 1 2 2X H —X X 3 3 10 2 2~~'- ---- X H 兀+ 13 -- 3 10 ° 10 20 X --- 兀 3 9 916~~1T —X ------ 9 927 441 --------- X ---------------16 256-3x 2+—x1649一一539 兀+ --- 27 256(2)利用辗转相除法，可以得到/(x) = 2xg(x)-(6x 2 +3兀-9),(\ 1Ag(x) = —(6x_ + 3兀一9) ——% + — — (% — 1), —(6x - + 3x — 9) = —(x —1)(6% + 9).因而，(/⑴,g(Q) = x-1，并且(1 1 …厶— —X + _ f (x) + _兀_—x~\ I 3 3丿 (3 3丿] 1 2 7 2fi/f 以 W (X )= X H —, V (X )= — --- X — \ •3 3 3 3(3) 利用辗转相除法，可以得到fM = X —3)g(x) + (x — 2),g(x) = (x+l)(x-2) + l ・因而( f(x), g(x)) = 1 ,并且(/(兀),g(x)) = 1 = g(x) - (x+1)(兀一 2)=g (兀)-(兀+1)(/(兀)-(x 2 一3)gCr))—(—兀―1) f (x) + (兀'+ 兀2 — 3兀—2)g(x),所以u (兀)= -x-l, v(x) = x 3 +x 2 -3x-2.9.设/(x) = %3+ (14-t)x 2+ 2x + 2w, g(x)二F+zx + u 的最大公因式是一个二次多项式，求/，凤的值.解利用辗转相除法，可以得到/(%) = g(x) + (l + /)兀2 +(2-/)兀 + « ,(/(x), g(x)) = x-l = -(6x 2+ 3x-9)+ | _g(x)I d J J(I ] \= (/(x)-2xg(x)) --x+- -g(x)\ 3丿 <2 o 2 d ,、 U 3 广—---- 兀+ (1 + r t-2(l +r)2(尸 + r—w)(i+r) + (t— 2)~u[(l + t)2 — (r —2)]由题意，/(x)与g(Q的最大公因式是一个二次多项式，所以(广 + / —w)(l + /) + (f— 2)~(T H?皿(l + r)2-(r-2)] A ；=0,(l + O2解得u = o^t = -4.10.设(x —I)[(A/+ B F+I),求A和B.由题意要求知解用(兀一1)2 去除f\x) = Ar4 + Bx2 +1 ,得余式”(x) = (4A + 2B)兀+1 -3人一B,斤（兀）=0,即4A + 2B = 0,1-3A-B = O,解得A = l,B = -2.11.证明:如果(/(x),g(x)) = l, (/(x),/z(x)) = l,那么(/(x), g(x)/z(x)) = l. 证明由条件可知,存在络(兀)和片⑴ 使得旳(兀)/(兀)+岭⑴g(x) = l,存在如(兀)和卩2(兀)使得u2(x)f(x) + v2(x)h(x) = 1.用/?(兀)乘以第一式得坷(x)f(x)h(x) + V, (x)g(x)h(x) = h(x),代入第二式得u2(x)f(x) + v2 (x) [u t (x)f(x)h(x) 4-Vj (x)g(x)/z(x)] = 1, 即[w2(兀)+ u\ (x)v2(x)h(x)]f(x) + [v, (x)v2(x)]g(x)h(x) = 1,所以(/(x),g(x)/z(x)) = l.12.证明：如果/(x)与g(x)不全为零，且/心)/(兀)+ 咻)g(兀)=(/(%), g(Q)，证明由于w(x)/(x) + v(x)g(x) = (/(x),g(x)), /(X )与 g(x)不全为零，所以(/(x)，g(x))HO.两边同时除以(/(Hg(Q)HO,有所以(弘(兀),咻)) = 1 .13.证明：如果〃(兀)|/(兀),〃(兀)|g(x),且〃(兀)为/(兀)与g(x)的一个组合，那么〃(兀)是/G)与 g(x)的一个最大公因式.证明由题意知d(x)是/(X )与g(x)的公因式.再由条件设d(x) = w(x)/(x) + v(x)^(x) •又设h(x) 为/(x)与g(x)的任一公因式，即/z(x)|/(x), h(x)\g(x),则由上式有h(x)\d(x).故而”(兀)是/(兀)与 g(x)的一个最大公因式.14.证明：(.fO)/2(X ), gO)/2(X )) = (.f(X ), g(x))〃(x),其中力(兀)的首项系数为 1.证明显然(/(x), g(x))/?(x)是f{x)h{x)与g(x)h(x)的一个公因式.下面來证明它是最大公因式. 设 /心),v(x)满足 w(x)/(x) + v(x)g(x) = (/(x), g(X>),贝iJu(x)f(x)h(x) + v(x)g(x)h(x) = (/(x),g(x))/z(x).由上题结果知，(/(兀),g(X ))/7(X )是/(X )/?(X )与g(JC”7(X )的一个最大公因式，又首项系数为1,所以(/(x)A(x), ^(%)/?(%)) = (/(x), ^(x))/i(x)・/⑴ g (兀)、(/(兀),g (兀))’(f(x),g(x))丿证明设〃(兀)=(/(兀),g(x)),则存在多项式M (x), v(x),使d(x) = u(x)f(x) + v(x)g(x)・因为/(X )与g (尢)不全为零，所以d(x)HO.上式两边同时除以〃(兀)，有故 /(兀) _____________ g (x)l (/(x),g(x))‘(/(x),g(x))‘u(x) /(X ) (/(%), g(x)) + v(x) g(x) (y (x ),^(x ))15.设多项式/(x)与gS)不全为零，证明1 = u(x)/(兀)(/(兀),g(x))+咻)g(x) (/(兀),g(x))=1成立.16. 分别在复数域、实数域和有理数域上分解兀4+ 1为不可约因式之积.在有理数域上兀°+1是不可约多项式.否则，若+ +1可约，有以下两种可能.(1) 兀4+1有一次因式，从而它有有理根，但/(±1)工0,所以卍+1无有理根.(2) x 4+ 1 无一次因式，设x 4+1 = (x 2+处 +方)(F +cx + d),其中 a,b y c,cl 为整数.于是a + c = O, b+ 〃 + ac = O, cut + be = 0 , bd = \,又分两种情况：① b = d = \,又 a = —c,从而由 b + 〃 + ac = O,得 a 2=2,矛盾； ② b = d = — \,则 a 2= —2 ,矛盾.综合以上情况，即证.17. 求下列多项式的有理根： (1) /(x) = x 3-6x 2+15兀一 14 ；(2) ^(X ) = 4X 4-7X 2-5X -1；(3) /z(x) = x 5+ %4— 6x^ — 14x~ — 1 lx — 3 ・解(1)由于/(x)是首项系数为1的整系数多项式，所以有理根必为整数根，且为-14的因数.-14的因数有：±1, ±2, ±7, ±14,计算得到:/(D = -4, /(-1) = -36, /(2) = 0, /(-2) = -72,/(7) = 140, /(-7) = -756, /(14) = 1764, /(一 14) = —4144,故x = 2是/(兀)的有理根.再由多项式除法可知，x = 2是于(兀)的单根.⑵ 类似(1)的讨论可知，g(x)的可能的有理根为:故x = --是巩兀)的有理根.再由多项式除法可知，兀二-丄是/(劝的2重根.2 2⑶ 类似地，加兀)的可能的有理根为：±1,±3,计算得到解在实数域上的分解式为X4+ 1 = (X 2 + 1)2-2X 2 =(X 2+V2X + 1)(X 2-V2X +1).在复数域上的分解式为x + ----------1 2 2%4+ 1 = f亠迈亠近、X ---------- 12 2/±1, ±1 ±?计算得到g(l) = -9,g(-1) = 1, g(］、r 、171=-5, g —=0, g — 一 —‘ g —〔2< 264 ,4丿11A(l) = -28, /?(-l) = 0,(3) = 0,加一3) = -96.故x = -l, x = 3是//(兀)的有理根.再由多项式除法可知，x = -\是/z(x)的4重根，兀=3是//(兀)的单根.18.若实系数方程x34- px + q = 0有一根a + bi (a,b为实数，/?工0),则方程x3 + px-q = 0有实根2—证明设原方程有三个根不失一般性，令=a + bi,从而有a2 =a-bi,由根与系数的关系可知0 = $ + 冬 + 他=(° + 勿)+ (a - bi) + ,所以冬二-2d，即(-2a)‘ + /?(-2a) + g = 0，故(2a)' + p(2a)-q = 0.这说明x3 + /zr-g = 0有实根2a .19.证明：如果(%-i)|/(r),那么证明因为u-i)|/(z),所以/(r)= /(i)= 0.因此，令y(x)=(x-i)g(x),则有E =(*-i)g(;),即(伙-1)|/(疋).20.下列多项式在有理数域上是否可约？(1)土 (%) = F+1；(2)/；(X)= X4-8?+12X2+2；(3)人(x) = x" +『+1 ；(4)厶(无)=* + "； + 1,门为奇素数；(5)厶(兀)=兀°+4尬+ 1, A为整数.解(1) ./；(兀)的可能的有理根为：±1,而/(±1) = 2,所以它在有理数域上不可约.(2)由Eisenstein判别法，取素数p = 2,则2不能整除1,而2|(-8), 2|12, 2|2,但是2?不能整除2,所以该多项式在有理数域上不可约.(3)令x=y + l,代入厶(x) = P+x'+l有^(y) = ^(y + l) = / + 6/+15/+21/+18y24-9y4-3.取素数0 = 3,由Eisenstein判别法知，g(y)在有理数域上不可约，所以/(兀)在有理数域上不可约.(4)令兀= y_l,代入f4(x) = x p 4-px + 1,得g(y)=厶(y j) = -+ cy~2——C；-2y2 + (Cf* + p)y-p，取素数p,由Eisenstein判别法知，g(y)在有理数域上不可约，所以£(兀)在有理数域上不可约.(5)令x=y + l,代入农(兀)=兀4+4Ax+l,得g(.y)=厶(y +1) = y" + 4y‘ + 6y2 + (4k + 4)y + 4R + 2 ,収素数p = 2,由Eisenstein判别法知，g(y)在有理数域上不可约，所以点(兀)在有理数域上不可约.1•设/(X)，g(X)，加兀)是实数域上的多项式，(1)若/2U) = xg2(x) + x/z2(x),则/(x) = g(x) = h{x) = 0 .(2)在复数域上，上述命题是否成立？证明(1)当g(兀)=/2(兀)=0时，有严⑴=0,所以/(%) = 0 ,命题成立.如果g(x), /z(x)不全为零，不妨设g(x)H0・当h(x) = 0时，a(xg2(x) + x/i2U)) = l + 2a^(x)为奇数；当加兀)工0时，因为g(x),瓜兀)都是实系数多项式，所以Xg2(x)与兀胪(兀)都是首项系数为正实数的奇次多项式，于是也有d(xg2(x) + x/『(x))为奇数.而这时均有/2(x)^0 ,且df\x) = 2df(x)为偶数,矛盾.因此有g(兀)=力(兀) = 0,从而有f(x) = 0 .(2)在复数域上，上述命题不成立.例如，设f(x) = 0 , g(x) = x\ h(x) = ix,1,其中斤为自然数, 有/2 (x) = xg2 (x)xh2 (x),但g(x) / 0 ,力(兀)工0.2.设/(x), g(x)9 h(x)e P[x],满足(x2 4-l)h(x)4-(x-l)/(x) + (x+2)g(x) = 0,(x2 + l)/?(x) + (x+ l)/(x) + (x - 2)^(%) = 0.证明(X2+1)|(/U), g(X))・证明两式相加得到2(x2 + l)h(x) + 2x(/(x) + g(兀))=0.由(x2+l,兀)=1可知(x2 + l)|(/(x) + g(x)).两式相减得到-2f(x) + 4g(x) = 0, f(x) = 2g(x).故(x2 + l)|/(x), (x2+l)|g(x), BP(X2+1)|(/(X),g(x)).3・设gi(x)g2(x)\f{(x)f2(x),证明(1)若/(x)|g](x)，/(X)H0,则g2(x)\f2(x)；(2)若g2(x)|/；(x)/；(x),是否有g2(x)\f2(x)?解(1)因为gi(兀)g2(兀)庞(兀)£(兀)，/O)|gi(X)，故存在多项式h(x), h}(x)使得fl(x)f 2(x) = g](x)g 2(x)h(x\ g](兀)=Z (x)h }(x).于是/；(兀)£(兀)=/(兀)人(兀)g2(x)力(兀)•由于土(兀)工0,故有 f 2(x) = h l (x)g 2(x)h(x),即g 2(x)\f 2(x).(2)否•例如取 g {(x) = x-2 , ^2(X ) = X 2-1 , (x) = (x-l)(x-2), (x) = (x + l)(x4-2).虽然 gSx)g 2(x)\f^x)f 2(x)且 g 2(x)\f {(x)f 2(x),但 g 2(x)不能整除 f 2(x).4.当R 为何值时，/(x) = X 2 +伙+ 6)x + 4k + 2和g(x) = F+(£ + 2)x + 2R 的最大公因式是一次的？并求出此吋的最大公因式.解显然 g(x) = (x + £)(x+2).当(/(x),g(Q) = x + 2时'/(一2) = 4 — 2伙+ 6) + 4£ + 2 = 0‘ 则k = 3.当(于(兀),g(Q )=兀 + £ 时’/(一灯=k 2 - k(k + 6) + 4Z: + 2 = 0 ‘ 则 k = l.这时(/(x), g(x))=兀+1. 5.证明：对于任意正整数斤，都有(/(x),g(Q)"=(/"(x),g"(x))・证明由题意可知/(%)与&(兀)不全为零.令(/(x), g(x)) = d(x),Z 、” g(x) 、d(x)丿/心)/"(兀)+ 咚)g"(兀)=d\x).又由 d(x)\f(x), d(x)|g(x),有 d n (x) f l \x), d"(x) g"(x),因此 d"(x)是厂(x)与 g"(x)的首项系数为1的最大公因式，从而有(广(x),g"(x))= 〃"(兀)=(/(x),g(x))" •6.设 / (x) = af(x) + bg(x), g[ (x) = c/(x) + dg(x),且 ad - be H 0 ,证明(/(x),g(x)) = (/](x), g](X ))・证明设(/(x), g(x)) = d(x),则 d(x)\f(x\d(x)\g(x).由于 “所以对任意正整如，有爲J 寫〕"卜于是有u{x) +咻) 则〃(兀)工0,从而fi (兀)=妙(x) + bg(x) , g] (x) = (x) + dg (x),故d (x)| (x), d (x)|g t (x).又设h(x)\ (x), /z(x)|(x),由上式及ad-bc^O ,可得从而/?(x)|/(x), h(x)\g(x),于是h(x)\d(x),即〃(兀)也是/；(兀)和g|(x)的最大公因式，即(/(x), g(x)) = (/；(x),&(兀))・7.设 /(x) = t/(x)/(x), g(Q 二 dCr)g](x),且/O)与 gd)不全为零，证明〃(兀)是/O)与 gCO的一个最大公因式的充分必要条件是(/(劝,g|(x)) = 1.证明必要性.若〃(x)是/(兀)与g (兀)的一个最大公因式，则存在多项式w(x),v(x)使W (x)/(x) +v(x)g(x) = d(x),于是u(x)d(x)f t (x) + v(x)d(x)g l (x) = d(x).由/(力与g (兀)不全为零知如工0,因此有u(x)f l (x) + v(x)g l (x) = l f 即(土(兀),g©))i •充分性.若(f l (x),g l (x)) = l ,则存在多项式u(x),v(x),使 u(x)f l (x)+ v(x)g l (x) = l. 两边同吋乘〃(兀)有u(x)f(x) + v(x)g(x) = d(x)・由d(x)是/(x)与g(x)的一个公因式知，d(x)是f(x)与g(x)的一个最大公因式.8.设于(兀)和g(x)是两个多项式，证明(f(x), g(x)) = l 当且仅当(f(x)-l-g(x), f(x)g(x)) = l. 证明必要性.设(f(x)9g(x)) = l,若f(x) + g(x)与/⑴g(x)不互素，则有不可约公因式p(x), 使p(x)lf(x)g(x)f所以 p(x)| /(X )或 0(x)|g(x).不妨设 p(x)\ /(x),由 P (x)|(/(x) + g (兀))可知 p(x)|g(x),因此 P (兀)是 /(兀)和g“)的公因式,与/(%), g (x)互素矛盾,故蚀+g (兀)与蚀g (兀)互素.充分性.设(/(兀)+ gO) J(x)g (兀)) = 1,则存在w(x), v(x)使(/(兀)+ g (兀))心)+ /(x)g(x)v(x) = 1 , f(x)u(x) + g (兀)(臥兀)+d ad-be zw- h ad 一gi (兀)， g(x) -c ad -be a ad -be g](x)，/(x)v(x)) = 1, 上式说明(/(兀),g(兀)) = 1.9.如果(x2 +x + l)|/j(x3) + x/^(x3),那么(x-l)|/；(x), 0 — 1)|/；(兀)・T；®所以，^3=£23 = 1.证明X2+X + l的两个根为£\= 士护和£2=因为U2+x+l)|(/；(^3) + x/；(^3)),所以(兀一£|)(x - £2)|/；(X')+/(F)，故有y 窗)+ £/(郃)=0,［爪哥)+ £2£(哥)=0,即解得/(l) = /；(l) = o,从而(兀—1)|久(兀)，(x-1)|/；(%).10.若f(x)\f(x H),则/(x)的根只能是零或单位根.证明因为f(x)\f(x n),故存在多项式g(x),使/(x n) = /(x)^(x).设。

线性代数(含全部课后题详细答案5-1.

当 r( A) r n 时，求得基础解系是1 ,2 , ,nr , 则 x k11 k22 knr nr 是AX 0 的解，
称为通解。
4. 解的结构
AX 0 的通解是 x k11 k22 knr n。
4 7
4 7

0 0
0 0
0 0
0 0
0 0

x1

13 7

3 7
x3

13 7
x4

x2

4 7

2 7
x3

4 7
x4
25

13 7

令 x3 x4 0,
得

4 7
0
0
又原方程组对应的齐次方程组的通解是

x1 x3

2x2 3
10 x4
1 5
x4
令

x2 x4

1
0

得

1

1 0 0

1

5

令

x2
x4

0

1

得
2

0 3

10 1
2
举例说明消元法具体步骤：
例1：解线性方程组

2 4
x1 x1
2 x1
2 1 3
解：(
A,
b)

4 2

(完整版)线性代数课后习题答案第1——5章习题详解

第一章行列式4.计算下列各行列式：（1）⎥⎥⎥⎥⎦⎥⎢⎢⎢⎢⎣⎢7110025*********4；（2）⎥⎥⎥⎥⎦⎥⎢⎢⎢⎢⎣⎢-265232112131412；（3）⎥⎥⎦⎥⎢⎢⎣⎢---ef cf bf de cd bd ae ac ab ；（4）⎥⎥⎥⎥⎦⎥⎢⎢⎢⎢⎣⎢---d c b a100110011001解(1)7110025102021421434327c c c c --0100142310202110214---=34)1(143102211014+-⨯---=143102211014-- 321132c c c c ++1417172001099-=0(2)2605232112131412-24c c -2605032122130412-24r r -0412032122130412- 14r r -0000032122130412-=0(3)ef cf bf de cd bd ae ac ab ---=ec b e c b ec b adf ---=111111111---adfbce =abcdef 4(4)d c b a 100110011001---21ar r +dc b a ab 100110011010---+=12)1)(1(+--dc a ab 10111--+23dc c +010111-+-+cd c ada ab =23)1)(1(+--cdadab +-+111=1++++ad cd ab abcd5.证明: (1)1112222b b a a b ab a +=3)(b a -; (2)bz ay by ax bx az by ax bx az bz ay bx az bz ay by ax +++++++++=y x z x z y z y x b a )(33+;(3)0)3()2()1()3()2()1()3()2()1()3()2()1(2222222222222222=++++++++++++d d d d c c c c b b b b a a a a ;(4)444422221111d c b a d c b a d c b a ))()()()((d b c b d a c a b a -----=))((d c b a d c +++-⋅;(5)1221100000100001a x a a a a x x x n n n +-----n n n n a x a x a x ++++=--111 . 证明(1)00122222221312a b a b a a b a ab a c c c c ------=左边a b a b a b a ab 22)1(22213-----=+21))((a b a a b a b +--=右边=-=3)(b a(2)bz ay by ax z by ax bx az y bx az bz ay x a ++++++分开按第一列左边bzay by ax x by ax bx az z bxaz bz ay y b +++++++ ++++++002y by ax z x bx az y z bz ay x a 分别再分bzay y x by ax x z bxaz z y b +++z y x y x z x z y b y x z x z y z y x a 33+分别再分右边=-+=233)1(yx z x z y zy x b y x z x z y z y x a(3) 2222222222222222)3()2()12()3()2()12()3()2()12()3()2()12(++++++++++++++++=d d d d d c c c c c b b b b b a a a a a 左边9644129644129644129644122222141312++++++++++++---d d d d c c c c b b b b a a a a c c c c c c 964496449644964422222++++++++d d d d c c c c b b b b a a a a 分成二项按第二列964419644196441964412222+++++++++d d d c c c b b b a a a 949494949464222224232423d d c c b b a a c c c c c c c c ----第二项第一项06416416416412222=+ddd c c c bb b a a a (4) 444444422222220001ad a c a b a ad a c a b a ad a c a b a ---------=左边=)()()(222222222222222a d d a c c a b b a d a c a b ad a c a b --------- =)()()(111))()((222a d d a c c a b b a d a c ab a d ac a b ++++++--- =⨯---))()((ad a c a b )()()()()(00122222a b b a d d a b b a c c a b b bd b c a b +-++-++--+ =⨯-----))()()()((b d b c a d a c a b )()()()(112222b d a b bd d b c a b bc c ++++++++=))()()()((d b c b d a c a b a -----))((d c b a d c +++-(5) 用数学归纳法证明.,1,2212122命题成立时当a x a x a x a x D n ++=+-==假设对于)1(-n 阶行列式命题成立，即 ,122111-----++++=n n n n n a x a x a x D:1列展开按第则n D1110010001)1(11----+=+-x xa xD D n n n n 右边=+=-n n a xD 1 所以，对于n 阶行列式命题成立.6.设n 阶行列式)det(ij a D =,把D 上下翻转、或逆时针旋转 90、或依副对角线翻转，依次得n nn n a a a a D 11111 =， 11112n nn n a a a a D = ，11113a a a a D n nnn =，证明D D D D D n n =-==-32)1(21,)1(.证明 )det(ij a D =nnn n nn n nn n a a a a a a a a a a D 2211111111111)1(--==∴ =--=--nnn n nnn n a a a a a a a a 331122111121)1()1( nnn n n n a a a a 111121)1()1()1(---=--D D n n n n 2)1()1()2(21)1()1(--+-+++-=-=同理可证nnn n n n a a a a D 11112)1(2)1(--=D D n n T n n 2)1(2)1()1()1(---=-= D D D D D n n n n n n n n =-=--=-=----)1(2)1(2)1(22)1(3)1()1()1()1(7.计算下列各行列式（阶行列式为k D k ）：(1)a aD n 11=,其中对角线上元素都是a ，未写出的元素都是0；(2)xaaax aa a x D n=; (3) 1111)()1()()1(1111n a a a n a a a n a a a D n n n n n n n ------=---+; 提示：利用范德蒙德行列式的结果． (4) nnnnn d c d c b a b a D000011112=; (5)j i a a D ij ij n -==其中),det(;(6)nn a a a D +++=11111111121 ,021≠n a a a 其中.解(1) aa a a a D n 00010000000000001000 =按最后一行展开)1()1(100000000000010000)1(-⨯-+-n n n aa a)1)(1(2)1(--⋅-+n n na aa(再按第一行展开)n n n nn a a a+-⋅-=--+)2)(2(1)1()1(2--=n n a a )1(22-=-a a n(2)将第一行乘)1(-分别加到其余各行，得ax x a ax x a a x x a aa a x D n ------=0000000 再将各列都加到第一列上，得ax ax a x aaa a n x D n ----+=000000000)1( )(])1([1a x a n x n --+=- (3) 从第1+n 行开始，第1+n 行经过n 次相邻对换，换到第1行，第n 行经)1(-n 次对换换到第2行…，经2)1(1)1(+=++-+n n n n 次行交换，得 nn n n n n n n n n a a a n a a a n a a aD )()1()()1(1111)1(1112)1(1-------=---++此行列式为范德蒙德行列式∏≥>≥++++--+--=112)1(1)]1()1[()1(j i n n n n j a i a D∏∏≥>≥+++-++≥>≥++-•-•-=---=111)1(2)1(112)1()][()1()1()]([)1(j i n n n n n j i n n n j i j i∏≥>≥+-=11)(j i n j i(4) nn nnn d c d c b a b a D 011112=nn n n n nd d c d c b a b a a 0000000011111111----展开按第一行0000)1(1111111112c d c d c b a b a b nn n n n nn ----+-+2222 ---n n n n n n D c b D d a 都按最后一行展开由此得递推公式：222)(--=n n n n n n D c b d a D即 ∏=-=ni i i iin D c b da D 222)(而 111111112c b d a d c b a D -==得 ∏=-=ni i i i i n c b d a D 12)((5)j i a ij -=432140123310122210113210)det( --------==n n n n n n n n a D ij n ,3221r r r r --0432111111111111111111111 --------------n n n n,,141312c c c c c c +++152423210222102210002100001---------------n n n n n =212)1()1(----n n n(6)nn a a D a +++=11111111121,,433221c c c c c c ---n n n n a a a a a a a a a a +-------10000100010000100010001000011433221 展开（由下往上）按最后一列))(1(121-+n n a a a a nn n a a a a a a a a a --------00000000000000000000000000022433221 nn n a a a a a a a a ----+--000000000000000001133221 ++ nn n a a a a a a a a -------000000000000000001143322n n n n n n a a a a a a a a a a a a 322321121))(1(++++=---)11)((121∑=+=ni in a a a a8.用克莱姆法则解下列方程组：⎪⎪⎩⎪⎪⎨⎧=+++-=----=+-+=+++;01123,2532,242,5)1(4321432143214321x x x x x x x x x x x x x x x x ⎪⎪⎪⎩⎪⎪⎪⎨⎧=+=++=++=++=+.15,065,065,065,165)2(5454343232121x x x x x x x x x x x x x解 (1)11213513241211111----=D 8120735032101111------=145008130032101111---=1421420005410032101111-=---= 112105132412211151------=D 11210513290501115----=1121023313090509151------=2331309050112109151------=1202300461000112109151-----=14200038100112109151----=142-=112035122412111512-----=D 811507312032701151-------=3139011230023101151-=2842840001910023101151-=----=426110135232422115113-=----=D ; 14202132132212151114=-----=D1,3,2,144332211-========∴DDx D D x D D x D D x (2) 510006510006510006510065=D 展开按最后一行61000510065100655-'D D D ''-'=65 D D D ''-'''-''=6)65(5D D '''-''=3019D D ''''-'''=1146566551141965=⨯-⨯=(,11的余子式中为行列式a D D ',11的余子式中为a D D ''''类推D D ''''''',) 51001651000651000650000611=D 展开按第一列6510065100650006+'D 46+'=D 460319+''''-'''=D 1507=51010651000650000601000152=D 展开按第二列5100651006500061-6510065000610005-365510651065⨯-= 1145108065-=--=51100650000601000051001653=D 展开按第三列51006500061000516500061000510065+6100510656510650061+= 703114619=⨯+=51000601000051000651010654=D 展开按第四列61000510065100655000610005100651--51065106565--=395-= 110051000651000651100655=D 展开按最后一列D '+10005100651006512122111=+= 665212;665395;665703;6651145;665150744321=-==-==∴x x x x x ． 9.齐次线性方程组取何值时问,,μλ⎪⎩⎪⎨⎧=++=++=++0200321321321x x x x x x x x x μμλ有非零解？解 μλμμμλ-==12111113D ，齐次线性方程组有非零解，则03=D即 0=-μλμ 得 10==λμ或不难验证，当,10时或==λμ该齐次线性方程组确有非零解.10.齐次线性方程组取何值时问,λ⎪⎩⎪⎨⎧=-++=+-+=+--0)1(0)3(2042)1(321321321x x x x x x x x x λλλ 有非零解？解λλλ----=111132421D λλλλ--+--=101112431)3)(1(2)1(4)3()1(3λλλλλ-------+-=3)1(2)1(23-+-+-=λλλ 齐次线性方程组有非零解，则0=D得 32,0===λλλ或不难验证，当32,0===λλλ或时，该齐次线性方程组确有非零解.第二章矩阵及其运算1. 已知线性变换:⎪⎩⎪⎨⎧++=++=++=3213321232113235322y y y x y y y x y y y x ,求从变量x 1, x 2, x 3到变量y 1, y 2, y 3的线性变换.解由已知:⎪⎪⎭⎫ ⎝⎛⎪⎪⎭⎫ ⎝⎛=⎪⎪⎭⎫ ⎝⎛221321323513122y y y x x x , 故 ⎪⎪⎭⎫ ⎝⎛⎪⎪⎭⎫ ⎝⎛=⎪⎪⎭⎫ ⎝⎛-3211221323513122x x x y y y ⎪⎪⎭⎫ ⎝⎛⎪⎪⎭⎫ ⎝⎛----=321423736947y y y , ⎪⎩⎪⎨⎧-+=-+=+--=321332123211423736947x x x y x x x y x x x y .2. 已知两个线性变换⎪⎩⎪⎨⎧++=++-=+=32133212311542322y y y x y y y x y y x , ⎪⎩⎪⎨⎧+-=+=+-=323312211323z z y z z y z z y , 求从z 1, z 2, z 3到x 1, x 2, x 3的线性变换.解由已知⎪⎪⎭⎫ ⎝⎛⎪⎪⎭⎫ ⎝⎛-=⎪⎪⎭⎫ ⎝⎛221321514232102y y y x x x ⎪⎪⎭⎫ ⎝⎛⎪⎪⎭⎫ ⎝⎛--⎪⎪⎭⎫ ⎝⎛-=321310102013514232102z z z⎪⎪⎭⎫ ⎝⎛⎪⎪⎭⎫ ⎝⎛----=321161109412316z z z , 所以有⎪⎩⎪⎨⎧+--=+-=++-=3213321232111610941236z z z x z z z x z z z x .3. 设⎪⎪⎭⎫ ⎝⎛--=111111111A , ⎪⎪⎭⎫⎝⎛--=150421321B , 求3AB -2A 及A T B . 解 ⎪⎪⎭⎫⎝⎛---⎪⎪⎭⎫ ⎝⎛--⎪⎪⎭⎫ ⎝⎛--=-1111111112150421321111111111323A AB ⎪⎪⎭⎫⎝⎛----=⎪⎪⎭⎫ ⎝⎛---⎪⎪⎭⎫ ⎝⎛-=2294201722213211111111120926508503, ⎪⎪⎭⎫ ⎝⎛-=⎪⎪⎭⎫ ⎝⎛--⎪⎪⎭⎫ ⎝⎛--=092650850150421321111111111B A T . 4. 计算下列乘积:(1)⎪⎪⎭⎫⎝⎛⎪⎪⎭⎫ ⎝⎛-127075321134; 解 ⎪⎪⎭⎫ ⎝⎛⎪⎪⎭⎫ ⎝⎛-127075321134⎪⎪⎭⎫ ⎝⎛⨯+⨯+⨯⨯+⨯-+⨯⨯+⨯+⨯=102775132)2(71112374⎪⎪⎭⎫ ⎝⎛=49635. (2)⎪⎪⎭⎫⎝⎛123)321(; 解 ⎪⎪⎭⎫⎝⎛123)321(=(1⨯3+2⨯2+3⨯1)=(10).(3))21(312-⎪⎪⎭⎫⎝⎛; 解 )21(312-⎪⎪⎭⎫ ⎝⎛⎪⎪⎭⎫ ⎝⎛⨯-⨯⨯-⨯⨯-⨯=23)1(321)1(122)1(2⎪⎪⎭⎫ ⎝⎛---=632142. (4)⎪⎪⎪⎭⎫ ⎝⎛---⎪⎭⎫ ⎝⎛-20413121013143110412 ; 解 ⎪⎪⎪⎭⎫ ⎝⎛---⎪⎭⎫ ⎝⎛-20413121013143110412⎪⎭⎫ ⎝⎛---=6520876. (5)⎪⎪⎭⎫ ⎝⎛⎪⎪⎭⎫ ⎝⎛321332313232212131211321)(x x x a a a a a a a a a x x x ; 解⎪⎪⎭⎫ ⎝⎛⎪⎪⎭⎫ ⎝⎛321332313232212131211321)(x x x a a a a a a a a a x x x =(a 11x 1+a 12x 2+a 13x 3 a 12x 1+a 22x 2+a 23x 3 a 13x 1+a 23x 2+a 33x 3)⎪⎪⎭⎫ ⎝⎛321x x x322331132112233322222111222x x a x x a x x a x a x a x a +++++=.5. 设⎪⎭⎫ ⎝⎛=3121A , ⎪⎭⎫ ⎝⎛=2101B , 问: (1)AB =BA 吗?解 AB ≠BA .因为⎪⎭⎫ ⎝⎛=6443AB , ⎪⎭⎫ ⎝⎛=8321BA , 所以AB ≠BA .(2)(A +B)2=A 2+2AB +B 2吗?解 (A +B)2≠A 2+2AB +B 2.因为⎪⎭⎫ ⎝⎛=+5222B A , ⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛=+52225222)(2B A ⎪⎭⎫ ⎝⎛=2914148, 但 ⎪⎭⎫ ⎝⎛+⎪⎭⎫ ⎝⎛+⎪⎭⎫ ⎝⎛=++43011288611483222B AB A ⎪⎭⎫ ⎝⎛=27151610, 所以(A +B)2≠A 2+2AB +B 2.(3)(A +B)(A -B)=A 2-B 2吗?解 (A +B)(A -B)≠A 2-B 2.因为⎪⎭⎫ ⎝⎛=+5222B A , ⎪⎭⎫ ⎝⎛=-1020B A , ⎪⎭⎫ ⎝⎛=⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛=-+906010205222))((B A B A , 而 ⎪⎭⎫ ⎝⎛=⎪⎭⎫ ⎝⎛-⎪⎭⎫ ⎝⎛=-718243011148322B A , 故(A +B)(A -B)≠A 2-B 2.6. 举反列说明下列命题是错误的:(1)若A 2=0, 则A =0;解取⎪⎭⎫ ⎝⎛=0010A , 则A 2=0, 但A ≠0. (2)若A 2=A , 则A =0或A =E ;解取⎪⎭⎫ ⎝⎛=0011A , 则A 2=A , 但A ≠0且A ≠E . (3)若AX =AY , 且A ≠0, 则X =Y .解取⎪⎭⎫ ⎝⎛=0001A , ⎪⎭⎫ ⎝⎛-=1111X , ⎪⎭⎫ ⎝⎛=1011Y , 则AX =AY , 且A ≠0, 但X ≠Y .7. 设⎪⎭⎫ ⎝⎛=101λA , 求A 2, A 3, ⋅ ⋅ ⋅, A k . 解 ⎪⎭⎫ ⎝⎛=⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛=12011011012λλλA , ⎪⎭⎫ ⎝⎛=⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛==1301101120123λλλA A A , ⋅ ⋅ ⋅ ⋅ ⋅ ⋅,⎪⎭⎫ ⎝⎛=101λk A k . 8. 设⎪⎪⎭⎫⎝⎛=λλλ001001A , 求A k . 解首先观察⎪⎪⎭⎫ ⎝⎛⎪⎪⎭⎫ ⎝⎛=λλλλλλ0010010010012A ⎪⎪⎭⎫ ⎝⎛=222002012λλλλλ, ⎪⎪⎭⎫ ⎝⎛=⋅=3232323003033λλλλλλA A A , ⎪⎪⎭⎫ ⎝⎛=⋅=43423434004064λλλλλλA A A ,⎪⎪⎭⎫ ⎝⎛=⋅=545345450050105λλλλλλA A A , ⋅ ⋅ ⋅ ⋅ ⋅ ⋅,⎝⎛=k A k k k k k k k k k k λλλλλλ0002)1(121----⎪⎪⎪⎭⎫ . 用数学归纳法证明:当k =2时, 显然成立.假设k 时成立，则k +1时，⎪⎪⎭⎫ ⎝⎛⎪⎪⎪⎪⎭⎫ ⎝⎛-=⋅=---+λλλλλλλλλ0010010002)1(1211k k k k k k k k k k k k A A A ⎪⎪⎪⎪⎭⎫ ⎝⎛+++=+-+--+11111100)1(02)1()1(k k k k k k k k k k λλλλλλ, 由数学归纳法原理知:⎪⎪⎪⎪⎭⎫ ⎝⎛-=---k k k k k k k k k k k A λλλλλλ0002)1(121. 9. 设A , B 为n 阶矩阵，且A 为对称矩阵，证明B T AB 也是对称矩阵. 证明因为A T =A , 所以(B T AB)T =B T (B T A)T =B T A T B =B T AB ,从而B T AB 是对称矩阵.10. 设A , B 都是n 阶对称矩阵，证明AB 是对称矩阵的充分必要条件是AB =BA . 证明充分性: 因为A T =A , B T =B , 且AB =BA , 所以(AB)T =(BA)T =A T B T =AB ,即AB 是对称矩阵.必要性: 因为A T =A , B T =B , 且(AB)T =AB , 所以AB =(AB)T =B T A T =BA .11. 求下列矩阵的逆矩阵:(1)⎪⎭⎫ ⎝⎛5221; 解 ⎪⎭⎫ ⎝⎛=5221A . |A|=1, 故A -1存在. 因为 ⎪⎭⎫ ⎝⎛--=⎪⎭⎫ ⎝⎛=1225*22122111A A A A A , 故*||11A A A =-⎪⎭⎫ ⎝⎛--=1225. (2)⎪⎭⎫ ⎝⎛-θθθθcos sin sin cos ; 解⎪⎭⎫ ⎝⎛-=θθθθcos sin sin cos A . |A|=1≠0, 故A -1存在. 因为 ⎪⎭⎫ ⎝⎛-=⎪⎭⎫ ⎝⎛=θθθθcos sin sin cos *22122111A A A A A , 所以*||11A A A =-⎪⎭⎫ ⎝⎛-=θθθθcos sin sin cos . (3)⎪⎪⎭⎫⎝⎛---145243121; 解 ⎪⎪⎭⎫ ⎝⎛---=145243121A . |A|=2≠0, 故A -1存在. 因为 ⎪⎪⎭⎫ ⎝⎛-----=⎪⎪⎭⎫ ⎝⎛=214321613024*332313322212312111A A A A A A A A A A , 所以 *||11A A A =-⎪⎪⎪⎭⎫ ⎝⎛-----=1716213213012.(4)⎪⎪⎪⎭⎫ ⎝⎛n a a a 0021(a 1a 2⋅ ⋅ ⋅a n≠0) .解 ⎪⎪⎪⎭⎫ ⎝⎛=n a a a A 0021, 由对角矩阵的性质知⎪⎪⎪⎪⎪⎪⎭⎫ ⎝⎛=-n a a a A 10011211 . 12. 解下列矩阵方程:(1)⎪⎭⎫ ⎝⎛-=⎪⎭⎫ ⎝⎛12643152X ; 解 ⎪⎭⎫ ⎝⎛-⎪⎭⎫ ⎝⎛=-126431521X ⎪⎭⎫ ⎝⎛-⎪⎭⎫ ⎝⎛--=12642153⎪⎭⎫ ⎝⎛-=80232. (2)⎪⎭⎫ ⎝⎛-=⎪⎪⎭⎫ ⎝⎛--234311*********X ; 解 1111012112234311-⎪⎪⎭⎫ ⎝⎛--⎪⎭⎫ ⎝⎛-=X⎪⎪⎭⎫ ⎝⎛---⎪⎭⎫ ⎝⎛-=03323210123431131 ⎪⎪⎭⎫ ⎝⎛---=32538122. (3)⎪⎭⎫ ⎝⎛-=⎪⎭⎫ ⎝⎛-⎪⎭⎫ ⎝⎛-101311022141X ;解 11110210132141--⎪⎭⎫ ⎝⎛-⎪⎭⎫ ⎝⎛-⎪⎭⎫ ⎝⎛-=X⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-⎪⎭⎫ ⎝⎛-=210110131142121 ⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛=21010366121⎪⎪⎭⎫ ⎝⎛=04111. (4)⎪⎪⎭⎫⎝⎛---=⎪⎪⎭⎫ ⎝⎛⎪⎪⎭⎫ ⎝⎛021102341010100001100001010X . 解 11010100001021102341100001010--⎪⎪⎭⎫ ⎝⎛⎪⎪⎭⎫ ⎝⎛---⎪⎪⎭⎫ ⎝⎛=X⎪⎪⎭⎫ ⎝⎛⎪⎪⎭⎫ ⎝⎛---⎪⎪⎭⎫ ⎝⎛=010100001021102341100001010⎪⎪⎭⎫ ⎝⎛---=201431012. 13. 利用逆矩阵解下列线性方程组:(1)⎪⎩⎪⎨⎧=++=++=++3532522132321321321x x x x x x x x x ;解方程组可表示为⎪⎪⎭⎫ ⎝⎛=⎪⎪⎭⎫ ⎝⎛⎪⎪⎭⎫ ⎝⎛321153522321321x x x , 故 ⎪⎪⎭⎫ ⎝⎛=⎪⎪⎭⎫ ⎝⎛⎪⎪⎭⎫ ⎝⎛=⎪⎪⎭⎫ ⎝⎛-0013211535223211321x x x , 从而有 ⎪⎩⎪⎨⎧===001321x x x .(2)⎪⎩⎪⎨⎧=-+=--=--05231322321321321x x x x x x x x x .解方程组可表示为⎪⎪⎭⎫⎝⎛=⎪⎪⎭⎫ ⎝⎛⎪⎪⎭⎫ ⎝⎛-----012523312111321x x x ,故 ⎪⎪⎭⎫ ⎝⎛=⎪⎪⎭⎫ ⎝⎛⎪⎪⎭⎫ ⎝⎛-----=⎪⎪⎭⎫ ⎝⎛-3050125233121111321x x x , 故有 ⎪⎩⎪⎨⎧===305321x x x .14. 设A k =O (k 为正整数), 证明(E -A)-1=E +A +A 2+⋅ ⋅ ⋅+A k -1. 证明因为A k =O , 所以E -A k =E . 又因为 E -A k =(E -A)(E +A +A 2+⋅ ⋅ ⋅+A k -1),所以 (E -A)(E +A +A 2+⋅ ⋅ ⋅+A k -1)=E , 由定理2推论知(E -A)可逆, 且(E -A)-1=E +A +A 2+⋅ ⋅ ⋅+A k -1.证明一方面, 有E =(E -A)-1(E -A). 另一方面, 由A k =O , 有E =(E -A)+(A -A 2)+A 2-⋅ ⋅ ⋅-A k -1+(A k -1-A k ) =(E +A +A 2+⋅ ⋅ ⋅+A k -1)(E -A),故 (E -A)-1(E -A)=(E +A +A 2+⋅ ⋅ ⋅+A k -1)(E -A), 两端同时右乘(E -A)-1, 就有(E -A)-1(E -A)=E +A +A 2+⋅ ⋅ ⋅+A k -1.15. 设方阵A 满足A 2-A -2E =O , 证明A 及A +2E 都可逆, 并求A -1及(A +2E)-1.证明由A 2-A -2E =O 得 A 2-A =2E , 即A(A -E)=2E ,或E E A A =-⋅)(21,由定理2推论知A 可逆, 且)(211E A A -=-.由A 2-A -2E =O 得A 2-A -6E =-4E , 即(A +2E)(A -3E)=-4E ,或E A E E A =-⋅+)3(41)2(由定理2推论知(A +2E)可逆, 且)3(41)2(1A E E A -=+-.证明由A 2-A -2E =O 得A 2-A =2E , 两端同时取行列式得 |A 2-A|=2,即 |A||A -E|=2, 故 |A|≠0,所以A 可逆, 而A +2E =A 2, |A +2E|=|A 2|=|A|2≠0, 故A +2E 也可逆. 由 A 2-A -2E =O ⇒A(A -E)=2E⇒A -1A(A -E)=2A -1E ⇒)(211E A A -=-,又由 A 2-A -2E =O ⇒(A +2E)A -3(A +2E)=-4E⇒ (A +2E)(A -3E)=-4 E ,所以 (A +2E)-1(A +2E)(A -3E)=-4(A +2 E)-1,)3(41)2(1A E E A -=+-.16. 设A 为3阶矩阵,21||=A , 求|(2A)-1-5A*|.解因为*||11A A A =-, 所以|||521||*5)2(|111----=-A A A A A |2521|11---=A A=|-2A -1|=(-2)3|A -1|=-8|A|-1=-8⨯2=-16. 17. 设矩阵A 可逆, 证明其伴随阵A*也可逆, 且(A*)-1=(A -1)*.证明由*||11A A A =-, 得A*=|A|A -1, 所以当A 可逆时, 有|A*|=|A|n |A -1|=|A|n -1≠0,从而A*也可逆.因为A*=|A|A -1, 所以 (A*)-1=|A|-1A .又*)(||)*(||1111---==A A A A A , 所以(A*)-1=|A|-1A =|A|-1|A|(A -1)*=(A -1)*. 18. 设n 阶矩阵A 的伴随矩阵为A*, 证明: (1)若|A|=0, 则|A*|=0; (2)|A*|=|A|n -1. 证明(1)用反证法证明. 假设|A*|≠0, 则有A*(A*)-1=E , 由此得 A =A A*(A*)-1=|A|E(A*)-1=O ,所以A*=O , 这与|A*|≠0矛盾，故当|A|=0时, 有|A*|=0.(2)由于*||11A A A =-, 则AA*=|A|E , 取行列式得到|A||A*|=|A|n . 若|A|≠0, 则|A*|=|A|n -1;若|A|=0, 由(1)知|A*|=0, 此时命题也成立. 因此|A*|=|A|n -1.19. 设⎪⎪⎭⎫⎝⎛-=321011330A , AB =A +2B , 求B .解由AB =A +2E 可得(A -2E)B =A , 故⎪⎪⎭⎫ ⎝⎛-⎪⎪⎭⎫ ⎝⎛---=-=--321011330121011332)2(11A E A B ⎪⎪⎭⎫ ⎝⎛-=011321330.20. 设⎪⎪⎭⎫⎝⎛=101020101A , 且AB +E =A 2+B , 求B .解由AB +E =A 2+B 得 (A -E)B =A 2-E ,即 (A -E)B =(A -E)(A +E).因为01001010100||≠-==-E A , 所以(A -E)可逆, 从而⎪⎪⎭⎫⎝⎛=+=201030102E A B .21. 设A =diag(1, -2, 1), A*BA =2BA -8E , 求B . 解由A*BA =2BA -8E 得 (A*-2E)BA =-8E , B =-8(A*-2E)-1A -1 =-8[A(A*-2E)]-1 =-8(AA*-2A)-1 =-8(|A|E -2A)-1 =-8(-2E -2A)-1 =4(E +A)-1=4[diag(2, -1, 2)]-1)21 ,1 ,21(diag 4-==2diag(1, -2, 1).22. 已知矩阵A 的伴随阵⎪⎪⎪⎭⎫⎝⎛-=8030010100100001*A , 且ABA -1=BA -1+3E , 求B .解由|A*|=|A|3=8, 得|A|=2. 由ABA -1=BA -1+3E 得 AB =B +3A ,B =3(A -E)-1A =3[A(E -A -1)]-1A11*)2(6*)21(3---=-=A E A E⎪⎪⎪⎭⎫ ⎝⎛-=⎪⎪⎪⎭⎫⎝⎛--=-1030060600600006603001010010000161. 23. 设P -1AP =Λ, 其中⎪⎭⎫ ⎝⎛--=1141P , ⎪⎭⎫ ⎝⎛-=Λ2001, 求A 11.解由P -1AP =Λ, 得A =P ΛP -1, 所以A 11= A=P Λ11P -1.|P|=3,⎪⎭⎫ ⎝⎛-=1141*P , ⎪⎭⎫ ⎝⎛--=-1141311P ,而 ⎪⎭⎫ ⎝⎛-=⎪⎭⎫⎝⎛-=Λ11111120 012001,故⎪⎪⎪⎭⎫⎝⎛--⎪⎭⎫ ⎝⎛-⎪⎭⎫ ⎝⎛--=31313431200111411111A ⎪⎭⎫ ⎝⎛--=68468327322731.24. 设AP =P Λ, 其中⎪⎪⎭⎫⎝⎛--=111201111P , ⎪⎪⎭⎫ ⎝⎛-=Λ511, 求ϕ(A)=A 8(5E -6A +A 2).解 ϕ(Λ)=Λ8(5E -6Λ+Λ2)=diag(1,1,58)[diag(5,5,5)-diag(-6,6,30)+diag(1,1,25)] =diag(1,1,58)diag(12,0,0)=12diag(1,0,0). ϕ(A)=P ϕ(Λ)P -1*)(||1P P P Λ=ϕ ⎪⎪⎭⎫⎝⎛------⎪⎪⎭⎫ ⎝⎛⎪⎪⎭⎫ ⎝⎛---=1213032220000000011112011112⎪⎪⎭⎫⎝⎛=1111111114.25. 设矩阵A 、B 及A +B 都可逆, 证明A -1+B -1也可逆, 并求其逆阵. 证明因为A -1(A +B)B -1=B -1+A -1=A -1+B -1,而A -1(A +B)B -1是三个可逆矩阵的乘积, 所以A -1(A +B)B -1可逆, 即A -1+B -1可逆.(A -1+B -1)-1=[A -1(A +B)B -1]-1=B(A +B)-1A .26. 计算⎪⎪⎪⎭⎫ ⎝⎛---⎪⎪⎪⎭⎫⎝⎛30003200121013013000120010100121. 解设⎪⎭⎫ ⎝⎛=10211A , ⎪⎭⎫ ⎝⎛=30122A , ⎪⎭⎫ ⎝⎛-=12131B , ⎪⎭⎫ ⎝⎛--=30322B ,则⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛2121B O B E A O E A ⎪⎭⎫ ⎝⎛+=222111B A O B B A A ,而⎪⎭⎫ ⎝⎛-=⎪⎭⎫ ⎝⎛--+⎪⎭⎫ ⎝⎛-⎪⎭⎫ ⎝⎛=+4225303212131021211B B A ,⎪⎭⎫ ⎝⎛--=⎪⎭⎫ ⎝⎛--⎪⎭⎫ ⎝⎛=90343032301222B A ,所以 ⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛2121B O B E A O E A ⎪⎭⎫ ⎝⎛+=222111B A O B B A A ⎪⎪⎪⎭⎫ ⎝⎛---=9000340042102521, 即 ⎪⎪⎪⎭⎫ ⎝⎛---⎪⎪⎪⎭⎫⎝⎛30003200121013013000120010100121⎪⎪⎪⎭⎫ ⎝⎛---=9000340042102521. 27. 取⎪⎭⎫ ⎝⎛==-==1001D C B A , 验证|||||||| D C B A D C B A ≠.解4100120021100101002000021010010110100101==--=--=D C B A , 而01111|||||||| ==D C B A , 故|||||||| D C B A D C B A ≠. 28. 设⎪⎪⎪⎭⎫ ⎝⎛-=22023443O O A , 求|A 8|及A 4. 解令⎪⎭⎫ ⎝⎛-=34431A , ⎪⎭⎫ ⎝⎛=22022A ,则⎪⎭⎫⎝⎛=21A O O A A ,故8218⎪⎭⎫ ⎝⎛=A O O A A ⎪⎭⎫ ⎝⎛=8281A O O A ,1682818281810||||||||||===A A A A A .⎪⎪⎪⎭⎫ ⎝⎛=⎪⎭⎫ ⎝⎛=464444241422025005O O A O O A A . 29. 设n 阶矩阵A 及s 阶矩阵B 都可逆, 求(1)1-⎪⎭⎫ ⎝⎛O B A O ;解设⎪⎭⎫ ⎝⎛=⎪⎭⎫ ⎝⎛-43211C C C C O B A O , 则⎪⎭⎫ ⎝⎛O B A O ⎪⎭⎫ ⎝⎛4321C C C C ⎪⎭⎫ ⎝⎛=⎪⎭⎫ ⎝⎛=s n E O O E BC BC AC AC 2143. 由此得 ⎪⎩⎪⎨⎧====s n E BC O BC O AC E AC 2143⇒⎪⎩⎪⎨⎧====--121413B C O C O C A C ,所以 ⎪⎭⎫ ⎝⎛=⎪⎭⎫ ⎝⎛---O A B O O B A O 111. (2)1-⎪⎭⎫ ⎝⎛B C O A .解设⎪⎭⎫ ⎝⎛=⎪⎭⎫ ⎝⎛-43211D D D D B C O A , 则⎪⎭⎫ ⎝⎛=⎪⎭⎫ ⎝⎛++=⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛s n E O O E BD CD BD CD AD AD D D D D B C O A 4231214321. 由此得 ⎪⎩⎪⎨⎧=+=+==s n E BD CD O BD CD O AD E AD 423121⇒⎪⎩⎪⎨⎧=-===----14113211B D CA B D O D A D ,所以 ⎪⎭⎫ ⎝⎛-=⎪⎭⎫ ⎝⎛-----11111B CA B O A BC O A . 30. 求下列矩阵的逆阵:(1)⎪⎪⎪⎭⎫⎝⎛2500380000120025; 解设⎪⎭⎫ ⎝⎛=1225A , ⎪⎭⎫ ⎝⎛=2538B , 则⎪⎭⎫ ⎝⎛--=⎪⎭⎫ ⎝⎛=--5221122511A , ⎪⎭⎫ ⎝⎛--=⎪⎭⎫ ⎝⎛=--8532253811B .于是 ⎪⎪⎪⎭⎫ ⎝⎛----=⎪⎭⎫ ⎝⎛=⎪⎭⎫ ⎝⎛=⎪⎪⎪⎭⎫⎝⎛----850032000052002125003800001200251111B A B A .(2)⎪⎪⎪⎭⎫⎝⎛4121031200210001. 解设⎪⎭⎫ ⎝⎛=2101A , ⎪⎭⎫ ⎝⎛=4103B , ⎪⎭⎫ ⎝⎛=2112C , 则⎪⎭⎫ ⎝⎛-=⎪⎭⎫ ⎝⎛=⎪⎪⎪⎭⎫⎝⎛------1111114121031200210001B CA B O A BC O A⎪⎪⎪⎪⎪⎪⎭⎫⎝⎛-----=411212458103161210021210001.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) ⎪⎪⎪⎭⎫ ⎝⎛--340313*********2)3()2(~r r r r -+-+⎪⎪⎪⎭⎫ ⎝⎛---020*********)2()1(32~-÷-÷r r ⎪⎪⎪⎭⎫ ⎝⎛--01003100120123~r r -⎪⎪⎪⎭⎫⎝⎛--300031001201 33~÷r ⎪⎪⎪⎭⎫ ⎝⎛--100031001201323~r r +⎪⎪⎪⎭⎫ ⎝⎛-1000010012013121)2(~r r r r +-+⎪⎪⎪⎭⎫ ⎝⎛100001000001(2) ⎪⎪⎪⎭⎫ ⎝⎛----1740343013201312)2()3(2~r r r r -+-+⨯⎪⎪⎪⎭⎫ ⎝⎛---31003100132021233~r r r r ++⎪⎪⎪⎭⎫ ⎝⎛000031001002021~÷r ⎪⎪⎪⎭⎫ ⎝⎛000031005010 (3) ⎪⎪⎪⎪⎭⎫⎝⎛---------12433023221453334311 141312323~rr r r rr ---⎪⎪⎪⎪⎭⎫ ⎝⎛--------1010500663008840034311)5()3()4(432~-÷-÷-÷r r r ⎪⎪⎪⎪⎭⎫⎝⎛-----22100221002210034311 2423213~r r r r r r ---⎪⎪⎪⎪⎭⎫⎝⎛---00000000002210032011(4) ⎪⎪⎪⎪⎭⎫⎝⎛------34732038234202173132 242321232~rr r r rr ---⎪⎪⎪⎪⎭⎫ ⎝⎛-----1187701298804202111110141312782~rr r r r r --+⎪⎪⎪⎪⎭⎫⎝⎛--410004100020201111134221)1(~r r r r r --⨯↔⎪⎪⎪⎪⎭⎫ ⎝⎛----0000041000111102020132~rr +⎪⎪⎪⎪⎭⎫⎝⎛--000004100030110202012.设⎪⎪⎪⎭⎫⎝⎛=⎪⎪⎪⎭⎫ ⎝⎛⎪⎪⎪⎭⎫ ⎝⎛987654321100010101100001010A ，求A 。

（精选）线性代数课后作业及参考答案

（精选）线性代数课后作业及参考答案《线性代数》作业及参考答案一．单项选择题1.设行列式a aa a11122122=m，a aa a13112321=n，则行列式a a aa a a111213212223++等于（）A. m+nB. -(m+n)C. n-mD. m-n2.设矩阵A=100020003，则A-1等于（）A.130012001B.100120013C. 1 3 00 010 00 1 2D. 1 2 00 10013.设矩阵A=312101214---，A*是A的伴随矩阵，则A *中位于（1，2）的元素是（）A. –6B. 6C. 24.设A是方阵，如有矩阵关系式AB=AC，则必有（）A. A =0B. B≠C时A=0C. A≠0时B=CD. |A|≠0时B=C5.已知3×4矩阵A的行向量组线性无关，则秩（A T）等于（）A. 1B. 2C. 3D. 46.设两个向量组α1，α2，…，αs和β1，β2，…，βs均线性相关，则（）A.有不全为0的数λ1，λ2，…，λs使λ1α1+λ2α2+…+λsαs=0和λ1β1+λ2β2+…λsβs=0B.有不全为0的数λ1，λ2，…，λs使λ1（α1+β1）+λ2（α2+β2）+…+λs（αs+βs）=0C.有不全为0的数λ1，λ2，…，λs使λ1（α1-β1）+λ2（α2-β2）+…+λs（αs-βs）=0D.有不全为0的数λ1，λ2，…，λs和不全为0的数μ1，μ2，…，μs使λ1α1+λ2α2+…+λsαs=0和μ1β1+μ2β2+…+μsβs=07.设矩阵A的秩为r，则A中（）A.所有r-1阶子式都不为0B.所有r-1阶子式全为0C.至少有一个r阶子式不等于0D.所有r阶子式都不为08.设Ax=b是一非齐次线性方程组，η1，η2是其任意2个解，则下列结论错误的是（）A.η1+η2是Ax=0的一个解2η1+12η2是Ax=b的一个解C.η1-η2是Ax=0的一个解D.2η1-η2是Ax=b的一个解9.设n阶方阵A不可逆，则必有（）A.秩(A)<n< bdsfid="226" p=""></n<>B.秩(A)=n-1C.A=0D.方程组Ax=0只有零解10.设A是一个n(≥3)阶方阵，下列陈述中正确的是（）A.如存在数λ和向量α使Aα=λα，则α是A的属于特征值λ的特征向量B.如存在数λ和非零向量α，使(λE-A)α=0，则λ是A的特征值C.A的2个不同的特征值可以有同一个特征向量D.如λ1，λ2，λ3是A的3个互不相同的特征值，α1，α2，α3依次是A的属于λ1，λ2，λ3的特征向量，则α1，α2，α3有可能线性相关11.设λ0是矩阵A的特征方程的3重根，A的属于λ0的线性无关的特征向量的个数为k，则必有（）A. k≤3B. k<3C. k=3D. k>312.设A是正交矩阵，则下列结论错误的是（）A.|A|2必为1B.|A|必为1C.A-1=A TD.A的行（列）向量组是正交单位向量组13.设A是实对称矩阵，C是实可逆矩阵，B=C T AC.则（）A.A与B相似B. A与B不等价C. A与B有相同的特征值D. A与B合同15.设有矩阵Ａｍ×ｎ，Ｂｍ×ｓ，Ｃｓ×ｍ，则下列运算有意义的是（）。

线性代数课后习题答案第1――5章习题详解

第一章行列式4.计算下列各行列式：（1）⎥⎥⎥⎥⎦⎥⎢⎢⎢⎢⎣⎢7110025*********4；（2）⎥⎥⎥⎥⎦⎥⎢⎢⎢⎢⎣⎢-265232112131412；（3）⎥⎥⎦⎥⎢⎢⎣⎢---ef cf bf de cd bd ae ac ab ；（4）⎥⎥⎥⎥⎦⎥⎢⎢⎢⎢⎣⎢---d c b a100110011001解(1)7110025102021421434327c c c c --0100142310202110214---=34)1(143102211014+-⨯---=143102211014-- 321132c c c c ++1417172001099-=0(2)2605232112131412-24c c -2605032122130412-24r r -0412032122130412- 14r r -0000032122130412-=0(3)ef cf bf de cd bd ae ac ab ---=e c b e c b e c b adf ---=111111111---adfbce =abcdef 4(4)d c b a 100110011001---21ar r +dc b a ab 100110011010---+=12)1)(1(+--dc a ab 10111--+23dc c +010111-+-+cd c ada ab =23)1)(1(+--cdadab +-+111=1++++ad cd ab abcd5.证明: (1)1112222b b a a b ab a +=3)(b a -; (2)bz ay by ax bx az by ax bx az bz ay bx az bz ay by ax +++++++++=y x z x z y z y x b a )(33+;(3)0)3()2()1()3()2()1()3()2()1()3()2()1(2222222222222222=++++++++++++d d d d c c c c b b b b a a a a ;(4)444422221111d c b a d c b a d c b a ))()()()((d b c b d a c a b a -----=))((d c b a d c +++-⋅;(5)1221100000100001a x a a a a x x x n n n +-----ΛΛΛΛΛΛΛΛΛΛn n n n a x a x a x ++++=--111Λ. 证明(1)00122222221312a b a b a a b a ab a c c c c ------=左边a b a b a b a ab 22)1(22213-----=+21))((a b a a b a b +--=右边=-=3)(b a(2)bz ay by ax z by ax bx az y bx az bz ay x a ++++++分开按第一列左边bzay by ax x by ax bx az z bxaz bz ay y b +++++++ ++++++002y by ax z x bx az y z bz ay x a 分别再分bz ay y x by ax x z bx az z y b +++zy x y x z xz y b y x z x z y z y x a 33+分别再分右边=-+=233)1(yx z x z y zy x b y x z x z y z y x a(3) 2222222222222222)3()2()12()3()2()12()3()2()12()3()2()12(++++++++++++++++=d d d d d c c c c c b b b b b a a a a a 左边9644129644129644129644122222141312++++++++++++---d d d d c c c c b b b b a a a a c c c c c c 964496449644964422222++++++++d d d d c c c c b b b b a a a a 分成二项按第二列964419644196441964412222+++++++++d d d c c c b b b a a a949494949464222224232423d d c c b b a a c c c c c c c c ----第二项第一项06416416416412222=+ddd c c c bb b a a a (4) 444444422222220001ad a c a b a ad a c a b a ad a c a b a ---------=左边=)()()(222222222222222a d d a c c a b b a d a c a b ad a c a b --------- =)()()(111))()((222a d d a c c a b b a d a c ab a d ac a b ++++++--- =⨯---))()((ad a c a b )()()()()(00122222a b b a d d a b b a c c a b b bd b c a b +-++-++--+ =⨯-----))()()()((b d b c a d a c a b )()()()(112222b d a b bd d b c a b bc c ++++++++=))()()()((d b c b d a c a b a -----))((d c b a d c +++-(5) 用数学归纳法证明.,1,2212122命题成立时当a x a x a x a x D n ++=+-==假设对于)1(-n 阶行列式命题成立，即 ,122111-----++++=n n n n n a x a x a x D Λ:1列展开按第则n D1110010001)1(11----+=+-x xa xD D n n n n ΛΛΛΛΛΛΛΛ右边=+=-n n a xD 1所以，对于n 阶行列式命题成立.6.设n 阶行列式)det(ij a D =,把D 上下翻转、或逆时针旋转ο90、或依副对角线翻转，依次得n nn n a a a a D 11111ΛM M Λ=， 11112n nn n a a a a D ΛM MΛ= ，11113a a a a D n nnn ΛM M Λ=，证明D D D D D n n =-==-32)1(21,)1(.证明 )det(ij a D =Θnnnn nn n nn n a a a a a a a a a a D 2211111111111)1(ΛMM ΛΛΛM M Λ--==∴ΛΛM M ΛΛ=--=--nnn n nnn n a a a a a a a a 331122111121)1()1( nnn n n n a a a a ΛM M ΛΛ111121)1()1()1(---=--D D n n n n 2)1()1()2(21)1()1(--+-+++-=-=Λ同理可证nnn n n n a a a a D ΛM M Λ11112)1(2)1(--=D D n n T n n 2)1(2)1()1()1(---=-= D D D D D n n n n n n n n =-=--=-=----)1(2)1(2)1(22)1(3)1()1()1()1(7.计算下列各行列式（阶行列式为k D k ）：(1)aaD n 11O=,其中对角线上元素都是a ，未写出的元素都是0；(2)xa a ax a a a x D n ΛΛΛΛΛΛΛ=;(3) 1111)()1()()1(1111ΛΛΛΛΛΛΛΛn a a a n a a a n a a a D n n n nn n n ------=---+; 提示：利用范德蒙德行列式的结果． (4) nnnnn d c d c b a b a D ON NO000011112=; (5)j i a a D ij ij n -==其中),det(;(6)nn a a a D +++=11111111121ΛΛΛΛΛΛΛ,021≠n a a a Λ其中.解(1) aa a a a D n 00010000000000001000ΛΛΛΛΛΛΛΛΛΛΛ=按最后一行展开)1()1(100000000000010000)1(-⨯-+-n n n a a a ΛΛΛΛΛΛΛΛΛΛ)1)(1(2)1(--⋅-+n n n a aaO(再按第一行展开)n n n nn a a a+-⋅-=--+)2)(2(1)1()1(O2--=n n a a )1(22-=-a a n(2)将第一行乘)1(-分别加到其余各行，得ax x a ax x a a x x a aa a x D n ------=0000000ΛΛΛΛΛΛΛΛ 再将各列都加到第一列上，得ax ax a x aaa a n x D n ----+=000000000)1(ΛΛΛΛΛΛΛΛ)(])1([1a x a n x n --+=- (3) 从第1+n 行开始，第1+n 行经过n 次相邻对换，换到第1行，第n 行经)1(-n 次对换换到第2行…，经2)1(1)1(+=++-+n n n n Λ次行交换，得 nn n n n n n n n n a a a n a a a n a a a D )()1()()1(1111)1(1112)1(1-------=---++ΛΛΛΛΛΛΛΛ 此行列式为范德蒙德行列式∏≥>≥++++--+--=112)1(1)]1()1[()1(j i n n n n j a i a D∏∏≥>≥+++-++≥>≥++-•-•-=---=111)1(2)1(112)1()][()1()1()]([)1(j i n n n n n j i n n n j i j i Λ∏≥>≥+-=11)(j i n j i(4) nn nnn d c d c b a b a D 011112ONN O=nn n n n nd d c d c b a b a a 0000000011111111----ΛONM NO展开按第一行0000)1(1111111112c d c d c b a b a b nn n n n nn ----+-+ON N O2222 ---n n n n n n D c b D d a 都按最后一行展开由此得递推公式：222)(--=n n n n n n D c b d a D即 ∏=-=ni i i iin D c b da D 222)(而 111111112c b d a d c b a D -==得 ∏=-=ni i i i i n c b d a D 12)((5)j i a ij -=0432********0122210113210)det(ΛΛΛΛΛΛΛΛΛΛΛ--------==n n n n n n n n a D ij n Λ,3221r r r r --0432111111111111111111111ΛΛΛΛΛΛΛΛΛΛΛ--------------n n n nΛ,,141312c c c c c c +++152423210222102210002100001---------------n n n n n ΛΛΛΛΛΛΛΛΛΛΛ=212)1()1(----n n n(6)nn a a D a +++=11111111121ΛΛΛΛΛΛΛΛ,,433221c c c c c c ---n n n n a a a a a a a a a a +-------10000100010000100010001000011433221ΛΛΛΛΛΛΛΛΛΛΛΛΛ 展开（由下往上）按最后一列))(1(121-+n n a a a a Λnn n a a a a a a a a a --------00000000000000000000000000022433221ΛΛΛΛΛΛΛΛΛΛΛΛΛ nn n a a a a a a a a ----+--000000000000000001133221ΛΛΛΛΛΛΛΛΛΛΛ++Λnn n a a a a a a a a -------000000000000000001143322ΛΛΛΛΛΛΛΛΛΛΛn n n n n n a a a a a a a a a a a a ΛΛΛΛ322321121))(1(++++=---)11)((121∑=+=ni in a a a a Λ8.用克莱姆法则解下列方程组：⎪⎪⎩⎪⎪⎨⎧=+++-=----=+-+=+++;01123,2532,242,5)1(4321432143214321x x x x x x x x x x x x x x x x ⎪⎪⎪⎩⎪⎪⎪⎨⎧=+=++=++=++=+.15,065,065,065,165)2(5454343232121x x x x x x x x x x x x x解 (1)11213513241211111----=D 8120735032101111------=145008130032101111---=1421420005410032101111-=---= 112105132412211151------=D 11210513290501115----=1121023313090509151------=2331309050112109151------=1202300461000112109151-----=14200038100112109151----=142-= 112035122412111512-----=D 811507312032701151-------=3139011230023101151-=2842840001910023101151-=----=426110135232422115113-=----=D ; 14202132132212151114=-----=D1,3,2,144332211-========∴DDx D D x D D x D D x (2) 510006510006510006510065=D 展开按最后一行61000510065100655-'D D D ''-'=65 D D D ''-'''-''=6)65(5D D '''-''=3019D D ''''-'''=1146566551141965=⨯-⨯=(,11的余子式中为行列式a D D ',11的余子式中为a D D ''''类推D D ''''''',) 51001651000651000650000611=D 展开按第一列6510065100650006+'D 46+'=D 460319+''''-'''=D 1507=51010651000650000601000152=D 展开按第二列5100651006500061-6510065000610005-365510651065⨯-= 1145108065-=--=51100650000601000051001653=D 展开按第三列51006500061000516500061000510065+6100510656510650061+= 703114619=⨯+=51000601000051000651010654=D 展开按第四列61000510065100655000610005100651--51065106565--=395-= 110051000651000651100655=D 展开按最后一列D '+10005100651006512122111=+= 665212;665395;665703;6651145;665150744321=-==-==∴x x x x x ． 9.齐次线性方程组取何值时问,,μλ⎪⎩⎪⎨⎧=++=++=++0200321321321x x x x x x x x x μμλ有非零解？解 μλμμμλ-==12111113D ，齐次线性方程组有非零解，则03=D即 0=-μλμ 得 10==λμ或不难验证，当,10时或==λμ该齐次线性方程组确有非零解.10.齐次线性方程组取何值时问,λ⎪⎩⎪⎨⎧=-++=+-+=+--0)1(0)3(2042)1(321321321x x x x x x x x x λλλ 有非零解？解λλλ----=111132421D λλλλ--+--=101112431)3)(1(2)1(4)3()1(3λλλλλ-------+-=3)1(2)1(23-+-+-=λλλ 齐次线性方程组有非零解，则0=D得 32,0===λλλ或不难验证，当32,0===λλλ或时，该齐次线性方程组确有非零解.第二章矩阵及其运算1. 已知线性变换:⎪⎩⎪⎨⎧++=++=++=3213321232113235322y y y x y y y x y y y x ,求从变量x 1, x 2, x 3到变量y 1, y 2, y 3的线性变换.解由已知:⎪⎪⎭⎫ ⎝⎛⎪⎪⎭⎫ ⎝⎛=⎪⎪⎭⎫ ⎝⎛221321323513122y y y x x x , 故 ⎪⎪⎭⎫ ⎝⎛⎪⎪⎭⎫ ⎝⎛=⎪⎪⎭⎫ ⎝⎛-3211221323513122x x x y y y ⎪⎪⎭⎫ ⎝⎛⎪⎪⎭⎫ ⎝⎛----=321423736947y y y , ⎪⎩⎪⎨⎧-+=-+=+--=321332123211423736947x x x y x x x y x x x y .2. 已知两个线性变换⎪⎩⎪⎨⎧++=++-=+=32133212311542322y y y x y y y x y y x , ⎪⎩⎪⎨⎧+-=+=+-=323312211323z z y z z y z z y , 求从z 1, z 2, z 3到x 1, x 2, x 3的线性变换.解由已知⎪⎪⎭⎫ ⎝⎛⎪⎪⎭⎫ ⎝⎛-=⎪⎪⎭⎫ ⎝⎛221321514232102y y y x x x ⎪⎪⎭⎫ ⎝⎛⎪⎪⎭⎫ ⎝⎛--⎪⎪⎭⎫ ⎝⎛-=321310102013514232102z z z⎪⎪⎭⎫ ⎝⎛⎪⎪⎭⎫ ⎝⎛----=321161109412316z z z , 所以有⎪⎩⎪⎨⎧+--=+-=++-=3213321232111610941236z z z x z z z x z z z x .3. 设⎪⎪⎭⎫ ⎝⎛--=111111111A , ⎪⎪⎭⎫⎝⎛--=150421321B , 求3AB -2A 及A T B . 解 ⎪⎪⎭⎫⎝⎛---⎪⎪⎭⎫ ⎝⎛--⎪⎪⎭⎫ ⎝⎛--=-1111111112150421321111111111323A AB ⎪⎪⎭⎫⎝⎛----=⎪⎪⎭⎫ ⎝⎛---⎪⎪⎭⎫ ⎝⎛-=2294201722213211111111120926508503, ⎪⎪⎭⎫ ⎝⎛-=⎪⎪⎭⎫ ⎝⎛--⎪⎪⎭⎫ ⎝⎛--=092650850150421321111111111B A T . 4. 计算下列乘积:(1)⎪⎪⎭⎫⎝⎛⎪⎪⎭⎫ ⎝⎛-127075321134; 解 ⎪⎪⎭⎫ ⎝⎛⎪⎪⎭⎫ ⎝⎛-127075321134⎪⎪⎭⎫ ⎝⎛⨯+⨯+⨯⨯+⨯-+⨯⨯+⨯+⨯=102775132)2(71112374⎪⎪⎭⎫ ⎝⎛=49635. (2)⎪⎪⎭⎫⎝⎛123)321(; 解 ⎪⎪⎭⎫⎝⎛123)321(=(1⨯3+2⨯2+3⨯1)=(10).(3))21(312-⎪⎪⎭⎫⎝⎛; 解 )21(312-⎪⎪⎭⎫ ⎝⎛⎪⎪⎭⎫ ⎝⎛⨯-⨯⨯-⨯⨯-⨯=23)1(321)1(122)1(2⎪⎪⎭⎫ ⎝⎛---=632142. (4)⎪⎪⎪⎭⎫ ⎝⎛---⎪⎭⎫ ⎝⎛-20413121013143110412 ; 解 ⎪⎪⎪⎭⎫ ⎝⎛---⎪⎭⎫ ⎝⎛-20413121013143110412⎪⎭⎫ ⎝⎛---=6520876. (5)⎪⎪⎭⎫ ⎝⎛⎪⎪⎭⎫ ⎝⎛321332313232212131211321)(x x x a a a a a a a a a x x x ; 解⎪⎪⎭⎫ ⎝⎛⎪⎪⎭⎫ ⎝⎛321332313232212131211321)(x x x a a a a a a a a a x x x =(a 11x 1+a 12x 2+a 13x 3 a 12x 1+a 22x 2+a 23x 3 a 13x 1+a 23x 2+a 33x 3)⎪⎪⎭⎫ ⎝⎛321x x x322331132112233322222111222x x a x x a x x a x a x a x a +++++=.5. 设⎪⎭⎫ ⎝⎛=3121A , ⎪⎭⎫ ⎝⎛=2101B , 问: (1)AB =BA 吗?解 AB ≠BA .因为⎪⎭⎫ ⎝⎛=6443AB , ⎪⎭⎫ ⎝⎛=8321BA , 所以AB ≠BA .(2)(A +B)2=A 2+2AB +B 2吗?解 (A +B)2≠A 2+2AB +B 2.因为⎪⎭⎫ ⎝⎛=+5222B A , ⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛=+52225222)(2B A ⎪⎭⎫ ⎝⎛=2914148, 但 ⎪⎭⎫ ⎝⎛+⎪⎭⎫ ⎝⎛+⎪⎭⎫ ⎝⎛=++43011288611483222B AB A ⎪⎭⎫ ⎝⎛=27151610, 所以(A +B)2≠A 2+2AB +B 2.(3)(A +B)(A -B)=A 2-B 2吗?解 (A +B)(A -B)≠A 2-B 2.因为⎪⎭⎫ ⎝⎛=+5222B A , ⎪⎭⎫ ⎝⎛=-1020B A , ⎪⎭⎫ ⎝⎛=⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛=-+906010205222))((B A B A , 而 ⎪⎭⎫ ⎝⎛=⎪⎭⎫ ⎝⎛-⎪⎭⎫ ⎝⎛=-718243011148322B A , 故(A +B)(A -B)≠A 2-B 2.6. 举反列说明下列命题是错误的:(1)若A 2=0, 则A =0;解取⎪⎭⎫ ⎝⎛=0010A , 则A 2=0, 但A ≠0. (2)若A 2=A , 则A =0或A =E ;解取⎪⎭⎫ ⎝⎛=0011A , 则A 2=A , 但A ≠0且A ≠E . (3)若AX =AY , 且A ≠0, 则X =Y .解取⎪⎭⎫ ⎝⎛=0001A , ⎪⎭⎫ ⎝⎛-=1111X , ⎪⎭⎫ ⎝⎛=1011Y , 则AX =AY , 且A ≠0, 但X ≠Y .7. 设⎪⎭⎫ ⎝⎛=101λA , 求A 2, A 3, ⋅ ⋅ ⋅, A k . 解 ⎪⎭⎫ ⎝⎛=⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛=12011011012λλλA , ⎪⎭⎫ ⎝⎛=⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛==1301101120123λλλA A A , ⋅ ⋅ ⋅ ⋅ ⋅ ⋅,⎪⎭⎫ ⎝⎛=101λk A k . 8. 设⎪⎪⎭⎫⎝⎛=λλλ001001A , 求A k . 解首先观察⎪⎪⎭⎫ ⎝⎛⎪⎪⎭⎫ ⎝⎛=λλλλλλ0010010010012A ⎪⎪⎭⎫ ⎝⎛=222002012λλλλλ, ⎪⎪⎭⎫ ⎝⎛=⋅=3232323003033λλλλλλA A A , ⎪⎪⎭⎫ ⎝⎛=⋅=43423434004064λλλλλλA A A ,⎪⎪⎭⎫ ⎝⎛=⋅=545345450050105λλλλλλA A A , ⋅ ⋅ ⋅ ⋅ ⋅ ⋅,⎝⎛=k A k k k k k k k k k k λλλλλλ0002)1(121----⎪⎪⎪⎭⎫ . 用数学归纳法证明:当k =2时, 显然成立.假设k 时成立，则k +1时，⎪⎪⎭⎫ ⎝⎛⎪⎪⎪⎪⎭⎫ ⎝⎛-=⋅=---+λλλλλλλλλ0010010002)1(1211k k k k k k k k k k k k A A A ⎪⎪⎪⎪⎭⎫ ⎝⎛+++=+-+--+11111100)1(02)1()1(k k k k k k k k k k λλλλλλ, 由数学归纳法原理知:⎪⎪⎪⎪⎭⎫ ⎝⎛-=---k k k k k k k k k k k A λλλλλλ0002)1(121. 9. 设A , B 为n 阶矩阵，且A 为对称矩阵，证明B T AB 也是对称矩阵. 证明因为A T =A , 所以(B T AB)T =B T (B T A)T =B T A T B =B T AB ,从而B T AB 是对称矩阵.10. 设A , B 都是n 阶对称矩阵，证明AB 是对称矩阵的充分必要条件是AB =BA . 证明充分性: 因为A T =A , B T =B , 且AB =BA , 所以(AB)T =(BA)T =A T B T =AB ,即AB 是对称矩阵.必要性: 因为A T =A , B T =B , 且(AB)T =AB , 所以AB =(AB)T =B T A T =BA .11. 求下列矩阵的逆矩阵:(1)⎪⎭⎫ ⎝⎛5221; 解 ⎪⎭⎫ ⎝⎛=5221A . |A|=1, 故A -1存在. 因为 ⎪⎭⎫ ⎝⎛--=⎪⎭⎫ ⎝⎛=1225*22122111A A A A A , 故*||11A A A =-⎪⎭⎫ ⎝⎛--=1225. (2)⎪⎭⎫ ⎝⎛-θθθθcos sin sin cos ; 解⎪⎭⎫ ⎝⎛-=θθθθcos sin sin cos A . |A|=1≠0, 故A -1存在. 因为 ⎪⎭⎫ ⎝⎛-=⎪⎭⎫ ⎝⎛=θθθθcos sin sin cos *22122111A A A A A , 所以*||11A A A =-⎪⎭⎫ ⎝⎛-=θθθθcos sin sin cos . (3)⎪⎪⎭⎫⎝⎛---145243121; 解 ⎪⎪⎭⎫ ⎝⎛---=145243121A . |A|=2≠0, 故A -1存在. 因为 ⎪⎪⎭⎫ ⎝⎛-----=⎪⎪⎭⎫ ⎝⎛=214321613024*332313322212312111A A A A A A A A A A , 所以*||11A A A =-⎪⎪⎪⎭⎫ ⎝⎛-----=1716213213012.(4)⎪⎪⎪⎭⎫ ⎝⎛n a a a O 0021(a 1a 2⋅ ⋅ ⋅a n≠0) .解 ⎪⎪⎪⎭⎫ ⎝⎛=n a a a A O 0021, 由对角矩阵的性质知⎪⎪⎪⎪⎪⎪⎭⎫ ⎝⎛=-n a a a A 10011211O . 12. 解下列矩阵方程:(1)⎪⎭⎫ ⎝⎛-=⎪⎭⎫ ⎝⎛12643152X ; 解 ⎪⎭⎫ ⎝⎛-⎪⎭⎫ ⎝⎛=-126431521X ⎪⎭⎫ ⎝⎛-⎪⎭⎫ ⎝⎛--=12642153⎪⎭⎫ ⎝⎛-=80232. (2)⎪⎭⎫ ⎝⎛-=⎪⎪⎭⎫ ⎝⎛--234311*********X ; 解 1111012112234311-⎪⎪⎭⎫ ⎝⎛--⎪⎭⎫ ⎝⎛-=X⎪⎪⎭⎫ ⎝⎛---⎪⎭⎫ ⎝⎛-=03323210123431131 ⎪⎪⎭⎫ ⎝⎛---=32538122. (3)⎪⎭⎫ ⎝⎛-=⎪⎭⎫ ⎝⎛-⎪⎭⎫ ⎝⎛-101311022141X ;解 11110210132141--⎪⎭⎫ ⎝⎛-⎪⎭⎫ ⎝⎛-⎪⎭⎫ ⎝⎛-=X⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-⎪⎭⎫ ⎝⎛-=210110131142121 ⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛=21010366121⎪⎪⎭⎫ ⎝⎛=04111. (4)⎪⎪⎭⎫⎝⎛---=⎪⎪⎭⎫ ⎝⎛⎪⎪⎭⎫ ⎝⎛021102341010100001100001010X . 解 11010100001021102341100001010--⎪⎪⎭⎫ ⎝⎛⎪⎪⎭⎫ ⎝⎛---⎪⎪⎭⎫ ⎝⎛=X⎪⎪⎭⎫ ⎝⎛⎪⎪⎭⎫ ⎝⎛---⎪⎪⎭⎫ ⎝⎛=010100001021102341100001010⎪⎪⎭⎫ ⎝⎛---=201431012. 13. 利用逆矩阵解下列线性方程组:(1)⎪⎩⎪⎨⎧=++=++=++3532522132321321321x x x x x x x x x ;解方程组可表示为⎪⎪⎭⎫ ⎝⎛=⎪⎪⎭⎫ ⎝⎛⎪⎪⎭⎫ ⎝⎛321153522321321x x x , 故 ⎪⎪⎭⎫ ⎝⎛=⎪⎪⎭⎫ ⎝⎛⎪⎪⎭⎫ ⎝⎛=⎪⎪⎭⎫ ⎝⎛-0013211535223211321x x x , 从而有 ⎪⎩⎪⎨⎧===001321x x x .(2)⎪⎩⎪⎨⎧=-+=--=--05231322321321321x x x x x x x x x .解方程组可表示为⎪⎪⎭⎫⎝⎛=⎪⎪⎭⎫ ⎝⎛⎪⎪⎭⎫ ⎝⎛-----012523312111321x x x ,故 ⎪⎪⎭⎫ ⎝⎛=⎪⎪⎭⎫ ⎝⎛⎪⎪⎭⎫ ⎝⎛-----=⎪⎪⎭⎫ ⎝⎛-3050125233121111321x x x , 故有 ⎪⎩⎪⎨⎧===305321x x x .14. 设A k =O (k 为正整数), 证明(E -A)-1=E +A +A 2+⋅ ⋅ ⋅+A k -1. 证明因为A k =O , 所以E -A k =E . 又因为 E -A k =(E -A)(E +A +A 2+⋅ ⋅ ⋅+A k -1),所以 (E -A)(E +A +A 2+⋅ ⋅ ⋅+A k -1)=E , 由定理2推论知(E -A)可逆, 且(E -A)-1=E +A +A 2+⋅ ⋅ ⋅+A k -1.证明一方面, 有E =(E -A)-1(E -A). 另一方面, 由A k =O , 有E =(E -A)+(A -A 2)+A 2-⋅ ⋅ ⋅-A k -1+(A k -1-A k ) =(E +A +A 2+⋅ ⋅ ⋅+A k -1)(E -A),故 (E -A)-1(E -A)=(E +A +A 2+⋅ ⋅ ⋅+A k -1)(E -A), 两端同时右乘(E -A)-1, 就有(E -A)-1(E -A)=E +A +A 2+⋅ ⋅ ⋅+A k -1.15. 设方阵A 满足A 2-A -2E =O , 证明A 及A +2E 都可逆, 并求A -1及(A +2E)-1. 证明由A 2-A -2E =O 得A 2-A =2E , 即A(A -E)=2E ,或E E A A =-⋅)(21,由定理2推论知A 可逆, 且)(211E A A -=-.由A 2-A -2E =O 得A 2-A -6E =-4E , 即(A +2E)(A -3E)=-4E ,或E A E E A =-⋅+)3(41)2(由定理2推论知(A +2E)可逆, 且)3(41)2(1A E E A -=+-.证明由A 2-A -2E =O 得A 2-A =2E , 两端同时取行列式得 |A 2-A|=2,即 |A||A -E|=2, 故 |A|≠0,所以A 可逆, 而A +2E =A 2, |A +2E|=|A 2|=|A|2≠0, 故A +2E 也可逆. 由 A 2-A -2E =O ⇒A(A -E)=2E⇒A -1A(A -E)=2A -1E ⇒)(211E A A -=-,又由 A 2-A -2E =O ⇒(A +2E)A -3(A +2E)=-4E⇒ (A +2E)(A -3E)=-4 E ,所以 (A +2E)-1(A +2E)(A -3E)=-4(A +2 E)-1,)3(41)2(1A E E A -=+-.16. 设A 为3阶矩阵,21||=A , 求|(2A)-1-5A*|.解因为*||11A A A =-, 所以|||521||*5)2(|111----=-A A A A A |2521|11---=A A=|-2A -1|=(-2)3|A -1|=-8|A|-1=-8⨯2=-16. 17. 设矩阵A 可逆, 证明其伴随阵A*也可逆, 且(A*)-1=(A -1)*.证明由*||11A A A =-, 得A*=|A|A -1, 所以当A 可逆时, 有|A*|=|A|n |A -1|=|A|n -1≠0,从而A*也可逆.因为A*=|A|A -1, 所以 (A*)-1=|A|-1A .又*)(||)*(||1111---==A A A A A , 所以(A*)-1=|A|-1A =|A|-1|A|(A -1)*=(A -1)*. 18. 设n 阶矩阵A 的伴随矩阵为A*, 证明: (1)若|A|=0, 则|A*|=0; (2)|A*|=|A|n -1. 证明(1)用反证法证明. 假设|A*|≠0, 则有A*(A*)-1=E , 由此得 A =A A*(A*)-1=|A|E(A*)-1=O ,所以A*=O , 这与|A*|≠0矛盾，故当|A|=0时, 有|A*|=0.(2)由于*||11A A A =-, 则AA*=|A|E , 取行列式得到|A||A*|=|A|n . 若|A|≠0, 则|A*|=|A|n -1;若|A|=0, 由(1)知|A*|=0, 此时命题也成立. 因此|A*|=|A|n -1.19. 设⎪⎪⎭⎫⎝⎛-=321011330A , AB =A +2B , 求B .解由AB =A +2E 可得(A -2E)B =A , 故⎪⎪⎭⎫ ⎝⎛-⎪⎪⎭⎫ ⎝⎛---=-=--321011330121011332)2(11A E A B ⎪⎪⎭⎫ ⎝⎛-=011321330.20. 设⎪⎪⎭⎫⎝⎛=101020101A , 且AB +E =A 2+B , 求B .解由AB +E =A 2+B 得 (A -E)B =A 2-E ,即 (A -E)B =(A -E)(A +E).因为01001010100||≠-==-E A , 所以(A -E)可逆, 从而⎪⎪⎭⎫⎝⎛=+=201030102E A B .21. 设A =diag(1, -2, 1), A*BA =2BA -8E , 求B . 解由A*BA =2BA -8E 得 (A*-2E)BA =-8E , B =-8(A*-2E)-1A -1 =-8[A(A*-2E)]-1 =-8(AA*-2A)-1 =-8(|A|E -2A)-1 =-8(-2E -2A)-1 =4(E +A)-1=4[diag(2, -1, 2)]-1)21 ,1 ,21(diag 4-==2diag(1, -2, 1).22. 已知矩阵A 的伴随阵⎪⎪⎪⎭⎫⎝⎛-=8030010100100001*A , 且ABA -1=BA -1+3E , 求B .解由|A*|=|A|3=8, 得|A|=2. 由ABA -1=BA -1+3E 得 AB =B +3A ,B =3(A -E)-1A =3[A(E -A -1)]-1A11*)2(6*)21(3---=-=A E A E⎪⎪⎪⎭⎫ ⎝⎛-=⎪⎪⎪⎭⎫⎝⎛--=-1030060600600006603001010010000161. 23. 设P -1AP =Λ, 其中⎪⎭⎫ ⎝⎛--=1141P , ⎪⎭⎫ ⎝⎛-=Λ2001, 求A 11.解由P -1AP =Λ, 得A =P ΛP -1, 所以A 11= A=P Λ11P -1.|P|=3,⎪⎭⎫ ⎝⎛-=1141*P , ⎪⎭⎫ ⎝⎛--=-1141311P ,而 ⎪⎭⎫ ⎝⎛-=⎪⎭⎫⎝⎛-=Λ11111120 012001,故⎪⎪⎪⎭⎫⎝⎛--⎪⎭⎫ ⎝⎛-⎪⎭⎫ ⎝⎛--=31313431200111411111A ⎪⎭⎫ ⎝⎛--=68468327322731.24. 设AP =P Λ, 其中⎪⎪⎭⎫⎝⎛--=111201111P , ⎪⎪⎭⎫ ⎝⎛-=Λ511, 求ϕ(A)=A 8(5E -6A +A 2).解 ϕ(Λ)=Λ8(5E -6Λ+Λ2)=diag(1,1,58)[diag(5,5,5)-diag(-6,6,30)+diag(1,1,25)] =diag(1,1,58)diag(12,0,0)=12diag(1,0,0). ϕ(A)=P ϕ(Λ)P -1*)(||1P P P Λ=ϕ ⎪⎪⎭⎫⎝⎛------⎪⎪⎭⎫ ⎝⎛⎪⎪⎭⎫ ⎝⎛---=1213032220000000011112011112⎪⎪⎭⎫⎝⎛=1111111114.25. 设矩阵A 、B 及A +B 都可逆, 证明A -1+B -1也可逆, 并求其逆阵. 证明因为A -1(A +B)B -1=B -1+A -1=A -1+B -1,而A -1(A +B)B -1是三个可逆矩阵的乘积, 所以A -1(A +B)B -1可逆, 即A -1+B -1可逆.(A -1+B -1)-1=[A -1(A +B)B -1]-1=B(A +B)-1A .26. 计算⎪⎪⎪⎭⎫ ⎝⎛---⎪⎪⎪⎭⎫⎝⎛30003200121013013000120010100121. 解设⎪⎭⎫ ⎝⎛=10211A , ⎪⎭⎫ ⎝⎛=30122A , ⎪⎭⎫ ⎝⎛-=12131B , ⎪⎭⎫ ⎝⎛--=30322B ,则⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛2121B O B E A O E A ⎪⎭⎫ ⎝⎛+=222111B A O B B A A ,而⎪⎭⎫ ⎝⎛-=⎪⎭⎫ ⎝⎛--+⎪⎭⎫ ⎝⎛-⎪⎭⎫ ⎝⎛=+4225303212131021211B B A ,⎪⎭⎫ ⎝⎛--=⎪⎭⎫ ⎝⎛--⎪⎭⎫ ⎝⎛=90343032301222B A ,所以 ⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛2121B O B E A O E A ⎪⎭⎫ ⎝⎛+=222111B A O B B A A ⎪⎪⎪⎭⎫ ⎝⎛---=9000340042102521, 即 ⎪⎪⎪⎭⎫ ⎝⎛---⎪⎪⎪⎭⎫⎝⎛30003200121013013000120010100121⎪⎪⎪⎭⎫ ⎝⎛---=9000340042102521. 27. 取⎪⎭⎫ ⎝⎛==-==1001D C B A , 验证|||||||| D C B A D C B A ≠.解4100120021100101002000021010010110100101==--=--=D C B A , 而01111|||||||| ==D C B A , 故|||||||| D C B A D C B A ≠. 28. 设⎪⎪⎪⎭⎫ ⎝⎛-=22023443O O A , 求|A 8|及A 4. 解令⎪⎭⎫ ⎝⎛-=34431A , ⎪⎭⎫ ⎝⎛=22022A ,则⎪⎭⎫⎝⎛=21A O O A A ,故8218⎪⎭⎫ ⎝⎛=A O O A A ⎪⎭⎫ ⎝⎛=8281A O O A ,1682818281810||||||||||===A A A A A .⎪⎪⎪⎭⎫ ⎝⎛=⎪⎭⎫ ⎝⎛=464444241422025005O O A O O A A . 29. 设n 阶矩阵A 及s 阶矩阵B 都可逆, 求(1)1-⎪⎭⎫ ⎝⎛O B A O ;解设⎪⎭⎫ ⎝⎛=⎪⎭⎫ ⎝⎛-43211C C C C O B A O , 则⎪⎭⎫ ⎝⎛O B A O ⎪⎭⎫ ⎝⎛4321C C C C ⎪⎭⎫ ⎝⎛=⎪⎭⎫ ⎝⎛=s n E O O E BC BC AC AC 2143. 由此得 ⎪⎩⎪⎨⎧====s n E BC O BC O AC E AC 2143⇒⎪⎩⎪⎨⎧====--121413B C O C O C A C ,所以 ⎪⎭⎫ ⎝⎛=⎪⎭⎫ ⎝⎛---O A B O O B A O 111. (2)1-⎪⎭⎫ ⎝⎛B C O A .解设⎪⎭⎫ ⎝⎛=⎪⎭⎫ ⎝⎛-43211D D D D B C O A , 则⎪⎭⎫ ⎝⎛=⎪⎭⎫ ⎝⎛++=⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛s n E O O E BD CD BD CD AD AD D D D D B C O A 4231214321. 由此得 ⎪⎩⎪⎨⎧=+=+==s n E BD CD O BD CD O AD E AD 423121⇒⎪⎩⎪⎨⎧=-===----14113211B D CA B D O D A D ,所以 ⎪⎭⎫ ⎝⎛-=⎪⎭⎫ ⎝⎛-----11111B CA B O A BC O A . 30. 求下列矩阵的逆阵:(1)⎪⎪⎪⎭⎫⎝⎛2500380000120025; 解设⎪⎭⎫ ⎝⎛=1225A , ⎪⎭⎫ ⎝⎛=2538B , 则⎪⎭⎫ ⎝⎛--=⎪⎭⎫ ⎝⎛=--5221122511A , ⎪⎭⎫ ⎝⎛--=⎪⎭⎫ ⎝⎛=--8532253811B .于是 ⎪⎪⎪⎭⎫ ⎝⎛----=⎪⎭⎫ ⎝⎛=⎪⎭⎫ ⎝⎛=⎪⎪⎪⎭⎫⎝⎛----850032000052002125003800001200251111B A B A .(2)⎪⎪⎪⎭⎫⎝⎛4121031200210001. 解设⎪⎭⎫ ⎝⎛=2101A , ⎪⎭⎫ ⎝⎛=4103B , ⎪⎭⎫ ⎝⎛=2112C , 则⎪⎭⎫ ⎝⎛-=⎪⎭⎫ ⎝⎛=⎪⎪⎪⎭⎫⎝⎛------1111114121031200210001B CA B O A BC O A⎪⎪⎪⎪⎪⎪⎭⎫⎝⎛-----=411212458103161210021210001.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) ⎪⎪⎪⎭⎫ ⎝⎛--340313*********2)3()2(~r r r r -+-+⎪⎪⎪⎭⎫ ⎝⎛---020*********)2()1(32~-÷-÷r r ⎪⎪⎪⎭⎫ ⎝⎛--01003100120123~r r -⎪⎪⎪⎭⎫⎝⎛--300031001201 33~÷r ⎪⎪⎪⎭⎫ ⎝⎛--100031001201323~r r +⎪⎪⎪⎭⎫ ⎝⎛-1000010012013121)2(~r r r r +-+⎪⎪⎪⎭⎫ ⎝⎛100001000001(2) ⎪⎪⎪⎭⎫ ⎝⎛----1740343013201312)2()3(2~r r r r -+-+⨯⎪⎪⎪⎭⎫ ⎝⎛---31003100132021233~r r r r ++⎪⎪⎪⎭⎫ ⎝⎛000031001002021~÷r ⎪⎪⎪⎭⎫ ⎝⎛000031005010 (3) ⎪⎪⎪⎪⎭⎫⎝⎛---------12433023221453334311 141312323~rr r r rr ---⎪⎪⎪⎪⎭⎫ ⎝⎛--------1010500663008840034311)5()3()4(432~-÷-÷-÷r r r ⎪⎪⎪⎪⎭⎫⎝⎛-----22100221002210034311 2423213~r r r r r r ---⎪⎪⎪⎪⎭⎫⎝⎛---00000000002210032011(4) ⎪⎪⎪⎪⎭⎫⎝⎛------34732038234202173132 242321232~rr r r rr ---⎪⎪⎪⎪⎭⎫ ⎝⎛-----1187701298804202111110141312782~rr r r r r --+⎪⎪⎪⎪⎭⎫⎝⎛--410004100020201111134221)1(~r r r r r --⨯↔⎪⎪⎪⎪⎭⎫ ⎝⎛----0000041000111102020132~rr +⎪⎪⎪⎪⎭⎫⎝⎛--000004100030110202012.设⎪⎪⎪⎭⎫⎝⎛=⎪⎪⎪⎭⎫ ⎝⎛⎪⎪⎪⎭⎫ ⎝⎛987654321100010101100001010A ，求A 。

线性代数课后习题1-4作业答案(高等教育出版社)

第一章行列式1. 利用对角线法则计算下列三阶行列式: (1)381141102---;解 381141102---=2⨯(-4)⨯3+0⨯(-1)⨯(-1)+1⨯1⨯8 -0⨯1⨯3-2⨯(-1)⨯8-1⨯(-4)⨯(-1) =-24+8+16-4=-4. (3)222111c b a c b a ; 解 222111c b a c b a=bc 2+ca 2+ab 2-ac 2-ba 2-cb 2 =(a -b )(b -c )(c -a ).4. 计算下列各行列式:(1)71100251020214214; 解 7110251020214214010014231020211021473234-----======c c c c 34)1(143102211014+-⨯---= 143102211014--=01417172001099323211=-++======c c c c .(2)2605232112131412-;解 2605232112131412-260503212213041224--=====c c 041203212213041224--=====r r 000003212213041214=--=====r r . (3)efcf bf de cd bd aeac ab ---;解 ef cf bf de cd bd ae ac ab ---e c b e c b ec b adf ---=abcdef adfbce 4111111111=---=.(4)dc b a 100110011001---. 解d c b a 100110011001---dc b aab ar r 10011001101021---++===== dc a ab 101101)1)(1(12--+--=+01011123-+-++=====cd c ada ab dc ccdad ab +-+--=+111)1)(1(23=abcd +ab +cd +ad +1. 6. 证明:(1)1112222b b a a b ab a +=(a -b )3;证明1112222b b a a b ab a +00122222221213a b a b a a b a ab a c c c c ------=====ab a b a b a ab 22)1(22213-----=+21))((a b a a b a b +--==(a -b )3 . (2)y x z x z y zy x b a bz ay by ax bx az by ax bx az bz ay bx az bz ay by ax )(33+=+++++++++;证明bzay by ax bx az by ax bx az bz ay bxaz bz ay by ax +++++++++bz ay by ax x by ax bx az z bxaz bz ay y b bz ay by ax z by ax bx az y bx az bz ay x a +++++++++++++=bz ay y x by ax x z bxaz z y b y by ax z x bx az y z bz ay x a +++++++=22z y x y x z xz y b y x z x z y z y x a 33+=y x z x z y zy x b y x z x z y z y x a 33+=yx z x z y zy x b a )(33+=.8. 计算下列各行列式(D k 为k 阶行列式): (1)aa D n 1 1⋅⋅⋅=, 其中对角线上元素都是a , 未写出的元素都是0; 解aa a a a D n 0 0010 000 00 000 0010 00⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅=(按第n 行展开) )1()1(10 000 00 000 0010 000)1(-⨯-+⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅-=n n n aa a )1()1(2 )1(-⨯-⋅⋅⋅⋅-+n n n a a an n n nn a a a+⋅⋅⋅-⋅-=--+)2)(2(1)1()1(=a n -a n -2=a n -2(a 2-1).(2)xa a a x a a a xD n ⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅= ; 解将第一行乘(-1)分别加到其余各行, 得 ax x a ax x a a x x a a a a x D n --⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅--⋅⋅⋅--⋅⋅⋅=000 0 00 0, 再将各列都加到第一列上, 得ax ax a x aaa a n x D n -⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅-⋅⋅⋅-⋅⋅⋅-+=0000 0 0000 )1(=[x +(n -1)a ](x -a )n 第二章矩阵及其运算 1. 计算下列乘积:(5)⎪⎪⎭⎫⎝⎛⎪⎪⎭⎫ ⎝⎛321332313232212131211321)(x x x a a a a a a a a a x x x ;解⎪⎪⎭⎫⎝⎛⎪⎪⎭⎫ ⎝⎛321332313232212131211321)(x x x a a a a a a a a a x x x=(a 11x 1+a 12x 2+a 13x 3 a 12x 1+a 22x 2+a 23x 3 a 13x 1+a 23x 2+a 33x 3)⎪⎪⎭⎫⎝⎛321x x x322331132112233322222111222x x a x x a x x a x a x a x a +++++=.2. 设⎪⎪⎭⎫ ⎝⎛--=111111111A , ⎪⎪⎭⎫⎝⎛--=150421321B , 求3AB -2A 及A TB .解 ⎪⎪⎭⎫⎝⎛---⎪⎪⎭⎫ ⎝⎛--⎪⎪⎭⎫ ⎝⎛--=-1111111112150421321111111111323A AB⎪⎪⎭⎫⎝⎛----=⎪⎪⎭⎫ ⎝⎛---⎪⎪⎭⎫ ⎝⎛-=2294201722213211111111120926508503, ⎪⎪⎭⎫ ⎝⎛-=⎪⎪⎭⎫ ⎝⎛--⎪⎪⎭⎫ ⎝⎛--=092650850150421321111111111B A T . 3.已知两个线性变换⎪⎩⎪⎨⎧++=++-=+=32133212311542322y y y x y y y x y y x ,⎪⎩⎪⎨⎧+-=+=+-=323312211323z z y z z y z z y , 求从z 1, z 2, z 3到x 1, x 2, x 3的线性变换.解由已知⎪⎪⎭⎫ ⎝⎛⎪⎪⎭⎫ ⎝⎛-=⎪⎪⎭⎫ ⎝⎛221321514232102y y y x x x ⎪⎪⎭⎫⎝⎛⎪⎪⎭⎫ ⎝⎛--⎪⎪⎭⎫ ⎝⎛-=321310102013514232102z z z ⎪⎪⎭⎫⎝⎛⎪⎪⎭⎫ ⎝⎛----=321161109412316z z z ,所以有⎪⎩⎪⎨⎧+--=+-=++-=3213321232111610941236z z z x z z z x z z z x .4.设⎪⎭⎫ ⎝⎛=3121A , ⎪⎭⎫ ⎝⎛=2101B , 问: (1)AB =BA 吗? 解 AB ≠BA .因为⎪⎭⎫ ⎝⎛=6443AB , ⎪⎭⎫ ⎝⎛=8321BA , 所以AB ≠BA .(3)(A +B )(A -B )=A 2-B 2吗? 解 (A +B )(A -B )≠A 2-B 2.因为⎪⎭⎫ ⎝⎛=+5222B A , ⎪⎭⎫ ⎝⎛=-1020B A ,⎪⎭⎫ ⎝⎛=⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛=-+906010205222))((B A B A ,而 ⎪⎭⎫ ⎝⎛=⎪⎭⎫ ⎝⎛-⎪⎭⎫ ⎝⎛=-718243011148322B A ,故(A +B )(A -B )≠A 2-B 2.5. 举反列说明下列命题是错误的: (1)若A 2=0, 则A =0;解取⎪⎭⎫ ⎝⎛=0010A , 则A 2=0, 但A ≠0. (2)若A 2=A , 则A =0或A =E ;解取⎪⎭⎫ ⎝⎛=0011A , 则A 2=A , 但A ≠0且A ≠E . (3)若AX =AY , 且A ≠0, 则X =Y .解取⎪⎭⎫ ⎝⎛=0001A , ⎪⎭⎫ ⎝⎛-=1111X , ⎪⎭⎫ ⎝⎛=1011Y ,则AX =AY , 且A ≠0, 但X ≠Y .7. 设⎪⎪⎭⎫⎝⎛=λλλ001001A , 求A k .解首先观察⎪⎪⎭⎫ ⎝⎛⎪⎪⎭⎫ ⎝⎛=λλλλλλ0010010010012A ⎪⎪⎭⎫⎝⎛=222002012λλλλλ,⎪⎪⎭⎫⎝⎛=⋅=3232323003033λλλλλλA A A ,⎪⎪⎭⎫⎝⎛=⋅=43423434004064λλλλλλA A A ,⎪⎪⎭⎫⎝⎛=⋅=545345450050105λλλλλλA A A ,⋅ ⋅ ⋅ ⋅ ⋅ ⋅,⎝⎛=kA kk kk k k k k k k λλλλλλ0002)1(121----⎪⎪⎪⎭⎫. 用数学归纳法证明: 当k =2时, 显然成立. 假设k 时成立，则k +1时，⎪⎪⎭⎫ ⎝⎛⎪⎪⎪⎪⎭⎫ ⎝⎛-=⋅=---+λλλλλλλλλ0010010002)1(1211k k k k k k k k k k k k A A A⎪⎪⎪⎪⎭⎫⎝⎛+++=+-+--+11111100)1(02)1()1(k k k k k k k k k k λλλλλλ,由数学归纳法原理知:⎪⎪⎪⎪⎭⎫ ⎝⎛-=---k k k k k k k k k k k A λλλλλλ0002)1(121.8. 设A , B 为n 阶矩阵，且A 为对称矩阵，证明B T AB 也是对称矩阵.证明因为A T =A , 所以(B T AB )T =B T (B T A )T =B T A T B =B T AB , 从而B T AB 是对称矩阵. 11. 求下列矩阵的逆矩阵:(1)⎪⎭⎫ ⎝⎛5221; 解 ⎪⎭⎫ ⎝⎛=5221A . |A |=1, 故A -1存在. 因为 ⎪⎭⎫⎝⎛--=⎪⎭⎫ ⎝⎛=1225*22122111A A A A A ,故 *||11A A A =-⎪⎭⎫ ⎝⎛--=1225. (3)⎪⎪⎭⎫⎝⎛---145243121; 解 ⎪⎪⎭⎫⎝⎛---=145243121A . |A |=2≠0, 故A -1存在. 因为⎪⎪⎭⎫⎝⎛-----=⎪⎪⎭⎫ ⎝⎛=214321613024*332313322212312111A A A A A A A A A A ,所以 *||11A A A =-⎪⎪⎪⎭⎫ ⎝⎛-----=1716213213012.(4)⎪⎪⎪⎭⎫ ⎝⎛n a a a 0021(a 1a 2⋅ ⋅ ⋅a n ≠0) .解 ⎪⎪⎪⎭⎫ ⎝⎛=n a a a A 0021, 由对角矩阵的性质知 ⎪⎪⎪⎪⎪⎪⎭⎫ ⎝⎛=-n a a a A 10011211 . 12. 利用逆矩阵解下列线性方程组: (1)⎪⎩⎪⎨⎧=++=++=++3532522132321321321x x x x x x x x x ;解方程组可表示为 ⎪⎪⎭⎫⎝⎛=⎪⎪⎭⎫ ⎝⎛⎪⎪⎭⎫ ⎝⎛321153522321321x x x ,故 ⎪⎪⎭⎫⎝⎛=⎪⎪⎭⎫ ⎝⎛⎪⎪⎭⎫ ⎝⎛=⎪⎪⎭⎫ ⎝⎛-0013211535223211321x x x ,从而有 ⎪⎩⎪⎨⎧===001321x x x .19.设P -1AP =Λ, 其中⎪⎭⎫ ⎝⎛--=1141P , ⎪⎭⎫ ⎝⎛-=Λ2001, 求A 11. 解由P -1AP =Λ, 得A =P ΛP -1, 所以A 11= A =P Λ11P -1.|P |=3, ⎪⎭⎫ ⎝⎛-=1141*P , ⎪⎭⎫ ⎝⎛--=-1141311P ,而 ⎪⎭⎫ ⎝⎛-=⎪⎭⎫ ⎝⎛-=Λ11111120 012001,故 ⎪⎪⎪⎭⎫⎝⎛--⎪⎭⎫ ⎝⎛-⎪⎭⎫ ⎝⎛--=31313431200111411111A ⎪⎭⎫ ⎝⎛--=68468327322731. 20. 设AP =P Λ, 其中⎪⎪⎭⎫⎝⎛--=111201111P , ⎪⎪⎭⎫ ⎝⎛-=Λ511, 求ϕ(A )=A 8(5E -6A +A 2). 解 ϕ(Λ)=Λ8(5E -6Λ+Λ2)=diag(1,1,58)[diag(5,5,5)-diag(-6,6,30)+diag(1,1,25)] =diag(1,1,58)diag(12,0,0)=12diag(1,0,0). ϕ(A )=P ϕ(Λ)P -1*)(||1P P P Λ=ϕ⎪⎪⎭⎫ ⎝⎛------⎪⎪⎭⎫ ⎝⎛⎪⎪⎭⎫ ⎝⎛---=1213032220000000011112011112⎪⎪⎭⎫⎝⎛=1111111114.21. 设A k =O (k 为正整数), 证明(E -A )-1=E +A +A 2+⋅ ⋅ ⋅+A k -1. 证明因为A k =O , 所以E -A k =E . 又因为 E -A k =(E -A )(E +A +A 2+⋅ ⋅ ⋅+A k -1), 所以 (E -A )(E +A +A 2+⋅ ⋅ ⋅+A k -1)=E , 由定理2推论知(E -A )可逆, 且 (E -A )-1=E +A +A 2+⋅ ⋅ ⋅+A k -1.证明一方面, 有E =(E -A )-1(E -A ).另一方面, 由A k =O , 有E =(E -A )+(A -A 2)+A 2-⋅ ⋅ ⋅-A k -1+(A k -1-A k ) =(E +A +A 2+⋅ ⋅ ⋅+A k -1)(E -A ), 故 (E -A )-1(E -A )=(E +A +A 2+⋅ ⋅ ⋅+A k -1)(E -A ), 两端同时右乘(E -A )-1, 就有(E -A )-1(E -A )=E +A +A 2+⋅ ⋅ ⋅+A k -1.22. 设方阵A 满足A 2-A -2E =O , 证明A 及A +2E 都可逆, 并求A -1及(A +2E )-1.证明由A 2-A -2E =O 得 A 2-A =2E , 即A (A -E )=2E , 或 E E A A =-⋅)(21,由定理2推论知A 可逆, 且)(211E A A -=-.由A 2-A -2E =O 得A 2-A -6E =-4E , 即(A +2E )(A -3E )=-4E , 或 E A E E A =-⋅+)3(41)2(由定理2推论知(A +2E )可逆, 且)3(41)2(1A E E A -=+-.证明由A 2-A -2E =O 得A 2-A =2E , 两端同时取行列式得 |A 2-A |=2, 即 |A ||A -E |=2, 故 |A |≠0,所以A 可逆, 而A +2E =A 2, |A +2E |=|A 2|=|A |2≠0, 故A +2E 也可逆. 由 A 2-A -2E =O ⇒A (A -E )=2E⇒A -1A (A -E )=2A -1E ⇒)(211E A A -=-,又由 A 2-A -2E =O ⇒(A +2E )A -3(A +2E )=-4E ⇒ (A +2E )(A -3E )=-4 E ,所以 (A +2E )-1(A +2E )(A -3E )=-4(A +2 E )-1, )3(41)2(1A E E A -=+-.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;解 ⎪⎪⎭⎫⎝⎛--340313021201(下一步: r 2+(-2)r 1, r 3+(-3)r 1. )~⎪⎪⎭⎫⎝⎛---020*********(下一步: r 2÷(-1), r 3÷(-2). )~⎪⎪⎭⎫⎝⎛--010*********(下一步: r 3-r 2. )~⎪⎪⎭⎫⎝⎛--300031001201(下一步: r 3÷3. )~⎪⎪⎭⎫⎝⎛--100031001201(下一步: r 2+3r 3. )~⎪⎪⎭⎫⎝⎛-100001001201(下一步: r 1+(-2)r 2, r 1+r 3. )~⎪⎪⎭⎫⎝⎛100001000001.(3)⎪⎪⎪⎭⎫⎝⎛---------12433023221453334311; 解 ⎪⎪⎪⎭⎫⎝⎛---------12433023221453334311(下一步: r 2-3r 1, r 3-2r 1, r 4-3r 1. )~⎪⎪⎪⎭⎫⎝⎛--------1010500663008840034311(下一步: r 2÷(-4), r 3÷(-3) , r 4÷(-5). )~⎪⎪⎪⎭⎫⎝⎛-----22100221002210034311(下一步: r 1-3r 2, r 3-r 2, r 4-r 2. )~⎪⎪⎪⎭⎫⎝⎛---00000000002210032011. 3. 已知两个线性变换⎪⎩⎪⎨⎧++=++-=+=32133212311542322y y y x y y y x y y x ,⎪⎩⎪⎨⎧+-=+=+-=323312211323z z y z z y z z y , 求从z 1, z 2, z 3到x 1, x 2, x 3的线性变换.解由已知⎪⎪⎭⎫ ⎝⎛⎪⎪⎭⎫ ⎝⎛-=⎪⎪⎭⎫ ⎝⎛221321514232102y y y x x x ⎪⎪⎭⎫⎝⎛⎪⎪⎭⎫ ⎝⎛--⎪⎪⎭⎫ ⎝⎛-=321310102013514232102z z z ⎪⎪⎭⎫⎝⎛⎪⎪⎭⎫ ⎝⎛----=321161109412316z z z ,所以有⎪⎩⎪⎨⎧+--=+-=++-=3213321232111610941236z z z x z z z x z z z x .4. 试利用矩阵的初等变换, 求下列方阵的逆矩阵:(1)⎪⎪⎭⎫⎝⎛323513123;解 ⎪⎪⎭⎫ ⎝⎛100010001323513123~⎪⎪⎭⎫⎝⎛---101011001200410123~⎪⎪⎭⎫ ⎝⎛----1012002110102/102/3023~⎪⎪⎭⎫⎝⎛----2/102/11002110102/922/7003~⎪⎪⎭⎫⎝⎛----2/102/11002110102/33/26/7001故逆矩阵为⎪⎪⎪⎪⎭⎫ ⎝⎛----21021211233267.(2)⎪⎪⎪⎭⎫ ⎝⎛-----1210232112201023.解 ⎪⎪⎪⎭⎫ ⎝⎛-----10000100001000011210232112201023~⎪⎪⎪⎭⎫ ⎝⎛----00100301100001001220594012102321~⎪⎪⎪⎭⎫ ⎝⎛--------20104301100001001200110012102321~⎪⎪⎪⎭⎫ ⎝⎛-------106124301100001001000110012102321 ~⎪⎪⎪⎭⎫⎝⎛----------10612631110`1022111000010000100021 ~⎪⎪⎪⎭⎫⎝⎛-------106126311101042111000010000100001 故逆矩阵为⎪⎪⎪⎭⎫⎝⎛-------10612631110104211. 5. (2)设⎪⎪⎭⎫ ⎝⎛---=433312120A , ⎪⎭⎫ ⎝⎛-=132321B , 求X 使XA =B . 解考虑A T X T =B T . 因为⎪⎪⎭⎫ ⎝⎛----=134313*********) ,(T T B A ⎪⎪⎭⎫⎝⎛---411007101042001 ~r ,所以 ⎪⎪⎭⎫⎝⎛---==-417142)(1T T T B A X ,从而 ⎪⎭⎫ ⎝⎛---==-4741121BA X . 9. 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).解用已知向量容易构成一个有4个非零行的5阶下三角矩阵:⎪⎪⎪⎪⎭⎫ ⎝⎛-0000001000001010001100001, 此矩阵的秩为4, 其第2行和第3行是已知向量.12. 设⎪⎪⎭⎫⎝⎛----=32321321k k k A , 问k 为何值, 可使(1)R (A )=1; (2)R (A )=2; (3)R (A )=3.解 ⎪⎪⎭⎫ ⎝⎛----=32321321k k k A ⎪⎪⎭⎫ ⎝⎛+-----)2)(1(0011011 ~k k k k k r . (1)当k =1时, R (A )=1; (2)当k =-2且k ≠1时, R (A )=2; (3)当k ≠1且k ≠-2时, R (A )=3. P106/ 1.已知向量组A : a 1=(0, 1, 2, 3)T , a 2=(3, 0, 1, 2)T , a 3=(2, 3, 0, 1)T ;B : b 1=(2, 1, 1, 2)T , b 2=(0, -2, 1, 1)T , b 3=(4, 4, 1, 3)T , 证明B 组能由A 组线性表示, 但A 组不能由B 组线性表示.证明由 ⎪⎪⎪⎭⎫⎝⎛-=312123111012421301402230) ,(B A ⎪⎪⎪⎭⎫ ⎝⎛-------971820751610402230421301~r ⎪⎪⎪⎭⎫⎝⎛------531400251552000751610421301 ~r ⎪⎪⎪⎭⎫ ⎝⎛-----000000531400751610421301~r 知R (A )=R (A , B )=3, 所以B 组能由A 组线性表示.由⎪⎪⎪⎭⎫ ⎝⎛-⎪⎪⎪⎭⎫ ⎝⎛---⎪⎪⎪⎭⎫ ⎝⎛-=000000110201110110220201312111421402~~r r B 知R (B )=2. 因为R (B )≠R (B , A ), 所以A 组不能由B 组线性表示. 4. 判定下列向量组是线性相关还是线性无关: (1) (-1, 3, 1)T , (2, 1, 0)T , (1, 4, 1)T ; (2) (2, 3, 0)T , (-1, 4, 0)T , (0, 0, 2)T .解 (1)以所给向量为列向量的矩阵记为A . 因为⎪⎪⎭⎫⎝⎛-⎪⎪⎭⎫ ⎝⎛-⎪⎪⎭⎫ ⎝⎛-=000110121220770121101413121~~r r A ,所以R (A )=2小于向量的个数, 从而所给向量组线性相关. (2)以所给向量为列向量的矩阵记为B . 因为022200043012||≠=-=B ,所以R (B )=3等于向量的个数, 从而所给向量组线性相无关.5. 问a 取什么值时下列向量组线性相关？ a 1=(a , 1, 1)T , a 2=(1, a , -1)T , a 3=(1, -1, a )T . 解以所给向量为列向量的矩阵记为A . 由)1)(1(111111||+-=--=a a a aa a A知, 当a =-1、0、1时, R (A )<3, 此时向量组线性相关.9.设b 1=a 1+a 2, b 2=a 2+a 3, b 3=a 3+a 4, b 4=a 4+a 1, 证明向量组b 1, b 2, b 3, b 4线性相关.证明由已知条件得a 1=b 1-a 2, a 2=b 2-a 3, a 3=b 3-a 4, a 4=b 4-a 1,于是 a 1 =b 1-b 2+a 3 =b 1-b 2+b 3-a 4 =b 1-b 2+b 3-b 4+a 1, 从而 b 1-b 2+b 3-b 4=0,这说明向量组b 1, b 2, b 3, b 4线性相关.11.(1) 求下列向量组的秩, 并求一个最大无关组:(1)a 1=(1, 2, -1, 4)T , a 2=(9, 100, 10, 4)T , a 3=(-2, -4, 2, -8)T ; 解由⎪⎪⎪⎭⎫⎝⎛-⎪⎪⎪⎭⎫ ⎝⎛--⎪⎪⎪⎭⎫ ⎝⎛----=000000010291032001900820291844210141002291) , ,(~~321r r a a a , 知R (a 1, a 2, a 3)=2. 因为向量a 1与a 2的分量不成比例, 故a 1, a 2线性无关, 所以a 1, a 2是一个最大无关组.12.利用初等行变换求下列矩阵的列向量组的一个最大无关组:(1)⎪⎪⎪⎭⎫⎝⎛4820322513454947513253947543173125;解因为⎪⎪⎪⎭⎫ ⎝⎛482032251345494751325394754317312513121433~r r r r r r ---⎪⎪⎪⎭⎫ ⎝⎛531053103210431731253423~rr r r --⎪⎪⎪⎭⎫ ⎝⎛00003100321043173125, 所以第1、2、3列构成一个最大无关组.(2)⎪⎪⎪⎭⎫⎝⎛---140113*********12211. 解因为⎪⎪⎪⎭⎫ ⎝⎛---1401131302151201221113142~rr r r --⎪⎪⎪⎭⎫ ⎝⎛------222001512015120122112343~rr r r +↔⎪⎪⎪⎭⎫ ⎝⎛---00000222001512012211, 所以第1、2、3列构成一个最大无关组. 13. 设向量组(a , 3, 1)T , (2, b , 3)T , (1, 2, 1)T , (2, 3, 1)T的秩为2, 求a , b .解设a 1=(a , 3, 1)T , a 2=(2, b , 3)T , a 3=(1, 2, 1)T , a 4=(2, 3, 1)T . 因为⎪⎪⎭⎫ ⎝⎛----⎪⎪⎭⎫ ⎝⎛---⎪⎪⎭⎫ ⎝⎛=52001110311161101110311131********) , , ,(~~2143b a a b a b a r r a a a a ,而R (a 1, a 2, a 3, a 4)=2, 所以a =2, b =5. 20.求下列齐次线性方程组的基础解系: (1)⎪⎩⎪⎨⎧=-++=-++=++-02683054202108432143214321x x x x x x x x x x x x ;解对系数矩阵进行初等行变换, 有⎪⎪⎭⎫ ⎝⎛--⎪⎪⎭⎫ ⎝⎛---=00004/14/3100401 2683154221081~r A ,于是得⎩⎨⎧+=-=43231)4/1()4/3(4x x x x x .取(x 3, x 4)T =(4, 0)T , 得(x 1, x 2)T =(-16, 3)T ; 取(x 3, x 4)T =(0, 4)T , 得(x 1, x 2)T =(0, 1)T . 因此方程组的基础解系为ξ1=(-16, 3, 4, 0)T , ξ2=(0, 1, 0, 4)T .(2)⎪⎩⎪⎨⎧=-++=-++=+--03678024530232432143214321x x x x x x x x x x x x .解对系数矩阵进行初等行变换, 有⎪⎪⎭⎫⎝⎛--⎪⎪⎭⎫ ⎝⎛----=000019/719/141019/119/201 367824531232~r A ,于是得⎩⎨⎧+-=+-=432431)19/7()19/14()19/1()19/2(x x x x x x .取(x 3, x 4)T =(19, 0)T , 得(x 1, x 2)T =(-2, 14)T ; 取(x 3, x 4)T =(0, 19)T , 得(x 1, x 2)T =(1, 7)T . 因此方程组的基础解系为ξ1=(-2, 14, 19, 0)T , ξ2=(1, 7, 0, 19)T .26. 求下列非齐次方程组的一个解及对应的齐次线性方程组的基础解系:(1)⎪⎩⎪⎨⎧=+++=+++=+3223512254321432121x x x x x x x x x x ;解对增广矩阵进行初等行变换, 有⎪⎪⎭⎫⎝⎛--⎪⎪⎭⎫ ⎝⎛=2100013011080101 322351211250011~r B . 与所给方程组同解的方程为⎪⎩⎪⎨⎧=+=--=213 843231x x x x x . 当x 3=0时, 得所给方程组的一个解η=(-8, 13, 0, 2)T . 与对应的齐次方程组同解的方程为⎪⎩⎪⎨⎧==-=043231x x x x x . 当x 3=1时, 得对应的齐次方程组的基础解系ξ=(-1, 1, 1, 0)T .(2)⎪⎩⎪⎨⎧-=+++-=-++=-+-6242163511325432143214321x x x x x x x x x x x x . 解对增广矩阵进行初等行变换, 有⎪⎪⎭⎫ ⎝⎛---⎪⎪⎭⎫ ⎝⎛-----=0000022/17/11012/17/901 6124211635113251~r B . 与所给方程组同解的方程为⎩⎨⎧--=++-=2)2/1((1/7)1)2/1()7/9(432431x x x x x x . 当x 3=x 4=0时, 得所给方程组的一个解η=(1, -2, 0, 0)T .与对应的齐次方程组同解的方程为⎩⎨⎧-=+-=432431)2/1((1/7))2/1()7/9(x x x x x x . 分别取(x 3, x 4)T =(1, 0)T , (0, 1)T , 得对应的齐次方程组的基础解系ξ1=(-9, 1, 7, 0)T . ξ2=(1, -1, 0, 2)T .。

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