《线性代数》课程教学资源（习题解答）第一章

aita12a,ana21a2证：设A=antan2anA=A,..ag=aj2a,4又A==0(i=l...,n)=)Za=..a=0i=l...,ni=l...,n.:A=0(B+I)，证明：A=A当且仅当B?=I12.设A=2证明“=”4=！(B+I)?=4A=B2+2B+I=2B+2I=B2+2B+I=B2=I1A!(B2 +2B+ I)= (21 +2B)=(B+I)= A42413.设X是n×1矩阵，且XT×=1，证明：S=I-2Xx是对称矩阵，且s=I证明：由于ST=(I-2XX"）=I-2XXT=S:.S是对称阵又xx=1知，S? =(I-2XXT) =(I-2XXT)(-2XXT)=1-2XXT-2XXT+4X(XX)X=1-4XXT+4XXT=1.14.利用等式(17-6)-6 9 ) ( -6 9+()35-12( 36)-6 9.解：注意到63 ]-6 6 9 66 9? )-6()因此6

6 12. 设 1 ( ) 2 A = B I + ，证明： 2 A = A 当且仅当 2 B = I . 13. 设 X 是 n1 矩阵, 且 1 T X X = , 证明: 2 T S I XX = − 是对称矩阵, 且 2 S I = . 14. 利用等式 17 6 2 3 2 0 7 3 35 12 5 7 0 3 5 2       − −       =       − − , 7 3 2 3 1 0 5 2 5 7 0 1      −      =      − 计算 5 17 6 35 12   −     − . 解：注意到 7 3 2 3 1 0 5 2 5 7 0 1      −      =      − ， 2 2 17 6 2 3 2 0 7 3 2 3 2 0 7 3 35 12 5 7 0 3 5 2 5 7 0 3 5 2 2 3 2 0 7 3 . 5 7 0 3 5 2               − − − =                             − − −       − =             − ， 因此 1 1 1 0 0 0 0 : 0 0 0 ( ) 0 0 0 0 0 1 1 6 (1, 2, 3), (1, , ), , . 2 3 : ( )( ) ( ) ( )( ) ( ) 3 1 1 1 2 3 2 3 3 3 2 1 3 3 3 1 2 7 n n n n T n n T T T T T T T n n T n n a a b b c c A A A A                     − − −         = − = −             = = = = = =        = = =             解 原式 、已知 且 计算 解 、举 2 2 2 2 : (1) , ; 0 0 : , 0, 1 0 (2) , , ; 1 0 : , , 0 0 , . (3) , , . 0 0 0 0 0 0 0 0 : 0, , , 1 0 2 0 3 0 0 0 A O A O A A A O A A A O A I A A A A O A I AX AY A O X Y A X Y AX AY = =   = =      = = =   = =       =  =         =  = = = =                 反例说明下列命题是错误的 若 则 解 取 但 若 则 若 解 取 但 且 若 且 则 解 取 但 2 11 12 1 21 22 2 1 2 2 1 1 2 2 1 2 1 . 8. , , : . : , , * 0, 0( 1, , ) * 0 1, , ; 1, , n n n n nn T ij ji n j n ij j n nj j ij X Y A A O A O a a a a a a A a a a A A a a a A a i n a a i n j n = = =  = =       =       =  =       = =  = =            = = =     然而, 如果 是实对称矩阵 且 证明 证 设 又 A = 0 2 2 2 2 2 2 2 2 2 2 1 9. ( ), : . 2 1 ( ) 4 2 2 2 2 4 1 1 1 ( 2 ) (2 2 ) ( ) 4 4 2 10. 1 , 1, : 2 , . : ( 2 ) 2 , T T T T T T A B I A A B I A B I A B B I B I B B I B I A B B I I B B I A X n X X S I XX S I S I XX I XX S S = + = =  = +  = + +  + = + +  =  = + + = + = + =  = = − = = − = − =  设 证明 当且仅当 证明 “ ” 设 是 矩阵 且 证明 是对称矩阵 且 证明 由于 是对 2 2 . 1 , ( 2 ) ( 2 )( 2 ) 2 2 4 ( ) 4 4 . T T T T T T T T T T X X S I XX I XX I XX I XX XX X X X X I XX XX I = = − = − − = − − + = − + = 称阵 又 知 2 2 2 2 2 2 2 2 2 2 1 9. ( ), : . 2 1 ( ) 4 2 2 2 2 4 1 1 1 ( 2 ) (2 2 ) ( ) 4 4 2 10. 1 , 1, : 2 , . : ( 2 ) 2 , T T T T T T A B I A A B I A B I A B B I B I B B I B I A B B I I B B I A X n X X S I XX S I S I XX I XX S S = + = =  = +  = + +  + = + +  =  = + + = + = + =  = = − = = − = − =  设 证明 当且仅当 证明 “ ” 设 是 矩阵 且 证明 是对称矩阵 且 证明 由于 是对 2 2 . 1 , ( 2 ) ( 2 )( 2 ) 2 2 4 ( ) 4 4 . T T T T T T T T T T X X S I XX I XX I XX I XX XX X X X X I XX XX I = = − = − − = − − + = − + = 称阵 又 知

63-6 9 93-(69 -)-69)(3197-1266(7385-292215.求平方等于零矩阵的所有二阶矩阵ab由A=0，即解：不妨设此二阶矩阵为A=Lcda’+bc(a+d)bab-[8]=I(a+d)ccb+d?dcdc因此a2+bc=0,[a’+bc=0(a+d)b=0,(a+d)c=0,Id=-acb+d2=0.习题1.21.解下列线性方程组：[x +2x +3x,=8(1) 32x, +5x, + 9x, =163x) -4x2-5x,=327

7 5 5 5 5 17 6 2 3 2 0 7 3 2 3 2 0 7 3 35 12 5 7 0 3 5 2 5 7 0 3 5 2 2 3 2 0 7 3 5 7 0 3 5 2 2 3 7 3 2 0 5 7 5 2 0 3 3197 1266 . 7385 2922               − − − =                             − − −       − =             −       − =           −     − =    − 15. 求平方等于零矩阵的所有二阶矩阵. 解：不妨设此二阶矩阵为 , a b c d       A = 由 2 A = 0，即 2 2 2 2 ( ) , ( ) a b a b a b a bc a d b c d c d c d a d c cb d         + +       = =           + + A = 因此 2 2 2 0, ( ) 0, 0, ( ) 0, . 0. a bc a d b a bc a d c d a cb d  + =   + = + =      + = = −    + = 习题 1.2 1. 解下列线性方程组: (1) 1 2 3 1 2 3 1 2 3 2 3 8 2 5 9 16 3 4 5 32 x x x x x x x x x  + + =   + + =   − − =

解：对增广矩阵作初等行变换213872381A=52916001[34-532]01021得同解方程组：X +2x2+3x =8x, +3x, = 02x, =1由最后一个方程逐个回代,得方程组的解为=/~专3-2X=XX,1-2x +2x+3x=4(2) 33x+5x+7x,=9;5x, +8x, +11x, =14解：对增广矩阵作初等行变换[1234][1234A=3507[581114000/0得同解方程组[x + 2x +3x, = 4x, +2x, =3取x，=k,由最后一个方程逐个回代,得方程组的解为x-2+k3-2kX=x.X其中k为任意常数2x + x2+3x=6(3) 33x +2x + x =1[5x, +2x, +4x, =276

8 (2) 1 2 3 1 2 3 1 2 3 2 3 4 3 5 7 9 5 8 11 14 x x x x x x x x x  + + =   + + =   + + = ; (3) 1 2 3 1 2 3 1 2 3 2 3 6 3 2 1 5 2 4 27 x x x x x x x x x  + + =   + + =   ++= ; 1 2 3 1 2 3 1 2 3 1 2 1.2 1. : 2 3 8, (1) 2 5 9 16, 3 4 5 32; : 1 2 3 8 1 2 3 8 2 5 9 16 0 1 3 0 3 4 5 32 0 0 2 1 : 2 3 x x x x x x x x x A x x  + + =   + + =   − − =         = →     − − + + 习题 解下列线性方程组 解 对增广矩阵作初等行变换 得同解方程组 3 2 3 3 1 2 3 1 2 3 1 2 3 1 2 3 8 3 0 2 1 , 11 2 3 2 1 2 2 3 4, (2) 3 5 7 9, 5 8 11 14; : 1 2 3 4 3 5 7 9 5 8 11 14 x x x x x X x x x x x x x x x x x A  =   + =   =             = = −                  + + =   + + =   + + =    =    由最后一个方程逐个回代 得方程组的解为 解 对增广矩阵作初等行变换 1 2 3 2 3 3 1 2 3 1 2 3 4 0 1 2 3 0 0 0 0 : 2 3 4 2 3 , , : 2 2 1 3 2 3 2 0 1 x x x x x x k x k X x k k x k         →        + + =   + = =         − + −         = = − = + −             得同解方程组 取 由最后一个方程逐个回代 得方程组的解为 . 其中 为任意常数 k 1 2 3 1 2 3 1 2 3 1 2 1.2 1. : 2 3 8, (1) 2 5 9 16, 3 4 5 32; : 1 2 3 8 1 2 3 8 2 5 9 16 0 1 3 0 3 4 5 32 0 0 2 1 : 2 3 x x x x x x x x x A x x  + + =   + + =   − − =         = →     − − + + 习题 解下列线性方程组 解 对增广矩阵作初等行变换 得同解方程组 3 2 3 3 1 2 3 1 2 3 1 2 3 1 2 3 8 3 0 2 1 , 11 2 3 2 1 2 2 3 4, (2) 3 5 7 9, 5 8 11 14; : 1 2 3 4 3 5 7 9 5 8 11 14 x x x x x X x x x x x x x x x x x A  =   + =   =             = = −                  + + =   + + =   + + =    =    由最后一个方程逐个回代 得方程组的解为 解 对增广矩阵作初等行变换 1 2 3 2 3 3 1 2 3 1 2 3 4 0 1 2 3 0 0 0 0 : 2 3 4 2 3 , , : 2 2 1 3 2 3 2 0 1 x x x x x x k x k X x k k x k         →        + + =   + = =         − + −         = = − = + −             得同解方程组 取 由最后一个方程逐个回代 得方程组的解为 . 其中 为任意常数 k

解：对增广矩阵作初等行变换[21362136A=21130-115[s342700020得同解方程组：[2x +x, + 3x, =6- 5x, =-11-x0-x=20显然，该方程组无解[x+x,+x =](4) x +2x2 -5x =222x, +3x2-4x, = 5解：对增广矩阵作初等行变换：1:1:111112:2-501-6A=1:[25lo00：23：-4类似上述(3)，此对应矛盾方程.因此该方程组无解+2x,+Xs=0X-x22+7x=03x, - 3x2(5)X-x+2x+3x4+2x,=02x -2x, +2x, +7x, -3x,=0

9 (4) 1 2 3 1 2 3 1 2 3 1 2 5 2 2 3 4 5 x x x x x x x x x  + + =   + − =   + − = ; 解：对增广矩阵作初等行变换： 1 1 1 1 1 1 1 1 1 2 5 2 0 1 6 1 2 3 4 5 0 0 0 2 A         = − → −             − 类似上述(3), 此对应矛盾方程. 因此该方程组无解. (5) 1 2 4 5 1 2 4 1 2 3 4 5 1 2 3 4 5 2 0 3 3 7 0 2 3 2 0 2 2 2 7 3 0 x x x x x x x x x x x x x x x x x  − + + =   − + =  − + + + =    − + + − = . 1 2 3 1 2 3 1 2 3 1 2 3 1 3 3 2 3 6, (3) 3 2 1, 5 3 4 27; : : 2 1 3 6 2 1 3 6 3 2 1 1 1 0 5 11 5 3 4 27 0 0 0 20 : 2 3 6 5 11 0 20 x x x x x x x x x A x x x x x x  + + =   + + =   + + =         = → − − −      + + =  − − = −    = 解 对增广矩阵作初等行变换 得同解方程组 显然 1 2 4 5 1 2 4 1 2 3 4 5 1 2 3 4 5 , . 2 0, 3 3 7 0, (4) 2 3 2 0, 2 2 2 7 3 0. : : 1 1 0 2 1 0 1 1 0 0 7 0 3 3 0 7 0 0 0 0 1 0 2 0 1 1 2 3 2 0 0 0 0 1 3 0 2 2 2 7 3 0 0 0 0 0 0 0 x x x x x x x x x x x x x x x x x A  − + + =   − + =  − + + + =    − + + − =     − −     − = → − −     − − 该方程组无解 解 对增广矩阵作初等行变换 得同解 1 2 5 3 5 4 5 2 1 5 2 1 1 2 2 1 3 1 2 2 4 2 5 2 : 7 0 2 0 3 0 , , : 7 1 7 1 0 2 0 2 3 0 3 0 1 x x x x x x x x k x k x k k x k X x k k k x k x k  − + =   + =   − = = =     −     −                 = = = +     −     −                               方程组 取 ,由最后一个方程逐个回代 得方程组的解为 1 2 , . k k   其中 为任意常数

解：对增广矩阵作初等行变换L门1-102/071001-171000S0700-3200A=00022-13100[2 -2 207-3/0]00000得同解方程组：7x, = 0[X -X, +2x, = 0+X4 -3x, = 0取x,=k,x,=k，由最后一个方程逐个回代,得方程组的解为：X-7k2[-7]k10kx2X=-2k,=k,0I+k.-2133k20x40/k2[x, ]其中k,k,为任意常数2.用行初等变换将矩阵A变为单位矩阵：000)1100(1) A=011111解：(1)对A施行行初等变换如下[1000]000.00-0010000U-rAi=2,3,44-520110000000111000001(2)A10

10 2. 用行初等变换将矩阵 A 变为单位矩阵: (1) 1 0 0 0 1 1 0 0 1 1 1 0 1 1 1 1       =       A ; (2) 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1     − −   =   − −     − − A ; 1 2 3 1 2 3 1 2 3 1 2 3 1 3 3 2 3 6, (3) 3 2 1, 5 3 4 27; : : 2 1 3 6 2 1 3 6 3 2 1 1 1 0 5 11 5 3 4 27 0 0 0 20 : 2 3 6 5 11 0 20 x x x x x x x x x A x x x x x x  + + =   + + =   + + =         = → − − −      + + =  − − = −    = 解 对增广矩阵作初等行变换 得同解方程组 显然 1 2 4 5 1 2 4 1 2 3 4 5 1 2 3 4 5 , . 2 0, 3 3 7 0, (4) 2 3 2 0, 2 2 2 7 3 0. : : 1 1 0 2 1 0 1 1 0 0 7 0 3 3 0 7 0 0 0 0 1 0 2 0 1 1 2 3 2 0 0 0 0 1 3 0 2 2 2 7 3 0 0 0 0 0 0 0 x x x x x x x x x x x x x x x x x A  − + + =   − + =  − + + + =    − + + − =     − −     − = → − −     − − 该方程组无解 解 对增广矩阵作初等行变换 得同解 1 2 5 3 5 4 5 2 1 5 2 1 1 2 2 1 3 1 2 2 4 2 5 2 : 7 0 2 0 3 0 , , : 7 1 7 1 0 2 0 2 3 0 3 0 1 x x x x x x x x k x k x k k x k X x k k k x k x k  − + =   + =   − = = =     −     −                 = = = +     −     −                               方程组 取 ,由最后一个方程逐个回代 得方程组的解为 1 2 , . k k   其中 为任意常数 1 3 2 4 3 4 2 2,3,4 2. : 1 0 0 0 1 1 0 0 (1) ; 1 1 1 0 1 1 1 1 : (1) : 1 0 0 0 1 0 0 0 1 0 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 1 1 0 0 0 1 0 0 0 1 0 0 1 1 1 0 0 1 1 0 0 0 1 i r r r r r r i r r A A A A − − − = −       =                  ⎯⎯⎯→ ⎯⎯⎯→ ⎯⎯⎯→                      用行初等变换将矩阵 变为单位矩阵 解 对 施行行初等变换如下 1 2 2 3 2 1 3 1 4 3 4 2 4 1 ( ) 1 2 ( ) 2 1 ( ) 2 (2) : 1 1 1 1 1 1 1 1 1 1 0 0 0 0 2 2 0 0 1 1 0 0 1 1 1 1 1 1 0 2 0 2 0 1 0 1 0 0 2 2 0 0 1 1 0 0 0 2 r r r r r r r r r r r r r A A −  −  − − − − −  −                        − − ⎯⎯⎯→ ⎯⎯⎯⎯→ ⎯⎯⎯⎯→            − − − −                 − − − 对 施行行初等变换如下 1 3 4 4 1 4 2 4 2 3 3 4 1 2 1 1 ( 1) ( ) 2 1 ( ) 2 ( 1 0 0 1 1 0 0 0 0 0 1 0 0 1 0 0 0 1 0 0 0 0 1 0 0 0 0 1 0 0 0 1 2 (3) , 4. 2 :1. 0, 0 2 0 2 r r r r r r r r r r r r r r r a A ab b a b A b −  −  + +  +  −         −     ⎯⎯⎯→ ⎯⎯⎯→                     −   =      =    =     解 当 时 2 1 1 2 1 2 1 1 ( ) 2 ) 1 2 ( ) 4 1 2 ( ) 1 0 1 0 0 2 0 1 2. 0 , 0 , 2 1 0 0 2 0 1 3. 0, 4 , 1 2 2 2 0 2 2 r b r a r r r r r b a b a A ab ab a a A b ab   −   −      ⎯⎯⎯⎯→ ⎯⎯⎯→          =     = ⎯⎯⎯⎯→               = ⎯⎯⎯⎯→      −  当 时 时 当 且 时 2 1 2 2 4 2 1 0 0 1 r ab a r r  − −       ⎯⎯⎯⎯→        