1备考考试习题选解2

1、第二章矩阵及其运算1已知线性变换：= 2 y1 + 2 y2 + y3 ,x1x= 3 y + y + 5 y ,2123x= 3 y + 2 y + 3 y , 3123求从变量 x1 , x2 , x3 到变量 y1 , y2 , y3 的线性变换解 x1 22121 y15 y2 由已知： x2 = 3 x 33 y2 3 1-1 x- 7- 432 y29y2121 1 1 x2 =-7 y2故 y2 = 3563 y 33 x-4y 2 3 3 = -7 x1 - 4 x2 + 9 x3y1y= 6 x + 3 x - 7 x2123y= 3 x + 2 x - 4 x 31232已

2、知两个线性变换= 2 y1 += -3z1 + z2 ,x1y1y3 ,xy= -2 y + 3 y + 2 y= 2z + z ,2123213xy= 4 y + y + 5 y ,= - z + 3z , 3 312323求从z1 , z2 , z3 到 x1 , x2 , x3 的线性变换解由已知1 - 3 x1203110- 10z11 z220311 y1 2 y2 = -2 x2 =-22205 y 53 z x 442 3 z1 3 3 - 61- 4- 1= 129 z2-1016 z 3 = -6z1 + z2 + 3z3x1x= 12z - 4z + 9z所以有2123x=

3、 -10z - z + 16z 3123 111- 1112- 2533设 A = 1- 1 ,B = - 14, 1110求3 AB - 2 A及 AT B.解 111- 1112- 253111- 11-1 -14- 21-1 3AB - 2A = 31 11110122- 2 05- 598 111- 1113- 1729 = 306 - 21- 1 = - 22020 1 - 21411- 112- 25 5- 59 113 08 ATB = 1- 1-14 = 06 112010 4计算下列乘积：17 323- 27 4) ;(2)(3) 1 (- 1,2);(1) 132;1,2,

4、3 2 50 1 13 13- 1- 3012100(4) 2 11- 143;414- 2a13 x1 a23 x2 ; a11a12 a22(5)( x1, x2 , x3 )a12 a xaa 130 133 3 23121001020010032- 201- 1.(6) 01 001 0303 0- 3解317 4 7 + 3 2 + 1 1 4 35 3 - 2 2 =1 7 + (-2) 2 + 31 = 6(1)1 50 1 495 7 + 7 2 + 0 17 3)(2)(3 2= (1 3 + 2 2 + 3 1) = (10)12 1 2 (-1)22-2 2(3)1 (-

5、 1422)= 1(-1)1 2 = - 1 3-36 3 (-1)32 13- 1- 30121- 7- 5(4) 2 1001- 14368 =- 6414 20- 2 a11a13 x1 a12 a22(5)(x1x3) a 12x2a23 x2 x aaa 1333 3 23= (aax + a x + a x )+ ax + axxax + ax+ ax22 223 323 233 312 113 111 112 213 3 x1 x = ax 2+ ax 2+ ax 2+ 2ax x + 2ax x + 2ax x 211 122 233 312 1 213 1 323 2 3 x

6、 3 12100102001010032- 201 1210052- 402- 10- 4 =010(6)010 033030 0- 3- 9 12 105设 A = ,B = ，问： 13 12(1) AB = BA 吗?(2)( A + B)2 = A2 + 2AB + B2 吗? (3)( A + B)( A - B) = A2 - B2 吗?解 12 102(1) A = ,B = 13 1则 AB = 3 446BA =1 328 AB BA2225 214298 + 12 = 8(2) ( A + B)= 2251438 + 616270 = 10但 A + 2AB + B=22

7、4118123415故( A + B)2 A2 + 2AB + B222 05 08 - 12 = 06987(3) ( A + B)( A - B) = 21030 = 2A - B= 22而411341( A + B)( A - B) A2 - B2故6举反列说明下列命题是错误的:（）若 A2 = 0 ,则 A = 0 ;（）若 A2 = A ,则 A = 0 或 A = E ;（）若 AX = AY ,且 A 0 ,则 X = Y . 0101000但 X解 (1)取 A = A= 0 ,但 A 02 01(2)取 A = A= A ,但 A 0 且 A E2 0取 A =111Y =

8、11X = - 1(3)0101AX = AY 且 A 0 Y7设 A =10 , 求 A123k, A , A.l0 1 10 = 10解 A = l21l2l011 0 11101A = A A = 32 = 2l1l1 3l= 1 kl01 k利用数学归纳法证明: A当k = 1时,显然成立,假设k 时成立,则k + 1 时 = A kA = 0 10 = 1101Ak kl1 l1 (k + 1)l01= 1 kl由数学归纳法原理知: Ak l01l01 ,A =0求 Ak .8设 0l解 首先观察 l0 ll22ll201l01l0012l A2= 01 = 01 0l2 0l 0l

9、03l 3l2 l3= A2 A = 03l2l30A3l30k(k - 1) lk -2 lklk -1lk0k= 2klk -1lkAk(k 2)00由此推测用数学归纳法证明:当k = 2 时,显然成立.假设k 时成立，则k + 1 时，k(k - 1)lk-2 lklk -1lk0kl01l02klk -1lkAk +1 = Ak A = 0 01 00l(k + 1)k lk -1 lk +1(k + 1)lk-1lk +10= 2(k + 1)lk-1lk +100klk -1lk0k(k - 1) lk -2 lk= 2klk -1lkAk由数学归纳法原理知:009设 A, B 为

10、n 阶矩阵，且 A 为对称矩阵，证明BT AB 也是对称矩阵.已知： AT = A(BT AB)T = BT (BT A)T证明= BT AT B = BTAB则BT从而AB 也是对称矩阵.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 .由已知： AT = A= BBT证明 AB = ( AB)T充分性： AB = BA AB = BT即 AB 是对称矩阵.AT必要性：( AB)T = AB = AB BA = AB .BT AT11求下列矩阵的逆矩阵：- 1 124- 4(2) cosq sinq- sinq ； (1) 1 22 ；5- 2;(

11、3) 3cosq5110 521000085002120031(4) 10;(5) 20 ;20 03 14 02 a10 a2(6) (0)a aa 01 2n%an 解125(1) A = A = 12= 5, A21 = 2 (-1), A12 = 2 (-1), A22 = 1A115- 2 A11A21 1*-1*A = = A=A - 21 A12A22 5- 2-1= - 21A故(2) A = 1 0故 A-1 存在A21 = s= cosqinqA22 = cosqA11cosqsinq -1= - sinqcosq A从而故 A-1 存在A = 2 ,(3)= -4A21

12、= 2A31 = 0A11A12 A13= -13= -32= 6= 14= -1而A22A23A32= -2A33 - 20 137-1 1 A13A-1*=A = -故2 - 162 - 1A1021200310(4) A = 1020 14A21A = 24= A31 = A41 = A32 = A42 = A43 = 0= 24A22 = 12A33 = 8A44 = 6A111031012120= (-1)3 2= (-1)4 20 = -120 = -12A12A1314141212010120A14 = (-1)5 2= (-1)5 23 = 30 = -4A2311141012

13、010220= (-1)6 2= (-1)7 13 = -50 = -2A24A3411111AA-1A*=012- 16- 5 2400 10013- 1 12 1-2-1= - 1故 A02114 8= 1 0故 A-1 存在(5) A= 1= -2= 0= 0而A11A12 A13A21A31A32A41A42= -2= 0= 5= 0= 0A22= 0= 2= -3A23A33A43= 0= 0= -5= 8A14A24A34A44- 25001002- 50= - 20从而 A-1- 3008a10 n a2(6) A = 0%a1a0 1 1= a-1A2由对角矩阵的性质知%01a

14、n 12解下列矩阵方程：-1 211- 1- 6 ;- 133 ;25X1 2= 4 20 = 4(1)(2)X21 13 114 X 2110 = 3-1-1;(3)- 1210010 0 - 40- 2100010 13 1 = 2-1.0 X0(4)1 00 1100 解5- 1 4- 54- 62- 232- 63(1)X = = = 32-1221 0811- 1-1211- 11033133- 13- 13(2)X = 1 4113 422=- 22- 2011- 30 - 2251= - 8- 23 34- 1 3-1- 4310 21021121= 2 0(3)X = - 1-

15、 1 1- 1 - 112 11 01 111 66 10 = 10 1=0123240-1 10-1- 40- 2 010031001 (4)X = 1- 10021 01 1000 01- 40- 2- 130 010031001020= 1- 1 0 1=- 4 102 0- 21100 10 13利用逆矩阵解下列线性方程组: x1 + 2 x2 + 3 x3 = 1,x1 - x2 - x3 = 2,(1) 2 x + 2 x + 5 x = 2, 2 x - x - 3x = 1,(2)123123 3 x + 5 x + x = 3;3 x + 2 x - 5 x = 0.1123

16、233 x11 1225 5 x2 = 2解(1)方程组可表示为231 x 3 3 x1 13-1 11225 x2 = 2 x 3 2 = 05故130 3 = 1= 0= 0x1x从而有2x 3 1 2- 1- 12-1 x1 2 = - 3 x 21(2) 方程组可表示为3- 5 x03 x1 1- 1- 12- 1-1 25 x2 = 2 x 3- 31 = 0故- 503 3 = 5= 0= 3x1x故有2x 314设 Ak = O ( k 为正整数),证明 (E - A)-1 = E + A + A2 + + Ak-1.一方面， E = (E - A)-1 (E - A)证明另一方

17、面，由 Ak = O 有 E = (E - A) + (A - A2 ) + A2 - - Ak-1= (E + A + A2 + + Ak-1 )(E - A)+ ( Ak -1- Ak )故 (E - A)-1(E - A) = (E + A + A2 + 两端同时右乘(E - A)-1就有(E - A)-1 = E + A + A2 + + Ak-1+ Ak-1)(E - A)15设方阵 A 满足 A2 - A - 2E = O ,证明 A 及 A + 2E 都可逆,并求 A-1及( A + 2E )-1 .由 A2 - A - 2E = O 得 A2 - A = 2E证明- A = 2

18、A2两端同时取行列式:A A - E= 2 ,故 0即A所以 A 可逆,而 A + 2E = A2A 2A + 2E= 0故 A + 2E 也可逆.A2由 A2 - A - 2E = O A( A - E ) = 2E A-1 A( A - E ) = 2 A-1 E A-1 = 1 ( A - E)2又由 A2 - A - 2E = O ( A + 2E )A - 3( A + 2E ) = -4E ( A + 2E )( A - 3E ) = -4E( A + 2E )-1( A + 2E )( A - 3E ) = -4( A + 2E )-1= 1 (3E - A)( A + 2E )

19、-143031216设 A = 0 , AB = A + 2B ,求B .1-13由 AB = A + 2B 可得( A - 2E )B = A解3-1 - 233- 120303213 0 10=-1-1B = ( A - 2E )A =11故 3 - 1- 113021 17设 P -1 AP = L ,其中 P = -1- 4-10,L = ,求 A11.1102= PL11 P-1P -1 AP = L 故 A = PLP -1 所以 A11解111441-1*= -1-1=3 - 1-1 = 3PPP011- 10= L112= 而211 13004-1-4-1 27312732 0

20、2113= =故 A11- 68411- 683110- 331记8 设m 次多项式 f (x)a0a1 xa2 x2ax ,mmf ( A) = a0 E + a1 A + a2 A2+ + amAmf ( A) 称为方阵 A 的m 次多项式. l1 f (l1)l0 0 k0= , f (L) =(1)设L= ,证明:;Lk1l2 f (l2 )l 000k2 (2)设 A = PLP -1 ,证明:= PLk P -1 , f ( A) = Pf (L)P -1 .Ak证明(1)i)利用数学归纳法.当k = 2 时 l10 l10 l0 2 = L2= 1l2 0l2 l 0202 命题

21、成立,假设k 时成立,则k + 1时 0 l1l0 lk +1k0= Lk L = = Lk+1110lllk +1k002 2 2故命题成立.ii)左边= f (L) = a0 E + a1L + a2L2 + + am Lml1l100 m0= a0 +a 1 +a1m ll1 0m02 02 = a + a l + a l+ + al2m001 12 1m 1a + a l + al + + a l2m0= f (l1)m 2 01 22 20 =右边 f (l2 )0i)利用数学归纳法.当k = 2 时(2)A2= PLP -1 PLP -1 = PL2 P -1 成立假设k 时成立,则k + 1时 = A = PLk P -1 PLP -1 = PLk+1 P -1 成立,故命题成立,Ak +1Ak= PLk P -1Ak即ii)证明右边= Pf (L)P -1m-12= P(a0 E + a1L + a2 L + + am L )P-1-12-1m-1= a0 PEP+ a1PLP+ a2PL P+ + am PL P= a0 E + a1 A + a2 A2 + + am Am = f ( A)=左边 19设n 阶矩阵 A 的伴随矩阵为 A* ，证明：= 0 ,则 A*= 0 ;(1)若AA n