《线性代数(2-5章答案)》由会员分享,可在线阅读,更多相关《线性代数(2-5章答案)(22页珍藏版)》请在金锄头文库上搜索。
1、习题二1、设, 111111111A 150421321B有 22942017222132222222222092650850 3111111111 2 150421321111111111 323AAB 092650850150421321111111111BAT2、求下列矩阵的乘积AB(1) 7 201321 (2) 121125147103121012132(3) 119912943110231101420121301(4) 0000 2121 1111(5) 0000002412122412(6) nnnnncbcbcbcbacbacba202020200010000221122211
2、13、求下列矩阵的乘积(1) niiinnbabbbaaa12121(2)nnnnnnnbabababababababababbbaaa22122212121112121(3))222(3223311321122 3332 2222 111321332313232212131211321xxaxxaxxaxaxaxaxxxaaaaaaaaaxxx 6、设,求与 A 可交换的矩阵;即 100110011A 333231232221131211bbbbbbbbbBBAAB BAbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbAB 3332323131232222212113121211
3、11333231332332223121231322122111得 为任意数13121133223221312312221121,00bbbbbbbbbbbbb 111211131211000bbbbbbB7、略 8、计算矩阵幂(1) 22211413 4321 4321 4321 43213(2) 2cos2sin2sin2cos1401104410013401102410010110 nnnnknknknknn(3)n 2312, 2 , 1 , 0122312210012312231223121001100123122312 kknknn 因(4)k nkkkn2121(5) 10001
4、01011000101011000101011000101011000100110001010110001030110001010110001020110001010110001020110001010110001010113kkkkk(6) kkkkkkkkkkk0002) 1(00100100303300100100201200100100201200100100100100100112132323222322229、设,43214/13/12/1113/4244/312/332/13/2124/13/12/114)()()4(43214/13/12/1113/4244/312/332/1
5、3/2124/13/12/114/13/12/11432111nnTTnTnTTAA10、分块计算(略) ,11、12、13、14(略) 15、求逆矩阵(1) acbdbcaddcba11(2) cossinsincos cossinsincos1(3),02 14524312132,13, 4131211AAA2, 1, 0,14, 6, 2333231232221AAAAAA2143216130242111AAA(4)11 21 11naaaA16.解矩阵方程(1) 3211 9553 2/12/312 9553 43211 X(2) 86122221576821 1091614 3512
6、1118765 1091614 251311 X(3) 98765432112523113501520950381X(4)BAEXBXAEBAXX1)()( 1102133502113/13/103/13/213/13/203502112011010111X17、1111)(66)(6EABABAEABAABAA 1236/13/12/1 6)(66/13/12/1 )( , 632 , 743111111EABEAEAA18、 9122692683321011324461351341321011324121011322)2()2()2(2111AEAAEABABEABAAB19、A 为 3
7、阶方阵,有;aA 0mammA320、A 为 3 阶方阵,;,2, 2/11AA1AAA41311112222323AAAAAAA21、略22、112)(212)(02EAAEEAAEAAAAEEAAEEAA21)(2)(0212因020)(2(EAEAEAEA23、)2(51)4(05)2)(4(03212EAEAEEAEAEAA24、因0mA有1221)(mmmmmmmAEAAEEAEAEEE所以121)(mAAAEAE25、 CACACCBmmm11)(26、199991PPBAPBPAPBAP27、28、略29、; 22112121,BAOOBAABBOOBBAOOAA30、(1)设
8、214321 EOOEAAAAOCBO有 1 2141 3212143 00CAAABAEOOE CACABABA即逆矩阵为 OBCO11(2)设 214231214321 EOOECAAACAAABABAAAAACAOB得逆阵为 1111CABCOB31、32、略33、求迭(1)2 00001140432122801140432121101542143211312 rrrrr(2)4211103000044000100112111011110022201001110011111100222021110r34、求逆阵(用软件算的与书后答案有些不同,请大家验证) (1)A =3 2 13 1 53 2 3det(A)= -6 inv(A) ans =1.1667 0.6667 -1.5000-1.0000 -1.0000 2.0000-0.5000 0 0.5000(2)B =2 3 11 2 0-1 2 -2det(B)=2 inv(B) ans =-2.0000 4.0000 -1.00001.0000 -1.5000 0.5000 2.0000 -3.5000 0.5000(3)C =3 -2