工程数学线性代数课后习题答案(第二单元)

1、精品文档0583 0 5 62900580 5 61 1 1 0 5 1290此答案从网上找的，仅供参考。习题二答案1、设 A1 11111233AB 2A 3 111124111051有1 1 12 1 1 11 1 12 13 222 17 204 29 21 1 1 1 2 3AT B 1 1 1 1 2 44)12、( 1) 1 2 3 0 72231 2 12) 1 23 0 19219 9 11101 0 3 1 1 1 3 3)2 1 0 2 2 0 11 1 1 21 1 1 20000精品文档210021（5） 4200422100a1b1c10b12c1000a2b2c20

2、b22c2（6）2010002anbncn0bn2cn3、求下列矩阵的乘积b1b2r11) a1 a2anai bii1bn2）a1a1b1a2b1a1b2a2b2a1bna2bna2b1b2bnananb1a2b2anbn3）a11a12a13x1x1x2x3a12a22a23x22(a11x122a22x2 a33x3 2a12x1x2 2a13 x1 x3 2a23x2x3 )a13a23a33x3110b11b12 b136、设A01 1 ，求与A 可交换的矩阵 Bb21 b22 b23；即AB BA001b31b32 b33b11 b21 b12 b22b13 b23 b11b11b

3、12 b12b13ABb21 b31 b22 b32b23 b33 b21b21b22 b22b23BAb31b32b33 b31b31b32 b32b33b21 0b11 b22 b12b23 b11b12b13得b31 0b21 b32 b22b33 B 0b11b12b11,b12,b13为任意数00b118、计算矩阵幂13212121213141）343434342122精品文档1 00 1(2)0 11 01 00 10 11 0(3)n 4k 2n 4k 3n 4k 4n 4k 1n.ncossin22nnsin -cos2221211 0 32 320 1因1 0 21 _ 21

4、0 1 32 _ 321 021 n_ 0 132 _ 213 2n 2kk 0,1,2,n 2k 1kk1k(4)(5)210 110 1 0 10 0 1010 2 10 10 1 0 0 1 0 0 1 0 0 1 0 0 0 1 0 0 1 0 0 1 0 0 1k10110k010010001001k 1k1 0 1 1 0 1 1 0 10 1 0 0 1 0 0 1 0 0 0 1 0 0 1 0 0 1精品文档2A212, A226, A2314, A310, A321,A332k(k1)2k k1111/21/31/42212/31/2AT1 1/21/31/4333/213

5、/44424/311T2T 11/21/31/4(4)3411/21/31/4n“ T、nT “ n 1212/31/2A ()(T)n 1433/213/4424/319、设12 3 4 ,11/2 1/3 1/4 ,10、分块计算（略）,11、12、13、14 (略)15、求逆矩阵1a bc d1 d b ad be c acos1sin.sincos(1)cossinsincos1 2(3) 34541220 ,A1114, A1213, A1332 ,精品文档4201a11(4) Aa2113 61232 142181an16解矩阵方程1123521351 1(1)X3459:3/21

6、/2592331114 165 61X529107 8(2)1211416 1861153910 2751 2 1222 6 81830531123(3)X5 9013245 62 15052 178 9(4)XAX B(EA)X B X(EA) 1B11101102/31/31131X1012012/31/320201025301/31/3531117、A1BA6A BA(A1E)BA 6AB6( A1E)1321/2A 14, A 1 E3,(A1 E) 11/3761/61/23B6( A 1E) 161/321/61精品文档AB A 2B(A2E)B(A2E) 1 A(A 2E) 1A

7、19、20、3AA为3阶方阵,A为3阶方阵,1 2A3A121/2,2AA 12A 123 A2421、22、A2 A2EA(AE)2E1(AE)1A2 A2E(EA)A2E(EA)1A因(A2E)(AE)2E A23、24、A2Am2A3E(A(A 4E)(A1 14E)(A52E)2E)5E所以(EA)1 EA A2Am125、Bm(C1AC)mC 1AmC26、APPBAPBP 1A99PB99P 127、28、略AOB,OA1B1O29、A,BABOA2O b2OA2 B2有EAmEm(E A)(Em1Em2AEm30、(1)设 OBA1A2C OA3A4E1OEAm2 Am1O E2

8、A3B1有 BA3BA4E1OA40riCA1CA2OE2A10A2C1OC1即逆矩阵为AB1OBA12)设A1A3A2A4BA2E1 OAA1 CA3 AA2 CA4O E2得逆阵为 1 1C 1 AB 1 C0 1 1 1 20 2 2 2 00 1 1 1 11 1 0 0 11 1 0 0 10 2 2 2 00 1 1 1 10 1 1 1 21 1 0 0 10 0 0 4 4 r40 0 0 0 30 1 1 1 231、 32、略33、求迭1234r2 r1123412341) 1245r3 r104110411r211012082200002)34、求逆阵(用软件算的与书后答

9、案有些不同，请大家验证)(1)A =321315323det(A)= -6 inv(A)ans =1.16670.6667 -1.5000-1.0000 -1.0000 2.0000-0.5000 00.50002)B =2 3 11 2 0 -1 2 -2det(B)=2 inv(B) ans =-2.00004.0000-1.00001.0000-1.50000.50002.0000-3.50000.5000（3）C =3-20-102211-2-3-20121det(C)=1 inv(C)ans =1.00001.0000-2.0000-4.000001.00000-1.0000-1.00001.00003.00006.00002.00001.0000-6.0000-1