工程數(shù)學(線性代數(shù)與概率統(tǒng)計)答案

1、工程數(shù)學（線性代數(shù)與概率統(tǒng)計）習題一2 1 1.1 22 2 1(1)5；x 11222.22(X 1)( X X 1) XX X X 13.ab2 a2b1 1 14.3 1 4 5 32 27 8 15 36 58 9 50a05. b0c0,（第一行與第三行對應成比例）0d01 2 36. 3 1 21 8 27 6 6 6 18。2 3 1.求逆序數(shù)342 15即221004312即3200n (n 1)2 1(n 1) (n 2)1055即 (n 1) (n 2)n(n 1)1 3014.(2n 1) (2n) (2n 2)n 1n 1n 21 2 (n 1) (n 1) (n 2)

2、2 12*n(n 1)2三.四階行列式中含有ana23的項為ana23a32a44ana23a34a42四.計算行列式值4 12 407 240 0 94512 0 2r1 4r21202r17 r412 0210 5 2 0 10r2015 220B 15r40 0 17 850 11701170 11701.2.01101111113.11004abcdef5.b2b2ca2b2cab2aa2c2aa2c2a2b(ab1)(cd1)ad2a2b2aa2c2a2b2aa2c2a2b4.按第一行展開其中a2a2a2bbac2b2c2ccaba 2a 2a2b b 2b2c 2c c1 2 2a

3、bc2 1 22 2 1a 2a2b b2c 2ca 2a 02bb 02c 2c a(ab)2b2ab25abc 3ab(a b) (a c)(a其余同法可求。方法2:a b c 2a 2a2b b a c 2b2c 2ccabb a c 2b(a b c)2c cab2 2416.31201222227. 2232228.a 00 a0 01 02a2bcaba 2a 02b b 02c 2c b(aa02a2ba2b2c0cabc)2c2acaba 2a(a c)2c c a bab 3ac)2b 2b2a2c c a b4 0 r2 2rl2 23530 323r3 2r1 0 451

4、14 %0 222212。2)a 02a2b c2b2c 0 c a b2b b a c2a2c 2c5 27030 10 0按第一行展開a五.證明下列等式a 00 a0 00 00202 (n 2)!n 2nn 1 n n 2 na ( 1) ( 1) a a0 a0 00 01 02 a2aab b102ab22ba(b a)a b2 ab22 cd2(a (b (c (dc3C42 ab22 cd22 )(a2)22 )(b2)22(c2)2)2(d2)22a14a2b14b2c14c2d14d11)444(a (b (c (d3)3)3)3)6a6b6cC2C16d2b1a(b bb2

5、b22 cd22c2d(b (c (d2)2)2)2)a)a(a (b (c (db22(b12aa)(ab)33)3)3)3)991c3 2c29 c4 2c292 ab22 cd22a2b2c2d3.xDnan(1)n11 (1)n 2xDn 1ananan 2an 3a2同理Dk1xDkak,返回代入得DnxDn 1anx(xDn 2 an 1)ann 1a1x1x anx1000 x10000 x1000 x10000 x00按第一行展開x00 x00000 x1000 x1ana n 1an 2a2x a1a n 1an 2an 3a2x1000Dn六、計算下列各式1.設X1,X2,

6、X3為方程3x px q 0的二個根,則由三次方程根的性質(zhì)3,2ax bxcxx1x2x3b/a,Xi x2 x3x1x2 x2x3 x1x3 c/ a得 x3px q 。的三個根滿足：x1 x2 x3 0, *1x2x3qx1X2X3所以x3xix2333 。./xx2 x3 3x1x2x3p(x1x2 x3) 3q 3x1x2x30。X2x3xi2. f(x)3.求D3中x3的系數(shù)為-1.A14A24A34的 A4 A24A34A44 的值,4.求D的An1An2Ann的值，A.1入2rian(ia0000 x a0000 x a01111xn)七.計算行列式值1.nx1mx2xn為 mx

7、2xni 1 nx1x2mxnnDnc1c jj 2x1x2xnmX m x2 mxni 1 nX mx2xn mi 1Xim X2i 1riri(i2, n) 0 mxnn0(xim)( m)n 1i 12.123110Dn0 220 00n 1n0000各行提出公因子11(n 1)!0021103010n 1 n00001nnk kk 2 k 310012n 1 n0000nkk 105Cj 1(j n 1,1) (n 1)!0000n(n 1)!( 1)n 1 k（1尸寸1）1（1尸寸D2n按第一行展開adD2n所以D2n4.Dnb(1)2n2n2 bcD2n 2,同理(adbc)na2a111a20a2D2n(adbc)D2nD2a1adrir1(i2,n)a1a1ana22n 1八.克萊姆法則解方程1.2x2y yyz2z解:1210211011202118 D11114 D22114 D3211112312132113D即 x 1/2, y 1/2,z 3/2；Xi2x23x34x42.x1乂23x27x2X3x4X4解: