工程数学(线性代数与概率统计)答案(1章)

工程数学(线性代数与概率统计)答案(1章)工程数学（线性代数与概率统计）习题一一、21；1.12221(1)52.x11xx2xx2x3x2；x2x2x1(1)(1)1abab2a2b3.b2a21114.314532278153658950a05.b0c0,(第一行与第三行对应成比率）0d01236.312182766618。231二．求逆序数1.34215即5221002.4312即532003.n(n1)21即(n1)(n2)n(n1)21(n1)(n2)10213(2n1)(2n)(2n2)4201n1n1n2104.n(n1)[12(n1)][(n1)(n2)21]2*2三．四阶行列式中含有a11a23的项为a11a23a32a44a11a23a34a42四．计算行列式值1/8工程数学(线性代数与概率统计)答案(1章)41240724009451.1202r14r21202r17r41202010520r310r2015220r315r40017850117011701172.011131111111111110113011101101001101c1c2c3c43101311013001031110311011100001abacae1113.bdcddeadfbce1114abcdefbfcfef111a100b101100按第一行睁开a4.1b11c10c101c101d01d001dabc111c1(ab1)(cd1)ada1d0d1d5.abc2a2a2bbac2b2c2ccaba2a2ab2a2ac2a2a2bbac2b0bac2b0bac2b2c2ccab02ccab02ccab此中2/8工程数学(线性代数与概率统计)答案(1章)a2a2aa2a2aa02aa02a2bbac2b2bb2b2ba2b2bc2b2c2ccab2c2ccab2c0cab2c0caba2a2aa2a0a2a0a2a2bb2b2bb02bb0(ac)2c2cc2c2ca2c2cb2ccab122a2aa2aabc212(a(ab)c)2212bb2ccab5abc3ab(ab)(ac)(a2ab3ac)其他同法可求。方法2:abc2a2a2bbac2b2c2ccab(abac2b2b2b2bbacbc)2cca2acab2a2cb2c2c2240r22r122404135r303556.3r12703123204832051r4r1021112221000222222227.2232rir2(i2)00102(n2)!222n000n28.a001a0000a000a000a000000按第一行睁开a(1)n100a000a0000a100a000an11000n1an(1)n1(1)nan2anan2五．证明以下等式3/8工程数学(线性代数与概率统计)答案(1章)a2abb20a(ba)b2a20a(ba)b2a22aab2br1a2r32aab2br22ar30ba2(ba)(ab)3111111111a2(a1)2(a2)2(a3)2a22a1(a2)2(a3)2b2(b1)2(b2)2(b3)2b22b1(b2)2(b3)2.2222c2c1222c(c1)(c2)(c3)c2c1(c2)(c3)d2(d1)2(d2)2(d3)2d22d1(d2)2(d3)2a22a14a46a9a22a123c3cb22b14b46b9c2cb22b1231320c4c1c22c14c46c9c4222c1232ccd22d14d46d9d22d1233.x1000x10000x1000x100Dn00x00按第一行睁开x00x00000x1000x1anaaa2xa1aaaa2xa1n1n2n1n2n3010000x10000x00an(1)n11(1)n2xDn1anxDn1000x1anaaa2xa1n2n3同理Dk1xDkak1，返回代入得DnxDn1anx(xDn2an1)anxna1xn1an1xan六、计算以下各式1.设x1,x2,x3为方程x3pxq0的三个根，则由三次方程根的性质ax3bx2cxd0x1x2x3b/a,x1x2x3d/a，4/8工程数学(线性代数与概率统计)答案(1章)x1x2x2x3x1x3c/a得x3pxq0的三个根知足：x1x2x30,x1x2x3qx1x2x3因此x3x1x2x13x23x333x1x2x3p(x1x2x3)3q3x1x2x30。x2x3x1xx102.f(x)1x233的系数为-1.中x23x2112xabcdAA3.求DcbdaA44的A142434的值，dbcaabdcAAAabc144cbd10142434Adbc1abd1xaaaxaA4.求D的An1Ann的值，n2aaxxaaaxa000AAAaxaa0xa00riarn(in)(xa)n1n1n2nnaaxa00xa011111111七.计算行列式值1.5/8工程数学(线性代数与概率统计)答案(1章)nx1mx2xnximx2xni1nnx1x2mxnximx2mxnDnc1cji1j2x1x2xnmnximx2xnmi1nximx2xni1nrir1(i2,,n)0m0(xim)(m)n1i100m2.123n1n123n1n11000各行提出公因子（11000）Dn02200n1!01100000n11n00011nnnkkk2n1nk1k2k3（01000）cjcj1(jn1,,1)n1!0010000001（）n1nn1(n1)n（）n1(n1)!k(1)n1!(1)k12n1!(1)23.6/8工程数学(线性代数与概率统计)答案(1章)abDab2ncdcdab00ababab按第一行睁开acdb(1)2n1cdcdcdd002n1

c2n1abadD2n2bcD2n2,同理D2n2(adbc)D2n4，，D2adbc，cd因此D2n(adbc)n4.1a1111a11111a21a1a2Dnrir1(i2,,n)111ana10an111a1110a2a1a200ana10an八．克莱姆法规解方程x2yz01.2xyz1xy2z3解：121021101120D2118D11114D22114D3211121123121321137/8工程数学(线性代数与概率统计)答案(1章)即x1/2,y1/2,z3/2；x12x23x34x442.x2x3x43x13x23xx4137x3x24解：12344234011116，D13111D128130113010731