16.计算下列行列式：111111111111211?30?1?1?50?1?1?51).???1；12250114000?143211?32）.13?11??10212111173112?10111h3?h20?1?1?13?312123212110113232?0?11?13121212110321111按第三列展开??2230111321h4?h102102001023124173413??(??)??.32323123012012?14?13521231?110?28110h1?h322h?h2133524212?13512?135122013).?135320?0001231105200010012012?1400121?0311023150012?140161145?0121671055790012?140?110?190?8170?412?41?19?130?1?19??63?57?111?28??483;57?1210012?10?11031?1?52?14?10123230?332?205?352?63111?1114).32121?101213010?2320?210020?2123?3?2305212??00?3212165?1855?2
155?21133??(??16?18)???.24248
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17.计算下列行列式：x01).?0yyx?000?00xy?000????按第一列展开x?0?x0?0yxy?00yx?00x?????(?1)1?ny?yx000?0y?0???0?y0?x00?0y00?x00?0?x?xn?1?(?1)1?ny?yn?1?xn?(?1)n?1yn;a1?b1a1?b2?a1?bn2).a2?b1a2?b2?a2?bn????an?b1an?b2?an?bn当n?1时，行列式值D?aa1?b1a1?b21?b1;当n?2时，D?a2?b1a2?b??(自己算！)2a1?b1a1?b2?a1?bn?b1a1?b2?a1?bna1a1?b2当n?3时，a2?b1a2?b2?a2?bn?b1a2?b2?a2?bna2a2?b2????????????an?b1an?b2?an?bn?b1an?b2?an?bnanan?b2?b1a1?a1a1b2?bnb?b1a2?a2b2?bn1?0?????a2?????0?0?0(显然b1?0时，也为0);?b1an?ananb2?bn - 7 -
a1?bna2?bn?an?bn????
10?0x1?mx2x1x2?m3).??x1x2????0xn?xn做这样的一个加边0?易检验知等号成立0xn?m01?000?????m0??????00?mxnxn00m???1000x1?mx20x1x2?m?x1?x2????000010?0m01?00?????0?10?001?????0100x10x1??1x10?m???0x2x2?00????xn?m2n?2n?m选定前n行，?xn未写出项均为0?xn??按拉普拉斯定理展开0?0x110?01?0x10m?0?????????0?1x100?m00x2?xn1001??000m0??xn00m??xn??????000??xn100?00?m0100?x2000?x20m?0010?x210?0???001?x2?????????00?m000?x210?0m?(?1)n20?0nm?0n?1n?1?(?1)m?xi?(?1)nmn;???i?101?00?????00?100?m111?200?10l?l4).223?2i1201?0按第二行展开即有(?2)(n?2)!;i?2,3,?n??????????222?n200?n?215)0?022?030?0?n?1??00?n00l1?li?122?2222?2?ii?1n2?12?030?0- 8 -
?n?1??00?n00??(?1)n?1(n?1)!. 21?1?2?00?0?2??n?11?n?n?11?n
18.证明：a011).1?11a10?010a2?0?1?0n1?0?a1a2?an(a0??)i?1ai???ana0??i?1n证明：用每一列将第一列的1消为0，可得：1ai1a10?010a2?0?1 00?0nn?01??ai(a0??)?0i?1aii?1???an得证。x0?1x0?12).??000000x?00??????000?x0a0a1a2?xn?an?1xn?1???a1x?a0?an?2x?an?1证明：当n?1时，x?a1?1?x?a0,成立；假设当n?k时，所证成立；现证，当n?k?1时也成立。原式x?10按第一行展开x?000x0?00?0a1a2a3?ak?1x?ak?1x00?1x?(?1)1?na000?1???000?0?0?0
????1?1x?0????00?x00?0由假设可得x(xk?akxk?1???a2x?a1)?a0?xk?1?akxk???a2x2?a1x?a0.由数学归纳法知所证成立。 - 9 -
???10?00?????1?000?000?10?000?3).????????000???n?1??n?1?????????????000?000??(???)Dn?1???Dn?2;证明：按第一列展开有 ??10(???)Dn?1??00???1?00????????00??????1?????(n?1)*(n?1)于是Dn?(???)Dn?1???Dn?2?Dn??Dn?1??(Dn?1??Dn?2) ??2(Dn?2??Dn?3) ????n?2(D2??D1) ??n?Dn??Dn?1??n*???n?1??n?1同理可得??Dn??Dn?1??**；解*式及**式可得Dn??.1?????nn?n?n??1??cos?14)0?0012cos?1?0001??000?1000?12cos?cos?11cos2??cosn?.2cos????00??2cos?证明：显然有 D1?cos??cos?,D2??2cos2??1?cos2?;假设对于级数小于n的行列式所证都成立，则： Dn按第n行展开2cos?Dn?1?Dn?2由假设有： Dn?2?cos(n?2)??cos[(n?1)???]?cos(n?1)?cos??sin(n?1)?sin?于是： Dn?2cos?Dn?1?Dn?2 ?2cos?cos(n?1)??[cos(n?1)?cos??sin(n?1)?sin?] ?cos?cos(n?1)??sin(n?1)?sin??cosn?.
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