3)135..(2n-1)(2n)(2n-2)42 τ(135..(2n-1)(2n)(2n-2)42)2+4+6+.+2(m-1)=m(n-1) 4在五阶行列式中,下列各项之前应取什么符号? 1)a3a24a244a5解:由于τ(34215)=2+3=5,所以-a1224a24145 2)a211y4a542解:由于(21354)(134522+3=5,所以-a21a13414a5a42 5写出四阶行列式中含有因子a1a23的各项 F:-a1a23a32a4, a1a23a3a42 6计算下列各行列式 3-12 056-2 0 506 056 056 41 105 2021 4 12021202 202 10520105201-152-200-152-20 011 01170117 1202 1202 001785 0015 0acca 9ddbb3) 1 3 5 … (2n-1) (2n) (2n-2)… 4 2 t(1 3 5 … (2n-1) (2n) (2n-2)… 4 2)=2+4+6+… +2(n-1)=n(n-1) 4.೼Ѩ䰊㸠߫ᓣЁ,ϟ߫৘乍ПࠡᑨপҔМヺো? 1) 13 24 32 41 55 aaaaa 㾷˖⬅Ѣt(34215)=2+3=5ˈ᠔ҹ 13 24 32 41 55 -aaaaa 2) 21 13 34 55 42 aaaaa 㾷˖⬅Ѣt(21354)+ t(13452)=2+3=5ˈ᠔ҹ 21 13 34 55 42 -aaaaa 5.ߎݭಯ䰊㸠߫ᓣЁ৿᳝಴ᄤ 11 23 a a ⱘ৘乍. 㾷˖ 11 23 32 44 11 23 34 42 -aaaa aaaa , 6.䅵ㅫϟ߫৘㸠߫ᓣ 1) 214 1 3 12 1 123 2 506 2 - - - = - - 㾷˖ 2 1 4 1 1 24 1 1 2 4 1 3 12 1 132 1 0 5 6 2 0 1 2 3 2 2 13 2 0 3 5 0 5 0 6 2 0 56 2 0 5 6 2 -- - - -- - - =- =- = - - -- -- - 2) 4 124 1 202 10 5 2 0 0 117 = 㾷˖ 4 124 1 202 1 2 0 2 1 2 0 2 1 202 4 124 0 7 2 4 0 7 2 4 10 5 2 0 10 5 2 0 0 15 2 20 0 15 2 20 0 117 0 117 0 1 1 7 0 1 1 7 12 0 2 1202 01 1 7 0117 17 9 0 0 0 17 85 0 0 1 5 0 0 9 45 0 0 1 5 -- -- =- =- =- -- -- = =´ = 3) cadb acdb acbd cabd =