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1、[线性代数课后习题答案]线性代数课后习题答案全解篇一:线性代数课后习题答案全解第一章行列式1.利用对角线法则计算下列三阶行列式:2011?4?;?183201解1?4??183=2××3+0××+1×1×8?0×1×3?2××8?1××=?24+8+16?4=?4.abcbca;cababc解bcacab=acb+bac+cba?bbb?aaa?ccc=3abc?a3?b3?c3.111abc;a2b2c2111解abca2b2c2=bc2+ca2+ab2?ac2?ba2?cb2=.xyx+yyx+yx.x+yxyxyx+
2、y解yx+yxx+yxy=xy+yx+yx?y3?3?x3=3xy?y3?3x2y?x3?y3?x3=?2.2.按自然数从小到大为标准次序,求下列各排列的逆序数:1234;解逆序数为04132;解逆序数为4:41,43,42,32.3421;解逆序数为5:32,31,42,41,21.2413;解逆序数为3:21,41,43.13???24???;n:解逆序数为23252,5472,74,76??????2,4,6,???,13??????2.解逆序数为n:3252,54??????2,4,6,???,4262,64???
3、???2,4,6,???,3.写出四阶行列式中含有因子a11a23的项.解含因子a11a23的项的一般形式为ta11a23a3ra4s,其中rs是2和4构成的排列,这种排列共有两个,即24和42.所以含因子a11a23的项分别是ta11a23a32a44=1a11a23a32a44=?a11a23a32a44,ta11a23a34a42=2a11a23a34a42=a11a23a34a42.4.计算下列各行列式:4101251202142;0741解1252024c2?c342======10?123202?104?1?1
4、02=122×4+3?140117c4?7c300103?14=4?110c2+c39910123?142======c00?2=0.1+2c31714231?114152203162;2242114解31?121=c=4?3=c22?140?15062==32r4?r222152120360=====3212212021340r4?r4====123?1=1200210320=0.00?bdabacbf?cfcdae?deef;解?bdabbf?accfcdae?deef=adf?bbb?ccce?ee=adfbce?1
5、1111?11?1=4abcdef.a1?001b?1001c?100.1da1解?001b?1001c?10r1+ar201+ab0=====?1b10?1d00a1c?1001d+aba0c3+dc2+abaad=2+1?1c1=====?1c1+cd0?1d0?105.证明:abad=abcd+ab+cd+ad+1.=3+2+?11+cda2abb22aa+b2b=3;111证明a2abb2c2?c1a2ab?a2b2?a22aa+b2b=====2ab?a2b?2a00111c3?c11222ab?ab?aab+a
6、=3.=1=2b?a2b?2a3+1ax+byay+bzaz+bxxyzay+bzaz+bxax+by=yzx;az+bxax+byay+bzzxy证明ax+byay+bzaz+bxay+bzaz+bxax+byaz+bxax+byay+bz篇二:线性代数答案线性代数浙大出版社主编杨浩波我要课后习题答案,全面的线性代数答案线性浙大出版社主编杨浩波我要课后习题答案,全面的下这个可作参考:篇三:线性代数课后习题答案全解第一章行列式1.利用对角线法则计算下列三阶行列式:2011?4?;?183201解1?4??183=2××3+
7、0××+1×1×8?0×1×3?2××8?1××=?24+8+16?4=?4.abcbca;cababc解bcacab=acb+bac+cba?bbb?aaa?ccc=3abc?a3?b3?c3.111abc;a2b2c2111解abca2b2c2=bc2+ca2+ab2?ac2?ba2?cb2=.xyx+yyx+yx.x+yxyxyx+y解yx+yxx+yxy=xy+yx+yx?y3?3?x3=3xy?y3?3x2y?x3?y3?x3=?2.2.按自然数从小到大为标准次序,求下列各排列的逆序数:1234;解逆序数为041
8、32;解逆序数为4:41,43,42,32.3421;解逆序数为5:32,31,42,41,21.2413;解逆序数为3:21,41,43.13???24???;n:解逆序数为23252,5472,74,76??????2,4,6,???,13??????2.解逆序数为n:3252,54??????2,