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时间:2018-03-08
《考研数学 线性代数 内部习题讲义 金典》由会员上传分享,免费在线阅读,更多相关内容在教育资源-天天文库。
1、YouStupidCunt!cunnilinguspenisvagina第一章行列式1.利用对角线法则计算下列三阶行列式:201abc(1)1−4−1;(2)bca−183cab111xyx+y(3)abc;(4)yx+yx.222abcx+yxy201解(1)1−4−1=2×(−4)×3+0×(−1)×(−1)+1×1×8−183−0×1×3−2×(−1)×8−1×(−4)×(−1)=−24+8+16−4=−4abc(2)bca=acb+bac+cba−bbb−aaa−ccccab333=3abc−a−b−c111222222(3)
2、abc=bc+ca+ab−ac−ba−cb222abc=(a−b)(b−c)(c−a)xyx+y(4)yx+yxx+yxy333=x(x+y)y+yx(x+y)+(x+y)yx−y−(x+y)−x322333=3xy(x+y)−y−3xy−3yx−x−y−x1YouStupidCunt!cunnilinguspenisvagina33=−2(x+y)2.按自然数从小到大为标准次序,求下列各排列的逆序数:(1)1234;(2)4132;(3)3421;(4)2413;(5)13…(2n−1)24…(2n);(6)13…(2n−1)(2n
3、)(2n−2)…2.解(1)逆序数为0(2)逆序数为4:41,43,42,32(3)逆序数为5:32,31,42,41,21(4)逆序数为3:21,41,43n(n−1)(5)逆序数为:2321个52,542个72,74,763个…………………(2n−1)2,(2n−1)4,(2n−1)6,…,(2n−1)(2n−2)(n−1)个(6)逆序数为n(n−1)321个52,542个…………………(2n−1)2,(2n−1)4,(2n−1)6,…,(2n−1)(2n−2)(n−1)个421个62,642个…………………(2n)2,(2n)4
4、,(2n)6,…,(2n)(2n−2)(n−1)个3.写出四阶行列式中含有因子aa的项.1123解由定义知,四阶行列式的一般项为t(−1)aaaa,其中t为pppp的逆序数.由于p=1,p=31p12p23p34p4123412已固定,pppp只能形如13□□,即1324或1342.对应的t分别为12342YouStupidCunt!cunnilinguspenisvagina0+0+1+0=1或0+0+0+2=2∴−aaaa和aaaa为所求.11233244112334424.计算下列各行列式:⎢4124⎥⎢2141⎥⎢⎥⎢⎥120
5、23−121(1)⎢⎥;(2)⎢⎥;⎢10520⎥⎢1232⎥⎢⎥⎢⎥⎣0117⎦⎣5062⎦⎢a100⎥⎢−abacae⎥⎢⎥⎢⎥⎢−1b10⎥(3)bd−cdde;(4)⎢⎥⎢0−1c1⎥⎢⎣bfcf−ef⎥⎦⎢⎥⎣00−1d⎦解41244−12−101202c2−c31202(1)10520c4−7c31032−14011700104−1−104+3=122×(−1)103−144−1109910c+c23=12−200−2=0c+1c12310314171714214121403−121c−c3−12242(2)1232123
6、0506250623YouStupidCunt!cunnilinguspenisvagina21402140r−r3−122r−r3−1224241=01230123021400000−abacae−bce(3)bd−cdde=adfb−cebfcf−efbc−e−111=adfbce1−11=4abcdef11−1a10001+aba0−1b10r+ar−1b1012(4)0−1c10−1c100−1d00−1d1+aba01+abaadc+dc2+132=(−1)(−1)−1c1−1c1+cd0−1d0−101+abad3+2=(
7、−1)(−1)=abcd+ab+cd+ad+1−11+cd5.证明:22aabb3(1)2aa+b2b=(a−b);111ax+byay+bzaz+bxxyz33(2)ay+bzaz+bxax+by=(a+b)yzx;az+bxax+byay+bzzxy4YouStupidCunt!cunnilinguspenisvagina2222a(a+1)(a+2)(a+3)2222b(b+1)(b+2)(b+3)(3)=0;2222c(c+1)(c+2)(c+3)2222d(d+1)(d+2)(d+3)1111abcd(4)2222abcd4
8、444abcd=(a−b)(a−c)(a−d)(b−c)(b−d)⋅(c−d)(a+b+c+d);x−10"000x−1"00nn−1(5)""""""=x+ax+"+ax+a.1n−1n000"x−1aaa"ax+an
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