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1、高等代数(北大版第三版)习题答案I 篇一：高等代数(北大版)第3章习题参考第三章线性方程组1用消元法解下列线性方程组：?x1?x?1?1)?x1x?1x13x2?5x3?4x4?1?3x2?2x3?2x42x2?x3?x4?x5?4x2?x3?x4?x5?2x2?x3?x4?x5x1?2x2?3x4?2x5?1x51?x1?x2?3x3?x4?3x5?232)?2x?3x?4x?5x?2x?72345?139x?9x?6x?16x?2x?252345?11x3?x7?0?3x1?4x2?5?x1?2x2?3x3?4x4?44x3?x2?0?x2?x3?x43?2x1?3x2?343)?4)?4
2、x?11x?13x?16x?0x?3xx?123424?1?17x?3x?x3?7x?2x?x?3x02342341?x1?2x2?3x3?x4?1?2x1?x2?x3?x4?1?3x1?2x2?x3?x4?13x1?2x2?2x3?3x4?25)?6)?2x1?3x2?x3?x4?12x?2x?2x?x?1?5x1?x2?x3?2x412341?2x?x?x?3x?4234?15x1?5x2?2x3?2解1 对方程组得增广矩阵作行初等变换，有11111?1?0?00033?2?420210?1521112?3?20?1?4?2?11?1?1202101?1?110100011103?0301
3、?011?200?0000?030?5?7?10000?15?3?4?4?400?20042358?1202101?1?11010001?2?2221?2?00?0因为rank(A)?rank(B)?4?5。因此方程组有没有穷多解，其同解方程组为x1?x4?1?2x1?x52?2x?03?x?x?0?24解得x1?x?2x3?x?4x51?k?k?0?k2?2k其中k为任意常数。2 对方程组德增广矩阵作行初等变换，有1?1291?0?0?02?1?3?920?34631?51632?3221120?07?25022?3?7?27120?346341110?2?5?2?1631?1?516?1?
4、3?34?512529?8?011333033?252972?10334?512529?8001?1?333000001?因为rank(A)?4?rank(A)?3。因此原方程无解。3 对方程组德增广矩阵作行初等变换，有1?010213?73?103411141?30?0130215?73?1?3341514?33?3?1?0?0001001?12?4210821300122400100002000088?3120?因为rank(A)?rank(A)?4。因此方程组有惟一解，且其解为x1x2x3?x?48?3?6?04 对方程组的增广矩阵作行初等变换，有3?247?1?0?004?311?27
5、?1717?3453?131?819?193871?22?416?37912000204007?311?27?1700?83?131?819009?2?163?9?2000?即原方程组德同解方程组为x1?7x2?8x3?9x4?0?17x2?19x3?20x4?0由此可解得x1?x2x3?x?43171917k1?k1?13172021k2k2。k1?k2其中k1,k2是任意常数。5 对方程组的增广矩阵作行初等变换，有?2?352?2?7?10?101?21?11000?12?11?10001?32?31?1001227?3?1?441247?1023010001000?1000?10001?
6、11?21?1001?425?14?2?1?因为rank(A)?4?rank(A)?3。因此原方程组无解。6 对方程组的增广矩阵作行初等变换，有?1?3?22522325311?1111?1112?3224555355421320?0?10?021222?122?5?1?1?0?0500022?15001001000201?050?100?0500007?6500100100?2?1?，5?0?0即原方程组的同解方程组为5x2?7x3?21?6?x?x?345?5x1?x3?0解之得x1x2?x3x4?k?25?75kk15?65k其中k是任意常数。2.把向量?表成?1,?2,?3,?4的线性
7、组合.。1)(1,2,1,1)1?(1,1,1,1),?2?(1,1,?1,?1)?3?(1,?1,1,?1),?4?(1,?1,?1,1)2)(0,0,0,1)1?(1,1,0,1),?2?(2,1,3,1)?3?(1,1,0,0),?4?(0,1,?1,?1)解1 设有线性关系k1?1?k2?2?k3?3?k4?4代入所给向量，可得线性方程组篇二：高等代数(北大版)第5章习题参考答案第五章二次型1用非退化线性替换化下列二次型为标准形，并利用矩阵验算所得结果。1 ?4x1x2?2x1x3?2x2x3；2222 x1；?2x1x2?2x2?4x2x3?4x3223 x1?3x2?2x1x2?2
8、x1x3?6x2x3；4 8x1x4?2x3x4?2x2x3?8x2x4；5 x1x2?x1x3?x1x4?x2x3?x2x4?x3x4；2226 x1?2x2?x4?4x1x2?4x1x3?2x1x4?2x2x3?2x2x4?2x3x4；22227 x1?x2?x3?x4?2x1x2?2x2x3?2x3x4。解1 已知f?x1,x2,x3?4x1x2?2x1x3?2x2x3，先作非退化线性替换x1?y1?y2x2?y1?y2 1x?y3?3则22f?x1,x2,x3?4y1?4y2?4y1y322224y1?4y1y3?y3?y3?4y222?2y1?y3y3，?4y23再作非退化线性替换1
9、1?y?z1212z3?y2?z2 2y?z33?则原二次型的标准形为f?x1,x2,x3?z1?4z2?z3。222最终将 2 代入 1 ，可得非退化线性替换为11?x?z?z?z32?121211?x2?z1?z2?z3 322x3?z3?于是对应的替换矩阵为110?11101?02?T?1?10?2012201?11?001?2?，?0012?001?且有100?T?AT?040?0012 已知f?x2221,x2,x3x1?2x1x2?2x2?4x2x3?4x3由配方法可得f?xx2221,2,x3?x1?2x1x2?x2?x2?4x2x3x21?x2?x2?2x23?。于是可令y1?
10、x1?x2y2?x2?2x3。y3x3则原二次型的标准形为f?x,x221,x23y1?y2且非退化线性替换为x1?y1?y2?2y3x2?y2?2y3。x3y3对应的替换矩阵为1?12?T?01?2?0014x23?且有001101?121001T?AT?11012201?2?010?。2?21024001000?223 已知f?x1,x2,x3x1?3x2?2x1x2?2x1x3?6x2x3。由配方法可得22222f?x1,x2,x3x1?2x1x2?2x1x3?2x2x3?x2?x3?4x2?4x2x3?x3x221?x2?x3?2x2?x3?。于是可令y1?x1?x2?x?3y2?2x
11、2?x3。y3x3则原二次型的标准形为f?xy221,x2,x31?y2且非退化线性替换为x1y1?1y2?3y3?22x11?2?2y2?2y3。x3?y3?对应的替换矩阵为11?322T1?01?2?0012?且有1?10112?3?T?AT110?11?012?232?1?3?302?1111?30?02?0121200?4 已知f?x1,x2,x3,x48x1x2?2x3x4?2x2x3?8x2x4，00?10。00先作非退化线性替换x1?y1?y4?x?y?22x?y3?3x4?y4则2f?x1,x2,x3,x48y1y4?8y4?2y3y4?2y2y3?8y2y42211111?18?y4?2y4?y1?y2?y3?y1?y2?y328228?2?1118?y1?y2?y32y2y328211118?y1?y2?y3?y42?y1?y2?y32y2y3。2842再作非退化线性替换222y1?z1y?z?z?223y3?z2?z3y4?z4则53531f?x1,x2,x3,x48?z1?z2?z3?z42?z1?z2?z3?88442222z2，?2z322再令53?w?z?x?x312?144w2?z2w3?z3?153w4?z1?z2?z3?z4
