20xx高等代數(shù)習(xí)題及答案1)精選(編輯修改稿)
2025-03-26 03:07 本頁(yè)面
【文章內(nèi)容簡(jiǎn)介】 1%）那么由p(x)|f(x)g(x)知p(x)|f(x)或p(x)|g(x)。(1%)不妨設(shè)p(x)|f(x)，再由p(x)|(f(x)?g(x))得p(x)|g(x)。故p(x)|1矛盾。(2%)充分性. 由(f(x)?g(x),f(x)g(x))?1知存在多項(xiàng)式u(x),v(x)使u(x)(f(x)?g(x))?v(x)f(x)g(x)?1,(2%)從而u(x)f(x)?g(x)(u(x)?v(x)f(x))?1,(2%) 故(f(x),g(x))?1。(1%)三、(16分)a,b取何值時(shí)，線性方程組ax1?bx2?2x3?1???ax1?(2b?1)x2?3x3?1 ?ax?bx?(b?3)x?2b?123?1有唯一解、沒(méi)有解、有無(wú)窮解？在有解情況下求其解。解：b21??ab21??a????a2b?131?0b?110?????a??bb?32b?1??00b?12b?2???(5%)01??a2?b????0b?110??00b?12b?2???當(dāng)a(b?1)?0時(shí)，有唯一解：x1?25?b?22b?2(4%) ， x2?, x3?；a(b?1)b+1b?1當(dāng)b?1時(shí)，有無(wú)窮解：x3?0,x2?1?ax1,x1任意取值；當(dāng)a?0,b?5時(shí)，有無(wú)窮解：x1?k,x2??,x3?,k任意取值；(3%)當(dāng)b??1或a?0 且 b??1且 b?5時(shí)，無(wú)解。(4%)四、(10分)設(shè)a1,a2,...,an都是非零實(shí)數(shù)，證明?a111..111?a21..111..1...1...1.....111..?a1a2...an(1??i?1n1?a3...11) ai...11?an1a1證: 對(duì)n用數(shù)學(xué)歸納法。當(dāng)n=1時(shí) , D1?1?a1?a1(1?), 結(jié)論成立(2%)。假設(shè)n1時(shí)成立。那么n時(shí)?a1111?a21..111...11...11?a111..111?a21..111..1...1...1.....000 ..Dn=1..11?a3...11?.........1...111?a3...1...1an=a1a2...an?1?anDn?1 (4%)?n?11?現(xiàn)由歸納假設(shè)Dn?1?a1a2...an?1?1???有?i?1ai??n?11?Dn=a1a2...an?1?anDn?1=a1a2...an?1?a1a2...an?1an?1????i?1ai?n?1?=a1a2...an?1an?1???,(3%)?i?1ai?故由歸納原理結(jié)論成立。(1%)五、(10分)證明f(x)?x4?1在有理數(shù)域上不可約。 證: 令x?y?1得(1%)g(y)?f(x)?y4?4y3?6y2?4y?2。(3%)取素?cái)?shù)p=2滿足2|2,2|4,2|6,2|4,且2不整除1, 4不整除2. (2%)再據(jù)艾茵斯坦茵判別法知g(y)?y?4y?6y?4y?2在有理數(shù)域上不可約，（2%）432從而f(x)?x4?1在有理數(shù)域上不可約（2%）六、（9分）令A(yù)為數(shù)域F上秩為r的m?n矩陣，r?0。求證：存在秩 為r的m?r矩陣F和秩為r的r?n矩陣G, 使得A?FG。證: A為數(shù)域F上秩為r的m?n矩陣，r?0, 那么存在m?m可逆陣使P和n?n可逆陣Q?IA?P?r?0進(jìn)而令0??Q.（3%） 0??I?F?P?r?,G??Ir?0?就得A?FG（2%） .0?Q（4%）七、（10分）設(shè)A, B是n?n矩陣, 且A?B,A?B可逆。求證2n?2n矩陣P??證:|P|??AB??1?可逆, 且求P。?BA?ABA?BBA?B??BAB?AA0B?|A?B||A?B|?0， A?B故P可逆 （5%）令P?1???X?TY??有 S??AB??X????BA??TY??In???S??00??.（1%） In?1??1?1??X?(A?B)?(A?B)????AX?BT?In2??AY?BS?01???1?1?(A?B)?(A?B)進(jìn)而?（1%），解得?Y??（3%）??
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