2022考研数学全程班作业答案——《高分强化521》新浪微博@考研数学周洋鑫12022考研数学全程班作业答案——《高分强化521》作业1第1章线代·行列式1.1代数余子式线性和问题【1】设3112513420111533D−−−=−−−,D的(),ij元的代数余子式记作ijA,求31323334+322AAAA−+.解析:代数余子式的线性和问题.313233343112132251345134+3221322311215331533AAAA−−−−−−−+==−−−−−−−1322132213220167608540854240854016760032085508550001−−−−−−−−=−===−−−−−−−−−.1.2数值型行列式计算【2】计算行列式1000100014321−−=−+.解析:43210010100010=01+4110++2+3+4.00132+101432+1−−−−−−=−−(-)【3】120naaa,111222111nnnaaaaaaaaa++=+.一笑而过考研数学2022考研数学全程班作业答案——《高分强化521》新浪微博@考研数学周洋鑫2解析:111122221111110101nnnnaaaaaaaaaaaa++−−+=+1221001001nnaaaaa+++=121naaa=+++.【4】1111111naaaaDaaa−−−==−−−−.解析:将行列式按照第一行展开可得:()()()1212111nnnDaDaD+−−=−+−−即()121nnnDaDaD−−=−+,其中()22111,111aaDaaDaa−==−+=−−−112211nnnnDaDDaDDaD−−−+=+==+=故11nnDaD−=−+,所以11111nnDaDaa−−=−−++,即11nDa−+是以111Da−+为首项,以a−为公比的等比数列,所以()21111nnaDaaa−−−=−++,即()1111nnnaDa+−+=+.【5】计算()11111111011111111xxDxyyy+−=+−.解析:221111111000000000000000xxxxxDxyxyyxyy+−−−===−−−−.一笑而过考研数学2022考研数学全程班作业答案——《高分强化521》新浪微博@考研数学周洋鑫31.3抽象型行列式计算【6】设A是三阶方阵,A是A的伴随矩阵,A的行列式12=A.求行列式()132AA−−=.解析:()111111123222333−−−−−−=−=−=−AAAAAAAA,所以()31122116323327−−−=−=−=−AAAA.【7】设A是n阶矩阵,满足T=AAE(E是n阶单位阵,TA是A的转置矩阵),0A,求+AE=.解析:根据T=AAE有TTTT|||||()||()||||()|+=+=+=+=+AEAAAAEAAEAAEA||||||||=+=+AEAAAE移项得(1||)||0−+=AAE.因为0A,故(1||)0−A,所以||0+=AE.【8】已知实矩阵()33ija=A满足条件:(1)(),1,2,3ijijAaij==,其中ijA是ija的代数余子式;(2)110a.计算行列式A.解析:由于(),1,2,3ijijAaij==,所以*T=AA,两边取行列式得2=AA,解得0=A或1,A沿第1行展开得1112132221111121213130aAaAaAaaa=++=++A,所以1=A.【9】设()T1,01=−,,矩阵T=A,n为正整数,则na−=EA.解析:显然T=A的特征值为0,0,2.故na−EA的特征值为,,2naaa−,所以()22nnaaa−=−EA.一笑而过考研数学