2025年專升本數(shù)學(xué)專業(yè)高等代數(shù)測試試卷（含答案）考試時(shí)間：______分鐘總分：______分姓名：______一、選擇題：本大題共5小題，每小題3分，共15分。在每小題給出的四個(gè)選項(xiàng)中，只有一項(xiàng)是符合題目要求的。1.設(shè)向量組α?,α?,α?線性無關(guān)，向量β?=α?+α?,β?=α?+α?,β?=α?+α?，則向量組β?,β?,β?的秩為（）。(A)1(B)2(C)3(D)不能確定2.設(shè)A是n階可逆矩陣，B是n階矩陣，則下列運(yùn)算中（）一定正確。(A)AB=BA(B)(AB)?=A?B?(C)|AB|=|BA|(D)(AB)?1=A?1B?13.設(shè)A是m×n矩陣，B是n×s矩陣，若AB=O（零矩陣），則必有（）。(A)A=O或B=O(B)r(A)+r(B)=n(C)r(A)≤n(D)s=04.n階矩陣A滿足A2-2A-3I=O，則A的特征值可能是（）。(A)-3(B)-1(C)1(D)35.二次型f(x?,x?,x?)=x?2+x?2+ax?2+2x?x?+4x?x?+2x?x?的正慣性指數(shù)為2，則實(shí)數(shù)a的取值范圍是（）。(A)a>1(B)a=1(C)a<-2(D)a≥-2二、填空題：本大題共5小題，每小題3分，共15分。6.設(shè)A=[aij]是三階矩陣，其中aij=i+j，則|A|=_______。7.若線性方程組Ax=b有無窮多解，其中A是4×3矩陣，則其增廣矩陣(A|b)的秩r(A|b)=_______。8.設(shè)向量α=(1,k,2)?與β=(0,1,-1)?正交，則實(shí)數(shù)k=_______。9.設(shè)A是三階矩陣，其特征值為1,2,-1，則|A|=_______，tr(A)=_______。10.設(shè)二次型f(x?,x?,x?)=x?x?+x?x?通過正交變換x=Pz化為標(biāo)準(zhǔn)形f=z?2-z?2+2z?2，則原二次型的矩陣A的特征值為_______，_______，_______。三、計(jì)算題：本大題共4小題，共55分。11.（10分）計(jì)算行列式|A|，其中A=???1234???。12.（15分）設(shè)線性方程組\[\begin{cases}x?+2x?+3x?=1\\2x?+3x?+ax?=3\\x?+3x?+4x?=b\end{cases}\](1)討論該方程組何時(shí)有解、無解；(2)若有解，求其通解。13.（15分）設(shè)向量組α?=(1,1,1)?,α?=(1,2,3)?,α?=(1,3,t)?。(1)當(dāng)t為何值時(shí)，向量組α?,α?,α?線性無關(guān)？(2)當(dāng)t為何值時(shí)，向量組α?,α?,α?線性相關(guān)？并在此時(shí)求出它的一個(gè)極大無關(guān)組。14.（15分）設(shè)矩陣A=???011???，求A的特征值與特征向量，并判斷A是否可對角化。若可對角化，求可逆矩陣P，使得P?1AP為對角矩陣。試卷答案一、選擇題1.C2.B3.C4.B5.D二、填空題6.67.38.-19.-2,310.1,-1,2三、計(jì)算題11.解：原式=|1234|（按第一行展開）=1×|234|-2×|134|+3×|124|-4×|123|=1×(2×4-3×4)-2×(1×4-3×1)+3×(1×3-2×1)-4×(1×2-2×1)=1×(-4)-2×(1)+3×(1)-4×(0)=-4-2+3=-3.12.解：對增廣矩陣(A|b)進(jìn)行行變換：(123|1)→(123|1)(23a|3)→(123|1)R?-2R?→(01a-6|1)(134|b)→(123|1)R?-R?→(011|b-1)→(123|1)→(01a-6|1)→(011|b-1)→(01a-6|1)R?-R?→(00a-7|b-2)(123|1)(01a-6|1)(00a-7|b-2)(1)當(dāng)a=7且b≠2時(shí)，r(A)=2,r(A|b)=3，方程組無解。當(dāng)a=7且b=2時(shí)，r(A)=r(A|b)=2<3，方程組有無窮多解。當(dāng)a≠7時(shí)，r(A)=r(A|b)=3，方程組有唯一解。(2)若方程組有解，即a≠7或(a=7且b=2)。若a≠7，則增廣矩陣可化為(123|1)(01a-6|1)(001|(b-2)/(a-7))，解得x?=(b-2)/(a-7)，x?=1-(a-6)x?=1-(a-6)(b-2)/(a-7)，x?=1-2x?-3x?=1-2[1-(a-6)(b-2)/(a-7)]-3(b-2)/(a-7)=(a+b-5)/(a-7)。通解為(x?,x?,x?)=[(a+b-5)/(a-7),1-(a-6)(b-2)/(a-7),(b-2)/(a-7)]。若a=7且b=2，則增廣矩陣可化為(123|1)(011|1)(000|0)，解得x?=1-x?，x?=1-2x?-3x?=1-2(1-x?)-3x?=-1+x?。通解為(x?,x?,x?)=(-1+x?,1-x?,x?)，即x=(-1,1,0)?+x?(1,-1,1)?，其中x?為任意常數(shù)。13.解：構(gòu)成矩陣A=[α?,α?,α?]=[(111),(123),(13t)]。(1)對A進(jìn)行行變換：(111)→(111)(123)→(123)R?-R?→(012)(13t)→(13t)R?-R?→(02t-1)→(111)→(012)→(02t-1)→(012)R?-2R?→(00t-5)當(dāng)t≠5時(shí)，r(A)=3，向量組α?,α?,α?線性無關(guān)。當(dāng)t=5時(shí)，r(A)=2<3，向量組α?,α?,α?線性相關(guān)。(2)當(dāng)t=5時(shí)，向量組線性相關(guān)。由行簡化階梯形(111)(012)(000)可知，α?,α?線性無關(guān)，α?=-2α?+α?。極大無關(guān)組為{α?,α?}。14.解：設(shè)A=[a??]，其中a??=0,a??=1,a??=1,a??=0,a??=0,a??=1,a??=0,a??=1,a??=0。|λI-A|=|λ0-1|=|λ0-1|λ|-(-1)|(-1)|+0|=λ[λ(λ)-0]+1[-(λ)-0]=λ3-λ=λ(λ2-1)=λ(λ-1)(λ+1)。令|λI-A|=0，得特征值λ?=0,λ?=1,λ?=-1。(1)對于λ?=0，解(0I-A)x=0，即(-A)x=0，得(0-1-1)(x?)=(0)(x?)=(0)(x?)(00-1)()=()()(010)()=()()→(0-1-1)(x?)=(0)(x?)=(0)(x?)→(00-1)(x?)=(0)(00)→(010)(x?)=(0)(00)→(001)(x?)=(0)(00)→(000)()=(0)()→(010)(x?)=(0)(00)→(001)(x?)=(0)(00)得x?=0,x?=0,x?任意。特征向量為k?(1,0,0)?，k?≠0。(2)對于λ?=1，解(1I-A)x=0，即(10-1)(x?)=(0)(x?)=(0)(x?)(01-1)()=()()(0-11)()=()()→(10-1)(x?)=(0)(x?)=(0)(x?)→(01-1)(x?)=(0)(00)→(0-11)(x?)=(0)(00)→(000)()=(0)()→(10-1)(x?)=(0)(x?)=(0)(x?)→(01-1)(x?)=(0)(00)→(000)()=(0)()得x?=x?,x?=x?。特征向量為k?(1,1,1)?，k?≠0。(3)對于λ?=-1，解(-1I-A)x=0，即(-10-1)(x?)=(0)(x?)=(0)(x?)(0-1-1)()=()()(01-1)()=()()→(-10-1)(x?)=(0)(x?)=(0)(x?)→(0-1-1)(x?)=(0)(00)→(01-1)(x?)=(0)(00)→(-10-1)(x?)=(0)(x?)=(0)(x?)→(0-1-1)(x?)=(0)(00)→(01-1)(x?)=(0)(00)→(-10-1)(x?)=(0)(x?)=(0)(x?)→(0-1-1)(x?)=(0)(00)→(002)(x?)=(0)(00)→(001)(x?)=(0)(00)→(000)()=(0)()→(101)(x?)=(0)(x?)=(0)(x?)→(011)(x?)=(0)(00)→(000)()=(0