2012年4月自考线性代数真题及答案[五篇范文]

第一篇：2012年4月自考线性代数真题及答案

全国2012年4月高等教育自学考试线性代数(经管类)试题课程代码：04184

一、单项选择题（本大题共10小题，每小题2分，共20分）

a111.设行列式a21a12a22a32a13a112a122a222a323a133a23=（）3a33D.12 a31A.-12 a23=2,则a21a33a31B.-6

C.6 1202.设矩阵A=120,则A*中位于第1行第2列的元素是（）003A.-6 B.-3

C.3

D.6

3.设A为3阶矩阵，且|A|=3，则(A)1=（）A.3 B.1 3C.1 3D.3 4.已知43矩阵A的列向量组线性无关，则AT的秩等于（）A.1 B.2

C.3

D.4 1005.设A为3阶矩阵,P =210,则用P左乘A，相当于将A（）001A.第1行的2倍加到第2行

B.第1列的2倍加到第2列 C.第2行的2倍加到第1行

D.第2列的2倍加到第1列 6.齐次线性方程组A.1

0x12x23x3的基础解系所含解向量的个数为（）x2+x3x4= 0B.2

C.3

D.4 7.设4阶矩阵A的秩为3，1，2为非齐次线性方程组Ax =b的两个不同的解，c为任意常数，则该方程组的通解为（）A.1c122 B.1223 5c1 C.1c122 D.1225 3c1

8.设A是n阶方阵，且|5A+3E|=0，则A必有一个特征值为（）A.5 3B.C.5D.1009.若矩阵A与对角矩阵D=010相似，则A3=（）001A.E B.D

C.A

D.-E

22210.二次型f(x1,x2,x3)=3x1是（）2x2x3A.正定的 B.负定的 C.半正定的 D.不定的

二、填空题（本大题共10小题，每小题2分，共20分）请在每小题的空格中填上正确答案。错填、不填均无分。

111.行列式21146=____________.4163600110012.设3阶矩阵A的秩为2，矩阵P =010，Q =010，若矩阵B=QAP，则r(B)=_____________.10010113.设矩阵A=1448，B=，则AB=_______________.141214.向量组1=(1,1,1,1),2=(1,2,3,4),3=(0,1,2,3)的秩为______________.15.设1,2是5元齐次线性方程组Ax =0的基础解系，则r(A)=______________.1000216.非齐次线性方程组Ax =b的增广矩阵经初等行变换化为01002，则方程组的通解是______.0012-217.设A为3阶矩阵，若A的三个特征值分别为1，2，3，则|A|=___________.18.设A为3阶矩阵，且|A|=6，若A的一个特征值为2，则A*必有一个特征值为_________.22219.二次型f(x1,x2,x3)=x1的正惯性指数为_________.x23x322220.二次型f(x1,x2,x3)=x12x22x34x2x3经正交变换可化为标准形______________.三、计算题（本大题共6小题，每小题9分，共54分）

3512453321.计算行列式D =

1201203413022.设A=210，矩阵X满足关系式A+X=XA,求X.00223.设，，2，3，4均为4维列向量，A=（，2，3，4）和B=（，2，3，4）为4阶方阵.若行列式|A|=4，|B|=1，求行列式|A+B|的值.24.已知向量组1=(1,2,1,1)T,2=(2,0,t,0)T,3=(0,4,5,2)T,4=(3,2,t+4,-1)T（其中t为参数），求向量组的秩和一个极大无关组.x1x22x3x4325.求线性方程组x12x2x3x42的通解..（要求用它的一个特解和导出组的基础解系表示）

2xx5x4x7341226.已知向量1=(1,1,1)T，求向量2，3,使1，2，3两两正交.四、证明题（本题6分）

27.设A为mn实矩阵，ATA为正定矩阵.证明：线性方程组Ax=0只有零解.全国2012年4月高等教育自学考试线性代数(经管类)试题答案课程代码：04184

一、单项选择题（本大题共10小题，每小题2分，共20分）

1．D 2.A 3.B 4.C 5.B 6.B 7.A 8.B 9.D 10.D

二、填空题（本大题共10小题，每小题2分，共20分）请在每小题的空格中填上正确答案。错填、不填均无分。

 2 0 2 00011.16 12.213.14.2 15.3 16.k,k为任意常数 17.6 2200

 0 12218.3 19.220.y14y2

三、计算题（本大题共6小题，每小题9分，共54分）

3421.解：D =1251215334201303422011201533013315120111034043212010111

01331043212010111101248 00101216001622.解：由AXXA，可知X(AE)A，则XA(AE)1，00301且AE200，(AE)130010000 011200111300122110010故XA(AE)210133

00200100223.解: AB(,22,23,24)8[(,2,3,4)(,2,3,4)]8AB40

1224.解：（1,2,3,4）=111000203120312042044801t5t40t25t70t20210224002031112003t3t00000021112

03t3t0000312

5t700t3时，秩为2，一个极大无关组为1,2 t3时，秩为3，一个极大无关组为1,2,3.25.解：对增广矩阵作初等行变换

112131121311213A(A,b)121120112101121

21547011210000010334

01121

00000x13x33x44同解方程组为.x3，x4是自由未知量，特解*(4,1,0,0)T

x2x32x41x13x33x4导出组同解方程组为.x3，x4是自由未知量，xx2x342基础解系1(3,1,1,0)T，2(3,2,0,1)T，通解为*k11k22,k1,k2R.26.解：设2=(x1,x2,x3)T，2与1正交，则有x1x2x30，故可取2==(1,0,-1)T, 设3=(y1,y2,y3)T，3与1，2两两正交，则故可取3=(1,2,1).四、证明题（本题6分）

27.证明：由于ATA为正定矩阵，则秩（ATA）= n，又秩（A）= 秩（ATA）= n，则线性方程组Ax=0只有零解.Ty1y2y30.y1y3 = 0

第二篇：0907线性代数真题及答案

全国2009年7月高等教育自学考试

线性代数（经管类）试题

课程代码：04184 试卷说明：在本卷中，AT表示矩阵A的转置矩阵；A*表示A的伴随矩阵;R(A)表示矩阵A的秩；|A|表示A的行列式；E表示单位矩阵。

一、单项选择题(本大题共10小题，每小题2分，共20分)在每小题列出的四个备选项中只有一个是符合题目要求的，请将其代码填写在题后的 括号内。错选、多选或未选均无分。

1.设A,B,C为同阶方阵，下面矩阵的运算中不成立的是(C)．．．A.（A+B）T=AT+BT C.A(B+C)=BA+CA a112.已知a21a31a12a22a32a132a11a23=3，那么a21a332a312a12a222a32B.|AB|=|A||B| D.(AB)T=BTAT 2a13a23=(B)2a33A.-24 C.-6

B.-12 D.12

3.若矩阵A可逆，则下列等式成立的是（C)A.A=1A* AB.A0

C.(A2)1(A1)2 D.(3A)13A1

41312021234.若A=，B=，C=2矩阵的312，则下列矩阵运算的结果为3×15221是(D)A.ABC C.CBA

B.ACTBT D.CTBTAT

5.设有向量组A：1，2，3，4，其中1，2，3线性无关，则(A)A.1，3线性无关

C.1，2，3，4线性相关

B.1，2，3，4线性无关 D.2，3，4线性相关

浙04184# 线性代数（经管类）试题

6.若四阶方阵的秩为3，则（B)A.A为可逆阵

C.齐次方程组Ax=0只有零解

B.齐次方程组Ax=0有非零解 D.非齐次方程组Ax=b必有解

7.设A为m×n矩阵，则n元齐次线性方程Ax=0存在非零解的充要条件是(B)A.A的行向量组线性相关 C.A的行向量组线性无关

8.下列矩阵是正交矩阵的是（A)010010A.

0011011110B.2011B.A的列向量组线性相关 D.A的列向量组线性无关

cosC.sinsin

cosD.220221666106333

3339.二次型fxTAx(A为实对称阵)正定的充要条件是（D)A.A可逆

C.A的特征值之和大于0

0k010.设矩阵A=0k2正定，则(C)024B.|A|>0

D.A的特征值全部大于0

A.k>0 C.k>1 1:D1k0;2:D2k00k0k20;0kB.k0 D.k1 3D3k1k20k2kk4k44kk10

2402

4二、填空题(本大题共10小题，每小题2分，共20分)请在每小题的空格中填上正确答案。错填、不填均无分。

浙04184# 线性代数（经管类）试题

1211263.11.设A=（1，3，-1），B=（2，1），则ATB=63。

ATB32,11221121012.若1310,则k-1。

k21210210211311312k1k10.k1.k11k21k110120060*630。20013.设A=，则A=214013120121*A2003120.A可逆.A1A20A0131A*AA1.而A,E320012011②401012-14013001③3201001②+(-2)①0001000130010①+(-2)②1000③+(-1)②001012003121201004021013001120-140141

12012010000060101012-140.A112-14063012001161121316112132140600601A*AA112630630.1221421414.已知A2-2A-8E=0，则（A+E）-1=A22A8E0.1A3E。5A22A3E5E015AEA3E5E.11AEA3EE5A3EAEE.AE 1A3E5浙04184# 线性代数（经管类）试题

15.向量组1(1,1,0,2),2(1,0,1,0),3(0,1,1,2)的秩为___2。

将三个向量的转置向量拼成一个43的矩阵，化简此矩阵11A021011011③+1②+(-1)①01②01101④+(-2)①④+(-2)②0110011020220001.rA2.0016.设齐次线性方程Ax=0有解，而非齐次线性方程且Ax=b有解,则是方程组____Ax=b的解。

x1x2017.方程组的基础解系为11。

x2x301___1。18.向量(3,2,t,1),(t,1,2,1)正交,则t__________t1,3,2,t,13t22t1t10.t1.211

103b119.若矩阵A=与矩阵B=相似，则x=4ab。304axAB103b143xabx4ab.04ax31232122220。20.二次型f(x1,x2,x3)x12x23x3x1x23x1x3对应的对称矩阵是123203

三、计算题（本大题共6小题，每小题9分，共54分）1340403521.求行列式D=的值。

20227622

浙04184# 线性代数（经管类）试题

13404035解D20227622④+2①13402090403522624351213222按

解：把所有行向量转置为列向量形成44的矩阵，并将其化为简化阶梯形矩阵.2354115302640264③①TTTA1T,2,3,41153235431953195115311531②③+2①02640132④+3①20515100515100264026410211021①+1②1①③+5②01320132④+2②1②记为,,,B.13420000000000000000显然B是A的简化阶梯形矩阵.易见：B的秩为2，从而A的秩为2,原向量组的秩为2.易见：B的列向量组的一个极大线性无关组为1,2.TA的列向量组的一个极大线性无关组为1T,2;从而1,2是原向量组的一个极大线性无关组.24.求取何值时,齐次方程组 (4)x13x20

4x1x30

5xxx0123

有非零解？并在有非零解时求出方程组的通解。

浙04184# 线性代数（经管类）试题

解方程的个数与未知量的个数相同，考察系数矩阵A是否为可逆矩阵.43A450011430③+1②41010按

631解A的特征方阵EnA053.A的特征方程为0641EnA006353253=-1-1-122-1205-4186464得1=2=1，3=-2，为A的两个特征值.用来求特征向量的矩阵方程为63x101x16x23x301053x0,即齐次线性方程组E3Ax5x23x30.20x0646x24x303属于121的特征向量满足线性方程组6x23x30,即x32x23个未知量1个方程,必有2个自由未知量,不妨取x1、x2为自由未知量，10x110令或，则x30或2，于是得2个线性无关的特征向量p10，p21.x201023x16x23x30属于32的特征向量满足线性方程组为3x23x30.,即x1x2x3,6x26x303个未知量2个方程,必有1个自由未知量,不妨取x1为自由未知量，令x11,则x21,x31,1于是得1个线性无关的特征向量p31.122226.用配方法求二次型f(x1,x2,x3)x14x2x32x1x34x2x3的标准形，并写出相应的线性变换。

222解二次型f(x1,x2,x3)x124x2x32x1x34x2x3x122x1x3x34x224x2x3x32x322x1x32x2x3x322x1y1y3y1x1x31设y22x2x3，即x2y2y3，2yx33x3y3222可使得f(x1,x2,x3)x1x32x2x3x3g(y1,y2,y3)y12y2y3.即二次型的标准形；221y1x110此时相应的线性变换xPy为x201212y2.x0013y3

浙04184# 线性代数（经管类）试题

四、证明题（本大题共1小题，6分）

27.证明：若向量组1,2,n线性无关,而11n,212,323,，nn1+n，则向量组1,2,,n线性无关的充要条件是n为奇数。

证设k11k22knn0.将已知条件代入得k11nk212k323knn1n0.整理得k1k21k2k32kn1knn1knk1n0.1,2,n线性无关,k1k2k2k3kn1knknk10.k1k2k3k4knk1,当n为奇数，则n1为偶数,则上式为k1k2k3k4kn1knknk1.由此knkn0,k1k2k3k4kn1kn0.因此,1,2,,n线性无关.反之,若1,2,,n线性无关,即当且仅当k1k2k3k4kn1kn0时,等式k11k22knn0才成立,k11nk212k323knn1n0k1k21k2k32kn1knn1knk1n01,2,n线性无关,k1k2k2k3kn1knknk10k1k2k3k4knk1,当n为偶数时,令k1k2k3k4kn1,则1234n1n0也成立,这与条件不符.当n为奇数时,则n1为偶数,则有k1k2k3k4kn1knknk1，立得k1k2k3k4kn1kn0,等式k11k22knn0才成立，这与条件完全相符.证毕.浙04184# 线性代数（经管类）试题

第三篇：2013年10月自考线性代数真题

2013年10月自考线性代数真题

说明：在本卷中，A表示矩阵A的转置矩阵，A表示矩阵A的伴随矩阵，E是单位矩阵，|A|表示方阵A的行列式，r(A)表示矩阵A的秩。T

*

选择题部分

注意事项：

1．答题前，考生务必将自己的考试课程名称、姓名、准考证号用黑色字迹的签字笔或钢笔填写在答题纸规定的位置上。

2．每小题选出答案后，用2B铅笔把答题纸上对应题目的答案标号涂黑。如需改动，用橡皮擦干净后，再选涂其他答案标号。不能答在试题卷上。

一、单项选择题（本大题共5小题，每小题2分，共10分）

在每小题列出的四个备选项中只有一个是符合题目要求的，请将其选出并将“答题纸”的相应代码涂黑。错涂、多涂或未涂均无分。1．设行列式a1a2b1b21，a1a2c1c22，则

a1a2b1c1b2c2

A.－3 C.1 2．设4阶矩阵A的元素均为3，则r(A)= A.1 C.3 3．设A为2阶可逆矩阵，若A1B.－1 D.3 B.2 D.4 13

2553C. 21A.13*，则A= 251B.25D.23 53 14．设A为m×n矩阵，A的秩为r，则 A.r=m时，Ax=0必有非零解 C.r

222B.r=n时，Ax=0必有非零解 D.r

5．二次型f(xl，x2,x3)=x12x23x38x1x312x2x3的矩阵为

1A.081C.04 0212026812 346 31B.001D.4008212 034026 63非选择题部分

注意事项：

用黑色字迹的签字笔或钢笔将答案写在答题纸上，不能答在试题卷上。

二、填空题（本大题共10小题，每小题2分，共20分）6．设A为3阶矩阵，且|A|=2，则|2A|=______．

127．设A为2阶矩阵，将A的第1行加到第2行得到B，若B=，则A=______.34a12a11a12a118．设矩阵A=，B=，且r(A)=1，则r（B）=______.aaaaaa2122112112229．设向量α=(1，0，1)，β=(3，5，1)，则β－2α=________．

T10．设向量α=(3，－4)，则α的长度||α||=______．

TT11．若向量αl=(1，k)，α2=(－1,1)线性无关，则数k的取值必满足______.12．齐次线性方程组xl+x2+x3=0的基础解系中所含解向量的个数为______．

T

T12210013．已知矩阵A=212与对角矩阵D=010相似，则数a=______ 22100a14．设3阶矩阵A的特征值为－1，0，2，则|A|=______．

15．已知二次型f(x1,x2,x3)=x1x2tx3正定，则实数t的取值范围是______．

三、计算题（本大题共7小题，每小题9分，共63分）

222abc16．计算行列式D=

2abac2c2a2bcabT

T2b2c.17．已知向量α=（1，2，k），β=1，，且βα=3，A=αβ，求(1)数k的值；

10(2)A．

11231231218．已知矩阵A=231，B=00，求矩阵X，使得AX=B.3401019．求向量组α1=(1,0,2,0), α2=(－1,－1,－2,0), α3=(－3,4,－4,l), α4=（－6,14,T－6,3）的秩和一个极大线性无关组，并将向量组中的其余向量由该极大线性无关组线性表出．

T

T

T2x3yz020．设线性方程组2xyz1，问：

xyz1(1)λ取何值时，方程组无解？

(2)λ取何值时，方程组有解？此时求出方程组的解．

00121．求矩阵A=010的全部特征值与特征向量． 1002222．用配方法化二次型f(x1,x2,x3)=2x12x24x1x38x2x3为标准形，并写出所用的可逆线性变换．

四、证明题（本题7分）

23．设向量组α1，α2线性无关，且β=clα1+c2α2，证明：当cl+c2≠1时，向量组β－α1,β－α2线性无关．

第四篇：线性代数02198自考2009年~2012年真题试题及答案(新)

2009年7月高等教育自学考试全国统一命题考试

线性代数试题

课程代码：02198 试卷说明：在本卷中，AT表示矩阵A的转置矩阵；A*表示A的伴随矩阵；R（A）表示矩阵A的秩；|A|表示A的行列式；E表示单位矩阵。

一、单项选择题（本大题共10小题，每小题2分，共20分）

在每小题列出的四个备选项中只有一个是符合题目要求的，请将其代码填写在题后的括号内。错选、多选或未选均无分。

1．设A，B，C为同阶方阵，下面矩阵的运算中不成立的是（）．．．A．（A+B）T=AT+BT C．A（B+C）=BA+CA a11a122．已知a21a22a31a32a132a11a23=3，那么a21a332a312a12a222a32B．|AB|=|A||B| D．（AB）T=BTAT 2a13a23=（）2a33A．-24 C．-6

B．-12 D．12 3．若矩阵A可逆，则下列等式成立的是（）

1*A．A=A B．|A|=0 |A|C．（A2）-1=（A-1）2

D．（3A）-1=3A-1

413124．若A=，B=23，C=021，则下列矩阵运算的结果为3×2的矩15231221阵的是（）A．ABC C．CBA

B．ACTBT D．CTBTAT

3线性无关，则（α2,α3,α4，其中α1，α2，α5．设有向量组A：α1,A．α1，α3线性无关

4线性相关）

B．α1，α2，α3，αD．α2，α3，α4线性无关

C．α1，α2，α3，α4线性无关

6．若四阶方阵的秩为3，则（）A．A为可逆阵

C．齐次方程组Ax=0只有零解

B．齐次方程组Ax=0有非零解 D．非齐次方程组Ax=b必有解

0207．已知方阵A与对角阵B=020相似，则A2=（）

020A．-64E B．-E

C．4E

8．下列矩阵是正交矩阵的是（）010A．010

001D．64E

1011110 B．

2011cosC．sinsin cosD．220221666106333 3339．二次型f=xTAx(A为实对称阵)正定的充要条件是（）A．A可逆

C．A的特征值之和大于0

B．|A|>0

D．A的特征值全部大于0

0k010．设矩阵A=0k2正定，则（）

024A．k＞0 C．k＞1

B．k≥0 D．k≥1

二、填空题（本大题共10小题，每小题2分，共20分）

请在每小题的空格中填上正确答案。错填、不填均无分。11．设A=（1，3，-1），B=（2，1），则ATB=__________.21012．若131=0，则k=__________.k2113．若ad≠bc，A=ab，则A-1=__________.cd14．已知A2-2A-8E=0，则（A+E）-1=__________.15．向量组α1=（1，1，0，2），α2=（1，0，1，0），α3=（0，1，-1，2）的秩为__________.16．两个向量α=（a,1,-1）和β=（b,-2,2）线性相关的充要条件是__________.x1x2017．方程组的基础解系为__________.xx03218．向量α=（3，2，t,1）β=(t,-1,2,1)正交，则t=__________.19．若矩阵A=10与矩阵B=3b相似，则x=__________.04ax22220．二次型f(x1,x2,x3)=x12x23x3x1x23x1x3对应的对称矩阵是__________.三、计算题（本大题共6小题，每小题9分，共54分）

11121．计算三阶行列式124.141622．已知A=23，B=31，C=011，D=120，矩阵X满足方程1021120101AX+BX=D-C，求X.23．设向量组为α1=（2，0，-1，3）

α2=（3，-2，1，-1）α3=（-5，6，-5，9）α4=（4，-4，3，-5）

求向量组的秩，并给出一个最大线性无关组.24．求λ取何值时，齐次方程组

(4)x13x20 4x1x305xxx0123有非零解？并在有非零解时求出方程组的结构式通解.16325．设矩阵A=053，求矩阵A的全部特征值和特征向量.40622226．用正交变换化二次型f(x1,x2,x3)=4x13x23x32x2x3为标准形，并求所用的正交矩阵P.四、证明题（本大题共1小题，6分）

27．若n阶方阵A的各列元素之和均为2，证明n维向量x=(1,1,…,1)T为AT的特征向量，并且相应的特征值为2.

2010年10月高等教育自学考试全国统一命题考试

线性代数试题

课程代码：02198 说明：在本卷中，AT表示矩阵A的转置矩阵，A*表示矩阵A的伴随矩阵，E是单位矩 阵，|A|表示方阵A的行列式，r（A)表示矩阵A的秩.一、单项选择题(本大题共10小题，每小题2分，共20分)在每小题列出的四个备选项中只有一个是符合题目要求的，请将其代码填写在题后的括号内。错选、多选或未选均无分。1．设矩阵A=11，B=（1，1）则AB=（）

A．0 B．（1，-1）C．11

D．1111

2．设A为3阶矩阵，|A|=1，则|-2AT|=（）

A．-8 B．-2 C．2 D．8 abcaabc3．设行列式D1=a1b1c1a1，D2=a1b1c1，则D1=（a2b2c2a2a2b2c2A．0 B．D2 C．2D2

D．3D2

*

124．设矩阵A的伴随矩阵A34

，则A-1=（）

A．143B．112221 234 C．121234 D．124231 5．设A,B均为n阶可逆矩阵，则必有（）

A．A+B可逆 B．AB可逆 C．A-B可逆 D．AB+ BA可逆1036．设A为3阶矩阵且r(A)=2，B=010，则r(AB)=（001A．0 B．1 C．2 D．3 7．设向量组α1=（1，2），α2=（0，2），β=（4，2），则（）））

A．αB．βC．βD．β线性无关

不能由α1，α2线性表示

可由α1，α2线性表示，但表示法不惟一 可由α1，α2线性表示，且表示法惟一 1，α2，β

2x1x2x308．设齐次线性方程组x1x2x30有非零解，则为（）

xxx0312A．-1 B．0 C．1 D．2 9．设A为3阶实对称矩阵，A的全部特征值为0，1，1，则齐次线性方程组(E-A)x=0的基础解系所含解向量的个数为（）A．0 B．1 C．2 D．3 10．二次型f(x1,x2,x3)=x12+x22+4x32-2tx2x3正定，则t满足（）A．-44

二、填空题（本大题共10小题，每小题2分，共20分）

请在每小题的空格中填上正确答案。错填、不填均无分。11．行列式0112的值为_________.1212．已知A=23，则|A|中 1219．与矩阵A=03相似的对角矩阵为_________.20．二次型f(x1，x2，x3)=2x1x2+2x1x3的秩为_________.三、计算题（本大题共6小题，每小题9分，共54分）0121．求行列式D=201012210102的值.1001012022．设矩阵A=100，B=210，求满足矩阵方程XA-B=2E的矩阵X.00100023．设向量组α1=(1,3,0,5)T，α2=(1,2,1,4)T，α3=(1,1,2,3)T，α4=(1,0,3,k)T，确定k的值，使向量组α1，α2，α3，α4的秩为2，并求该向量组的一个极大线性无关组.x12x2x3124．当数a为何值时，线性方程组2x1x2x31有无穷多解？并求出其通解.（要求用它

x1x2ax32的一个特解和导出组的基础解系表示）

25．已知3阶矩阵A的特征值为-1，1，2，设B=A2+2A-E，求（1）矩阵A的行列式及A的秩.（2）矩阵B的特征值及与B相似的对角矩阵.x12y12y2y326．求二次型f(x1,x2,x3)=-4x1x2+2x1x3+2x2x3经可逆线性变换x22y12y2y3所得的标准形.x2y3

3四、证明题（本题6分）

27．已知n阶矩阵A，B满足A2=A，B2=B及(A-B)2=A+B，证明AB=0.

第五篇：2011年4月自考英语二试题真题及答案

2011年4月高等教育自学考试全国统一命题考试

Ⅰ.Vocabulary and Structure(10points 1 point each)从下列各句四个选项中选出一个最佳答案，并在答题纸上将相应的字母涂黑。1.You may use bike________ you give it back to me tomorrow morning.A.unless

B.provided C.though D.because 2.He could not ________his tears on hearing that he was not admitted to the university A.hold up B.hold back C.get over D.get through 3.Good parents have the ability to communicate messages of love, trust, and self-worth ______their children A.on B.for

C.with

D.to 4.The teacher required that all errors should be _______eliminated before the students turn in their term paper.A.deported B.eliminated C.deprived D.implemented 5.A person is lucky if his career ________with his interest and hobby.A.concerns B.competes C.coinciders D.compares 6.Out sleep influences our mood.our mood, ________,affects our performance.A.in return B.in vain C.in short D.in turn 7.It was ________of you not to disturb us while we were sleeping.A.considerate B.considering C.considerable D.considered 8.I'd appreciate it very much if you could make some ________on my recent article at the conference.A.requests B.references C.remarks D.restrictions 9.He is ______absorbed in his own thoughts that he seems unaware of what's going on in the room.A.so

B.each

C.very

D.much 10.No sooner(选C)had I reached home than Michael arrived with Jane in his car.A.did I reach

B.I had reached

C.had I reached

D.I reached

Ⅱ.Cloze Test(10point,1 point each)下列短文中有十个空白，每个空白有四个选项，根据上下文要求选出最佳答案。并在答题纸上将相应的字母涂黑。

It is difficult to escape the influence of television.If you fit the statistical messages, by the age of 20 you will have been esposed___11___at least 20,000 hours of television.You can add 10,000 hours for each decade you have lived ___12___the age of 20.The only things Americans do more than them watch the television are work an sleep.Calculate for a moment what could be done with even a part of those ____13____.Five thousand, I am sold, are what a typical college undergraduate____14____working in a bachelor’s degree.In 10,000 hours you corded have learned several lauguages fluently, you could be reading Shakespeare in the ____15____,and you could have walked around the world the world and ____16____a book about these hours.The trouble with television is____17____it discourages concentrations.Almost anything interesting and rewarding in life____18____some constructive effort.The dultest , the least gifted of us can achieve things that seem remarkable to those____19____never concertrate on anything.But television encourages us not to make any___20___.It makes the time pass without gain.11.A.to

B.under 12.A.at

B.from 13.A.decades B.hours

14.A.spends B.casts 16.A.written B.write 17.A.what B.that

18.A.requests B.asks

19.A.what B.whose 20.A.money B.time

C.for

C.after

C.years C.takes C.wrote C.which C.orders C.which C.effort

D.at D.before D.things D.uses D.original D.writing D.why D.requires D.who D.skill 15.A.origin B.beginning C.source

Ⅲ.Reading Comprehension(30point,2points each)从下列每篇短文的问题后所给的四个选项中选出一个最佳答案，并在答题纸上将相应的字母涂黑。

Passage one Questions 21 to 25 are based on the following passage.Michael Stadtlander set Toronto’s restarant world on fire in the 1980s with his original food, and eastablished a national and international fame.In 1994,however,he decided to leave the city restaurant scene and seek a quiet life in the country, where he has been preparing meals on his farm two hours borth of Toronto and asks guests to bring their own wine.Not long ago, Stadlander was charged with selling liquer without a licence, He said when a guest asked for wine at a dinner in December, he provided two bottles from his private stock.But the customer turned out to be an undercover Ontario Provincial Police(OPP)officer who returned a few days later along with four armed officers to seize 83 bottles of wine,and to search the farmhouse for records.The punishment could run as high as $100,000 in fines and a year in jail.When Mr and Mrs Smisth arrived at the farm on a weekend befre Christmas, they said that it was their annual celebrstion and asked if they could get some wine.Nobuyo,Stadtlander’s wife,told them the policy was thar guests peovide their own..Given the occasion, though, Stadtlander agreed to provide a bottle of white wiee from own cellar,at cost,as a favor, Later, Mr Smith asked for a second bottle and a detailed receipt that included the price of thr wine.He turned out to be OPP Detective Paul Smith and ―his wife,‖an assistant who went as a witness.―I was shocked,‖Nobuyo Stadtlander recalls.‖How could thy do this to us when we did them a favor?We sold them our wine at our cost.‖.But Mr.Smith say,‖ No.they made $20.‖

Stadtlander is determined to prove his innocence and continue the business.―People who live in the area have been very supportive,‖ he says.‖And when this is over,I want my wine back.‖

21.Stadtlander was accused because________.A.be offered too much wine for a guest B.be asked guests to bring their own wine C.be charged too much for the wine offered D.be sold wise to his guests without a license 22.In the 1980s,Stadtlander________.A.threw his original restaurant menu into the fine B.made his restaurant well-known around the world C.sought for a quiet life in the central part of the city D.opened one resyaurant after another home and abroad.23.IT turned out that Mr.and Mrs.Smith were________.A.two farmers B.two witnesecs C.hasbend and wife D.police officers 24.Nobuy Studtlander claimed that in the wine case they made________.A.$1.20 B.no money C.$100,000 D.a lot of money 25.Which of the following could be used as evidence against Stadtlander? A.What Stadtlander’s wife said B.The second bottle of wine

C.The records from the farmhouse.D.A receipt with the wine price.Passage Two Questions26 to 30 are based on the following passage.A recent study shows that sixteen out of every 100 American couoles have violent confrontations of one sort or another during the course of a year.In six of these cases there is severe kicking, biting, punching or hitting with objects.Almost four of every 100 wives are seriously beaten by their husbands.three of every 100 children are kicked or punched by their parents.More than a third of all brothers and sisters severely attack each other.As expected, the incidence of violence is highest among the urban poor(many of them minoroties).blue-collar workers, people under 20 or without religious beliefs, families with a husband who is jobless and those with four to six children.But the study also showed that violence occurs among wealthy families as well.Indeed, the wife of a university president once quietly called Straus, one of the sociologists who conducted the study, to ask what she could do about her husband, who often beat her.Straus suggested seeking assistance from marriage advisors.Steaus and his colleagues found out that there are various root causes that give rise to such behavior.―The reason are mixed –psychological, sociological, situational,‖ says Straus.―The husband, for example, may feel under particular stress because be has been out of work too long.Violence may also be an echo of the past,‖ Straus explains.―When Mummy gives her two-year-old a slap（巴掌）for putting something dirty in his mouth, he is learning from infancy that those who love you hit you.‖ Another reason may be the worsening economic situations.―If we have a real economos decline, It’s going to get worse,‖ economic situations.―If we have a real economic declines,it’s going to get worse,‖ says Gelles, one of Straus‖ colleagues.These sociologists have no easy answer to violence in the American family.While they welcome such move as the opening of shelters for beaten wives and the establishment of a National Center for Child Abuse and Neglect, they belive that there must be more basic attack on vilolence,including the reduction of ―macho‖（大男子主义）themes on the television,the outlawing of physical punishment in schools and perhaps even the wlimination of death senernces.As Straus explains,―Volence is an acceptable solution to problems in American society,And that is how it is used in families.26.Which of the following statements is true? A.More than a third of brothers attack sisters in American families.B.Almost four percent of husbands are seriously beaten by their wives.C.Six out of one hundred couples experience severe domestic violence.D.Six percent of American couples have some kind of domestic violence.27.The example concerning a university president’s wife shows that_________ A.domestic violence is found in well-to do families as well B.the incidence of violence is highest among urban families C.marriage advisons’ assistance is the best solution to violence D.domestic vilence exists regardless of age, race,and social status.28.What Gelles says shows that________.A.vilence is responsible for the decline of economy B.violence may be a reflection of one’s past experience C.violence is related to the economic situation of the time D.violence is the best form of emotional release for a husband 29.The word ―outlawing‖ in line5 of the last paragraph means________.A.making something illegal B.freeing someone from prison C.learning something from law D.throwing someone out of court 30.According to the sociologists, it is impossible to curb domestic violence unless________.A.more shelters for beateb wives are opened B.children are protected by a National Center C.violence is not accepted as a solution to problems D.Americans get tougher with violent behaviors Passage Three Qestions 31to35are based on the following passage.American scientists are developing an ―intelligent‖ mabile phone capable of blocking incoming calls depending on the owner’s mood., Using ―context aware‖

Technology the ―Sensay‖ phone will monitor calls and send back polite messages saying the user may be contacted later.A research team at the Institute for Complex Engineering Systems at Carragie Mellon University in Pittsburg, Pennsylvania, are developing body temperature and electrical skin monitors to help the device understand the emotional state of its user.If the phone senses that the user is busy —for instance, involved in a conversation—it might block an incoming call and turn it onto voicemail.The phone would send back a text message saying the user is unavailable, but advising that id the matter is urgent the caller can try again in three minutes.If a call from the same person came in again, the phone would put it through.The researchers are interested in four basic different states—busy and not to be interrupted, physically active, idle,and ―normal.‖ Most people are said to change between these states,an average of 6 to 12 times a day.Professor Asim Smailagic, a leading member of the Carnegic Mellon team, told The Engineer magazine.‖Today’s computers are pretty dumb compared with the device.We got to work at the beginning of May and since then have been improving it.The next stage is to make it smarter, adding various intellingence systems so it can learn about the user.The phone also employs four primary sensons—two microphones to pick up conversations and monitor local noise.a light detector and an acceleromster（加速度计）.The light sensor shows if the phone is being carried in a bag or pocket ,while the accelerometer determines whether the user is walking,running or standing still, In the sensor box, phone is being carried in a bag or pocket, while the accelerometer determites whether the user is walking,running or standing still,In the future,the sensor box,phone and personal organizer will be combined into one device.‖

31.According to the passage, the ―Sensay‖ phone is capable of ________.A.showing body temperature B.blocking unwanted calls C.detecting the owner’s mood D.sensing the caller’s mood

32.The second paragraph mainly tells us ________.A.where the intelligent phone is developed B.how the new phone is going to function C.whether the researchers failed in their experiment D.why the researchers failed in their experiment 33.Accorfing to Professor Smailagic, the detector will show________.A.where the mobile phone is B.where the mobile phone user is C.whether the mobile phone user is busy D.whether the mobile phone is within reach 34.the word ―it‖ in line 5 of paragraph 4 refers to________.A.the phone B.the monitor C.the computer D.the light sensor 35.The best title for this passage is ________.A.How to Tell a Person’s Mood B.How to Block Incoming Calls C.Sensay Your Personal Organizer D.Sensay The Futuer Mobile Phone

Ⅳ.Word Spelling(10points for two words)将下列汉语单词译成英语，每个单词的词类、首字母及字母数目均已给出，请将完整的单词写在答题纸上。

36.乐意地，容易地adv.r________ 37.前景n.p________ 38.永恒的,不断的a.________ 39.有益的，有好处的a.b________ 40.进口n.i________ 41.预报n.f________ 42.元素，成分n.e________ 43.缩短，减少n.s________ 44.诚实n.h________ 45.有规律的a.r________ 46.改变，使不同v.v________ 47.坦率地adv.F________ 48.减轻，救济n.r________ 49.和谐，融洽n.h________ 50.赔偿v.c________ 51.有效率的a.________ 52.真诚的a.s________ 53.建设v.e________ 54.市长n.m________ 55.好奇心n.e________

Ⅴ.Word Form(10point each)将括号内的各词变为适当的形式填入空白，答案写在答题纸上。

56.It will be interesting to see what’s going to happen at the next________(elect).57.For those involved, the scandal has been a very________(pain)experience.58.________(safe)needs to be improved on the railways in that country.59.The present crisis bears some ________(similar)to the oil erisis of the 1970s.60.His mother becomes disappointed whenever he fails to live up to her________(expect).61.Being a ________(high)motivated lauguage learner, he takes every opportunity to improve his English.62.Our most ________(success)product is based on a very simple idea.63.local people are questioning the ________(wise)of spending so much money on the new road.64.Don’t worry.This is the ________(desire)weight range for you health.65.This scientist was awarded with several titles for his________(accomplish)in biology.Ⅵ.Translation from Chinese into English(15points,3points each)将下列各句译成英语并将答案写在答题纸上。66.尽管有困难，他们最终还是把工作完成了。67.这次会议在促进两国友谊方面起了重要作用。68.他长久以来梦想着称为一名足球运动员。

69.在过去的二十年中，用于科学实验的动物数量大大减少了。70.这次英语考试比我们预想的要容易得多。

Ⅶ.Translation from English into Chinese(15points)将下列短文译成汉语并将答案写在答题纸上。

Shy people are likely to be passive and easily influenced by other.They are very sensitive t criticism.They also find it difficult to be pleased by praises because they believe they are unworthy of them.A shy person may respond to a praise with a statement like this:―You’re saying that to make me feel good.I know it’s not me.‖It is clear that, while self-awareness is a good quality.overdoing it is harmful.Can shyness be completely gotten rid of ,or at least reduced? Fortunately ,people can overcome shyness with determined and patient effort in building self-confidence.