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2013年10月《线性代数(经管类)04184》试卷及标准答案

全国2013年10月高等教育自学考试

线性代数试题

课程代码：02198

选择题部分

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11221a b a b =，112

22a c a c =-，则1112

b c a b c +=+ B

A.－3

B.－

C.1

D.3

2．设4阶矩阵A 的元素均为3，则r(A )= A A.1 C.3

D.4

3．设A 为2阶可逆矩阵，若1

1325-??= ?

，则A *

= A.1325--??

?--??

B.1325?? ???

C.5321-?? ?-??

D.532

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

B. r =n 时，Ax =0必有非零解

C. r

D. r

5．二次型f (x l ，x 2,x 3)=2

123132323812x x x x x x x ++-+的矩阵为

A. 1080

2128123-?? ?

? ?-?? B. 1080212003-??

? ? ???

C. 104026463-?? ? ? ?-??

D. 140426063-??

?- ? ???

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 |=__16____．

7．设A 为2阶矩阵，将A 的第1行加到第2行得到B ，若B =1234??

???

，则A =_1 2

2 2.

8．设矩阵A =11122122a a a a ?? ???，B=11

11211222a a a a a a ??

?++??

，且r(A )=1，则r （B ）=__1____.

9．设向量α=(1，0，1)T ，β=(3，5，1)T ，则β－2α=_(1 5 -1)_______． 10．设向量α=(3，－4)T ，则α的长度||α||=_5_____．

11．若向量αl =(1，k )T ，α2=(－1,1)T 线性无关，则数k 的取值必满足_K 不等于-1____. 12．齐次线性方程组x l +x 2+x 3=0的基础解系中所含解向量的个数为__2____．

13．已知矩阵A =122212221?? ? ? ???与对角矩阵D =10001000a -??

- ? ???

相似，则数a =_5___

14．设3阶矩阵A 的特征值为－1，0，2，则|A |=_0____．

15．已知二次型f (x 1,x 2,

x 3)=222

123

x x tx ++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）

16．计算行列式D=

a b c a a

b b a

c b

c c c a b

.＝（a+b+c）3

步骤：1.将第二行和第三行全部加到第一行，使第一行全部变成：a+b+c,

2.提取公因子a+b+c, 第一行变成11 1

3.第一行的-2b倍加到第二行，变成：0－（a+b+c）0

4 第一行的-2c倍加到第三行，变成：00－（a+b+c）

5 行列式的值为：对角线元素之积，（a+b+c）3

17．已知向量α=（1，2，k），β=

1,,

，且βαT=3，A=αTβ，求

(1)数k的值；

(2)A10．

解：（１）：βαT=3矩阵的乘法：1+1+k/3＝3Ｋ=3

(2):αTβ不能乘吧？这个不会解，请老师帮忙

18．已知矩阵A=

123

231

340

，B=

-??

，求矩阵X，使得AX=B.

解：Ｘ＝Ａ-1*Ｂ

先求出Ａ-1利用初等行变换，将前面部分转为简化梯阶形式，后面部分就成为了Ａ的逆矩阵。

-4-127

-39-5

1-2 1

所以Ｘ＝Ａ-1*Ｂ

19．求向量组α1=(1,0,2,0)T, α2=(－1,－1,－2,0)T, α3=(－3,4,－4,l)T, α4=（－6,14,－6,3）T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．

解：就是初等行变换，化为简化阶梯形式，

100-1

010-2

001 3

0000

秩为：3

α1α2α3为一个极大无关组

α1－α4=0

α2－２α4=0

α3+3α4=0

线性表出：α4

＝

α1-2α2+3α3

20．设线性方程组

230

x y z

x y zλ

++=

?++=

，问：

(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．解：方程组无解，就是系数矩阵的秩不等于增广矩阵的秩λ＝1/2时，方程组无解

(2)有解，就是系数矩阵的秩等于增广矩阵的秩，

Λ不等于1/2时，方程组有解，是唯一解，

其解为：Ｚ=3/（4λ-2），y=-1/2, x=

21．求矩阵A =001010100?? ?

? ???

的全部特征值与特征向量．

解：λＥ-Ａ=0 代入求特征值 λ１＝λ２＝1 λ＝-1

求特征向量，将λ１＝λ２＝1 λ＝-1代入计算当λ＝-1时，矩阵为： -1 0 -1 0 -2 0 0 0 0

特征向量为（-1 -1/2 1）

当λ１＝λ２＝1时，矩阵为： 1 0 -1 0 0 0 0 0 0

特征向量为（-1 0 1）

22．用配方法化二次型f (x 1,x 2,x 3)=22

1213232248x x x x x x --+为标准形，并写出所用的可逆

线性变换．

解：标准形：只有平方项，没有交叉项的为标准形

f (x 1,x 2,x 3)= (2Ｘ12 – 4Ｘ1Ｘ3 –Ｘ32 ) + (Ｘ32 + 8Ｘ2Ｘ3 + 4Ｘ22 ) - 6Ｘ22 = (2Ｘ12

-Ｘ32)2

+ (Ｘ32

+2Ｘ22)2

- 6Ｘ22

可逆线性变换: Ｐ-1ＤＰ ?

四、证明题（本题7分）

23．设向量组α1，α2线性无关，且β=c lα1+c2α2，证明：当c l+c2≠1时，向量组β－α1,β－α2线性无关．

证明向量组β－α1,β－α2线性无关．用反证法，如果向量组β－α1,β－α2线性相关

因为：β=c lα1+c2α2

β－α1=c lα1+c2α2－α1

c lα1+c2α2－α２

β－α

２=

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