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11121-18191_线代_期末1
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2011 ~ 2012 学年第一学期期末考试试卷
《线性代数及其应用》
Æ
}
( 共 15 分,
x
8 #
3
)
1.
; "
9
f(X) = X
T
AX
'
K
[
`
%
T
X = SY
S
*
h :
y
2
1
+2y
2
2
+3y
2
3
,
U
; "
9
g(X) = X
T
A
−1
X
'
K
[
`
%
T
X = SY
S
*
h :
.
2
.
λ = −1
n
b
8 g
X
A
/
k
e " Y ^
,
U
g
X
−E − A
/
b
=
.
3
.
;
+ 4 <
k ℄
V
/
4 <
Æ T
σ
T
V
α
1
, α
2
2
/ g
X *
a
11
a
12
a
21
a
22
,
U
σ
T
V
α
2
, α
1
2
/ g
X *
.
4
.
D ℄
g
X
A =
0 5 3 0 0
0 2 1 0 0
5 0 0 0 0
0 0 0 3 2
0 0 0 5 4
,
U
A
−1
=
.
5
.
λ
1
N
λ
2
g
X
A
/ p D
(
/
" Y ^
,
α
1
N
α
2
* 5 J
/
"
Y 7
q
,
U
α
2
, A(α
1
+ 3α
2
)
4 < 5
H /
&
^
.
t
(
F
15
,
x
8 #
3
)
1.
2
r
( )
G
R
4
/
j
k ℄
.
(A) W
1
= {(x
1
, x
2
, x
3
, x
4
)
x
1
+ x
2
= x
3
+ x
4
}
(B) W
2
= {(x
1
, x
2
, x
3
, x
4
)
x
1
+ x
2
= 1}
(C) W
3
= {(x
1
, x
2
, x
3
, x
4
)
x
1
∈ Z}
(D) W
4
= {(x
1
, x
2
, x
3
, x
4
)
x
2
1
= x
2
}
2.
7
q
n
(I)α
1
, α
2
, . . . , α
s
j
L
(II)β
1
, β
2
, . . . , β
s
4 <
,
2
r
<
[
/
( ).
(A)
(I)
4 < -
H
,
U
(II)
4 < 5
H
(B)
(I)
4 < -
H
,
U
(II)
4 < -
H
(C)
(II)
4 < -
H
,
U
(I)
4 < 5
H
(D)
(II)
4 < -
H
,
U
(I)
4 < -
H
3.
g
X
A, B ∈ P
n×n
*
8 g
X
,
A
Q
B
P
(
,
U
( ).
(A) A
Q
B
M 5 ( " Y ^
(B) A
Q
B
M 5 (
/
b
(C) A
Q
B
M 5 ( " Y 7
q
(D) A
Q
B
M 5 (
/
;
r
4.
A
*
j ~ g
X
,A
/ 2
1
;
/
2
\ , 2
3
;
. g
X
B,
U
( ).
(A) A
∗
/ 2
1
;
/
-2
\ , 2
3
;
.
B
∗
1
(B) A
∗
/ 2
3
;
/
-2
\ , 2
1
;
.
B
∗
(C) A
∗
/ 2
1
r /
-2
\ , 2
3
r .
B
∗
(D) A
∗
/ 2
3
r /
-2
\ , 2
1
r .
B
∗
5.
2
r ; "
9 [
3 ; "
9
( ).
(A) f
1
= (x
1
− x
2
)
2
+ (x
2
− x
3
)
2
+ (x
3
− x
1
)
2
(B) f
2
= (x
1
+ x
2
)
2
+ (x
2
− x
3
)
2
+ (x
3
+ x
1
)
2
(C) f
3
= (x
1
+ x
2
)
2
+ (x
2
+ x
3
)
2
+ (x
3
− x
4
)
2
+ (x
4
− x
1
)
2
(D) f
4
= (x
1
+ x
2
)
2
+ (x
2
+ x
3
)
2
+ (x
3
+ x
4
)
2
+ (x
4
− x
1
)
2
(8
)
D ℄
b >
X
A
/
" Y ^ *
2, −2, 1,
N
b >
X
B
Q
A
5
,
(1)B
/
g
X
B
∗
/
" Y ^
(2)B
2
− 2B
/
;
r
.
(8
)
T
R[x]
3
3 V
(I) : 1, x, x
2
, x
3
N
(II) : 1 + x + x
2
, 1 + x, 1, 1 +
x + x
2
+ x
3
.
(1)
L
V
(I)
, V
(II)
/ K 6 g
X
(2)
f(x) = 1 + 2x + 3x
2
+ 4x
3
T
V
(II)
2
/
q
(3)
J
g(x) ∈ R[x]
3
T
V
(II)
2
/
q
*
[1, 2, 3, 4]
T
,
! T
V
(I)
2
/
q
.
/
(10
)
D ℄
ξ =
1
1
−1
b g
X
A =
2 −1 2
5 a 3
−1 b −2
/
" Y 7
q
,(1)
a, b
/
^
X
ξ
8
J
/
" Y ^
(2)
g
X
A
} A 8 a S
z m
L
.
u
(12
)
D ℄
α
1
= [1, −1, 0]
T
, α
2
= [−1, 0, 1]
T
, α
3
= [0, 1, 1]
T
R
3
/
B
D V
,σ
R
3
/
B
D
4 <
Æ T
,
σ
T
V
α
1
, α
2
, α
3
2
/ g
X *
A =
1 2 0
0 1 2
1 −2 0
.
(1)
\
z
β
1
= α
1
, β
2
= −α
1
+ α
2
, β
3
= −α
2
+ α
3
A
R
3
/
B
D V
(2)
σ
T
β
1
, β
2
, β
3
2
/ g
X
(3)
α = α
1
+ 2α
2
− 3α
3
,
σ(α)
T
V
α
1
, α
2
, α
3
2
/
q
.
(12
)
D ℄
?
"
4 <
>
n
x
1
+ x
2
+ x
3
+ x
4
= −1
4x
1
+ 3x
2
+ 5x
3
− x
4
= −1
ax
1
+ x
2
+ 3x
3
+ bx
4
= 1
M
D
4 < -
H
.
(1)
>
n 1
g
X
A
/
b
(2)
a, b
/
^
X >
n
/
'
(
K
+
n
/ V
1
).
2
(16
) (1)
B
D
[
`
%
T
,
_
; "
9
f(x
1
, x
2
, x
3
) = 3x
2
1
+3x
2
2
+6x
2
3
+
8x
1
x
2
− 4x
1
x
3
+ 4x
2
x
3
S
*
h :
(2)
; "
9
/
[
I
< _
N B M
.
e
(4
)
n
b
>
X
A
v
m
A
2
−4A + 3E = O,
\
z
(2E −A)
T
(2E −A)
* [
3 g
X
.
3
2011-2012(
)
F
A
C
1. y
2
1
+
1
2
y
2
2
+
1
3
y
2
3
2. n − k 3.
a
22
a
21
a
12
a
11
4.
0 0
1
5
0 0
−1 3 0 0 0
2 −5 0 0 0
0 0 0 2 −1
0 0 0 −
5
2
3
2
5. λ
1
= 0
z
z
!
C
ABBDD
N
(1)
Q
B
A
f 4
,
;
B
A
5
f I
D
A
/ 5
,
5
B
D
A
/
5
Q
2, −2, 1.
?
C
|B| = 2 × (−2) × 1 = −4,
B
∗
D
A
/ 5
Q
µ
1
=
|B|
2
= −2, µ
2
=
|B|
−2
= 2, µ
3
=
|B|
1
= −4.
(2)
W
(1)
3
,B
D
A
/ 5
Q
2, −2, 1.
λ
Q
B
D
A
/ 5
,
"
f(B) = B
2
− 2B
D
A
/ 5
Q
f(λ) = λ
2
−2λ.
5
f(B)
A
/ 5
Q
f(2), f(−2), f(1),
Q
0, 8, −1.
'
|f(B)| = | B
2
− 2B| = 0 × 8 × (−1) = 0.
W U
B
A
f 4
,
;
f(B)
f(A)
f 4
,
'
|f(B)| = |f(A)|.
λ
Q
A
D
A
/ 5
,
"
f(A) = A
2
− 2A
D
A
/ 5
Q
f(λ) = λ
2
− 2λ.
5
f(A)
A
/ 5
Q
f(2), f(−2), f(1),
Q
0, 8, −1.
'
|f(A)| = |A
2
− 2A| = 0 × 8 × (−1) = 0.
7 S
|f(B)| = | f ( A)| = 0.
\
(1)
Q
(1 + x + x
2
, 1 + x, 1, 1 + x + x
2
+ x
3
) = (1, x, x
2
, x
3
)
1 1 1 1
1 1 0 1
1 0 0 1
0 0 0 1
,
;
Æ
(I)
A
Æ
(II)
D { M
3
.
S =
1 1 1 1
1 1 0 1
1 0 0 1
0 0 0 1
.
(2)
3
f(x) = 1 + 2x + 3x
2
+ 4x
3
?
C
f(x)
Æ
(I)
`
D
Q
Q
X =
1
1
2
3
4
.
7 S
f(x)
Æ
(II)
`
D
Q
Q
Y = S
−1
X.
[S
.
.
.X] =
1 1 1 1 1
1 1 0 1 2
1 0 0 1 3
0 0 0 1 4
→
1 0 0 0 −1
0 1 0 0 −1
0 0 1 0 −1
0 0 0 1 4
,
;
Y =
−1
−1
−1
4
.
(3)
g(x)
Æ
(II)
`
D
Q
Q
Y =
1
2
3
4
,
?
C
g(x)
Æ
(I)
`
D
Q
Q
X = SY =
1 1 1 1
1 1 0 1
1 0 0 1
0 0 0 1
1
2
3
4
=
10
7
5
4
.
o
(1)
λ
'
A
D
f
Æ
A
/
k
P
ξ
D
A
/ 5
,
"
Aξ = λξ,
2 −1 2
5 a 3
−1 b −2
1
1
−1
= λ
1
1
−1
.
0
G
C
−1 = λ,
2 + a = λ,
1 + b = −λ,
)
C
λ = −1, a = −3, b = 0.
(2)
(1)
C
A =
2 −1 2
5 −3 3
−1 0 −2
.
A
D
A
/
R
j $ Q
|λE−A| =
λ − 2 1 −2
−5 λ + 3 −3
1 0 λ + 2
c
1
+c
2
−c
3
=======
λ + 1 1 −2
λ + 1 λ + 3 −3
−λ − 1 0 λ + 2
=
λ + 1 1 −2
0 λ + 2 −1
0 1 λ
= (λ + 1)
3
.
2
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