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同济大学线性代数答案详解.pdf
1
SK1-1
1. ˜A = (A, b)
¥
A =
1
3
5
3
5
ˇdk
1
1
2
1 −1 1
2
2
3 −1
4
1
, b =
1
x1
=
1
⁄5§|/“ x1 + x2 + 2x3 + 2x4 = 1,
2x1 + x2 + 3x3 − x4 = 3,
x1 − x2 + x3 + 4x4 = 5.
1
2
2
3 −1
4
1
2
1 −1 1
x2
x3
2.
1 2
a b
+
x y
3 4
1 + x 2 + y
a + 3 b + 4
=
=
3 −4
7
1
3 −4
1
7
1 + x = 3
2 + y = −4
a + 3 = 7
b + 4 = 1
⁄–kx = 2, y = −6, a = 4, b = −3.
3.
1
(1)
(2)
A + 2B =
=
=
3A − B = 3
+ 2
2
1 −2
3 −1
2
3 + 2 −1 + 10 2 + 2
2 − 4
−2
1 − 2
5
1
1
−2 −1 0
9
5
4
−2 −1 −2
5
1
1
−2 −1 0
−
2
1 −2
3 −1
2
9 − 1 −3 − 5 6 − 1
−6
6 + 2
8 −8
8
5
4 −6
3 + 1
=
=
ABT =
AT B =
=
=
3 −1
2
1 −2
2
1 −2
5 −1
1
0
3 − 5 + 2 −6 + 1
2 + 5 − 2 −4 − 1
0 −5
5 −5
=
=
3
3 − 4
−1
2
−1
1
2 −2
5
1
1
−2 −1 0
15 − 2
3
−1 − 2 −5 − 1 −1
2
2 + 4
10 + 2
13
3
−3 −6 −1
2
6
12
2
−
1 −3
2
1
2
0
3 −1
2
0
3 −1
+
1 −3
1
2
3 −3
1
4
−1 −3
−3 + 6 −9 − 9
−4 − 2 −12 + 3
3 −18
−6
9
−2
3
(A + B)(A − B) =
=
=
=
4.
5.
A2 + 3A − 2B =
+ 3
0 2 1
0 1 2
0 2 5
1 0 0
− 2
1 0 0
−2
+
3 0 0
0 5 2
0 3 6
0 6 3
4 + 6 − 10
5 + 3 − 4
0
0
0
0
0 −4 −10
0 −10 −4
5 + 3 − 4
4 + 6 − 10
=
6 −4 2
9 −6 3
3 −2 1
= 2 − 3 + 2 = 1
=
=
=
0
0
0 4 5
0 5 4
0 2 1
0 1 2
2
1 0 0
+
1 0 0
1 + 3 − 2
2 0 0
1
0 4 0
0 0 4
2
3
2 3 1
3 −2 1
1
−1
2
3
6. (1)
(2)
1
1 −1
1
2
1
−1 1
0
(3)
x y z
(4)
K
k > 2,Bk = 0
=
x + y − z x + 2y + z −x + y
x
y
z
x
y
z
= (x2 + xy − xz) + (xy + 2y2 + yz) + (−xz + yz)
= x2 + 2y2 + 2xy − 2xz + 2yz
B =
0 0 0
0 0 1
0 1 0
1 1 0
0 1 1
0 0 1
A = I + B =
An = (I + B)n = I + nB +
B2 + 0
n(n − 1)
0 1 0
1 0 0
+
1 n n(n−1)
0 0 1
0 1
2
n
0 0
1
2
0 0 n
0 n 0
0 0
0
=
=
+
0 0
0 0
0 0
n(n−1)
2
0
0
7. ABvAB = BA,
B =
a b
c d
4
a b
c d
=
=
1 1
0 1
a b
c d
1 1
0 1
a + c
b + d
d
a a + b
c
c + d
c
c = 0
a = d
,
a + c = a
b + d = a + b
c = c
⇒
d = c + d
B =
a b
0 a
=k
ˇd
A
¥a, b?¿Œ.
8.
y†
(AT + A)T = A + AT = AT + A
(AT − A)T = A − AT = −(AT − A)
⁄–AT + A·Ø¡,AT − A·Ø¡.
9.
SK1-2
1.
An =
λ1
λ2
n
. . .
λn
3 1
2 1
0 0
0 0
A =
0 0
0 0
1 4
2 5
5
=
=
λn
1
λn
2
. . .
λn
n
A1 O
O A2
B =
−1
0
0 −1
3
0
1 −1
1 0
0 1
2 1
1 2
C =
2 4
1 3
0 0
0 0
0 0
0 0
3 1
0 2
=
=
B1 B2
B3 B4
C1 O
O C2
AC =
=
C1 O
O C2
A1 O
O A2
A1C1 =
=
2 4
1 3
3 1
2 1
7 15
5 11
, A2C2 =
A1C1
O
O
A2C2
=
3 1
0 2
1 4
2 5
3
9
6 12
⁄–
AC =
AB − BT A =
=
=
A1 O
O A2
A1B1 A1B2
A2B3 A2B4
A1B1 − BT
A2B3 − BT
7 15 0
5 11 0
3
0
0
0
0
0
0
9
6 12
−
B1 B2
B3 B4
−
BT
BT
1 A1 BT
2 A1 BT
3 A2
4 A2
1 A1 A1B2 − BT
2 A1 A2B4 − BT
3 A2
4 A2
6
A1 O
O A2
BT
BT
1 BT
3
2 BT
4
A1B1 − BT
1 A1 =
=
=
−1
−3 −1
−
0
0 −1
−1
−
−2 −1
3 1
−3 −1
−2 −1
2 1
0 0
0
0 −1
3 1
2 1
1 4
2 5
3
1
0 −1
−
1 0
0 1
5
17
−2 −5
3 1
2 1
1 0
0 1
−
A1B2 − BT
3 A2 =
=
=
A2B3 − BT
2 A1 =
=
=
A2B4 − BT
4 A2 =
=
6
4
0 0
−
2 1
3 1
1 4
2 1
3 1
−2 −16
2 5
7 −4
11 −5
4 −5
9 −6
1 4
2 5
6
9
9 12
2 −4
4 −2
3
0
1 −1
−
3 1
2 1
−
2 1
1 2
−
4 13
5 14
1 4
2 5
2 1
1 2
=
⁄–
AB − BT A =
0
0 −2 −16
4
6
0
0
2 −4
4 −5
4 −2
9 −6
7
B =
Λ1 Λ2 Λ3
λ2
a21
a31
a12
a22
a32
a13
a23
a33
=
a11
λ1
=
λ1a11 λ2a12 λ3a13
λ1a21 λ2a22 λ3a23
λ1a31 λ2a32 λ3a33
= A1Λ1 + A2Λ2 + A3Λ3
A1 A2 A3
λ3
=
AB =
···
A1 A2 A3
Λ1 Λ2 Λ3
2.
A =
en−1
e1
e2
en
3. A =
duAekuA1k, ⁄–k = 1,Aek = ek−1,¿Ae1 = en.
i > k,Akei = Ak−1ei−1 = ··· = Aei−k+1 = ei−k,
i ≤ k,Akei = Ak−i+1Ai−1ei = Ak−i+1e1 = Ak−iAe1 = Ak−ien = en−k+i.
ˇd
Ak = AkE = Ak
···
en
··· Aken
en−k
=
=
=
···
e1
ek
ek+1
Ake1
en−k+1
··· Akek Akek+1
···
···
en
e1
0 En−k
Ek
0
.
4.
SK1-3
Dk =
A1 O ··· O
··· O
O A2
...
...
...
. . .
O O ··· As
1 O ··· O
Ak
··· O
O Ak
2
...
...
...
. . .
O O ··· Ak
s
k
=
8
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