Mô tả:
LỜI GIẢI MỘT SỐ BÀI TẬP
TOÁN CAO CẤP 2
Lời giải một số bài tập trong tài liệu này dùng để tham khảo. Có một số bài tập do
một số sinh viên giải. Khi học, sinh viên cần lựa chọn những phương pháp phù hợp và
đơn giản hơn. Chúc anh chị em sinh viên học tập tốt
BÀI TẬP VỀ HẠNG CỦA MA TRẬN
Bài 1:
Tính hạng của ma trận:
2 −4 3 1 0
1 −2 1 −4 2 η1↔
η2
A
=
→
1)
0 1 −1 3 1
1 −7 4 −4 5
1 −2 1 −4
2 −4 3 1
0 1 −1 3
1 −7 4 −4
2
0
1
5
→
1 −2 1 −4 2
0 0 1 9 −4 η
2↔ η3
→
0 1 −1 3 1
0 −5 3 0
3
1 −2 1 −4 2
0 1 −1 3 1
0 0 1 9 −4
0 −5 3 0
3
η2(5)+η4
→
1 −2 1 −4 2
0 1 −1 3 1 η3( 2)+ η4
→
0 0 1 9 −4
0 0 −2 15 8
1 −2 1 −4 2
0 1 −1 3 1
0 0 1 9 −4
0 0 0 33 0
h1(−2)+η2
η1(−1)+η4
⇒ ρ( Α) = 4
2)
A=
0 2 −4
−1 −4 5
1↔ η2
3 1
7 η
→
0 5 −10
2 3
0
1
η2
2
→
−1 −4
5
0
1
−2
0 −11 22
0
5 −10
0 −5 10
−1 −4 5
0 2 −4 η1(3)+ η3
η1( 2 )+ η4
3 1
7
→
0 5 −10
2 3
0
η2(11)+ η3
ηη22((5−5)+)+η5η4
→
−1 −4 5
0 1 −2
0 0 0
0 0 0
0 0 0
1
−1 −4
5
0
2
−4
0 −11 22
0
5 −10
0 −5 10
⇒ ρ( Α) = 2
2 −1 3 −2 4 η1(−2)+η2 2 −1 3 −2 4
1)+η3
2) A = 4 −2 5 1 7 η1(−
→ 0 0 −1 5 −1
2 −1 1 8 2
0 0 −2 10 −2
2 −1 3 −2 4
h2(2)+η3
→ 0 0 −1 5 −1 ⇒ ρ( Α) = 2
0 0 0 0 0
3)
A=
1 3 5 −1
2 −1 −5 4
5 1 1 7
7 7 9 −1
→
η1( −2)+η2
( −5)+η3
η1
η1( −7 )+η4
→
1 3
5 −1
0 −7 −15 6
0 0
1
0
0 0
4 −6
1
η3
6
1 3
5 −1
( −2 )+η3
η2
0 −7 −15 6 η2( −2)+η4
→
0 −14 −24 12
0 −14 −26 6
η4 −4 +η4
( )
→
1 3
5 −1
0 −7 −15 6
0 0
1
0
0 0
0 −6
1 3
5 −1
0 −7 −15 6
0 0
6
0
0 0
4 −6
⇒ ρ( Α) = 4
4)
A=
3
5
1
7
−1 3 2
−3 2 3
−3 −5 0
−5 1 4
5
4
7
1
η3
η1↔
→
1
5
3
7
−3 −5 0 7 η1( −5)+η2
)+η3
η1( −3
−3 2 3 4 η1
( −7 )+η4
→
−1 3 2 5
−5 1 4 1
→
1 −3 −5 0 7
( −3)+η3
η2
0 4 9 1 −8 η2
( −4 )+η4
→
0 12 27 3 −31
0 16 36 4 −48
→
5)
1 −3 −5 0 7
0 4 9 1 −8 ⇒ ρ Α = 3
( )
0 0 0 0 −7
0 0 0 0 0
1
η3 ↔ η2
2
16
+ η4
7
η3 −
2
1 −3 −5 0 7
0 4 9 1 −8
0 0 0 0 −7
0 0 0 0 −16
1 −3 −5 0 7
0 12 27 3 −31
0 8 18 2 −16
0 16 36 4 −48
A=
2 2 1 5 −1
1 0 4 −2 1
2 1 5 −2 1 η1↔ η2
→
−1 −2 2 −6 1
−3 −1 −8 1 −1
1 2 −3 7 −2
1 0 4 −2 1
2 2 1 5 −1
2 1 5 −2 1
−1 −2 2 −6 1
−3 −1 −8 1 −1
1 2 −3 7 −2
η1( −2)+ η2
η1( −2)+ η3
η1+ η4
η
→
1(3)+ η5
η1( −1)+ η6
1 0
4 −2 1
0 2 −7 9 −3
0 1 −3 2 −1 η
2↔ η3
→
0 −2 6 −8 2
0 −1 4 −5 2
0 2 −7 9 −3
η2( −2)+ η3
η2( 2)+ η4
η
→
2+ η5
η2(−2)+ η6
1
0
0
0
0
0
0 4 −2 1
1 −3 2 −1
0 −1 3 −1
0 0 −4 0
0 1 −3 1
0 −1 3 −1
η3+ η5
→
η3( −1)+ η6
1 0 4 −2 1
0 1 −3 2 −1
0 2 −7 9 −3
0 −2 6 −8 2
0 −1 4 −5 2
0 2 −7 9 −3
1
0
0
0
0
0
0 4 −2 1
1 −3 2 −1
0 −1 3 −1 ⇒ ρ Α = 4
( )
0 0 −4 0
0 0 0 0
0 0 0 0
6)
A=
1 −1 2 3
4
2 1 −1 2
0 η1(−2)+ η2
1+ η3
−1 2 1 1
3 ηη
→
1(−1)+ η4
η1(−3)+ η5
1 5 −8 −5 −12
3 −7 8 9 13
2↔ η3
η
→
1 −1 2
3
4
0 3 −5 −4 −8
0 1
1
3
7
0 6 −10 −8 −16
0 −4 2
0
1
1 −1 2
3
4
0 1
1
3
7 η2(−3)+ η3
2(−6)+ η4
0 3 −5 −4 −8 η
→
η2( 4)+ η5
0 6 −10 −8 −16
0 −4 2
0
1
h3( −1)+ η4
3+ η5
η
→
1 −1 2
3
4
0 1 1
3
7
5( −4)+ η3
0 0 −8 −13 −29 η
→
0 0 0
0
0
0 0 −2 −1
0
3
1 −1 2
3
4
0 1
1
3
7
0 0 −8 −13 −29
0 0 −16 −26 −58
0 0
6
12
29
1 −1 2 3
4
0 1 1 3
7
0 0 0 −9 −29
0 0 0 0
0
0 0 −2 −1 0
η4↔ η3
h5↔
→
1 −1 2
3
4
0 1 1
3
7
0 0 −2 −1 0 ⇒ ρ( Α) = 4
0 0 0 −9 −29
0 0 0 0
0
7)
A=
−3 2 −7 8
−1 0
5 −8 η1↔ η2
→
4 −2 2 0
1 0
3 7
2(−1)+ η3
η
→
−1
0
0
0
−1 0
5 −8
η1(−3)+ η2
−3 2 −7 8 η1(4)+ η3
→
η1+ η4
4 −2 2 0
1 0
3 7
0 5 −8
2 −22 32 η
3↔ η4
→
0 0
0
0 8
−1
−1
0
0
0
−1 0
5
−8
0 2 −22 32
0 −2 22 −32
0 0
8
−1
0 5 −8
2 −22 32 ⇒ ρ( Α) = 3
0 8 −1
0 0
0
8)
A=
1
−1 3 3 −4
η1( 4)+ η2
4 −7 −2 1 η
1( −3)+ η3
→
η1( −2)+ η4
−3 5 1 0
−2 3 0 1
2+ η3
ηη
→
2+ η4
−1
0
0
0
3
1
0
0
η2
5
−1 3 3 −4
η3 1
0 5 10 −15
4
→
1
η
4
0 −4 −8 12
3
0 −3 −6 9
−1 3 3 −4
0 1 2 −3
0 −1 −2 3
0 −1 −2 3
3 −4
2 −3 ⇒ ρ( Α) = 2
0 0
0 0
9)
A=
1
7
17
3
3
1
1
4
2(−1)+ η3
η
→
η2(−1) η4
10)
−1 6
η1(−7)+ η2
−3 10 η1(−17 )+ η3
→
η1(−3)+ η4
−7 22
−2 10
1 3 −1 6
1
0 −20 4 −32 η2 4
→
1
0 −50 10 −80 η3 10
0 −5 1 −8
1 3 −1 6
0 −5 1 −8 ⇒ ρ( Α) = 2
0 0 0 0
0 0 0 0
4
1 3 −1 6
0 −5 1 −8
0 −5 1 −8
0 −5 1 −8
A=
0 1 10 3
2 0 4 −1
16 4 52 9
8 −1 6 −7
1↔ η2
η
→
2 0 4 −1
η2( −4 )+ η3
0 1 10
3 ⇒ ρ( Α) = 3
→
η2+ η4
0 0 −20 5
0 0 0
0
Bài 2:
Biện luận theo tham số λ hạng của các ma trận:
3 1 1 4
3 1 1
λ 4 10 1 η2
2 2 4
↔ η4
→
1) A =
1 7 17 3
1 7 17
2 2 4 1
λ 4 10
η2
h1↔
→
1
4
3
1
↔ η3
η2
→
1 2
4
2
0 1
5
−5
0 −7 −15 −5
0 2
6 λ−2
→
Vậy :
Nếu
Nếu
1
η3 + η4
5
2) A =
3
λ
1
2
2 4 2
1 1 3
7 17 1
4 10 λ
1
0
0
0
η1 −8 + η3
( )
η
→
1( −4 )+ η4
2 0 4 −1
0 1 10 3
16 4 52 9
8 −1 6 −7
η1( −4)+η2
( −3)+η3
η1
η1( −1)+η4
→
4
1
3
1
χ4
χ1↔
→
1 2
4
2
0 −7 −15 −5
0 1
5
−5
0 2
6 λ−2
( 7 )+η3
η2
( −2 )+η4
η2
→
1
0
0
0
2 0
4 −1
0 1 10
3
0 4 20 17
0 −1 −10 −3
4
1
3
1
1 1 3
2 4 2
7 17 1
4 10 λ
2 4
2
1 5
−5
0 20 −40
0 −4 λ + 8
2 4
2
1 5 −5
0 20 −40
0 0
λ
= 0 thì r(A) = 3
0 thì r(A) = 4
1 1 4
4 10 1 η2 ↔ η4
→
7 17 3
2 4 3
3
2
1
λ
1 1 4
2 4 3
7 17 3
4 10 1
5
χ4
χ1↔
→
4
3
3
1
1 1 3
2 4 2
7 17 1
4 10 λ
χ2
c1↔
→
1
2
7
4
4 1 3
3 4 2
3 17 1
1 10 λ
η1( −2)+η2
( −7 )+η3
η1
η1( −4)+η4
→
1 4 1 3
0 −5 2 −4
→
0 0 0 0
0 0 0 λ
Vậy:
Nếu = 0 thì r(A) = 2
Nếu 0 thì r(A) = 3
η2( −5)+η3
η2( −3)+η4
3) A =
4 1 3 3
0 6 10 2 Χ2 ↔ Χ4
→
1 4 7 2
6 λ −8 2
→
h1( −4 )+η3
η1( −6 )+η4
η4
η3↔
→
4
0
1
6
1 2 7
4
0 1 5
3 η3↔ η4
→
→
0
0 0 0
0 0 0 λ + 6
Vậy:
Khi λ + 6 = 0 ⇔ λ = −6 thì r(A) = 2
Khi λ + 6 ≠ 0 ⇔ λ ≠ −6 thì r(A) = 3
−3
0
1
3
9 14 1
6 10 2 Χ2 ↔ Χ4
→
4 7 2
λ 1 2
→
h1(3)+η3
η1( −3)+η4
−3
0
1
3
1 2
7
4
0 2 10
6
0 7
35
21
0 −4 −20 λ − 12
η3
η1↔
→
1
η2 2
→
η2(5)+ η3
η2(10 )+ η4
4) A =
1 4 1 3
0 −5 2 −4
0 0 0 λ
0 0 0 0
3 3 1
2 10 6
2 7 4
2 −8 λ
1 2
7
4
0 2
10
6
0 −5 −25 −15
0 −10 −50 λ − 24
1
0
4
6
1 2 7
4
0 −1 −5 −3
0 0 0 λ +6
0 0 0
0
1
η2
2
→
2 7 4
2 10 6
3 3 1
2 −8 λ
1 2
7
4
0 1
5
3
0 −5 −25 −15
0 −10 −50 λ − 24
1 14 9
2 10 6 η1↔ η3
→
2 7 4
2 1 λ
6
1 4
1
3
0 −5 2
−4
0 −25 10 −20
0 −15 6 λ − 12
1
0
−3
3
2 7 4
2 10 6
1 14 9
2 1 λ
1 2
7
4
0 1
5
3
0 7
35
21
0 −4 −20 λ − 12
→
h2( −7 )+η3
η2( 4 )+η4
Vậy :
1
0
0
0
2
1
0
0
7 4
5 3 η3↔ η4
→
0 0
0 λ
Nếu = 0 thì r(A) = 2
Nếu 0 thì r(A) = 3
7
1
0
0
0
2
1
0
0
7 4
5 3
0 λ
0 0
BÀI TẬP VỀ MA TRẬN NGHỊCH ĐẢO
VÀ PHƯƠNG TRÌNH MA TRẬN
Bài 1:
Tìm ma trận nghịch đảo của các ma trân sau:
3 4
1) A =
5 7
Ta có:
1
5
η1
3 4 1 0 η1 − 3 + η2 3 4 1 0 η2(33) 1
→
(A I)=
→ 0 1 − 5 1
5 7 0 1
0
3
3
4 1
0
3 3
1 −5 3
4
η2 − + η1
3
1 0 7 −4
7 −4
−1
→
⇒ Α =
0 1 −5 3
−5 3
1 −2
2) A =
4 −9
Ta có:
−1
−9 2 9 −2
1 δ −β
1
A = 1 −2 =
=
=
αδ − βχ − χ α 1.(−9) − (−2).4 −4 1 4 −1
4 −9
−1
3) A =
Ta có:
(A I) =
3 −4 5
2 −3 1
3 −5 −1
3 −4 5 1 0 0
2 −3 1 0 1 0
3 −5 −1 0 0 1
1 −1 4 1 −1 0
η2(−1) + η1
→ 2 −3 1 0 1 0
3 −5 −1 0 0 1
1 −1 4
1 −1 4 1 −1 0
1 −1 0
η2(−2) + η3
→ 0 −1 −7 −2 3 0
→ 0 −1 −7 −2 3 0
0 −2 −13 −3 3 1
0 0 1 1 −3 1
η1( −2)+η2
η1( −3)+η3
1 −1 4 1 −1 0
1 −1 0 −3 11 −4
η3(−7)+η2
→ 0 1 7 2 −3 0 η3
→ 0 1 0 −5 18 −7
( −4)+η1
0 0 1 1 −3 1
0 0 1 1 −3 1
η2(−1)
1 0 0 −8 29 −11
η2+η1
→ 0 1 0 −5 18 −7
0 0 1 1 −3 1
8
8
Vậy ma trận A là ma trận khả nghịch và A =
1
29
11
5 18
7
1
2
4) A = 3
1
Ta có:
2
(A I)= 3
1
3
1
7 3
9 4
5 3
1 5 3 0 0 1
7 3 1 0 0
η3↔η1
9 4 0 1 0
→ 3 9 4 0 1 0
5 3 0 0 1
2 7 3 1 0 0
1 5 3 0 0 1
1 5 3 0 0 1
η3↔η2
→ 0 −6 −5 0 1 −3
→ 0 −3 −3 1 0 −2
0 −3 −3 1 0 −2
0 −6 −5 0 1 −3
η1( −3)+η2
η1( −2 )+η3
1 5 3 0 0 1 η2 − 1 1 5 3 0 0
3
1
h2(2)+η3
→ 0 −3 −3 1 0 −2 → 0 1 1 −
0
3
0 0 1 −2 1 1
0 0 1 −2 1
1
→ 0
0
7
−
3
5
⇒ Α−1 =
3
−2
h3( −1)+η2
η3( −3)+η1
5 0
6 −3 −2
5
1
1 0
−1 −
3
3
0 1 −2 1
1
1
3
1
−1 −
3
1
1
2
−
7
1
0
0
−
3
5
5)+η1
η2(−
→ 0 1 0
3
0 0 1 −2
1 2 2
5) A = 2 1 −2
2 −2 1
Ta có:
9
1
2
3
1
1
3
1
−1 −
3
1
1
2
−
1 2 2 1 0 0�
�
�
�
A = �2 1 −2 0 1 0 �
�2 −2 1 0 0 1 �
�
�
h 2( −2 ) + h 3
1 2 2
�
�
0 −3 −6
�
�
0 0 9
�
5
�
1 2 0
�
9
�
h 3( −2 ) + h 2
2
h 3( −2 ) + h1
�
0 1 0
�
9
�
2
�
0 0 1
�
9
�
2 �
�1 2
�9 9
9 �
�
�
2 1
2�
−1
�
�A =
−
�9 9
9�
�
�
2
2 1 �
�
−
�
�
9 9 �
�9
1 2 2 1 0 0�
�
�
�
0 −3 −6 −2 1 0 �
�
�
0 −6 −3 −2 0 1 �
�
�
�
�1�
h 2�
− � �
1 2 2 1 0
�3�
1 0 0� �
�
1�
h 3� �
2
1
�9 �
�
−2 1 0 �
0
1
2
−
�
�
3
3
2 −2 1 �
�
�
2
2
�
0 0 1
−
9
9
�
4
2�
1 2
�
− �
1 0 0
�
9
9
9 9
�
�
1
2
2 1
h 2 −2 + h1
�
− � ( )
0 1 0
�
�
9
9
9 9
�
�
2 1 �
2
2
�
−
0 0 1
−
�
�
9 9 �
9
9
�
h1( −2 ) + h 2
h1( −2 ) + h 3
Bài 2
Giải các phương trình ma trận sau
1 2� �
3 5�
�
1) � �X = � �
3 4� �
5 9�
�
1 2� �
3 5�
�
;B =� �
Đặt A = � �
3 4� �
5 9�
�
Ta có: AX = B � X = A−1 B
−1
�−2
1 2�
�
�4 −2 � �
1 �d −b �
1
A =� � =
�
�=
�
�= 3
3 4 � ad − bc �
−c a � 1.4 − 2.3 �
−3 1 � �
�
�2
�−2 1 �
3 5 � �−1 −1�
�
� X = �3 −1 �
�
�= �
�
�
�
5 9 � �2 3 �
�
�2
2 �
3 −2 � �−1 2 �
�
2) X �
�= �
�
5 −4 � �
−5 6 �
�
−1
10
1 �
−1 �
�
2 �
�
0�
�
0�
�
1�
�
9�
2 �
9 �
�
2�
−
9�
�
1 �
�
9 �
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