Khoá h c Toán cao c p:

i s tuy n tính (Th y Lê Bá Tr n Ph

ng)

nh th c – Ma tr n

MA TR N (PH N 01+ PH N 02)
ÁP ÁN BÀI T P T LUY N
Giáo viên: LÊ BÁ TR N PH

NG

(Tài li u dùng chung cho P1+P2)

Bài 1. Cho các ma tr
 1 0 2
A   1 1
2
 2 2 1
Tính :

n


1 5 3 
 1 2 4 
 , B   0 4 2  , C   3 5 8 






 2 3 5 
 5 7 6 


a) A-2B +5C

b)

1 t
A  Bt  3C .
2

Gi i

0 16 
 4

a) A 2 B  5C   16 18 38
 27 31 19 
11
 5
 2
2

1 t
23
A  Bt  3C   14

b)

2
2

 17 20



11 

22 

25 
2 

 1 8 4 3 
 5 1 4 0 
; B =
Bài 2. Cho A = 

 . Tìm ma tr n X, bi t :
 0 5 1 2 
2 1 3 7 
1
1) 3A  X  B ;
2

2) A 5 X  O ;


3) X  2B  O .

Gi i

32 22 16 6 
1) X  6 A 2 B  

12 4 20 38 


1
A 
2) X    
5  2

 5

1
4

5
5
1
3


5
5



0 

7

5 

 2 16 8 6 
3) X  2 B  

 0 10 2 4
a b 
Bài 3. Ch ng minh r ng X  
 tho mãn ph ng trình :
c d 

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Khoá h c Toán cao c p:

i s tuy n tính (Th y Lê Bá Tr n Ph

ng)


nh th c – Ma tr n

1 0 
X 2   a  d  X   ad  bc  I  0 , v i I  
.
0 1
Gi i

 a 2  bc ab  bd 
Ta có X  
2
ca  cd cb  d 
2

a b 
1 0 
  ad  bc  


c d 
0 1
0 
ab  bd   ad  bc



ad  bc 
ad  d 2   0

 a  d  X   ad  bc  I   a  d  

 a 2  ad

 ac  cd
 X2
=> PCM

1

1
2

Bài 4. Ch ng minh r ng A 
4  là nghi m c a đa th c f  x  x  5x  3 .


4 4

Gi i
5 

2

Ta có A 
4  thay vào ta đ


 20 17 
2

c f  A  A2  5 A 3.I 3  0


V y A là nghi m c u đa th c đã cho

1 2 3 


Bài 5. Cho f ( x)  3x2  2 x  5 và A   2 4 1  .Tìm f ( A)
 3 5 2 
Gi i

 6 9 7 
A   3 7 4  mai
 1 4 8 
2

 f  A  3 A2  2 A 5I 3
 6 9 7  1 2 3  1 0 0   21 23 15 
 3  3 7 4   2  2 4 1   5 0 1 0    13 34 10 
 1 4 8   3 5 2  0 0 1   9 22 25
 1 3
1
1 1


; C  0
Bài 6. Cho A   1 2 ; B  

 0 1
 3 4
1

Tính : 1)  AB  C  ;
t

2) Bt . At .C ;

1
1
0
2  ; D  
0
1 

0
3) A. At ;

2 1 3
1 2 2 
0 1 2

0 0 1
4) At . A ; 5) Dt .D

Gi i

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Khoá h c Toán cao c p:

i s tuy n tính (Th y Lê Bá Tr n Ph

ng)

nh th c – Ma tr n

 0 1 2 
t
1)  AB  C   

3 1 0
 4 0
2) Bt . At .C  

 3 5

10 5 15 


3) A. A   5 5 5 
 15 5 25 


t


 11 13 
4) At . A  

13 29 
1

2
t
5) D .D  
1

3

3

8
4 6 9

8 9 18 
2 1
5 4

Bài 7. Tìm ma tr n X th a mãn:
1)

1 3 2   2 5 6   0 6 6 
1
X   3 4 1  1 2 5    2 9 2 
2
 2 5 3  1 3 2   4 8 6 


 9 20 27   5 8 1  3 2 5
2) 4 X   11 29 30    6 9 5  4 1 3
 11 19 28   4 7 3 9 6 5
Gi i
1)

1 3 2   2 5 6   0 6 6 
1
X   3 4 1  1 2 5    2 9 2 
2
 2 5 3  1 3 2   4 8 6 

1 5 5   0 6 6 
1
 X   3 10 0    2 9 2 
2
 2 9 7   4 8 6 
 1 1 1 
 2 2
1


 X1
1
2   X   2
2
2
 2 17 1
 4 34


2
4 
2 

 9 20 27  5 8 1  3 2 5 38

 

 
2) 4 X   11 29 30   6 9 5  4 1 3   9
 11 19 28  4 7 3 9 6 5 13

 29
 4
 4 44 

1

27 32   X   
 2
 17 26 
 1

 2

4
1
2
1

2

17 
4 

1 
2 
1 

2 

Bài 8. Hãy th c hi n phép nhân các ma tr n sau :
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Khoá h c Toán cao c p:

i s tuy n tính (Th y Lê Bá Tr n Ph

2
2


1) 1  .1 2 3 ; 2) 1 2 3  4  ;

3 
1 
3 1 1  1 1 1
5)  2 1 2  .  2 1 1  ;
1 2 3  1 0 1 
5

1 2 3 4   2
7) 

3 2 4 1   0

1

ng)

1 
3 2 1   
3) 
 . 2 ;
 0 1 2  3 
 

nh th c – Ma tr n

3 1 
 2 1 1 
4) 
.  2 1  ;


3 0 1 1 0 



1 3 1  1 0 0   0
6)  2 2 1  0 2 0   1
 3 4 2  0 0 1   2

2
1
5

1 
1 
4  .

1
6 
2 1

0 2
3
4

Gi i

2
2



1) 1  .1 2 3 =13 2) 1 2 3  4   13
3 
1 

1 
3 2 1    10 
3) 
 . 2   
 0 1 2  3   8 
 

3 1 
 2 1 1 
 9 3
4) 
.  2 1  = 


3 0 1 1 0  10 3


3 1 1  1 1 1 6 2 1
5)  2 1 2  .  2 1 1  = 6 1 1 
1 2 3  1 0 1  8 1 4 
1 3 1  1 0 0   0 2 1  4 3 3



 


6)  2 2 1  0 2 0  1 1 1 =  2 3 2 
 3 4 2  0 0 1   2 5 4   4 4 3
Bài 9. Hãy th c hi n phép tính sau :
2

 3 2 
 3 2

1) 
;

4 8 
 4 2 
5

 2 1 1  7 4 4

2) 3 1 0   9 4 3 
 0 1 2   3 3 4 

6
1 0 0 1 0 0  1 0 0


3) 0 2 0   0 26 0   0 64 0  ;
0 0 1   0 0 16  0 0 1 
6

 4 3 3 190 189 189


 

4)  2 3 2   126 127 126
 4 4 3  252 252 251
6

5) V i n=2, 3, 4 ta có

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Khoá h c Toán cao c p:

i s tuy n tính (Th y Lê Bá Tr n Ph
3

2

ng)

nh th c – Ma tr n

4


1 1 1 2 1 1 1 3 1 1 1 4
 0 1  0 1  , 0 1  0 1 , 0 1  0 1 

 
 
 
 

 
n

1 1 1 n 
=>d đoán 
 

0 1 0 1 
CM: th t v y v i n=1, 2, 3, 4 ta đ u đúng
Gi s đúng v i n=k, ta c n ch ng minh đúng v i n=k+1
k

1 1
1 1 1 k 
T c là ta có 
c n CM 




0 1
0 1 0 1 


k 1

1 k  1

1 
0

Th t v y

1 1
0 1



k 1

1 k  1 1 1 k  1



1 
0 1  0 1 0
n

1 1 1 n 
=> 
 

0 1 0 1 

cos   sin  
6) 

sin  cos  

n

cos 
n  2  
 sin 

 sin   cos 2

cos    sin 2

 sin 2 
cos 2 

cos 
n  3  
 sin 

 sin   cos 2

cos    sin 2

 sin 2  cos 
cos 2   sin 

2


3

cos 
=>D đoán 
 sin 

 sin   cosn 

cos    sinn 
n

 sin   cos 3

cos    sin 3

 sin 3 
cos 3 

 sinn  
cosn  

Gi s đúng v i n =k ta c n CM đúng v i n=k+1 ta có:

cos 
 sin 


 sin  
cos  


k 1

cos 

 sin 

 sin   cos 
cos    sin 
k

 sin   cos  k  1 

cos    sin  k  1 

cosk 

 sink 

 sink   cos 
cosk    sin 

cos 
=> 
 sin 

 sin   cosn 

cos    sinn 


0 0 1 

7) 1 0 0 
 0 1 0 

n

 sin  
cos  
 sin  k  1  

cos  k  1  

 sinn  
cosn  

n

2

0 0 1  0 1 0 
n  2  1 0 0   0 0 1 
0 1 0  1 0 0 

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Khoá h c Toán cao c p:

i s tuy n tính (Th y Lê Bá Tr n Ph

ng)

nh th c – Ma tr n

3

 0 0 1   0 1 0   0 0 1  1 0 0 
n  3  1 0 0   0 0 1  1 0 0   0 1 0 
0 1 0  1 0 0  0 1 0  0 0 1 
4

 0 0 1  1 0 0   0 0 1 
0 0 1  0 0 1 






n  4  1 0 0  0 1 0 1 0 0  I 3 . 1 0 0  1 0 0 
0 1 0 0 0 1  0 1 0 
0 1 0  0 1 0 


=>nh v y t ng quát nên ta có:
n

 0 0 1  1 0 0 
+ V i n=3k thì  1 0 0   0 1 0 
 0 1 0  0 0 1 

 


k  N

n

 0 0 1  0 0 1 


+ V i n=3k+1 thì  1 0 0   1 0 0 
 0 1 0  0 1 0 

 

n

 0 0 1  0 1 0 


+ V i n=3k+2 thì  1 0 0   0 0 1 
 0 1 0  1 0 0 


 


a 0 0 
8) b a 0 
 0 0 a 

n

2
0
a 0 0   a

n  2  b a 0    2ab a 2
 0 0 a   0
0

0

0
a 2 

2
0
a 0 0   a



n  3  b a 0    2ab a 2
 0 0 a   0

0

0  a 0 0   a 3
0

 2

0  b a 0   3a b a 3
0
a 2   0 0 a   0

2

3

0

0
a 3 

D đoán
3
0
a 0 0   a
b a 0    na n 1b a 3

 
 0 0 a   0
0
n


0

0  Các b n CM t
a 3 

ng v n t nh các câu trên.

3 3
1 0 
, I 
Bài 10. Cho A  
.Tính A3  5 A2  6 A I .


1 3
0 1 

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Khoá h c Toán cao c p:

i s tuy n tính (Th y Lê Bá Tr n Ph


ng)

nh th c – Ma tr n

A3  5 A2  6 A I
54

30
11

6

90 
12 18 
3 3   1 0 
5
6




54 
 6 12 
1 3 0 1 
18
11

 0 1 1
1 0 0 



Bài 11. Cho A   2 1 3 , I  0 1 0  .Tính ( A I )2 .
 0 1 1
0 0 1 
 1 1 1  3 2 6 
  2 2 3   6 13 14 
1 
 0 1 2  2 4
2

 A I 

2

1 0 0 
Bài 12. Cho f ( x)  x3  x2  9 x  9 và A  0 1 0  . Tìm f ( A)
0 0 3 
1 0 0 
1 0 0 


3
A  0 1 0  ; A  0 1 0 
0 0 9 
0 0 27 
f  A  A3  A2  9 A 9 
2

1

=>  0

0
0
 0
0

0  1 0 0  9 0 0  9 0 0 
1 0   0 1 0   0 9 0    0 9 0 
0 27  0 0 9  0 0 27  0 0 9 
0 0
0 0 
0 0 
0

a 1 0 


Bài 13. Cho A   0 a 1  . Tính A100 .
 0 0 a 
2
 a 1 0   a 1 0   a 2a 1 


A2   0 a 1   0 a 1    0 a 2 2a 
 0 0 a   0 0 a   0 0 a 2 
 a 2 2a 1   a 1 0   a 3 3a 2 3a 





A3   0 a 2 2a   0 a 1    0 a 3 3a 2 
 0 0 a 2   0 0 a   0
a 3 
0



 a 3 3a 2 3a   a 1 0   a 4 4a 3 6a 2 




A4   0 a 3 3a 2   0 a 1    0 a 4 4a 3 
0
0
a 4 
a 3   0 0 a   0
0


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Khoá h c Toán cao c p:

a n

D đoán An   0
0


i s tuy n tính (Th y Lê Bá Tr n Ph

na n 1
an
0

ng)

nh th c – Ma tr n

2(n  1)a n 2 

na n 1  Các b n t CM.

an


Giáo viên : Lê Bá Tr n Ph
Ngu n

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T ng đài t v n: 1900 58-58-12

:

ng

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