5 Repaso De Matrices

5. Repaso de matrices

(© Chema Madoz, VEGAP, Madrid 2009)

1

Matrices
Elemento: aij
Tamaño: m  n
Matriz cuadrada: n  n
(orden n)
Elementos de la diagonal: ann

Vector columna
(matriz n x 1)

Vector fila
(matriz 1 x n)

 a11 a12  a1n 


 a21 a22  a2 n 
 



 am1 am 2  amn 

 a1 
 
 a2 
 
 
 an 
(a1 a2  an )

2

Suma:

3
 2 1
 4 7  8




A  0 4
6 , B  9
3
5
  6 10  5 
1  1

2





 24

A  B  0  9
  6 1


 1 7
43

3  ( 8)   6
 
6  5   9
10  ( 1)  5  2    5

6
7
9

5

11 
 3 

Multiplicación por un escalar:
 ka11 ka12  ka1n 


 ka21 ka22  ka2 n 
k A 

(
k
a
)
ij
mn





 kam1 kam 2  kamn 

3

Si A, B, C son matrices mn, k1 y k2 son escalares:
(i)
(ii)
(iii)(iv)
(v)
(vi)

A+B=B+A
A + (B + C) = (A + B) + C
(k1k2) A = k1(k2A)
1A=A
k1(A + B) = k1A + k1B
(k1 + k2) A = k1A + k2A

4

Multiplicación:
 4 7
 9  2
 , B 

(a) A 
8
 3 5
6
 4． 9  7． 6 4． ( 2)  7． 8   78 48 
AB 
 

 3． 9  5． 6 3． ( 2)  5． 6   57 34 
 5 8
(b) A  1 0  , B   4  3 


0
 2
 2 7



 5． ( 4)  8． 2 5． ( 3)  8． 0    4  15 



AB  1． ( 4)  0． 2 1． ( 3)  0． 0    4  3 
 2． ( 4)  7． 2 2． ( 3)  7． 0   6  6 

 

Nota: En general, AB  BA

5

Transpuesta de una matriz A:
 a11

 a12
T
A 


 a1n

a21  am1 

a22  am 2 


a2 n  amn 

(i) (AT)T = A
(ii) (A + B)T = AT + BT
(iii) (AB)T = BTAT
(iv) (kA)T = kAT
Nota:

(A + B + C)T = AT + BT + CT
(ABC)T = CTBTAT

6

Matriz cero
A+0=A
A +(–A) = 0

 0 0


 0
 0 0
0   , 0 
 , 0  0 0 
 0
 0 0
 0 0



Matrices triangulares
1

0
0

0


2 3 4

5 6 7
0 8 9

0 0 1 

Triangular superior

  2 0 0 0 0


 1 6 0 0 0
 8 9 3 0 0


 1 1 1 2 0
 15 2 3 4 1 


Triangular inferior

7

Matriz diagonal:
Matriz cuadrada n  n, i ≠ j, aij = 0

7

0
0


0
1

2

0

0

0
1 

 5 0
 1 0
 5 

 0 5
 0 1

Matriz identidad:
A: m  n, entonces
I m A = A In = A

1 0 0 

0 1 0 


0 0 0 

0

0


1
8

Una matiz A n × n es simétrica si AT = A.
1 2 7


A  2 5 6 
 7 6 4


1 2 7


T
A  2 5 6  A
 7 6 4



9

Resolución de sistemas de ecuaciones lineales:
a11 x1  a12 x2    a1n xn b1
a21 x1  a22 x2    a2 n xn b2


am1 x1  am 2 x2   amn xn bm

Matriz aumentada
asociada, para resolver
el sistema de ecuaciones
lineales.

 a11 a12  a1n

 a21 a22  a2 n
 


 am1 am 2  amn

b1 

b2 


bm 
10

2x1 + 6x2 + x3 = 7
x1 + 2x2 – x3 = –1
5x1 + 7x2 – 4x3 = 9
 2 R1  R2
 5 R1  R3



3 R2  R3



1 7
2 6


 1 2  1  1
 5 7  4 9



2  1  1
1


2
3 9
0
0  3

1
14



 1 2  1  1


3
90 1
2
2
 0 0 11 55 

2
2

1

2 R2


2

11 R3



R12



 1 2  1  1


1 7
2 6
5 7  4

9



2  1  1
1


3
9
1
0
2
2
0  3

1
14



 1 2  1  1


3
9
0 1
2
2
0 0

1
5


x1  2 x2  x3  1

x3 = 5, x2 = –3, x1 = 10

3
9
x2  x3 
2
2
x3 5

11

Resolver mediante el método de Gauss-Jordan
x1 + 3x2 – 2x3 = – 7
4x1 + x2 + 3x3 = 5
2x1 – 5x2 + 7x3 = 19

1

42

 111 R2
1
 111 R3 

 0
0


3  2  7

1
3
5
5
7 19 
3  2  7

1  1  3
1  1  3 

 4 R1  R2
 2 R1  R3



 3 R2  R1
 R2  R3



1

0
0

1

0
0


3  2  7

 11 11 33 
 11 11 33 
0
1
2

1  1  3
0
0
0 

Entonces:

x2 – x3 = –3
x1 + x3 = 2
Haciendo x3 = t, tenemos x2 = –3 + t, x1 = 2 – t.

12

Resolver:

x1 + x2 = 1
4x1 − x2 = −6
2x1 – 3x2 = 81
1
1


 4  1  6 
2  3

8



 1 0  1


 0 1 2
 0 0 16 



0 + 0 = 16 !!  No tiene soluciones.

13

 a11 a12  a1n 


 a21 a22  a2 n 
A 




 am1 am 2  amn 

Vectores fila:
u1 = (a11 a12 … a1n),
u2 = (a21 a22, … a2n),…,
um = (am1 am2 … amn)

Vectores columna:

 a11 
 a12 
 a1n 






 a21 
 a22 
 a2 n 
v1 
, v 2 
, , v n...