Matematicas

Ejercicios a Resolver:

I. Resuelve los siguientes sistemas de ecuaciones lineales.
a) Aplica el método de la matriz inversa
b) Comprueba los resultados obtenidos en a), aplicando laregla de Cramer.
1) [pic] 2) [pic]

Procedimientos:

1.
a)

1 2 3 5
2 3 2 2
-1 -2 -4 -1

1 2 3 1 2
2 3 2 2 3
-1 -2 -4 -1 -2

((1)(3)(4) + (2)(2)(-1)+ (3)(2)(-2)) – ((-1)(3)(3) + (-2)(2)(1) + (-4)(2)(2))

-12 + (-4)(-12)
-28 – (-9 + (-4) + (-16))
-28 + 29
29-28 = 1

Det = 1 ≠ 0 invertible

-8 -6 -1
MA -2 -1 0-7 -4 -1

-8 6 -1
Cof 2 -1 0
-5 4 -1

Adj = (Cof(A))^ t

Adj = -8 2 -5
6 -1 4
-1 0 -1

= -82 -5
6 -1 4
-1 0 -1
1

A^ -1= -8 2 -5
6 -1 4
-1 0 -1

Comprobación

Ax = B
X = BA^ -1

X -8 2 -5 5Y = 6 -1 4 2
Z -1 0 -1 -1

X -31
Y = 24
Z -4

X = -31 Y = 24 Z = -4

b)

1 2 3 x 5
2 3 2 y = 2
-1 -2 -4 z -1Det = 1 ≠ 0 invertible

5 2 3
X = 2 3 2
-1 -2 -4
1

5 2 3 5 2
2 3 2 2 3
-1 -2 -4 -1 -2

(-60 – 4 – 12) – (-9 – 20 – 16)
= - 76– (-45)
= -76 + 45
= -31

X = -31/1 = -31

1 5 3
Y = 2 2 2
-1 -1 -4
1

1 5 3 1 5
2 2 2 2 2
-1 -1 -4 -1 -1

(-8 – 10 – 6) – (- 6 – 2 – 40)
= - 24 – (-48)
= - 24 + 48
= 24
Y = 24/1 = 24

1 2 5
Z = 2 3 2
-1 -2 -1
1

1 2 5 1 2
2 3 2 2 3
-1 -2 -1 -1 -2(-3 – 4 – 20) – ( -15 – 4 – 4)
= - 27 – (-23)
= - 27 + 23
= - 4
Z = -4/1 = -4

X = -31 Y = 24 Z = -4

2.
a)

2 4 6
-1 -4 -3
0 1 -1

2 4 6 2 4
-1 -4...