Examen Julio 2015, preguntas y respuestas PDF

Preview

Summary

Examen Algebra Julio 2015...

Description

x, y, z x + 2y + z = 11.5 2x + 3y + 2z = 20.5



⇐⇒



1 2 1 11.5 2 3 2 20.5

3F1 + F2

[5 9 5 55]

2F2 −3F1

[1 0 1 6.5]



 x + 2y + z = 11.5  2x + 3y + 2z = 20.5  2y − z = 0

x = 1.5 y = 2.5 z = 5



 a b c A= 1 0 0  0 1 0



 0 0 −1 P =  0 −1 a  −1 a b

P AP −1 = AT . P P A = AT P

P

 0 −1 0 a 0  = AT P P A =  −1 0 0 −c 

P

1 −a 0  0 1 −a A =  0 0 1 0 0 0 

A  0 0   −a  1 a

A−1



1 0 = 0 0

 a a2 a3 1 a a2   0 1 a  0 0 1

LU Ax = b 

A

 4 3 −5 7 A =  −4 −5 8 6 −8

 4 0 0 L =  −4 −2 0  , 8 0 2 

•



 2 b =  −4  6



 1 3/4 −5/4 U = 0 1 −1  0 0 1

Ly = b 

   4 0 0 | 2 1/2  −4 −2 0 | −4  → y =  1  8 0 2 | 6 2

•

Ux = y     1 3/4 −5/4 | 1/2 1/4  0 1 −1 | 1  → x = 2  0 0 1 | 2 1  5 5 5 A=5 5 5  5 5 5 

A

A



1 1 1

T

      5 5 5 1 15 1  5 5 5   1  =  15  = 15  1  . 5 5 5 1 15 1 

−λ3 + 15λ2 15 •

λ=0    5 5 5 | 0 1 1 1 | 0  5 5 5 | 0 → 0 0 0 | 0  5 5 5 | 0 0 0 0 | 0 

     −1  −α − β  −1  = Env  1  ,  0  .  α   0 1 β 

•

λ = 15     −10 5 5 | 0 1 0 −1 | 0  5 −10 5 | 0  →  0 1 −1 | 0  5 5 −10 | 0 0 0 0 | 0     −α  1   α  = Env  1  .   α 1 

P −1 AP = D     −1 −1 1 0 0 0 D =  0 0 0 . 0 1 , P = 1 0 1 1 0 0 15

0...