Your SlideShare is downloading. ×
21533843 calculo-del-rango-de-una-matriz
Upcoming SlideShare
Loading in...5
×

Thanks for flagging this SlideShare!

Oops! An error has occurred.

×

Introducing the official SlideShare app

Stunning, full-screen experience for iPhone and Android

Text the download link to your phone

Standard text messaging rates apply

21533843 calculo-del-rango-de-una-matriz

142
views

Published on


0 Comments
0 Likes
Statistics
Notes
  • Be the first to comment

  • Be the first to like this

No Downloads
Views
Total Views
142
On Slideshare
0
From Embeds
0
Number of Embeds
0
Actions
Shares
0
Downloads
2
Comments
0
Likes
0
Embeds 0
No embeds

Report content
Flagged as inappropriate Flag as inappropriate
Flag as inappropriate

Select your reason for flagging this presentation as inappropriate.

Cancel
No notes for slide

Transcript

  • 1. Cálculo del rango de una matriz:Vamos a calcular el rango de las siguientes matrices: 1 4 -1 1 3 -1A = -1 3 2 B = 2 -1 5 2 2 0 1 10 - 8 1 0 2 1 -1 1 -2 0 -3 0 2 -1 1 2C = -1 3 1 4 D= -1 1 3 2 0 2 1 5 -1 0 8 7 9 4 F2 + F1 F3 - 2F2 1 4 -1 1 4 -1 1 4 -1A = -1 3 2 → 0 7 1 → 0 7 1 → rango(A) = 3 2 2 0 0 -6 2 0 - 20 0 F3 - 2F1 F2 - 2F1 F3 + F2 1 3 -1 1 3 -1 1 3 -1B = 2 -1 5 → 0 -7 7 → 0 - 7 7 → rango(A) = 2 1 10 - 8 0 7 -7 0 0 0 F3 - F1 F2 + F1 F3 - 5F2 1 -2 0 -3 1 -2 0 -3 1 -2 0 -3C = -1 3 1 4 → 0 1 1 1 → 0 1 1 1 → rango(C) = 2 2 1 5 -1 0 5 5 5 0 0 0 0 F3 - 2F1 F3 + F1 2F3 – F2 1 0 2 1 -1 1 0 2 1 -1 1 0 2 1 -1 0 2 -1 1 2 0 2 -1 1 2 0 2 -1 1 2D= → → → -1 1 3 2 0 0 1 5 3 -1 0 0 - 11 - 5 4 0 8 7 9 4 0 8 7 9 4 0 0 11 5 - 4 F4 – 4F2
  • 2. F4 + F3 1 0 2 1 -1 0 2 -1 1 2 → → rango(D) = 3 0 0 - 11 - 5 4 0 0 0 0 0Estudiar el rango de una matriz según los valores de un parámetro:Vamos a calcular el rango de las siguientes matrices según los valores delparámetro k: 1 -1 -1 2 -1 4M = 1 -1 2 N = -2 1 3 2 1 k 1 k 2 1 3 2 -1 -1 1 0 2P= 2 6 4 k Q= 1 3 1 0 4 12 8 - 4 2 10 3 k F2 - F1 1 -1 -1 1 -1 -1M = 1 -1 2 → 0 0 3 → rango(M) = 3 ∀k ∈ R 2 1 k 0 3 k+2 F3 - 2F1 F2 + F1 2 -1 4 2 -1 4 1N = -2 1 3 → 0 0 7 → 1+2k = 0 → k = - 2 1 k 2 0 1+ 2k 0 2F3 - F1Por tanto tenemos que: 1Si k = - rango(N) = 2 2 1Si k ≠ - rango(N) = 3 2
  • 3. F3/4 F2 – F1 1 3 2 -1 1 3 2 -1 1 3 2 -1P= 2 6 4 k → 1 3 2 - 1 → 0 0 0 0 → k+2=0→ k = – 2 4 12 8 - 4 2 6 4 k 0 0 0 k+2 F2 F3 F3 – 2F1Por tanto tenemos que:Si k = – 2 rango(P) = 1Si k ≠ – 2 rango(P) = 2 F2 + F1 F3 – 3F2 -1 1 0 2 -1 1 0 2 -1 1 0 2Q= 1 3 1 0 → 0 4 1 2 → 0 4 1 2 → k–2=0→k =2 2 10 3 k 0 12 3 k + 4 0 0 0 k-2 F3 + 2F1Por tanto tenemos que:Si k = 2 rango(Q) = 2Si k ≠ 2 rango(Q) = 3