Cls Notes 9-5-2011

Cls Notes 9-5-2011 - Cls Notes 9-5-2011 Consider x1 1 2 0 5...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
Cls Notes 9-5-2011 Consider 1 2 3 4 1205 1 41 31 8 3687 5 x x x x ⎛⎞ ⎡⎤ ⎜⎟ ⎢⎥ −− = ⎣⎦ ⎝⎠ A basic solution using () 123 0 xxx is 1 2 3 1 24 3 120 1 3 8 368 5 1 1, an d 0 1 x x x x xx x ⎜⎟⎜⎟ −= ⎛⎞⎛⎞ =− = ⎝⎠⎝⎠ Now if we want another basic solution we could include 4 x and replace one of the basic variables: 12 3 ,, o r. x Which one? Any of them would work since 1 2 3 1 2 3 5 3 1 7 1 2 1 λ = Thus, using 11 4 ,a n d x we have 1 2 4 1 23 4 12 5 1 411 8 36 7 5 2 1,a n d 0 . 1 x x x x x = ==
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 10/26/2011 for the course ISEN 620 taught by Professor Curry during the Fall '10 term at Texas A&M.

Ask a homework question - tutors are online