Jordan exchange-Independence

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mathmari
Posts: 81
Joined: Sat Apr 27, 2013 11:58 pm UTC

Jordan exchange-Independence

Postby mathmari » Sat Nov 30, 2013 7:04 pm UTC

Hello! I need some help using the Matlab function ljx.m for solving a system of linear equations. A= [1 , -1 ,0 , 1;1 , 0, 1 , 0; 1 ,1 ,2 , -1] and b=[1; 1;0].

When I call the function ljx(T,1,1), then ljx(2,2) I get that the row y3 is dependent and can be written as y3=-y1+2y2+1.

When I call the function ljx(T,2,1), then ljx(T,3,2) I get y1=2y2-y3+1 .
If I solve for y3 I get the same answer as at the first case. But the result y1=2y2-y3+1 doesn't mean that y1 is dependent and y3 independent?
Are the results at both cases equal??

flownt
Posts: 70
Joined: Sat Feb 09, 2013 5:24 pm UTC

Re: Jordan exchange-Independence

Postby flownt » Mon Dec 02, 2013 3:21 pm UTC

mathmari wrote:Are the results at both cases equal??


Yes. Your matrix only has rank 3 so your solution set is an entire line.

mathmari
Posts: 81
Joined: Sat Apr 27, 2013 11:58 pm UTC

Re: Jordan exchange-Independence

Postby mathmari » Thu Dec 05, 2013 11:08 pm UTC

Great! Thanks!


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