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Jordan exchange

mathmari

Well-known member
MHB Site Helper
Apr 14, 2013
4,047
Hey! I need some help using the Matlab function ljx.m for solving a system of linear equations. I found a solved example for A= [tex] \bigl(\begin{smallmatrix}
1 & -1 &0 & 1\\
1 & 0& 1 & 0\\
1 &1 &2 & -1
\end{smallmatrix}\bigr) [/tex] and b=[tex] \bigl(\begin{smallmatrix}
1\\
1\\
-1
\end{smallmatrix}\bigr) [/tex].
At the first time they call the function by ljx(T,2,1).
Why do they take as pivot this element, and not for example ljx(T,1,1)??
Is there a specific reason, or can we take as pivot any element we want as long as it's different from 0?
:confused:
 
Last edited:

mathmari

Well-known member
MHB Site Helper
Apr 14, 2013
4,047
When I call the function ljx(T,1,1), then ljx(2,2) I get that the row [tex] y_{3} [/tex] is dependent and can be written as [tex] y_{3}=-y_{1}+2y_{2}+1 [/tex].

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