# Jordan exchange

#### mathmari

##### Well-known member
MHB Site Helper
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= $$\bigl(\begin{smallmatrix} 1 & -1 &0 & 1\\ 1 & 0& 1 & 0\\ 1 &1 &2 & -1 \end{smallmatrix}\bigr)$$ and b=$$\bigl(\begin{smallmatrix} 1\\ 1\\ -1 \end{smallmatrix}\bigr)$$.
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?

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#### mathmari

##### Well-known member
MHB Site Helper
When I call the function ljx(T,1,1), then ljx(2,2) I get that the row $$y_{3}$$ is dependent and can be written as $$y_{3}=-y_{1}+2y_{2}+1$$.

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