Hi!! I need some help using the Matlab function ljx.m for solving a system of linear equations. Let 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 \(\displaystyle y_{3} \) is dependent and can be written as \(\displaystyle y_{3}=-y_{1}+2y_{2}+1 \).
When I call the function ljx(T,2,1), then ljx(T,3,2) I get \(\displaystyle y_{1}=2y_{2}-y_{3}+1 \).
If I solve for \(\displaystyle y_{3} \) I get the same answer as at the first case. But the result \(\displaystyle y_{1}=2y_{2}-y_{3}+1 \) doesn't mean that \(\displaystyle y_{1} \) is dependent and \(\displaystyle y_{3} \) independent?
Are the results at both cases equal??
How do I choose at each step the pivot? Can I take each element as long as it is different from zero?
When I call the function ljx(T,1,1), then ljx(2,2) I get that the row \(\displaystyle y_{3} \) is dependent and can be written as \(\displaystyle y_{3}=-y_{1}+2y_{2}+1 \).
When I call the function ljx(T,2,1), then ljx(T,3,2) I get \(\displaystyle y_{1}=2y_{2}-y_{3}+1 \).
If I solve for \(\displaystyle y_{3} \) I get the same answer as at the first case. But the result \(\displaystyle y_{1}=2y_{2}-y_{3}+1 \) doesn't mean that \(\displaystyle y_{1} \) is dependent and \(\displaystyle y_{3} \) independent?
Are the results at both cases equal??
How do I choose at each step the pivot? Can I take each element as long as it is different from zero?