why can we tackle a matrix with linearly dependent rows while driving artificial variables out of the basis?

Asked Jan 14 '19 at 21:16

Active Jan 15 '19 at 12:27

Viewed 42 times

in introduction to linear programming, section 3.5 driving artificial variables out of the basis, the authors consider the case where that when trying to drive the lth basic variable (which is artificial), we find out that the lth row of $B^{-1}A$ is zero, which means that A has linearly dependent rows (last paragraph of page 113)

How is that possible? I thought that our assumption is that A has full rank! and that we test for that before the optimization...

Thanks!

edited Jan 15 '19 at 12:27

asked Jan 14 '19 at 21:16

ihadanny

Please specify where you´ve read that the lth row is $0$. – callculus42 Jan 15 '19 at 06:26
@callculus - edited – ihadanny Jan 15 '19 at 12:27

why can we tackle a matrix with linearly dependent rows while driving artificial variables out of the basis?

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