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Assuming that A is already in row reduced echelon form

$$ A= \begin{pmatrix} 1 & -1 & 0 & 0 & -5/3 & 6\\ 0 & 0 & 1 & 0 & -1 & -3 \\0 & 0 & 0 &1 & -2/3 & -2 \end{pmatrix} $$

(this is a row reduced echelon form matrix that i obtained from a system of linear equations, through gaussian elimination, that used the variables x1, x2, x3, x4,x5)

from my understanding, the basic variables are x1, x3 ,and x4. the free variables is just x2. But then we have a missing variable x5 that does not correspond to any pivot in the matrix, does this mean that x5 is NEITHER a free variable nor a basic variable

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    $\begingroup$ Everything except $x_1, x_3$ and $x_4$ is a free variable. $\endgroup$ – user8277998 Dec 15 '17 at 20:30
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    $\begingroup$ ah i understand now. So we have to use the logic that if a variable is not basic, then it must be free. thank you $\endgroup$ – Soon_to_be_code_master Dec 15 '17 at 20:32
  • $\begingroup$ Is that last column sixth column of the matrix ? Or it is the augmented column ? $\endgroup$ – user8277998 Dec 15 '17 at 20:40
  • $\begingroup$ it is an augmented column, sorry i forgot to mention that. $\endgroup$ – Soon_to_be_code_master Dec 15 '17 at 21:16

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