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Find the free variable of the following linear equations system

\begin{cases} x + 2y + -3z=2 \\ 2x – 3y + z = 1\\ 5x -4y -z=4\end{cases}

Now the book says that because the system is not row reduced there is not free variables.

So I row reduced it to

$$ \left[\begin{array}{rrr|r} 1 & 2 & -3 & 2 \\ 0 & -7 & 7 & -3 \\ 0 & 0 & 0 & -1 \end{array}\right] $$

Now the system of course has no solution (As once I read, because there is a pivot element in the b column).

will it be write to say that there is a free variable namely $z$ or because there is not solution there isn't any?

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  • $\begingroup$ Are you sure that is correct? As the RREF, I get $$\begin{bmatrix} 1 & 0 & -1 & \dfrac{8}{7} \\ 0 & 1 & -1 & \dfrac{3}{7} \\ 0 & 0 & 0 & 0 \\ \end{bmatrix}$$ $\endgroup$ – Moo Nov 2 '17 at 16:59
  • $\begingroup$ @Moo sorry, typo $\endgroup$ – gbox Nov 2 '17 at 17:01
  • $\begingroup$ Updated RREF - looks like you made a slight error because our second rows match. This system does have a free variable, namely $z$. $\endgroup$ – Moo Nov 2 '17 at 17:03
  • $\begingroup$ @Moo Why do you say it is $z$? I can write $$y= x-\frac{5}{7};\;z= x-\frac{8}{7}$$ $\endgroup$ – Raffaele Nov 2 '17 at 18:13
  • $\begingroup$ @Raffaele: My comment is based on the typical convention of solving using bottoms-up because that makes calculations easier. I have no issue with your observation, I am just following convention. $\endgroup$ – Moo Nov 2 '17 at 18:15

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