The original system of equations, for reference, was $$x+2y=8$$ $$x-y=2$$ $$x+y=4$$ I put the resulting matrix into row echelon form getting $$\begin{bmatrix}1&0&0\\0&1&0\\0&0&1 \end{bmatrix}$$ I tried to put this back into equation form, where column 1 is x, column 2 is y, and column 3 is the constant on the right of the equals sign. I got $$x=0$$ $$y=0$$ $$0=1$$ which obviously doesn't make sense. This also doesn't fit the original equations. Did I do something wrong (I can add my work if anyone requests it)? Or does this simply mean that there are no solutions?

Thanks! Any help would be appreciated.

Edit: This is thought to be a duplicate of another question I asked. This question is with a different system of equations, and ended up being a very different question; the first suggested solving by putting the matrix into row echelon form, and this one said I needed to learn how to put augmented matrices into row echelon form. Please don't close this as a duplicate, they are not the same.


You have the augmented system $$\left[\begin{array}{cc|c}1&2&8\\1&-1&2\\1&1&4 \end{array}\right]$$ Then we get $$\left[\begin{array}{cc|c}1&2&8\\1&-1&2\\1&1&4 \end{array}\right] \leadsto \left[\begin{array}{cc|c}1&2&8\\0&-3&-6\\0&-1&-4 \end{array}\right]\leadsto\left[\begin{array}{cc|c}1&2&8\\0&1&2\\0&1&4 \end{array}\right]$$ But this is impossible to solve in two variables since from the third equation we get $y = 4$ and from the second $y = 2$.

  • $\begingroup$ Is there anything special I have to do to put an augmented matrix into row echelon form? $\endgroup$ – heather Sep 18 '16 at 22:44
  • $\begingroup$ @heather No, not really, just the evaluation is a bit more trickier. $\endgroup$ – TheGeekGreek Sep 18 '16 at 22:47
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    $\begingroup$ Alright, thanks, I'll figure it out. But the end result is that there are no solutions? $\endgroup$ – heather Sep 18 '16 at 22:51
  • $\begingroup$ @heather Sorry, I just saw, what you did in your post. I would recommend to use such a vertical line to separate the left and the right side of the equations. It makes things easier. But yes, your conclusion was right, there exists no solution to this system. $\endgroup$ – TheGeekGreek Sep 18 '16 at 22:53

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