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This question already has an answer here:

Finding the solution(s) for:

$$ \begin{align*} \begin{cases} 2x_1+x_2+3x_3+2x_4 &=5\\ x_1+x_2+x_3+2x_4&=3\\ -x_2+x_3+6x_4&=3 \end{cases} \end{align*} $$

I tried using elimination to rewrite the system in row echelon form, then back substituting. I have no idea what I did wrong, or if I was doing the elimination process correctly.. but I kept getting stuck. I am confused and not really sure how to proceed, I must be awful because I've been staring at this problem for hours.

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marked as duplicate by Namaste linear-algebra Jan 31 '17 at 21:24

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ pudge Why did you delete your first post math.stackexchange.com/questions/2122976/…, after this post here was identified as a duplicate of it? That's game-playing. And therefore I have flagged this question, and your behavior. $\endgroup$ – Namaste Jan 31 '17 at 21:19
  • $\begingroup$ I deleted it because it was a duplicate though? It was an issue so I got rid of it. On my first question it was reformatted and it appeared that the question I had was with the wrong equations I wanted. I was confused -> new correct post -> delete the older one. jeez $\endgroup$ – pudge Jan 31 '17 at 21:22
  • $\begingroup$ Well, you've got a duplicate to turn to and apply to this system of 4 variables and three equations. And, regardless, when you've posted a question, and want to edit it, you are to edit it, not leave it and ask the same question again. $\endgroup$ – Namaste Jan 31 '17 at 21:26
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Adding the third row to the first one and adding the third one to the second one you obtain \begin{array}{lcl} 2x_1+4x_3+8x_4&=&8\\ x_1+2x_3+8x_4&=&6\\ -x_2+x_3+6x_4&=&3 \end{array} Now multiply the second equation by $2$ and subtract it from the first one \begin{array}{lcl} -8x_4&=&-4\\ x_1+2x_3+8x_4&=&6\\ -x_2+x_3+6x_4&=&3 \end{array} Now you know that $x_4=2$ and hence \begin{array}{lcl} x_4&=&2\\ x_1+2x_3&=&-10\\ x_3&=&-9+x_2 \end{array} Finally subtract the last one multiplied by two from the second equation \begin{array}{lcl} x_4&=&2\\ x_1&=&8-2x_2\\ x_3&=&-9+x_2 \end{array} Now it's clear that your system has an infinite number of solutions (one per each value given to $x_2$). This is not surprising since you have four variables and just three equations.

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You can call $x_1=p$ and get the system:

$$ \begin{align*} \begin{cases} x_2+3x_3+2x_4 &=5-2p\\ x_2+x_3+2x_4&=3-p\\ -x_2+x_3+6x_4&=3 \end{cases} \end{align*} $$

If you sum the last equation at the first and second you get

$$ \begin{align*} \begin{cases} 4x_3+8x_4 &=8-2p\\ 2x_3+8x_4&=6-p\\ -x_2+x_3+6x_4&=3 \end{cases} \end{align*} $$

Now subtract the second at the first

$$ \begin{align*} \begin{cases} 2x_3 &=2-p\\ 2x_3+8x_4&=6-p\\ -x_2+x_3+6x_4&=3 \end{cases} \end{align*} $$

And solve as usual. Your solution will be a function of $p$.

Can you finish?

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  • $\begingroup$ When you call x1 = p, x1 would be considered a free variable, correct? $\endgroup$ – pudge Feb 1 '17 at 18:31
  • $\begingroup$ @pudge: yes, that is correct! $\endgroup$ – Arnaldo Feb 1 '17 at 19:41
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$\begin{matrix} 2& 1 &3 &2 &|&5\\ 1 &1 &1 &2 &|&3\\ 0 &-1& 1& -2&|&-1\\ 0&-1& 1& 6& |&3\\ 0& 0& 0& -8& |&-4\end{matrix}\\ x_4 = \frac 12\\ -x_2+x_3 =0\\ x_2 = x_3 = t\\ x_1 + 2t + 1 = 3\\ x_1 = 2-2t\\$

$(x_1,x_2,x_3,x_4) = (2,0,0,\frac 12) + (-2,1,1,0)t$

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  • $\begingroup$ Sorry, new to the subject and this made no sense to me! $\endgroup$ – pudge Jan 31 '17 at 21:51
  • $\begingroup$ There is a line of solutions (not a single solution). What I have there is the vector representation of that line. $\endgroup$ – Doug M Jan 31 '17 at 22:22
  • $\begingroup$ As for the top section, I have put the equation in to augmented matrix form, and then conducted row operations to reduce the problem. $\endgroup$ – Doug M Jan 31 '17 at 22:24
  • $\begingroup$ Thanks! I ended up learning this in the next section, so I understand now. $\endgroup$ – pudge Feb 1 '17 at 18:30

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