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I keep getting the wrong set of solutions can someone help me. I know that when using the Gauss-Jordan method, the rules that I must follow can be applied in a variety of different procedures then why do I keep getting a different result, I reduce this matrix from its previous form on the right to its reduced system also known as the reduced echelon form that is on the left side. $$ \left[\begin{matrix} 5 & -5 & -15 & 40 \\ 4 & -2 & -6 & 19\\ 3 & -6 & -17 & 41 \\ \end{matrix}\right] = \left[\begin{matrix} 1 & -1 & -3 & 8\\ 0 & 2 & 6 &-13\\ 0 & -3 & -8 & 17\\ \end{matrix}\right] $$

$$ \left[\begin{matrix} 1 & -1 & -3 & 8\\ 0 & 2 & 6 &-13\\ 0 & -3 & -8 & 17\\ \end{matrix}\right]= \left[\begin{matrix} 1 & -1 & -3 & 8\\ 0 & 1 & 3 &\frac{-13}{2}\\ 0 & -3 & -8 & 17\\ \end{matrix}\right] $$

$$ \left[\begin{matrix} 1 & -1 & -3 & 8\\ 0 & 1 & 3 &\frac{-13}{2}\\ 0 & -3 & -8 & 17\\ \end{matrix}\right]= \left[\begin{matrix} 1 & 0 & 0 & \frac{3}{2}\\ 0 & 1 & 3 &\frac{-13}{2}\\ 0 & 0 & 1 & \frac{-5}{2}\\ \end{matrix}\right] $$ $$ \left[\begin{matrix} 1 & 0 & 0 & \frac{3}{2}\\ 0 & 1 & 3 &\frac{-13}{2}\\ 0 & 0 & 1 & \frac{-5}{2}\\ \end{matrix}\right]= \left[\begin{matrix} 1 & 0 & 0 & \frac{3}{2}\\ 0 & 1 & 0 & 1\\ 0 & 0 & 1 & \frac{-5}{2}\\ \end{matrix}\right] $$ Therefore my solution set should be: $$(\frac{3}{2}, 1, \frac{-5}{2})$$ But the solution to the problem from the book I am studying from, states that this is the solution. $$(-\frac{5}{2}, −6, -\frac{11}{2})$$

What rules should I consider and how can I follow all the rules of the Gauss-Jordan method without getting an incorrect set of solutions.

Here is the equation:

5x − 5y − 15z = 40

4x − 2y − 6z = 19

3x − 6y − 17z = 41

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  • $\begingroup$ What rules did you apply? Did you mix row and column operations? $\endgroup$
    – M. Vinay
    Commented Jun 23, 2014 at 7:50
  • $\begingroup$ Well I started by first reducing the top row by multiplying it by -1/5 $\endgroup$ Commented Jun 23, 2014 at 7:53
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    $\begingroup$ Neither your solution, nor solution from the textbook seem to be correct. Unless you typed it wrong here. See this. $\endgroup$
    – Kaster
    Commented Jun 23, 2014 at 7:54
  • $\begingroup$ @user3593705 Nothing wrong in that. Could you update your answer with all the steps you used? $\endgroup$
    – M. Vinay
    Commented Jun 23, 2014 at 7:58
  • $\begingroup$ @M.Vinay Yeah just give me a couple of minutes and I'll rework it, I've been at this problem all day it's driving me crazy. $\endgroup$ Commented Jun 23, 2014 at 8:00

1 Answer 1

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The first mistake is in the fourth matrix. The entry in row 2, column 4 should be $-\frac{13}{2}$ and not $-5$.

EDIT Seems that you have silently corrected this in the meanwhile.

BTW: Formally, your equality signs are not correct. Use, for example, arrows instead.

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  • $\begingroup$ Okay I'll use arrows. $\endgroup$ Commented Jun 23, 2014 at 8:51

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