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 = 404x − 2y − 6z = 19
3x − 6y − 17z = 41
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.