In this homework problem, I'm asked to find the solution to the following augmented matrix that represents a system of linear equations. Extracting the solution from the matrix isn't a problem, so I'll just show you guys how I got to my reduced echelon form.
$\left[\begin{array}{c c c c c} 3 & 3 & 6 & 12 \\ -6 & -6 & -12 & -24 \\ 0 & -4 & -12 & 16 \\ \end{array}\right]$
Here is my thought process:
I see that the 2nd row is simply a scalar multiple of the first, so that will turn into a row of zeros and move to the bottom. Further, the first and third rows (in the original matrix) can be divided by scalars $3$ and $4$, respectively. So far, we have
$\left[\begin{array}{c c c c c} \color{red}3 & \color{red} 3 & \color{red} 6 & \color{red}{12} \\ \color{blue}{-6} & \color{blue}{-6} & \color{blue}{-12} & \color{blue}{-24} \\ \color{green}{0} & \color{green}{-4} & \color{green}{-12} & \color{green}{16} \\ \end{array}\right] \sim \left[\begin{array}{c c c c c} \color{red}1 & \color{red}{1} & \color{red}{2} & \color{red}{4} \\ \color{green}{0} & \color{green}{-1} & \color{green}{-3} & \color{green}{4} \\ \color{blue}0 & \color{blue}0 & \color{blue}0 & \color{blue}0 \end{array}\right]$
Now we replace row 2 with the sum of rows 1 and 2, and get
$\left[\begin{array}{c c c c c} 1 & 1 & 2 & 4 \\ 0 & 0 & -1 & 8 \\ 0 & 0 & 0 & 0 \\ \end{array}\right]$
Finally, we add replace row 1 with sum of twice row 2 and row 1 to get the following reduced echelon form:
$\left[\begin{array}{c c c c c} 1 & 1 & 0 & 20 \\ 0 & 0 & -1 & 8 \\ 0 & 0 & 0 & 0 \\ \end{array}\right]$
Apparently, the correct reduced echelon form is
$\left[\begin{array}{c c c c c} 1 & 0 & -1 & 8 \\ 0 & 1 & 3 & -4 \\ 0 & 0 & 0 & 0 \\ \end{array}\right]$
I can see that the discrepancy results from my second step: instead of replacing row 2 with the sum of rows 1 and 2, it seems it is correct to replace row 1 instead with the same row operation. But why is what I did wrong?
