When solving the matrix
$$\left(\begin{array}{ccc|c}
1 & 1 & 1 & 4\\
1 & 3 & 1 & 4\\
-1 & 2 & 3 & -2\end{array}\right)$$
I somehow made an error with the elementary row operations, but I cannot see it:
$$\left(\begin{array}{ccc|c}
1 & 1 & 1 & 4\\
0 & 5 & 4 & 2\\
0 & 5 & 4 & 2\end{array}\right)\begin{array}{c}
\\
R_2+R_3\\
R_3+R_2\end{array}
\Leftrightarrow\left(\begin{array}{ccc|c}
1 & 0 & 1/5 & 18/5\\
0 & 1 & 4/5 & 2/5\\
0 & 0 & 0 & 0\end{array}\right)\begin{array}{c}
R_1-R_2/5\\
R_2/5\\
R_2-R_3\end{array}$$
This solution is wrong because when substituting the set of solutions $\left\{\begin{array}{c}
x=\frac{18}{5}-\frac{1}{5}t\\
y=\frac{2}{5}-\frac{4}{5}t\\
z=t\end{array}\right.$ into row 2 or 3, I do not get 4 or -2 respectively. I do however note that I could have used a different elementary row operation in the third row of the first calculation:
$$\left(\begin{array}{ccc|c}
1 & 1 & 1 & 4\\
0 & 5 & 4 & 2\\
0 & 3 & 4 & 2\end{array}\right)\begin{array}{c}
\\
R_2+R_3\\
R_3+R_1\end{array}
\Leftrightarrow\left(\begin{array}{ccc|c}
1 & 0 & -1/3 & 10/3\\
0 & 1 & 4/5 & 2/5\\
0 & 0 & 8/5 & 4/5\end{array}\right)\begin{array}{c}
R_1-R_3/3\\
R_2/5\\
R_3-3R_2/5\end{array}
\Leftrightarrow\left(\begin{array}{ccc|c}
1 & 0 & 0 & 7/2\\
0 & 1 & 0 & 0\\
0 & 0 & 1 & 1/2\end{array}\right)\begin{array}{c}
R_1+5R_3/24\\
R_2-R_3/2\\
5R_3/8\end{array}$$
and this solution is correct.
So, what went wrong the first time round?
Best Answer
Both rows in your first expansion are the same, i.e. $R_2 + R_3$ and $R_3 + R_2$. This means that your first transformation is not reversible. Elementary operations are reversible, i.e. If you can transform $A$ to $A'$ using elementary operations, you can also transform $A'$ to $A$ using elementary operations.
In order to avoid doing such a mistake, always perform only one operation at a time. Thus, you would first transform the third row into $R_2 + R_3$, and in the second step, even if you transformed the second row into $R_2 + R_3$, it would not be equal to the third row.