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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?

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1 Answer 1

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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.

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