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I am a little confused about understanding when we consider a column to have free and pivot variables. So let's say I have the following matrix:

\begin{matrix} 1 & 2 & 0 & -1 & 8\\ 0 & 0 & 3 & 6 & 18 \\ -1 & -2 & 0 & 0 & -3 \end{matrix}

This in the reduced row-echelon form would be:

\begin{matrix} 1 & 2 & 0 & 0 & 3\\ 0 & 0 & 1 & 0 & 16\\ 0 & 0 & 0 & 1 & -5 \end{matrix}

Based on this we say, that columns 1,3 and 4 are pivot columns, and the rest are free. But if I switched columns 2 and 4, or any other columns the results that we would get would be completely different. That is why it feels as though the selection of the pivot and free variables is arbitrary.

Can someone please help me?

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  • $\begingroup$ If you switch columns two and four, you have a different matrix, so you should expect different results, no? $\endgroup$ Mar 19, 2020 at 5:23
  • $\begingroup$ For instance say I had: a+2b=2, 3a-b=3. On switching the columns I get, 2a+b = 2 and -a+3b = 3. This is just like solving what we had before, right? Just that our variable a is now b, and b is now a? $\endgroup$
    – user681443
    Mar 19, 2020 at 5:41

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The essential reason is that interchanging of columns is not a permitted operation when it comes to manipulation of matrices in row echelon form. As you know (hopefully!) there are (essentially) three permitted operations: (1) interchanging of rows; (2) adding/subtracting of rows; and (3) multiplication of a row by a scalar. Think about this in the context of solving a system of equations. If you think about it, interchanging of rows is simply moving the equations around. Whereas, interchanging columns changes the system of equations completely. For example, consider the system \begin{equation} \begin{split} x+y=2 \\ x-y=3 \end{split} \end{equation}

If I interchange rows, the system becomes \begin{equation} \begin{split} x-y=3 \\ x+y=2 \end{split} \end{equation}

which is the same system, whereas if I interchange columns, the system becomes \begin{equation} \begin{split} x+y=3 \\ -x+y=2 \end{split} \end{equation} which is a different system, with a different solution set.

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  • $\begingroup$ I guess I am confused because you could write x+y = 2 as y+x=2. And the same for. x-y=3 as y-x=3. In this case, I am switching the columns, aren't I? $\endgroup$
    – user681443
    Mar 19, 2020 at 5:37
  • $\begingroup$ Yes, but by switching columns you are changing the system. The first case is okay, because you are not changing the relationship between $x$ and $y$. The equations $x+y=2$ and $y+x=2$ are the same. However, in the second case, $x-y=3$ is not the same equation as $y-x=3$. The first has (e.g.) solution $x=6,y=3$, but this is not a solution to the second. $\endgroup$ Mar 19, 2020 at 5:44
  • $\begingroup$ Okay. I understand this. Say we had, x-y=2. Isn't this the same as -y+x=2.I am still switching the columns right? $\endgroup$
    – user681443
    Mar 19, 2020 at 5:49
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    $\begingroup$ Yes, you're right that $x-y=2$ is the same as $-y+x=2$. But 'switching the columns' would mean going from $x-y=2$ to $-x+y=2$, which are not the same. $\endgroup$ Mar 19, 2020 at 5:49
  • $\begingroup$ @OishikaChaudhury Remember that you’re expressing the system of equation in the “bulk” matrix form $A\mathbf x=\mathbf b$. If you “swap columns” by rewriting the system of equation so that the variables are in a different order, then not only does the order of the columns of $A$ change, but so does the order of the rows of $\mathbf x$ and $\mathbf b$. $\endgroup$
    – amd
    Mar 19, 2020 at 19:59

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