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Suppose we transform an augmented matrix ( originated form a system Ax=b ) into reduced row echelon form like in this example (original image here)

$$\begin{array}{lr} \left[\begin{array}{cccc|c} \color{red}{1}&2&0&3&2\\ 0&0&\color{red}{1}&-2&5\\ 0&0&0&0&0 \end{array}\right]=\operatorname{rref}(A)&&& \begin{align*} \color{red}{x_1}+2x_2\quad\,+3x_4&=2\\ \color{red}{x_3}-2x_4&=5 \end{align*}\\ \hline \text{Red shows pivot variables; the others are free.} \end{array}$$

Why we need to solve for $x_1$ and $x_3$ ? what so special aobut this variables ? Why cant we solve for $x_2$ and $x_3$ for example ? or for $x_2$ and $x_4$ ?

Why do some columns show up with pivot and we call the corresponding variables the free variables ? Why every variable doesn't have the same importance as the others ?

Thanks

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    $\begingroup$ That link's broken, I think... $\endgroup$ – DonAntonio Mar 29 '13 at 19:46
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    $\begingroup$ @DonAntonio: Click on the thumbnail. $\endgroup$ – Brian M. Scott Mar 29 '13 at 19:47
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    $\begingroup$ I’ve embodied the information from the image into the question, retaining the form as well as I could. $\endgroup$ – Brian M. Scott Mar 29 '13 at 19:57
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There is nothing special about $x_1$ and $x_3$. By elementary row operations you can change $x_2$ and $x_4$ to be pivot variables with $x_1$ and $x_3$ as free variables by the following: $$\begin{array}{lr} \left[\begin{array}{cccc|c} \frac{1}{2}&\color{red}{1}&\frac{3}{4}&0&\frac{19}{4}\\ 0&0&-\frac{1}{2}&\color{red}{1}&-\frac{5}{2}\\ 0&0&0&0&0 \end{array}\right]&&& \begin{align*} \frac{1}{2}x_1+\color{red}{x_2}+\frac{3}{4}x_3\quad\quad&=\frac{19}{4}\\ -\frac{1}{2}x_3+\color{red}{x_4}&=-\frac{5}{2} \end{align*}\\ \hline \end{array}$$ However it is not possible to choose arbitrary variables as pivot variables. We can write $A$ in the form $[a_1,...,a_4]$, where $a_i$'s are the column vectors of $A$. Solving they system $Ax=b$ is essentially to find a linear combination of $(a_1,...,a_4)$ such that $$ x_1a_1+...+x_4a_4=b. $$ The textbook method is to find a particular solution $x^p$ such that $$ x^p_1a_1+...+x^p_4a_4=b $$ and a homogeneous solution $x^h$ such that $$ x^h_1a_1+...+x^h_4a_4=0. $$ Summing them together we get $$ (x^p_1+x^h_1)a_1+...+(x^p_4+x^h_4)a_4=b $$ The general solution is given $x=x^p+x^h$. What determines the pivot variables is the homogeneous equation. The Gaussian elimination reduces the problem to the echelon form. Let $\text{rref}(A)=(a_1^*,...,a_4^*)$, solving the homogeneous equation is equivalent to finding $$ x^h_1a_1^*+...+x^h_4a_4^*=0. $$ Now we need to use linear combination of $a_1^*,...,a_4^*$ to form the $0$ vector. We can choose two vectors, say $a_2^*,a_3^*$, and allow their coefficients $x_2,x_3$ to vary. Now $$ x_2a_2^*+x_3a_3^*\neq 0 $$ in general. To get the homogeneous equation, we need to set the coefficients of $a_1^*$ and $a_4^*$ to vary accordingly such that $$ x_1a_1^*+x_4a_4^*=-(x_2a_2^*+x_3a_3^*) $$
Thus we have chosen $x_1$ and $x_4$ to be pivot variables and $x_2$ and $x_3$ to be free variables. Now you can see why $x_1$ and $x_2$ cannot be pivot variables. As in the reduced echelon form $$ x_1a_1^*+x_2a_2^*=-(x_3a_3^*+x_4a_4^*)\Leftrightarrow x_1\begin{bmatrix}1\\0\\0\end{bmatrix}+x_2\begin{bmatrix}2\\0\\0\end{bmatrix}=-x_3\begin{bmatrix}0\\1\\0\end{bmatrix}-x_4\begin{bmatrix}3\\-2\\0\end{bmatrix} $$ cannot be satisfied since the left hand side only has nonzero first coordinate and cannot be manipulated to match the right hand side.

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  • $\begingroup$ Can we also make x1 and x2 as the pivot variables ? $\endgroup$ – nerdy Mar 30 '13 at 1:01
  • $\begingroup$ No we can't. I made the edit to explain why not. $\endgroup$ – user60610 Mar 30 '13 at 10:25
  • $\begingroup$ Is the matrix in your answear in RREF form ? Is it true that we dont really need the first ( from the left ) non zero element to be 1 ? Can we say we only need columns with 1 and rest of 0's and one pivot for each row? $\endgroup$ – nerdy Mar 31 '13 at 0:40
  • $\begingroup$ I'm in terrible doubt because a lot of ppl said the first matrix you wrote is not in RREF but i think it should be since it has the same coeficients of a RREF matrix with columns swapped. $\endgroup$ – nerdy Mar 31 '13 at 1:36
  • $\begingroup$ @nerdy The first matrix I wrote is not in RREF. Check wikipedia Row echelon form term. RREF is unique and here it is what you wrote in you question. $\endgroup$ – user60610 Mar 31 '13 at 9:02
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There is nothing special about $x_1$ and $x_3$. If you want to make $x_2$ a pivot variable instead of $x_1$ then do the following: $$ x_1 + 2x_2 + 3x_4 = 2 $$ $$ 2x_2 + x_1 + 3x_4 = 2 $$ Divide by 2 $$ x_2 + \frac{1}{2}x_1 + \frac{3}{2}x_4 = 1 $$

Now $x_2$ and $x_3$ are the pivot variables.

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  • $\begingroup$ Okay.What about making x1 and x2 the pivot variables ? $\endgroup$ – nerdy Mar 30 '13 at 1:13
  • $\begingroup$ No, if you choose x_1 to be your pivot variable then you cannot choose x_2 to be the other pivot variable. This is because x_1 is dependant on x2 and x2 is dependant on x_1. If we rearrange the first equation to represent x_1 and x_2, we have: x_1 = 2 - 2x_2 - 3x_4 and x_2 = 1 - (3/2)x_4 - (1/2)x_1 So, the value of x_1 is partly determined by the value of x_2 and vice versa, so we must choose one or the other to be the independent variable. I hope that was understandable. $\endgroup$ – rafidhoda Mar 30 '13 at 5:43
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As I mentioned in chat, there's nothing particularly special about the pivot variables, other than they happen to be the terms that, by our construction, end up with unity coefficients.

This simplifies things in that you don't need to do any division to get the result you want. But in the end, $x_1, x_2, x_3, x_4$ are all independent variables in your system. Writing $x_3$ in terms of $x_4$ is no better or worse than writing $x_4$ in terms of $x_3$ in this case.

There is some notion that the columns that these variables appear in form a basis (indeed, in reduced row echelon form, the standard basis) for your vector space. But once you know how to change bases, then this is of minimal value, except in terms of economy of solution-finding.

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  • $\begingroup$ I will editorialize that the introduction of fairly meaningless terminology in introductory linear algebra is to the detriment of just about everyone involved. $\endgroup$ – Emily Mar 29 '13 at 20:08

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