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My prof gave us this definition of an Echelon system:

A system of m linear equations in n variables is called an echelon system if

1. m ≤ n.


2. Every variable is the leading variable of at most one equation.


3. Every leading variable is to the left of the leading variables of all lower equations.


4. Every equation has a leading variable.

I don't understand this definition at all.

Q1)If you have this system of equations:

$ax_1 + bx_2 + cx_3 = z$

$dx_1 + ex_2 + fx_3 = z$

its not an echelon system because of multiple equations having the same variable as the leading variable and therefore violating rule #2.

But if you switch around variables and rewrite the equation as:

$ax_1 + bx_2 + cx_3 = z$

$ex_2 + dx_1 + fx_3 = z$

It becomes an echelon system! Why would the way you write a system of equations affect its definition?

Q2)Why would point number three matter? Can't you just move around the equations so that they are in order of variables?

Q3)Why would point number four matter? If you have a equation in the system that doesn't have a leading variable can't you just disregard it? For example if you have the equation $0x = 0$ in the equation can't you just ignore it to have a echelon system since it doesn't affect the solution of the system?

Point no $1$ : please convince yourself that rewriting the equation does not change the nature of the system.

what exactly does he mean by this is... if you "fix some variable" as "leading variable" in first equation ($textbf{WARNING}$ : we fix some variable as leading variable if its coefficient is non zero, so, as you are fixing $x_1$ as leading variable we assume $aneq 0$) then, no other equation in the system should have that variable with non zero coefficient... So, what you are thinking is not quite correct

$$ax_1 + bx_2 + cx_3 = z$$

$$ex_2 + dx_1 + fx_3 = z$$

In this system, fixing $x_1$ as leading variable, we see that first equation has non zero coefficent $a$ of $x_1$ and second equation also has non zero coefficient $d$ (we do assume $dneq 0$)

i am sure you know how to relate linear system with matrices,
for the system
$$ax_1 + bx_2 + cx_3 = z$$

$$ex_2 + dx_1 + fx_3 = z$$

we have corresponding matrix form as
$ left( begin{array}{ccc}
a & b & c \
d & e & f \
end{array} right) $
$ left( begin{array}{c}
x_1 \
x_2\
x_3\
end{array} right) $ = $ left( begin{array}{c}
z \
z\
end{array} right) $

if you have any doubt in seeing this matrix form please let me know.

So, now you have coefficient matrix as $ left( begin{array}{ccc}
a & b & c \
d & e & f \
end{array} right) $

what does first condition for a system being echelon is, as we have assumed coefficient of $x_1$ i.e., $a$ is non zero in first equation, no other equation in the system has non zero coefficient of $x_1$ i.e., in second equation $d$ has to be zero for the system to pass test $1$ for it to be echelon...

So, i think now you are alright with your question no $1$.

Point No $2$ : what you have observed its partially true, after moving around equations we will get into that form.. but moving around equations should be careful enough. that is why we usually ask to $textbf{Transform}$ the system into $textbf{echelon form}$.

If after those rearrangements if the system can be changed to the given form we say that new form to be echelon form

On the whole, what actually we are looking is that the coefficient matrix looks like

$ left( begin{array}{cccc}
a & b & c &d \
0 & e & f &g\
0 & 0 & h &iend{array} right) $

Though this is not exactly "the way it looks" but it is one possibility....