Suppose you and I (and a few hundred other people) independently solve a system of linear equations, and we want to compare our answers. If we have taken Gaussian elimination to reduced row-echelon form, and have expressed our answers with the free variables as parameters, we can immediately compare our answers; if there is any difference at all, then (at least) one of us is wrong. If I used the free variables, and you used the leading variables, then we have some calculating to do before we can see whether we got the same answer.
So, it's really a matter of selecting a standard form so everyone can agree on an answer without having to do a lot more calculation to check.
EDIT: Lots of things that come later depend on solving systems of equations (e.g., finding a basis for the kernel of a linear transformation, finding a basis for an eigenspace of a matrix), and it continues to be useful to have a standard form.
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 $a\neq 0$) 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 $d\neq 0$)
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 $1$.
Best Answer
Echelon form helps up solve the system, pure and simple. If all these 4 are met, then we can successfully solve our system for our n variables. I'm assuming you will see the importance of putting a matrix in echelon form, and after this, into reduced row echelon form in the following classes...
Let's look at a random echelon $3\times{3}$ matrix $A$ and let $b=\left( \begin{matrix} 1\\ 1\\ 1\\ \end{matrix}\right)$. Then we have our $A\mathbf{x}=b$ $$ \left( \begin{matrix} 1 & 3 & 3\\ 0 & 1 & -1\\ 0 & 0 & 3\\ \end{matrix}\right) \left( \begin{matrix} x_1\\ x_2\\ x_3\\ \end{matrix}\right)= \left( \begin{matrix} 1\\ 1\\ 1\\ \end{matrix}\right) $$ We can turn this into an augmented $3\times{4}$ matrix by adjoining $b$ onto $A$. $$ \left( \begin{matrix} 1 & 3 & 3 & 1\\ 0 & 1 & -1 & 1\\ 0 & 0 & 3 & 1\\ \end{matrix}\right) $$ From here we can deduce that for our vector $\mathbf{x}=\left( \begin{matrix} x_1\\ x_2\\ x_3\\ \end{matrix}\right)$in $\mathbb{R}^3$ we have $$3x_3=1\rightarrow{x_3}=\frac{1}{3}$$ $$x_2-x_3=1\rightarrow{x_2}=\frac{4}{3}$$ $$x_1+3x_2+3x_3=1\rightarrow{x_1}=-4$$ Thus, the vector that solves this system is $\mathbf{x}=\left( \begin{matrix} -4\\ \frac{4}{3}\\ \frac{1}{3}\\ \end{matrix}\right)$
If $A$ were in this echelon form, $$\left( \begin{matrix} 1 & 3 & 3 & 1\\ 0 & 1 & -1 & 1\\ 0 & 0 & 0 & 0\\ \end{matrix}\right)$$ Then our system of equations looks like this $$x_1+3x_2+3x_3=1$$ $$x_2-x_3=1$$ What arises here is we have multiple values for both $x_2$ and $x_3$ that solve this equation. To generalize a solution we introduce a parameter, $t$, such that $x_3=t$ for example and see that $$x_3=t$$ $$x_2=t+1$$ $$x_1=-6t-2$$ Thus here for values of $t,$ where $t\in\mathbb{R}$ all solve $A\mathbf{x}=b$. Two vectors that are solutions are $\mathbf{x}=\left( \begin{matrix} -2\\ 1\\ 0\\ \end{matrix}\right)_{t=0}$ and $\mathbf{x}=\left( \begin{matrix} 4\\ 0\\ -1\\ \end{matrix}\right)_{t=-1}$
Now if the random echelon matrix looks like this, $$ \left( \begin{matrix} 1 & 3 & 3 & 1\\ 0 & 1 & -1 & 1\\ 0 & 0 & 0 & 1\\ \end{matrix}\right) $$ and you perform the elimination, you can see there is no solution to this system since you essentially get $0=1$ in row $3$. So echelon form expedites solving system of equations. There are other benefits as well such as discerning the rank, dimension of various subspaces, etc. You will learn much of this in the coming classes.