Why is echelon form important? My professor gave us this definition for a system of equations in echelon form:

A system of m linear equations in n variables is called an echelon system if
  
  
*
  
*m ≤ n.
  
*Every variable is the leading variable of at most one equation.
  
*Every leading variable is to the left of the leading variables of all lower equations.
  
*Every equation has a leading variable.
  

He stressed the importance of this form, but didn't actually give us a reason why it was important. Why is this form so important? It seems kinda random.
 A: 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.
