Relationship between the rows and columns of a matrix I am having trouble understanding the relatioship between rows and columns of a matrix.
Say, the following homogeneous system has a nontrivial solution.
$$
3x_1 + 5x_2 − 4x_3 = 0  \\
−3x_1 − 2x_2 + 4x_3 = 0  \\
6x_1 + x_2 − 8x_3 = 0\\
$$
Let A be the coefficient matrix and row reduce 
$\begin{bmatrix}
A & \mathbf 0
\end{bmatrix}$ 
to row-echelon form:  
$\begin{bmatrix}3&5&-4&0\\-3&-2&4&0\\6&1&-8&0\\ \end{bmatrix}
\rightarrow
\begin{bmatrix}3&5&-4&0\\0&3&0&0\\0&0&0&0\\ \end{bmatrix}$
$\quad a1 \quad a2 \quad \,a3$
Here, we see $x_3$ is a free variable and thus we can say 3rd column,$\,a_3$, is in $\text{span}(a_1, a_2)$
But what does it mean for an echelon form of a matrix to have a row of $0$'s?
Does that mean 3rd row can be generated by 1st & 2nd rows?
just like 3rd column can be generated by 1st & 2nd columns?
And this raises another question for me, why do we mostly focus on columns of a matrix?
because I get the impression that ,for vectors and other concepts, our only concern is
whether the columns span $\mathbb R^n$ or the columns are linearly independent and so on.
I thought linear algebra is all about solving a system of linear equations, 
and linear equations are rows of a matrix, thus i think it'd be logical to focus more on rows than columns. But why?

Having a row of $0$'s in the row-echelon form means that we were able to write the third row of $A$ as a linear combination of the second and first rows.  As it so happens for square matrices, this is true precisely when we can write the columns as a linear combination of each other (that is, when the columns are not linearly independent).  If you further reduce this to reduced row-echelon form, you get
$$\begin{bmatrix}
1 & 0 & -4/3 & 0\\
0 & 1 & 0 & 0 \\
0 & 0 & 0 & 0
\end{bmatrix}$$
Because the third column lacks a pivot, $x_3$ is our free variable, which means that we can write $a_3$ as a linear combination of the other two columns.
There's a very good reason for focusing on the columns of a matrix.  This comes out of using $A$ as a linear transformation, where the "column space" gives us the "range" of the function $f(\vec x) = A \vec x$.
