How do I know if the linear system has a line of intersection? I was wondering how can I determine if there is a line of intersection with any matrix?
For example, if I have the following matrix:
$$\left(\begin{array}{rrr|r}
1 & -3 & -2 & -9 \\
2 & -5 & 1 & 3 \\
-3 & 6 & 2 & 8  \\
\end{array} \right)$$
What does the solution have to look like for me to conclude that there is a line of intersection?
P.S. I know this matrix has a point of intersection but I used this as an example because I didn't know an example for a matrix that had a line of intersection. 
 A: The existence of a "point of intersection" is the existence of a point $(x, y, z)$ that satisfies the system of equations: a point lying on each line represented by the corresponding system of equations:
$$\begin{align} x - 3y -2z & = -9 \\ 2x - 5y + z & = 3 \\ -3x + 6y + 2z & = 8 \end{align}$$
A unique solution exists (a single point of intersection exists) if the augmented matrix does not reduce to a row of all zeros, and no row has all zeros, augmented by a non-zero entry. 
If you obtain a row of all zeros, through row reduction, then infinitely many points of intersection occur (two or more lines will be concurrent): the entries in one or more lines will be a scalar multiple of the entries of another. Put differently, there will be a line of intersection.
If the matrix reduces to a row of three zeros, with a non-zero entry in the last column of that row, no solution exists (i.e., no point of intersection exists.)
All these possibilities can be determined by reducing the matrix to row echelon form.
We can reduce your example matrix:
$$\left(\begin{array}{rrr|r}
1 & -3 & -2 & -9 \\
2 & -5 & 1 & 3 \\
-3 & 6 & 2 & 8  \\
\end{array}\right)$$
$$\left(\begin{array}{rrr|r}
1 & -3 & -2 & -9 \\
0 &  1 & 5 & 21 \\
0 & 0 & 11 & 42  \\
\end{array}\right)$$
If reduced further, you'd see that the system represented by the system of equations has a unique point of intersection, hence no common line of intersection. 
Note that if the last row were $(0\;\;2\;\;10\;\;42)$, we could "zero it" by taking $-2R_2 + R_3 \rightarrow R_3$, and obtain a row of all zeros, since the third row would be a scalar multiple of the second row. In that event, there would indeed be a line of intersection. If any row is a linear combination of the other rows, we have a linearly dependent system of equations: this shows when a row-reduced matrix has a row of all zeros. And in that case, there is at least a line of intersection.
A: If you are familiar with ranks and Rouché-Capelli (RC) theorem, the following reasoning can be used to answer. The system $A\mathbf x=\mathbf b$, where $A$ is an $n\times m$ matrix, has solutions if and only if 
$$
r(A)=r(A|\mathbf b)=r,
$$
and in this case you have "$\infty^{n-r}$" solutions. This really means that your solutions depend on $n-r$ free parameters. So, if the rank $r=n-1$, your solution set depends on $n-(n-1)=1$ free parameter or, in other words, it's monodimensional. This helps the intuition that the solution set is indeed a line in this case. Hence, you'll get a line of solutions when the rank of the complete matrix $r$ is equal to the number of unknowns $n$ less 1. As an example, consider the system
$$
\begin{align} x - y & = 2 \\ -2x +2y &=-4
\end{align}
$$
The rank $r(A)=r(A|\mathbf b)=1$ and you can indeed write the solutions set as $x=2+t,y=t$, which is equal to the set of points on the line $y=x-2$.
A more interesting example (I like a lot $A$!) is
$$
\begin{align} 
x +y +z& = 6 \\ 
4x +5y +6z&=15\\
7x+8y+10z&=24
\end{align}
$$
The reduced complete matrix is
$$
\left(\begin{array}{rrr|r}
1 & 2 & 3 & 6 \\
0 & -3 & -6 & -9 \\
0 & 0 & 0 & 0  \\
\end{array}\right)
$$
from which you get the solutions $(x=t,y=3-2t,z=t), \forall t$. The solution set can be described in terms of a unique parameter $t$ and it is a straight line in $\mathbf R^3$.
