How did my book see the rank and what is wrong with my null space 

Reduced Row Echleon form: 
$$\begin{bmatrix} 3 & 1 & 3 & -2 \\ 2 & -1 & 4 & -5 \\ 0 & 0 & 0 & \theta+6 \\ 0 & 0 & 0 & \theta +6 \end{bmatrix}$$
$$\begin{bmatrix} 1 & 2 & -1 & 3 \\ 0 & -5  & 6 & -11 \\ 0 & 0 & 0 & \theta+6 \\ 0 & 0 & 0 & 0 \end{bmatrix}$$
Surprisingly my book ends here and states : 

It is then a simple matter to deal with the cases θ ≠ −6 and θ = −6 .

How come? I don't see how the rank becomes in the cases at this stage. How did the book see it! How is it simple to see this?

Ofcourse, I paused the Row reduction
I had to continue by doing Column reduction: 
$$\begin{bmatrix} 1 & 0 & 1 & -3 \\ 0 & -7 & 1 & 4 \\ 0 & 0 & 0 & \theta+6 \\ 0 & 0 & 0 & 0 \end{bmatrix}$$
$$\begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & \theta+6 \\ 0 & 0 & 0 & 0 \end{bmatrix}$$
Only now is it clear to me.
How did the book see the rank before these further steps? And by not doing column reduction.

For the Null Space, which matrix do I use? I am confused because look at this matrix: 
$$\begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & \theta+6 \\ 0 & 0 & 0 & 0 \end{bmatrix}$$
Using this I get : 
$x=0, y= 0 , z= ? , (\theta+6)t=0$
But this is not obvious the Null Space. What is wrong with my Linear Algebra knowledge? Please help.
 A: The kicker, here, is that the rank of a matrix is equal to the number of rows with non-zero entries in row echelon form. In the case that $\theta=-6,$ there are only $2$ such rows, while for $\theta\ne-6,$ there are $3$.
How are the rank and the dimension of the null space related?
As for the second part, remember that $\theta\ne-6,$ so $\theta+6\ne0,$ and so from $(\theta+6)t=0,$ we can conclude that $t=0.$ Consequently, a basis vector for $K$ will have the form $[x_1,y_1,z_1,0]^\dagger,$ as mentioned in the problem.
However, you should not be using column-reduction to find the null-space. Instead, you should continue on to find the RREF, keeping in mind that $\theta\ne-6.$ You should find equations with $x$ and $y$ dependent on $z,$ $z$ is a free variable (since it doesn't correspond to a pivot column), and the equation $t=0$.
A: Hint: for certain values of $\theta$, you can just divide the third row by $(\theta+6)$ to get a $1$ in that row.  When is this impossible?
Also, column reduction does not preserve null-space.

We now have the system of equations
$$
x + 2y + z + 3t = 0\\
-5y + 6z - 11 t = 0\\
(\theta + 6) t = 0
$$
In either case, we have
$$
y = \frac{1}5(6z - 11t)\\
x = -2y - z - 3t = -\frac{2}5(6z - 11t) - z - 3t
$$
From there, you just need to make sense of the last equation.
A: For $\theta \neq -6$, you have three linearly independent rows. For $\theta = 6$, you have two linearly independent rows. Then you recall that the row rank and the column rank are the same things. Maybe this is how the author sees it. You can also think of it in terms of pivots ; for $\theta \neq -6$, you have three pivots, so the rank is $3$. For $\theta = -6$, you have two pivots, so the rank is $2$. Note that the rank plus the dimension of the null space equals $4$ by the rank-nullity theorem.
Hope that helps,
