Arrangements of Solutions to $n$ Linear Equations with $n$ Unknowns In a system of two linear equations with two unknowns, there are three "arrangements" that we can see when we graph the two lines in the $xy$ plane:

*

*The two lines intersect at a single point (one solution to the system).

*The two lines are parallel and never intersect (no solution to the system).

*The two equations describe the same line (infinitely many solutions to the system).

When we move up to a system of three linear equations with three unknowns, now we have three planes in space, and there are eight distinct arrangements of the three planes:

If we consider one linear equation with one unknown, I suppose it makes sense to say that there is one arrangement, which is a single point on the real number line.
So, for one, two and three unknowns, we have the start of a sequence: $1, 3, 8, ...$
I am interested in how this sequence continues. I have searched in vain on the OEIS. Alas, there are many sequences that have $1, 3, 8,...$ and I'm not sure which, if any, are the right one.
This one: https://oeis.org/A001792 looks like it might be it, because the comments say that sequence is related to matrices in certain ways. Also I expect a formula for this sequence should involve powers of 2. But, I wouldn't bet any money on it.
Is there a sequence with a simple formula to work out the number of arrangements?
 A: More than an answer, the following represent some considerations on how the problem could be tackle.
Start and consider just two planes (in $3$-D) with the relevant
matrix of coefficients $\bf C$ and the augmented matrix (coefficients plus known terms) $\bf A$
$$
\underbrace {\left( {\overbrace {\matrix{
   {a_1 } & {b_1 } & {c_1 }  \cr 
   {a_2 } & {b_2 } & {c_2 }  \cr 
 } }^{\bf C}\;
\left| {\;\matrix{
   {d_1 }  \cr 
   {d_2 }  \cr 
 } } \right.} \right)}_{\bf A}\;
$$
Now, if the matrix $\bf C$

*

*has full rank $2$, then the two normal vectors are independent, i.e.
the planes are incident;

*has rank $1$, then the two vectors are dependent, i.e. parallel;

*the rank $0$ corresponds to having all the coefficients null, which is a
degenerated situation.

The rank of $\bf A$ (call it $R$) cannot be less than that of $\bf C$ ($r$), nor it can be grater than $r+1$.
When $r=0$, $R$ can be $0$ or $1$ and it will be in any case a degenerated situation.
When $r=1$, $R$ can be $1$ (planes coincident), or $2$ (planes parallel).
When $r=2$ also $R=2$ and the planes are incident.
Passing to consider three $3$-D planes, your scheme of distinction
is unfortunately wider than that provided by considering  the ranks of the $3 \times 4$ matrices.
For the eight cases that you sketched in fact we have
$$
\begin{array}{c|cccccccc}
   {} &  &  {\left( 1 \right)} & {\left( 2 \right)} & {\left( 3 \right)}
 & {\left( 4 \right)} & {\left( 5 \right)} & {\left( 6 \right)}
 & {\left( 7 \right)} & {\left( 8 \right)}  \\
\hline
   r &  &  1 & 2 & 1 & 2 & 1 & 3 & 2 & 2  \\
   R &  &  1 & 2 & 2 & 3 & 2 & 3 & 2 & 3  \\
\end{array}
$$
We have that the three couples $[(2),\, (7)] , \; [(3),\, (5)], \; [(4),\, (8)]$ cannot be distinguished.
To be able to distinguish between them, for $r < 3$,  we could introduce a fourth plane, with coefficients
independent from the previous three, which means that it increases $r$ to $r+1$, and count
the crosses we are going to have.
Equivalently, we can consider both the ranks of the first two planes and how they change by
adding the third.
However this way would soon become impractical for higher dimensions.
