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:

enter image description here

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?

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    $\begingroup$ There is also a similar sequence if you just count the distinct arrangements of $n$ planes in $3$ dimensions. One plane has $1$ arrangement, two planes have $3$ arrangements (parallel, intersecting, coinciding), and three planes have the $8$ arrangements pictured. How does this sequence continue? Is it the same as the one in my original question? $\endgroup$ Oct 30, 2020 at 0:31
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    $\begingroup$ Can you explicitly define the arrangements? This wiki page may contain relevant information. $\endgroup$
    – Noah Tang
    Nov 5, 2020 at 12:40
  • $\begingroup$ What does this have to do with systems of equations? For example, in your image of arrangements of planes, (3), (4), (5) and (8) all have the same type of solution for the corresponding system of equations: that is, no solution. Should we just stop talking about systems of equations and just talk about hyperplanes? $\endgroup$ Nov 5, 2020 at 13:49
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    $\begingroup$ @BrianDrake I brought up systems because systems produce the graphs pictured above. I'm not worried whether a system has solution(s) or not. I guess it's not even necessary to consider hyperplanes. What if a 4th plane was added to each of the pictures above? How many more arrangements would there be? $\endgroup$ Nov 6, 2020 at 14:16

1 Answer 1


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.


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