This table, like the rest of the information on this topic, has been obtained by straightforward counting arguments.

The numbers of secants, the total number of tangents, and the total number of exterior lines are given by the parameters of the two geometries, as are the numbers of points in the three classes. The structure of the secants is likewise given, and the structures of one type of tangent and of one type of exterior line are given by considering the plane of type B0, discussed here. The structure of the second type of tangent, as is the number of such lines, is then determined. The information thus far obtained is then sufficient to get the rest of the matrix, using a set of simultaneous equations.

The points of type C₁ are on the secants only, while the points of type C₂ are in the space "between" the secants. A tangent of type 1, "originating" in a point of the sub-geometry, is embedded in one, and one only, plane of type B₀ and meets all the secants in that plane and not passing through that point of origin in a point of type C₁, as well as having some points of type C₂ in that plane. An exterior line of type 1 is also embedded in one, and one only, plane of type B₀, and intersects all the secants in that plane in points lying "between" the points of the sub-geometry.

There are q^3ν+1 planes through each secant: of these q^ν+1 are planes of type B₀, so there are q^3ν-q^ν of type B₁. There are (q^2ν+1)(q^2ν+q^ν+1) secants, so the number of planes of type B₁ is as shown. Supposing the sub-geometry points of this secant to be A,B,...,C, these planes then cut the planes of type B₀ not passing through that secant, but passing through one of A,B,...,C, in tangents of type 1, and the remaining planes of type B₀ in exterior lines of type 1.

Likewise, there are (q^6ν+q^3ν+1) planes through a point of the sub-geometry (say, A): of these, (q^2ν+q^ν+1) are planes of the sub-geometry (i.e. of type B₀) and a further (q^3ν-q^ν) pass through each of the secants through u, leaving the given number of planes of type B₂. Each of these planes meets the other planes through A in a tangent: so it must meet the planes of type B₀ in a tangent of type 1, of which there are (q^2ν+q^ν+1), which accounts for all the tangents of type 1 in a plane of type B₂.

Since every line in the plane must meet every other such line in a point, the tangents containing point A (say) of the sub-geometry must meet all the secants not containing A in a point: in other words, the C₁ points in the tangent comprise a transversal of the C₁ points in the q^2ν such secants. Likewise, the C₁ points in an exterior line comprise a transversal of the C₁ points in the q^2ν+q^ν+1 secants.

In 3-space, there are q^3ν+1 planes through a secant. Of these, only q^ν+1 pass through another secant, so the remainder have in addition just tangents and exterior lines.

There are no lines of type x₁ in a plane of type B₁. That this must be so may be seen as follows.

Suppose a plane α of type B₀, containing an exterior line x of type 1. Firstly, no plane of type B₁ containing a secant from α can contain x, obviously. So suppose a plane γ of type B₁ containing a secant t (not in α), where t is contained by a plane β of type B₀, distinct from α. The two planes α and β meet in a line, in particular in a secant, say s. Since s and t both lie in β, they must meet in a point, say A, belonging to the subgeometry. This must also belong to γ since it lies on t. If γ contains x, then it contains a point (of type C1) from each of the secants of α, so it contains two points from some secant of α, and therefore a whole secant from α, which is a contradiction.

The points of type C₁ which occur in the secant clearly cannot appear in any tangents, so the class has been split in two, C₁₁ and C₁₂, to accommodate the tangents, where C₁₁ comprises the points which lie on the secant. Clearly, no pair of points from C₁₁ can occur together elsewhere in the geometry. This gives the structure of the exterior lines of type 2, the C₁ points being split 1:q^ν between C₁₁ and C₁₂. The total number of C₁₂ and C₂ points is q^6ν. The total number of tangents is (q^ν+1)q^3ν, so the number of exterior lines (of type 2) is q^4ν(q^2ν-1).

This information is sufficient to determine successively: the occurrence of C₁₂ points and their total number, hence the total number of C₂ points, the occurrence of C₀ points in a tangent of type 1, and so in a tangent of type 2, and the total number of such tangents, therefore the occurrence of C₁₂ and C₂ points in those tangents, and therefore in the exterior lines.

The points of type C₂ have been split into two sub-classes. Each point of type C₁ and C₂ can appear only once in a tangent, so if there are tangents of both types, the C₂ points must be split, and each tangent type uses points from just one of the sub-classes.

The next step is to determine the number of type 1 tangents (the total number of tangents is (q^3ν+1), so the total number of exterior lines is q^6ν). Now, suppose the plane of type B₂ is α and the C₀ point in α is A. Through A there pass (q^2ν+q^ν+1) planes of type B₀: each of these meets α in a line, which must be a tangent of type 1. Conversely, each tangent of type 1 lies in one and one only plane of type B₀ . Hence, the number of tangents of type 1 is (q^2ν+q^ν+1), and that of type 2 tangents is (q^3ν-q^2ν-q^ν). This gives the number of C₁, C₂₁ and C₂₂ points. The rest of the matrix can be obtained by solving a set of simultaneous equations.