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For convenience, the generic multiplication table from the home page is repeated here:

generic multiplication	q	r	s	t	u	v
q	3r − 2t + 2v	q + s	r + t	s + u	t + v	u
r	q + s	2r − t + 2v	q + u	r + v	s	t − v
s	r + t	q + u	2r − 2t + 3v	q	r − v	s − u
t	s + u	r + v	q	2r − 2t + v	q − u	r − t
u	t + v	s	r − v	q − u	2r − 3t + 2v	q − s
v	u	t − v	s − u	r − t	q − s	r − 2t + 2v

There exist solutions employing 6-by-6 integer matrices and the usual tools of linear algebra. As an example, the six matrices below (here displayed as HTML tables) simultaneously satisfy the multiplicative relations above. They bear the subscript 1 to distinguish them from other matrices to appear later.

q1 0 +3 0 −2 0 +2 +1 0 +1 0 0 0 0 +1 0 +1 0 0 0 0 +1 0 +1 0 0 0 0 +1 0 +1 0 0 0 0 +1 0	s1 0 +1 0 +1 0 0 +1 0 0 0 +1 0 0 +2 0 −2 0 +3 +1 0 0 0 0 0 0 +1 0 0 0 −1 0 0 +1 0 −1 0	u1 0 0 0 +1 0 +1 0 0 +1 0 0 0 0 +1 0 0 0 −1 +1 0 0 0 −1 0 0 +2 0 −3 0 +2 +1 0 −1 0 0 0
r1 +1 0 +1 0 0 0 0 +2 0 −1 0 +2 +1 0 0 0 +1 0 0 +1 0 0 0 +1 0 0 +1 0 0 0 0 0 0 +1 0 −1	t1 0 0 +1 0 +1 0 0 +1 0 0 0 +1 +1 0 0 0 0 0 0 +2 0 −2 0 +1 +1 0 0 0 −1 0 0 +1 0 −1 0 0	v1 0 0 0 0 +1 0 0 0 0 +1 0 −1 0 0 +1 0 −1 0 0 +1 0 −1 0 0 +1 0 −1 0 0 0 0 +1 0 −2 0 +2

The table below, excerpting one column from the generic multiplication table above, shows how q₁ was obtained:

generic multiplication	q		q expanded		q1 matrix
q	3r − 2t + 2v	⇒	0q + 3r + 0s − 2t + 0u + 2v	⇒	0	+3	0	−2	0	+2
r	q + s		1q + 0r + 1s + 0t + 0u + 0v		+1	0	+1	0	0	0
s	r + t		0q + 1r + 0s + 1t + 0u + 0v		0	+1	0	+1	0	0
t	s + u		0q + 0r + 1s + 0t + 1u + 0v		0	0	+1	0	+1	0
u	t + v		0q + 0r + 0s + 1t + 0u + 1v		0	0	0	+1	0	+1
v	u		0q + 0r + 0s + 0t + 1u + 0v		0	0	0	0	+1	0

This simple rtv3 calculation produces the identity matrix as hoped:

r1 − t1 + v1 = r1 t1 v1
+1	0	0	0	0	0
0	+1	0	0	0	0
0	0	+1	0	0	0
0	0	0	+1	0	0
0	0	0	0	+1	0
0	0	0	0	0	+1

Here is the parallel qsu3 calculation:

q1 + s1 − u1 = q1 s1 u1
0	+4	0	−2	0	+1
+2	0	0	0	+1	0
0	+2	0	−1	0	+4
0	0	+1	0	+2	0
0	−1	0	+4	0	−2
−1	0	+2	0	0	0

Another solution results if q₁ through v₁ are all transposed.

In the following equation, let a through f be real numbers:

a · q₁ + b · r₁ + c · s₁ + d · t₁ + e · u₁ + f · v₁ = 0

The six matrices are linearly independent because the only solution is when a through f are all equal to zero. Observe:

• Only q₁ has a nonzero at row 4, column 3; whence a must be zero.
• r₁ at 1, 1; whence b = 0.
• s₁ at 2, 5; whence c = 0.
• t₁ at 5, 5; whence d = 0.
• u₁ at 6, 1; whence e = 0.
• v₁ at 3, 3; whence f = 0.

Matrices in this system tend to have many zeroes. This characteristic invites a search for a more condensed representation, which follows.

The following 3-by-3 matrices are one way to represent the R-T-V subdomain. They were formed by deleting the 1st, 3rd, and 5th rows and columns of the source 6-by-6 matrices.

r2 +2 −1 +2 +1 0 +1 0 +1 −1

t2 +1 0 +1 +2 −2 +1 +1 −1 0

v2 0 +1 −1 +1 −1 0 +1 −2 +2

There is a rationale for deleting those rows and columns. Here is the generic multiplication table reduced to the R-T-V subdomain:

generic subdomain multiplication	r	t	v
r	2r − t + 2v	r + v	t − v
t	r + v	2r − 2t + v	r − t
v	t − v	r − t	r − 2t + 2v

The table below, excerpting one column from the generic subdomain multiplication table immediately above, shows how r₂ was obtained:

generic subdomain multiplication	r		r expanded		r2 matrix
r	2r − t + 2v	⇒	2r − 1t + 2v	⇒	+2	−1	+2
t	r + v		1r + 0t + 1v		+1	0	+1
v	t − v		0r + 1t − 1v		0	+1	−1

Now 3-by-3 matrices q₂, s₂, u₂ can be defined in order to enable representation of the entire septright domain, although these matrices' components will not be integers. To find out what the components might be, define matrix a₂ = 2I + r₂, where "I" represents the 3-by-3 identity matrix. Because (q₂)² should equal a₂, a step toward finding a suitable q₂ is to examine the square roots of a₂. Toward that end, observe that the 3-by-3 matrix a₂ has 3 distinct positive eigenvalues:

• 2 + 2 cos (180° / 7)
• 2 − 2 cos (360° / 7)
• 2 + 2 cos (540° / 7)

Now define k = √(1/7) ≈ 0.377964. Then a convenient choice of √a₂ for establishing q₂ is the following:

q2
+ 6k	− 3k	+ 5k
+ k	+ 3k	+ 2k
− k	+ 4k	− 2k

(Except for − q₂, the components of the other square roots of a₂ do not fall into simple integer ratios.)

Simple matrix calculation yields the others:

s2 + 3k + 2k − k + 4k − 2k + k + 3k − 5k + 6k

u2 + 2k − k + 4k + 5k − 6k + 3k + 2k − k − 3k

Now q₂ through v₂ will satisfy the generic multiplication table at the top of this page.

When any two of q₂, s₂, u₂ are multiplied, the product is an integer matrix.

The following 3-by-3 matrices are another way to represent the R-T-V subdomain. These symmetric matrices were discovered by deleting the 2nd, 4th, and 6th rows and columns of the source 6-by-6 matrices:

r3 +1 +1 0 +1 0 +1 0 +1 0

t3 0 +1 +1 +1 0 0 +1 0 −1

v3 0 0 +1 0 +1 −1 +1 −1 0

By a diagonalization procedure similar to that above, the matrices q₃, s₃, u₃, also symmetrical, can be produced. Of the eight square roots of 2 + r₃, a convenient choice for q₃ is:

q3
+ 4k	+ 2k	− 1k
+ 2k	+ 1k	+ 3k
− 1k	+ 3k	+ 2k

where k = √(1/7) as before.

Simple matrix calculation yields the others:

s3 + 2k + 1k + 3k + 1k + 4k − 2k + 3k − 2k + 1k

u3 − 1k + 3k + 2k + 3k − 2k + 1k + 2k + 1k − 4k

Another way to manage the many zeroes in q₁ through v₁ is to make block matrices by rearranging rows and columns. There are of course many ways to do this, the one shown below being particularly suitable.

All six matrices are transformed the same way, described here with q₁ as an example. Rows of q₁ will be shifted to form intermediate result m, whose colums will be shifted to form q₄.

first copy …	from this row of q1	to this row of m	then copy …	from this column of m	to this column of q4
	1	4		1	4
	2	1		2	1
	3	5		3	5
	4	2		4	2
	5	6		5	6
	6	3		6	3

The following are yielded:

q4 0 +1 +1 0 0 +1 +1 0 0 +1 +3 −2 +2 0 +1 +1 0 0 +1 +1	s4 0 +1 0 +1 +1 0 0 0 +1 −1 +1 +1 0 0 +2 −2 +3 +1 0 −1	u4 0 0 +1 0 +1 0 −1 +1 −1 0 0 +1 +1 0 +1 0 −1 +2 −3 +2
r4 +2 −1 +2 0 +1 0 +1 0 +1 −1 0 +1 +1 0 +1 0 +1 0 +1 0	t4 +1 0 +1 0 +2 −2 +1 +1 −1 0 0 0 +1 +1 +1 0 0 +1 0 −1	v4 0 +1 −1 0 +1 −1 0 +1 −2 +2 0 0 0 +1 0 +1 −1 +1 −1 0

Three of these matrices are more concisely written thus:

r4 r2 0 0 r3

t4 t2 0 0 t3

v4 v2 0 0 v3

Familiar numbers appear among the determinants and eigenvalues of the matrices above:

matrices	determinant	eigenvalues		matrices	determinant	eigenvalues
q1, s1, u1 q4, s4, u4	− 7	+ 2 cos ( 90° / 7), − 2 cos ( 90° / 7), + 2 cos (270° / 7), − 2 cos (270° / 7), + 2 cos (450° / 7), − 2 cos (450° / 7)		r1, v1 r4, v4	+ 1	two instances of each: + 2 cos (180° / 7), − 2 cos (360° / 7), + 2 cos (540° / 7)
				t1 t4	+ 1	two instances of each: − 2 cos (180° / 7), + 2 cos (360° / 7), − 2 cos (540° / 7)
q2, s2 q3, s3	− √7	+ 2 cos ( 90° / 7), + 2 cos (270° / 7), − 2 cos (450° / 7)		r2, v2 r3, v3	− 1	+ 2 cos (180° / 7), − 2 cos (360° / 7), + 2 cos (540° / 7)
u2 u3	+ √7	− 2 cos ( 90° / 7), − 2 cos (270° / 7), + 2 cos (450° / 7)		t2 t3	+ 1	− 2 cos (180° / 7), + 2 cos (360° / 7), − 2 cos (540° / 7)