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

 genericmultiplication 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+30−20+2
+10+1000
0+10+100
00+10+10
000+10+1
0000+10
s1
0+10+100
+1000+10
0+20−20+3
+100000
0+1000−1
00+10−10
u1
000+10+1
00+1000
0+1000−1
+1000−10
0+20−30+2
+10−1000
r1
+10+1000
0+20−10+2
+1000+10
0+1000+1
00+1000
000+10−1
t1
00+10+10
0+1000+1
+100000
0+20−20+1
+1000−10
0+10−100
v1
0000+10
000+10−1
00+10−10
0+10−100
+10−1000
0+10−20+2

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

 genericmultiplication 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:

r1t1 + v1 = r1 t1 v1
+100000
0+10000
00+1000
000+100
0000+10
00000+1

Here is the parallel qsu3 calculation:

q1 + s1u1 = q1 s1 u1
0+40−20+1
+2000+10
0+20−10+4
00+10+20
0−10+40−2
−10+2000

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
+10+1
0+1−1
t2
+10+1
+2−2+1
+1−10
v2
0+1−1
+1−10
+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:

 genericsubdomainmultiplication 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:

 genericsubdomainmultiplication 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)
making it diagonalizable with 8 = 23 distinct square roots which can be written without the need for complex numbers. These facts greatly simplify the calculation.

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+ 2kk
+ 4k− 2k+ k
+ 3k− 5k+ 6k
u2
+ 2kk+ 4k
+ 5k− 6k+ 3k
+ 2kk− 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+10
+10+1
0+10
t3
0+1+1
+100
+10−1
v3
00+1
0+1−1
+1−10

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 … then copy … from thisrow of q1 to thisrow of m from thiscolumn of m to thiscolumn 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+10
0+1+1
00+1
+3−2+20
+1+10
0+1+1
s4
0+10+1
+100
0+1−1
+1+100
+2−2+3
+10−1
u4
00+10
+10−1
+1−10
0+1+10
+10−1
+2−3+2
r4
+2−1+20
+10+1
0+1−1
0+1+10
+10+1
0+10
t4
+10+10
+2−2+1
+1−10
00+1+1
+100
+10−1
v4
0+1−10
+1−10
+1−2+2
000+1
0+1−1
+1−10

Three of these matrices are more concisely written thus:

r4
r20
0r3
t4
t20
0t3
v4
v20
0v3

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) t1t4 + 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)