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Let A, B, C and D be eight-dimensional vectors. The problem here is to solve A × B × C = D for C, when A, B and D are furnished.

First define these 28 real numbers:

h10 = a3b2a2b3 + a5b4a4b5 + a7b6a6b7
h20 = a1b3a3b1 + a5b7a7b5 + a6b4a4b6
h30 = a2b1a1b2 + a4b7a7b4 + a5b6a6b5
h40 = a1b5a5b1 + a2b6a6b2 + a7b3a3b7
h50 = a4b1a1b4 + a6b3a3b6 + a7b2a2b7
h60 = a1b7a7b1 + a3b5a5b3 + a4b2a2b4
h70 = a2b5a5b2 + a3b4a4b3 + a6b1a1b6

h21 = a3b0a0b3 + a6b5a5b6 + a7b4a4b7
h31 = a0b2a2b0 + a5b7a7b5 + a6b4a4b6
h41 = a2b7a7b2 + a3b6a6b3 + a5b0a0b5
h51 = a0b4a4b0 + a2b6a6b2 + a7b3a3b7
h61 = a4b3a3b4 + a5b2a2b5 + a7b0a0b7
h71 = a0b6a6b0 + a3b5a5b3 + a4b2a2b4

h32 = a1b0a0b1 + a4b5a5b4 + a6b7a7b6
h42 = a5b3a3b5 + a6b0a0b6 + a7b1a1b7
h52 = a0b7a7b0 + a3b4a4b3 + a6b1a1b6
h62 = a0b4a4b0 + a1b5a5b1 + a7b3a3b7
h72 = a1b4a4b1 + a3b6a6b3 + a5b0a0b5

h43 = a0b7a7b0 + a2b5a5b2 + a6b1a1b6
h53 = a0b6a6b0 + a1b7a7b1 + a4b2a2b4
h63 = a1b4a4b1 + a2b7a7b2 + a5b0a0b5
h73 = a4b0a0b4 + a5b1a1b5 + a6b2a2b6

h54 = a1b0a0b1 + a2b3a3b2 + a6b7a7b6
h64 = a2b0a0b2 + a3b1a1b3 + a7b5a5b7
h74 = a0b3a3b0 + a2b1a1b2 + a5b6a6b5

h65 = a0b3a3b0 + a2b1a1b2 + a4b7a7b4
h75 = a0b2a2b0 + a1b3a3b1 + a6b4a4b6

h76 = a1b0a0b1 + a2b3a3b2 + a4b5a5b4

The pattern is that ai bj contributes to hkl when Ui × Uj × Uk = Ul accoring to the baseline cross product. Should hkl not be listed, it equals − hlk. Meanwhile, hkk had it been needed would have equalled zero.

Now form the skew-symmetric matrix H:

H =
 0 +h10 +h20 +h30 +h40 +h50 +h60 +h70 −h10 0 +h21 +h31 +h41 +h51 +h61 +h71 −h20 −h21 0 +h32 +h42 +h52 +h62 +h72 −h30 −h31 −h32 0 +h43 +h53 +h63 +h73 −h40 −h41 −h42 −h43 0 +h54 +h64 +h74 −h50 −h51 −h52 −h53 −h54 0 +h65 +h75 −h60 −h61 −h62 −h63 −h64 −h65 0 +h76 −h70 −h71 −h72 −h73 −h74 −h75 −h76 0

Then, with the double dot indicating matrix multiplication, seek values of C that satisfy H ·· C = D. This problem is equivalent to solving a system of eight linear equations in eight variables. Not yet established is whether H is invertible, therefore there may be zero, one or many solutions.

Here is an example where three vectors are given:

 A = [ +1, 0, +1, −1, 0, +1, 0, −1 ] B = [ −1, 0, 0, +1, −1, +1, −1, 0 ] D = [ +3, 0, −7, −4, −4, +2, −1, +2 ]

and the goal is to find C in A × B × C = D.

The coefficients in the following system correspond to the values in the H matrix above, and the right member of each equation comes from D:

 0c0 − 1c1 + 1c2 − 2c3 − 2c4 − 1c5 − 1c6 + 2c7 = +3 1c0 + 0c1 + 2c2 + 2c3 − 1c4 − 3c5 − 1c6 − 2c7 = 0 − 1c0 − 2c1 + 0c2 + 0c3 + 3c4 + 0c5 − 2c6 − 1c7 = −7 2c0 − 2c1 + 0c2 + 0c3 + 0c4 + 0c5 − 2c6 + 2c7 = −4 2c0 + 1c1 − 3c2 + 0c3 + 0c4 + 0c5 − 2c6 − 1c7 = −4 1c0 + 3c1 + 0c2 + 0c3 + 0c4 + 0c5 − 1c6 + 1c7 = +2 1c0 + 1c1 + 2c2 + 2c3 + 2c4 + 1c5 + 0c6 + 2c7 = −1 − 2c0 + 2c1 + 1c2 − 2c3 + 1c4 − 1c5 − 2c6 + 0c7 = +2

With application of Gaussian elimination, two equations drop out, leaving an underdetermined system:

 c3 + c6 − c7 = 0 c5 + c6 + c7 = +1 c4 − c6 = −2 c2 + c7 = +1 c0 − c6 + c7 = −1 c1 = +1

Among the many parameterizations of C is:

C = [ c6c7 − 1, 1, 1 − c7, c7c6, c6 − 2, 1 − c6c7, c6, c7 ]

which, being linear and having two independent variables, namely c6 and c7, denotes a plane. A particular solution is C = [ 0, +1, +1, −1, −1, 0, +1, 0 ].

If A × B × C = 0, but A and B are non-zero, then C must be a linear combination of A and B, also governed by two independent variables.

We prepared a computer program to evaluate the determinant of matrix H by symbolic manipulation (as opposed to numerical techniques). It turns out that for all complex values of an and bn, the determinant vashishes.

Although evaluating the determinant of an 8-by-8 matrix most generally requires calculating 8! = 40320 products of 8 factors each, we bypassed any product that would have incorporated a zero element on the diagonal, reducing the number of products to 14833 (see derangement). Beyond that, we did not take advantage of the skew symmetry of the matrix, preferring to keep the algorithm simple to facilitate checking.

An example of a product is:

h30h51h12h23h04h45h76h67

which is a polynomial in 214207 terms. (Recall that h67 = −h76) An example of a term in that product is:

−2a1(a2)3(a5)2a6a7b1b2(b3)2(b6)3b7

In all terms of all products, the sum of exponents of all the an is eight, and no exponent is greater than six. The same applies to the bn.

A standard result is that the rank of a skew-symmetric matrix must be an even number.

• Because H has a determinant of zero, its rank can never be 8.
• A rank of 6 is what usually results when A and B are filled with random numbers.
• The rank is zero if A = 0 or B = 0.
• An open question is whether a rank of 4 or 2 is possible.