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 = a₃b₂ − a₂b₃ + a₅b₄ − a₄b₅ + a₇b₆ − a₆b₇
h₂₀ = a₁b₃ − a₃b₁ + a₅b₇ − a₇b₅ + a₆b₄ − a₄b₆
h₃₀ = a₂b₁ − a₁b₂ + a₄b₇ − a₇b₄ + a₅b₆ − a₆b₅
h₄₀ = a₁b₅ − a₅b₁ + a₂b₆ − a₆b₂ + a₇b₃ − a₃b₇
h₅₀ = a₄b₁ − a₁b₄ + a₆b₃ − a₃b₆ + a₇b₂ − a₂b₇
h₆₀ = a₁b₇ − a₇b₁ + a₃b₅ − a₅b₃ + a₄b₂ − a₂b₄
h₇₀ = a₂b₅ − a₅b₂ + a₃b₄ − a₄b₃ + a₆b₁ − a₁b₆

h₂₁ = a₃b₀ − a₀b₃ + a₆b₅ − a₅b₆ + a₇b₄ − a₄b₇
h₃₁ = a₀b₂ − a₂b₀ + a₅b₇ − a₇b₅ + a₆b₄ − a₄b₆
h₄₁ = a₂b₇ − a₇b₂ + a₃b₆ − a₆b₃ + a₅b₀ − a₀b₅
h₅₁ = a₀b₄ − a₄b₀ + a₂b₆ − a₆b₂ + a₇b₃ − a₃b₇
h₆₁ = a₄b₃ − a₃b₄ + a₅b₂ − a₂b₅ + a₇b₀ − a₀b₇
h₇₁ = a₀b₆ − a₆b₀ + a₃b₅ − a₅b₃ + a₄b₂ − a₂b₄

h₃₂ = a₁b₀ − a₀b₁ + a₄b₅ − a₅b₄ + a₆b₇ − a₇b₆
h₄₂ = a₅b₃ − a₃b₅ + a₆b₀ − a₀b₆ + a₇b₁ − a₁b₇
h₅₂ = a₀b₇ − a₇b₀ + a₃b₄ − a₄b₃ + a₆b₁ − a₁b₆
h₆₂ = a₀b₄ − a₄b₀ + a₁b₅ − a₅b₁ + a₇b₃ − a₃b₇
h₇₂ = a₁b₄ − a₄b₁ + a₃b₆ − a₆b₃ + a₅b₀ − a₀b₅

h₄₃ = a₀b₇ − a₇b₀ + a₂b₅ − a₅b₂ + a₆b₁ − a₁b₆
h₅₃ = a₀b₆ − a₆b₀ + a₁b₇ − a₇b₁ + a₄b₂ − a₂b₄
h₆₃ = a₁b₄ − a₄b₁ + a₂b₇ − a₇b₂ + a₅b₀ − a₀b₅
h₇₃ = a₄b₀ − a₀b₄ + a₅b₁ − a₁b₅ + a₆b₂ − a₂b₆

h₅₄ = a₁b₀ − a₀b₁ + a₂b₃ − a₃b₂ + a₆b₇ − a₇b₆
h₆₄ = a₂b₀ − a₀b₂ + a₃b₁ − a₁b₃ + a₇b₅ − a₅b₇
h₇₄ = a₀b₃ − a₃b₀ + a₂b₁ − a₁b₂ + a₅b₆ − a₆b₅

h₆₅ = a₀b₃ − a₃b₀ + a₂b₁ − a₁b₂ + a₄b₇ − a₇b₄
h₇₅ = a₀b₂ − a₂b₀ + a₁b₃ − a₃b₁ + a₆b₄ − a₄b₆

h₇₆ = a₁b₀ − a₀b₁ + a₂b₃ − a₃b₂ + a₄b₅ − a₅b₄

The pattern is that a_i b_j contributes to h_kl when U_i × U_j × U_k = U_l accoring to the baseline cross product. Should h_kl not be listed, it equals − h_lk. Meanwhile, h_kk 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 = [ c₆ − c₇ − 1, 1, 1 − c₇, c₇ − c₆, c₆ − 2, 1 − c₆ − c₇, 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 a_n and b_n, 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 a_n is eight, and no exponent is greater than six. The same applies to the b_n.

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.