Note: I need to update this page: see this forum discussion with Manfred and Brombo14
Can we use the same trick that we used with complex numbers, that is multiply top and bottom of equation by the conjugate, which will convert the denominator into a scalar making the division possible.
Terminology:
input | result (1/input) | |
scalar | a.e | e |
vector | a.e1 , a.e2 , a.e3 | e1 , e2 , e3 |
bivector | a.e12 , a.e31 , a.e23 | e12 , e31 , e23 |
trivector | a.e123 | e123 |
Since I'm not yet sure how to calculate the inverse of a multivector, I'll first try an easier problem, how to calculate the inverse of parts of a multivector (scalar, vector, bivector and trivector).
inverse of scalar
e-1 = 1/a.e
inverse of vector
(e1 + e2 + e3 )-1
use conjugate (-e1 - e2 - e3 )
so multiplying a vector by its conjugate gives:
(e1 + e2 + e3 ) (-e1 -e2 -e3 ) = -e12 -e22 -e32
so (e1 + e2 + e3 )-1 = (-e1 - e2 - e3 )/(-e12 -e22 -e32)
e1 = a.e1/(a.e12 +a.e22 +a.e32)
e2 = a.e2/(a.e12 +a.e22 +a.e32)
e3 = a.e3/(a.e12 +a.e22 +a.e32)
inverse of bivector
(e12 + e31 + e23)-1
use conjugate (-e12 - e31 - e23)
so multiplying a bivector by its conjugate gives:
(e12 + e31 + e23) (-e12 - e31 - e23) = e122 + e312 + e232
so (e12 + e31 + e23)-1 = (-e12 - e31 - e23)/(e122 + e312 + e232)
e12 = - a.e12/(a.e12 +a.e22 +a.e32)
e31 = - a.e31/(a.e12 +a.e22 +a.e32)
e23 = - a.e23/(a.e12 +a.e22 +a.e32)
inverse of trivector
e123-1 = a.e123 / (a.e123 * a.e123) = -1/a.e123
inverse of multivector
If it follows the pattern then it would be:
(e1 + e2 + e3 + e12 + e31 + e23 + e123)-1 = (e -e1 -e2 -e3 - e12 - e31 - e23 + a.e123)/(e2 - ex2 - ey2 - ez2 + e122 + e312 + e232 - e1232)
but is it? this would make the conjugate e -e1 -e2 -e3 - e12 - e31 - e23 + a.e123
so if we multiply a multivector by this possible conjugate we get:
(a.e1 + a.e2 + a.e3 + a.e12 + a.e31 + a.e23 + a.e123)(a.e -a.e1 -a.e2 -a.e3 - a.e12 - a.e31 - a.e23 + a.e123)
which gives:
e = a.e2 - a.e12 - a.e22 - a.e32 + a.e122 + a.e312 + a.e232 - a.e1232
e123 = 2*a.e * a.e123 - 2*a.e1 *a.e23 -2*a.e2*a.e31 - 2*a.e3*a.e12
so due to this extra e123 term this will only work if:
a.e * a.e123 = a.e1 *a.e23 + a.e2*a.e31 + a.e3*a.e12
which would make the extra term zero.
Although not necessary to calculate the inverse, the calculation would be a lot simpler if:
1 = a.e2 - a.e12 - a.e22 - a.e32 + a.e122 + a.e312 + a.e232 - a.e1232
so if,
a.e * a.e123 = a.e1 *a.e23 + a.e2*a.e31 + a.e3*a.e12
and
1 = a.e2 - a.e12 - a.e22 - a.e32 + a.e122 + a.e312 + a.e232 - a.e1232
then,
(e1 + e2 + e3 + e12 + e31 + e23 + e123)-1 = e -e1 -e2 -e3 - e12 - e31 - e23 + a.e123