## Inverse

Here we look at the multiplicative inverse of multivector 'm', that is '1/m'. On this page we saw that the inverse of a pure vector is the same vector with a scaling factor.

On the same page we saw that the inverse of a bivector blade is calculated by replacing the bivector bases with their reverse (with a scaling factor). Taking the reverse of a bivector is the same as chaging its sign. There is a problem in 4D and above though, the non-scalar terms may not cancel outcancel out because bivectors like e12 and e34 don't have a common term. So do four dimensional bivectors allways have a multiplicative inverse?

So how can we calculate the inverse for a general bivector based in 4D which contains upto 6 terms? The inverse of any two terms with a common term is the reverse, but if there is no common term the result is more complicated as we have seen.

## General Bivector Case

We want to find the most general case for the inverse of a 4D bivector.

1/ (a e12 + b e31 + c e23 + d e41 + f e42 + g e43)

we start by multiplying the top and bottom by the conjugate:

(-a e12 - b e31 - c e23 - d e41 - f e42 - g e43)

multiplying all the terms out gives the following denominator:

a² + b² + c² + d² + f² + g² + 2*(a*g + b*f + c*d)e1234

this is quite long so lets substitute:

x = a² + b² + c² + d² + f² + g²

y =2*(a*g + b*f + c*d)

so the denominator is:

x + y e1234

to get rid of the e1234 term we will multiply top & bottom by,

x - y e1234

so the inverse is:

(-a e12 - b e31 - c e23 - d e41 - f e42 - g e43)(x - y e1234)/(x² - y²)

= ((g y -a x) e12

+ (f y - b x) e31

+ (d y - c x) e23

+ (c y - d x) e41

+ (b y - f x) e42

+ (a y - g x) e43)/(x² - y²)

where:

- x = a² + b² + c² + d² + f² + g²
- y =2*(a*g + b*f + c*d)

#### Inverse of 4D Trivector

1/(a e_{123} + b e_{142} + c e_{134} + d e_{324})

is equal to

(-a e_{123} - b e_{142} - c e_{134} - d e_{324})/(a² + b² + c² + d²)

So we don't have the same problem for trivectors that we had for bivectors, in 4d we can just use the reverse of all the terms, with the appropriate scaling factor.

#### Inverse of 4D Quadvector

1/(a e_{1234}) = (1/a) e_{1234}

There is only one quadvector in 4D which is its own inverse.

## Example

Lets try inverting the bivector: a e12 + b e34, this multiplying top & bottom by the reverse gives:

So the denominator still has a non-scalar term, we can make another attempt to canel it out by multipling by its reverse, so the numirator is,

(a e_{21} + b e_{43})(a² + b² + 2ab e_{1234})

= (a (a² + b²) - 2ab²) e_{21} + (b (a² + b²) - 2a²b) e_{43}

= (a (a² - b²)) e_{21} + (b (b² - a²)) e_{43}

= (a (a + b)(a - b)) e_{21} + (b (b - a)(b + a)) e_{43}

and the denominator is:

(a² + b² - 2ab e_{1234})(a² + b² + 2ab e_{1234})

= (a + b)²(a² + b² - 2 a b)

so the result is:

= (a (a - b)/((a + b)(a² + b² - 2 a b))) e_{21}

+ (b (b - a)/((a + b)(a² + b² - 2 a b))) e_{43}

This looks very messy. Can it be simplified and further?