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Aug 15, 2018, 11:54:10 PM8/15/18

to Geometric_Algebra

I am applying the Cramer's rule derivation using the wedge product to a system of two simultaneous equations in two variables.

The equations are:

x1 - x2 = 0

3x1 - 2x2 = -1

So that x1 = (c ^ b)/(a ^ b)

where a = e1 + 3e2, b = -e1 -2e2, and c = -e2

The variables have to be equal by the first equation. The answer is both x1 and x2 = -1.

For the numerator for the x1 term I get -e2 ^ -e1. I calculate the denominator as 1.

The only thing that seems to make sense to me is that to get -(e1 ^ e2) is to do three sign flips as follows:

-e2 ^ -e1 = e1 ^ -e2 = e2 ^ e1 = -(e1 ^ e2)

Is my reasoning correct?

Thanks.

All the best,

Matthew

The equations are:

x1 - x2 = 0

3x1 - 2x2 = -1

So that x1 = (c ^ b)/(a ^ b)

where a = e1 + 3e2, b = -e1 -2e2, and c = -e2

The variables have to be equal by the first equation. The answer is both x1 and x2 = -1.

For the numerator for the x1 term I get -e2 ^ -e1. I calculate the denominator as 1.

The only thing that seems to make sense to me is that to get -(e1 ^ e2) is to do three sign flips as follows:

-e2 ^ -e1 = e1 ^ -e2 = e2 ^ e1 = -(e1 ^ e2)

Is my reasoning correct?

Thanks.

All the best,

Matthew

Aug 16, 2018, 10:59:36 AM8/16/18

to Geometric_Algebra

Answered my own question. The sign flips are correct. NFCM pg 23.

Aug 16, 2018, 7:17:08 PM8/16/18

to Geometric_Algebra

Yes, correct. However, you do not need three applications of the anticommutativity property of the wedge product (on suffices), since scalar multipliers from any factor can be brought out with no other change (linearity of the wedge product).

You are implicitly using the linearity property anyway, by treating (-e1) ^ ... and -(e1 ^ ...) as the same thing before you cancel signs:

-e2 ^ -e1 = -(-e1 ^ -e2) (anti-commutativity)

-(-e1 ^ -e2) = e1 ^ -e2 (linearity)

Manfred

Aug 23, 2018, 11:21:12 AM8/23/18

to Geometric_Algebra

Thanks, Manfred. It is taking me a bit to get used to the operator, but I am gradually getting it.

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