/sci/ - Science & Math

[OP !!yzPI6QwWJCg No.5391144]

Hi, I'm currently trying to prove that G (hereby defined as the set which contains all 2x2 matrices) is a group under matrix multiplication.

Associativity and existence of the identity are done, but I'm stuck on the inverse element. I know it's just the inverse matrix, so that

<span class="math"> A \cdot A^{-1} = id [/spoiler]

(where A, its inverse and id are in G, and id is the unit matrix)

I'm sure it's really simple, but if I want to solve the equation above, how can I ensure that the variables encircled in the image always yield 1 when used in the multiplication? I'm not seeing it.
Thanks for any help!

[Anonymous No.5391153]

Firstly, the set of all such matrices doesn't form a group as the matrices with determinant 0 are not invertible.
As for your question, if you just make a general ansatz, you'll be lead to four equations which you solve.

E.g. here

http://www.wolframalpha.com/input/?i={{a%2Cb}%2C{c%2Cd}}.{{A%2CB}%2C{C%2CD}}%3D%3D{{1%2C0}%2C{0%2C1}
}&dataset=

from the entry 12 you deduce a·B=-b·D
and then you go on
If a \ne 0, then you can write B=-b·D/a,
plug it in ... etc.

[Anonymous No.5391159]

I could add this,
http://en.wikipedia.org/wiki/Invertible_matrix#Methods_of_matrix_inversion
might help, I don't know.
Furthermore you're dealing with
http://en.wikipedia.org/wiki/GL%28n%29
with n=2.

[OP !!yzPI6QwWJCg No.5391166]

>>5391153
>>5391153
Hah, my bad. I was in a little hurry and didn't exclude the ones with det(a) = 0, of course you're right.
I'll try out your approach now, then. Thanks!

>>5391159
Thanks. Didn't see that page before for some reason..

[Anonymous No.5391169]

A-1 = 1/det(A) * transp(comatrix(A))