Let . Show that G is a group under matrix multiplication. Explain why each element of G has an inverse even though the matrices have 0 determinant.

Why isn't my latex code working? I want to produce a matrix with two rows, each rows with [ a a ]...

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- Aug 24th 2007, 09:27 PMtttcomraderGroup problem with matrices
Let . Show that G is a group under matrix multiplication. Explain why each element of G has an inverse even though the matrices have 0 determinant.

Why isn't my latex code working? I want to produce a matrix with two rows, each rows with [ a a ]... - Aug 25th 2007, 12:07 AMred_dog

Let

We have

**Comutativity:**

**Asociativity:**

Then

**Identity element:**

Let be the identity element:

Then

We have

So the identity element is

**Reciprocal element:**

Let and the reciprocal element.

Then

So, the reciprocal element of is

Each element has an inverse because the identity element is not and we can't apply the same rule for an inverse of a matrix. - Aug 25th 2007, 12:52 AMCaptainBlack
- Aug 25th 2007, 02:51 PMtttcomrader
Thanks, that was really helpful, I just need to think a bit more about why each element has an inverse because of the different identity.