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Fearless Symmetry by Robert Gross, Avner Ash

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GROUPS OF MATRICES 129
we define the determinant of A, det(A) to be the number ad bc.
Then if det(A) = 0,
A
1
=
d
ad bc
b
ad bc
c
ad bc
a
ad bc
which you can check by matrix multiplication: Both AA
1
and A
1
A
should equal the identity matrix I =
10
01
. If det(A) = 0, then A
has no inverse.
You can now check that the six matrices listed in (11.2) all have
determinant 1, and you can figure out each one’s matrix inverse.
The General Linear Group of Invertible Matrices
THEOREM 11.3: The set of invertible n-by-nR-matrices form
a group.
PROOF: Well, we said that the associative law holds. We
have the neutral identity matrix I. And, by definition, every
element in this set is invertible, so we have inverses. Are we
done?
No. Now that we have limited ourselves to invertible
matrices, we have to check point (1) on page 125 again. Is this
new and smaller set of square invertible matrices still closed
under matrix multiplication? The answer is yes.
If you desire a proof of that, here it is: Suppose A and B are
invertible square matrices. Then A
1
and B
1
exist. We claim
that the inverse of AB is B
1
A
1
. To check this, we must use
the associative law several times: (AB)(B
1
A
1
) =
((AB)B
1
)A
1
) = (A(BB
1
))A
1
= (AI)A
1
= AA
1
= I. Similarly,
you can prove as an exercise that (B
1
A
1
)(AB) = I.
We need a symbol for this group: GL(n, R). The “GL stands for
the words “general linear, which are a historical artifact of the
theory.
5
5
The adjective “general” meant that the determinant is allowed to be general, that is,
not necessarily equal to 1.

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