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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