Section 9.1 Algebraic Structures
689
c
a 1,1 a 1,2
a 2,1 a 2,2
d + c
b 1,1 b 1,2
b 2,1 b 2,2
d = c
a 1,1 + b 1,1 a 1,2 + b 1,2
a 2,1 + b 2,1 a 2,2 + b 2,2
d
= c
b 1,1 + a 1,1 b 1,2 + a 1,2
b 2,1 + a 2,1 b 2,2 + a 2,2
d
= c
b 1,1 b 1,2
b 2,1 b 2,2
d + c
a 1,1 a 1,2
a 2,1 a 2,2
d
The matrix
c
0 0
0 0
d
is an identity. The matrix
c
1
2
−4
5
d
is an inverse of the matrix
c
−1
−2
4
−5
d
ReminDeR
It’s important to understand all this new terminology. You can’t prove
that P S Q if you don’t
know what you’re starting
with or where you want
to go.
example 3
Consider 3M 2 (Z), # 4 where # denotes matrix multiplication. Closure holds. It can be
shown (Exercise 2) that matrix multiplication is associative. The matrix
c
1
0
0
1
d
serves as an identity because
c
1
0
0
1
d # c
a
c
b
d
d = c
a
c
b
d
d # c
1
0
0
1
d = c
a
c
b
d
d
Thus 3M 2 (Z), # 4 is at least a monoid.
■
PrACtiCe 2 Prove that 3M 2 (Z), # 4 is not a commutative monoid.
PrACtiCe 3 Prove that 3M 2 (Z), # 4 is not a group.
■
A structure called a monoid results from dropping the inverse property in the
definition of a group; thus a monoid has an associative operation and an identity
element, but in a monoid that is not also a group, at least one element has no inverse. A semigroup results from dropping the identity property and the inverse
property in the definition of a group; thus a semigroup has an associative operation, but in a semigroup that is not also a monoid, no identity element exists. Many
familiar forms of arithmetic are instances of semigroups, monoids, and groups.