Answer to Question #105097 in Abstract Algebra for Sourav Mondal

Question

Let R be a ring. Show that M3(R) is a ring with respect to the usual matrix
addition and multiplication. Further, if R is commutative, will M3(R )be
commutative? Why, or why not?

Answer 1

$(M3, +,.)$ is a ring as :

1. (M3,+) form an abelian group. Since:

1.1. A+(B+C) = (A+B)+C, addition is associative over ME.

1.2 a + b = b + a for all a, b in M3  (that is, + is commutative).

1.3 There is an element 0 in M3 such that a + 0 = a for all a in M3(that is, 0 is the additive identity).

1.4 For each a in M3 there exists −a in M3 such that a + (−a) = 0  (that is, −a is the additive inverse of a).

2. M3 is associative over .

3. There exists a multiplicative identity $I=\begin{bmatrix} 1&0& 0 \\ 0&1& 0\\0&0&1 \end{bmatrix}$ , such that A.I=I.A=A, $\forall A\in M3$

4.The . is distributive over + in M3, i.e. A.(B+C)=A.B+A.C, for all A, B, C in M3.

All these conditions show that M3 is a ring.

However, M3 is not a commutative ring, since, matrix multiplication is not commutative in nature, ie, $A.B , B.A , \forall A, B \in M3$ are not always the same.

Thus, M3(R) is a ring, but not a commutative ring.