Question

(a) If $\mathit{R}$is a ring, show that the ring $\mathit{M}\left(R\right)$of all $\mathbf{2}\mathbf{\times }\mathbf{2}$matrices with entries in $\mathit{R}$is a ring.

(b) If $\mathit{R}$has an identity, show that $\mathit{M}\left(R\right)$also has an identity.

Short answer

It is proved that,$M\left(R\right)$ forms a ring with entries in$R$ and$M\left(R\right)$ has identity, $\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$.

Step-by-step solution

Step: 1: Ring

A ring is a non-empty set$R$ equipped with two operations (usually written

as addition and multiplication) that satisfy the following axioms.

For all $a,b,c\in R$:

1. If , $a\in R$and$b\in R$ then, $a+b\in R$.
2. $a+\left(b+c\right)=\left(a+b\right)+c$.
3. $a+b=b+a$.
4. There is an element,$0_R$ in $R$such that, $a+0_R=0_R+a=a$for every $a\in R$.
5. For each $a\in R$, the equation$a+x=0_R$ has a solution in $R$.
6. If , and then, .
7. $a\left(bc\right)=\left(ab\right)c$.
8. role="math" localid="1653627245805" $a\left(b+c\right)=ab+ac;$ and $\left(a+b\right)c=ac+bc$.

Check for the conditions

Since, elements in$M\left(R\right)$is$2\times 2$ matrices, we try to satisfy all the axioms,

a. Let, role="math" localid="1653628552533" $$\begin{gathered} \left[\begin{array}{cc}{a}_{11}& a_{12}\\ a_{21}& a_{22}\end{array}\right],\left[\begin{array}{cc}{b}_{11}& b_{12}\\ b_{21}& b_{22}\end{array}\right]\in M\left(R\right) \\ \left[\begin{array}{cc}{a}_{11}& a_{12}\\ a_{21}& a_{22}\end{array}\right]+\left[\begin{array}{cc}{b}_{11}& b_{12}\\ b_{21}& b_{22}\end{array}\right]=\left[\begin{array}{cccc}{a}_{11}& b_{11}& a_{12}& b_{12}\\ a_{21}& b_{21}& a_{21}& b_{22}\end{array}\right]\in M\left(R\right) \end{gathered}$$,

So addition is closed.

b. Let, $\left[\begin{array}{cc}{a}_{11}& a_{12}\\ a_{21}& a_{22}\end{array}\right],\left[\begin{array}{cc}{b}_{11}& b_{12}\\ b_{21}& b_{22}\end{array}\right]\left[\begin{array}{cc}{c}_{11}& c_{12}\\ c_{21}& c_{22}\end{array}\right]\in M\left(R\right)$, then,

$\left(\left[\begin{array}{cc}{a}_{11}& a_{12}\\ a_{21}& a_{22}\end{array}\right]+\left[\begin{array}{cc}{b}_{11}& b_{12}\\ b_{21}& b_{22}\end{array}\right]\right)+\left[\begin{array}{cc}{c}_{11}& c_{12}\\ c_{21}& c_{22}\end{array}\right]=\left[\begin{array}{cc}\left(a_{11}+b_{11}\right)+c_{11}& \left(a_{12}+b_{12}\right)+c_{12}\\ \left(a_{21}+b_{21}\right)+c_{21}& \left(a_{22}+b_{22}\right)+c_{22}\end{array}\right]=\left[\begin{array}{cc}{a}_{11}+\left(b_{11}+c_{11}\right)& a_{12}+\left(b_{12}+c_{12}\right)\\ a_{21}+\left(b_{21}+c_{21}\right)& a_{22}+\left(b_{22}+c_{22}\right)\end{array}\right]=\left[\begin{array}{cc}{a}_{11}& a_{12}\\ a_{21}& a_{22}\end{array}\right]+\left(\left[\begin{array}{cc}{b}_{11}& b_{12}\\ b_{21}& b_{22}\end{array}\right]+\left[\begin{array}{cc}{c}_{11}& c_{12}\\ c_{21}& c_{22}\end{array}\right]\right)$

Hence, addition is associative.

c. Now,

role="math" localid="1653629573902" $\left[\begin{array}{cc}{a}_{11}& a_{12}\\ a_{21}& a_{22}\end{array}\right]+\left[\begin{array}{cc}{b}_{11}& b_{12}\\ b_{21}& b_{22}\end{array}\right]=\left[\begin{array}{cccc}{a}_{11}& b_{11}& a_{12}& b_{12}\\ a_{21}& b_{21}& a_{21}& b_{22}\end{array}\right]=\left[\begin{array}{cc}{b}_{11}+a_{11}& b_{12}+a_{12}\\ b_{21}+a_{21}& b_{22}+a_{22}\end{array}\right]=\left[\begin{array}{cc}{b}_{11}& b_{12}\\ b_{21}& b_{22}\end{array}\right]+\left[\begin{array}{cc}{a}_{11}& a_{12}\\ a_{21}& a_{22}\end{array}\right]$

So, addition is commutative.

d. Here,

$\left[\begin{array}{cc}{a}_{11}& a_{12}\\ a_{21}& a_{22}\end{array}\right]+\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]+\left[\begin{array}{cc}{a}_{11}& a_{12}\\ a_{21}& a_{22}\end{array}\right]=\left[\begin{array}{cc}{a}_{11}& a_{12}\\ a_{21}& a_{22}\end{array}\right]$

.

Which shows,$\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]$ is the identity element.

e. Now,

$\left[\begin{array}{cc}{a}_{11}& a_{12}\\ a_{21}& a_{22}\end{array}\right]+\left[\begin{array}{cc}-a_{11}& -a_{12}\\ -a_{21}& -a_{22}\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]$, it shows that.

The equation, $a+x=0_R$, has a solution in $R$.

f.$\left[\begin{array}{cc}{a}_{11}& a_{12}\\ a_{21}& a_{22}\end{array}\right]\left[\begin{array}{cc}{b}_{11}& b_{12}\\ b_{21}& b_{22}\end{array}\right]=\left(\begin{array}{cc}{a}_{11}{b}_{11}+a_{12}{b}_{21}& a_{21}{b}_{11}+a_{22}{b}_{21}\\ a_{11}{b}_{12}+a_{12}{b}_{22}& a_{21}{b}_{12}+a_{22}{b}_{22}\end{array}\right)\in M\left(R\right)$

So, multiplication is closed

g. Since, multiplication is associative for matrices. So, multiplication is associative in $M\left(R\right)$

h. Now,

$\left[\begin{array}{cc}{a}_{11}& a_{12}\\ a_{21}& a_{22}\end{array}\right]\left(\left[\begin{array}{cc}{b}_{11}& b_{12}\\ b_{21}& b_{22}\end{array}\right]\left[\begin{array}{cc}{c}_{11}& c_{12}\\ c_{21}& c_{22}\end{array}\right]\right)=\left[\begin{array}{cc}{a}_{11}\left(b_{11}+c_{11}\right)+a_{12}\left(b_{12}+c_{12}\right)& a_{11}\left(b_{21}+c_{21}\right)+a_{12}\left(b_{22}+c_{22}\right)\\ a_{21}\left(b_{11}+c_{11}\right)+a_{22}\left(b_{12}+c_{12}\right)& a_{21}\left(b_{21}+c_{21}\right)+a_{22}\left(b_{22}+c_{22}\right)\end{array}\right]=\left[\begin{array}{cc}{a}_{11}& a_{12}\\ a_{21}& a_{22}\end{array}\right]\left[\begin{array}{cc}{b}_{11}& b_{12}\\ b_{21}& b_{22}\end{array}\right]+\left[\begin{array}{cc}{a}_{11}& a_{12}\\ a_{21}& a_{22}\end{array}\right]\left[\begin{array}{cc}{c}_{11}& c_{12}\\ c_{21}& c_{22}\end{array}\right]$

Also,

$$\begin{gathered} \left(\left[\begin{array}{cc}{a}_{11}& a_{12}\\ a_{21}& a_{22}\end{array}\right]+\left[\begin{array}{cc}{b}_{11}& b_{12}\\ b_{21}& b_{22}\end{array}\right]\right)\left[\begin{array}{cc}{c}_{11}& c_{12}\\ c_{21}& c_{22}\end{array}\right] \\ =\left[\begin{array}{cc}\left(a_{11}+b_{11}\right)c_{11}+\left(a_{21}+b_{21}\right)c_{12}& \left(a_{11}+b_{11}\right)c_{21}+\left(a_{21}+b_{21}\right)c_{22}\\ \left(a_{12}+b_{12}\right)c_{11}+\left(a_{22}+b_{22}\right)c_{12}& \left(a_{12}+b_{12}\right)c_{21}+\left(a_{22}+b_{22}\right)c_{22}\end{array}\right] \\ =\left[\begin{array}{cc}{a}_{11}& a_{12}\\ a_{21}& a_{22}\end{array}\right]\left[\begin{array}{cc}{c}_{11}& c_{12}\\ c_{21}& c_{22}\end{array}\right]+\left[\begin{array}{cc}{b}_{11}& b_{12}\\ b_{21}& b_{22}\end{array}\right]\left[\begin{array}{cc}{c}_{11}& c_{12}\\ c_{21}& c_{22}\end{array}\right] \end{gathered}$$

Hence, distributive property is proved.

Therefore, $M\left(R\right)$forms a ring with entries in $R$.

Conclusion

It is given that,$R$has an identity. Now, $\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\in M\left(R\right)$. We get,

$\left[\begin{array}{cc}{a}_{11}& a_{12}\\ a_{21}& a_{22}\end{array}\right]\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}{a}_{11}& a_{12}\\ a_{21}& a_{22}\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\left[\begin{array}{cc}{a}_{11}& a_{12}\\ a_{21}& a_{22}\end{array}\right]$.

Therefore,$M\left(R\right)$ has identity, $\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$.