Question

Let$S$ be the subset of$M\left(R\right)$ consisting of all matrices of the form$\left(\begin{array}{cc}a& a\\ b& b\end{array}\right)$

1. Prove that $S$ is a ring.
2. Show that $J=\left(\begin{array}{cc}1& 1\\ 0& 0\end{array}\right)$is a right identity in ${\mathrm{\mathbb{Z}}}_n$(meaning that $AJ=A$ for every $A$in $S$ ).
3. Show that $J$ is not a left identity in $S$by finding a matrix $B$ in $S$ such that$JB\ne B$ .

Short answer

1. It is proved that S is a ring.
2. It is proved that $\left(\begin{array}{cc}1& 1\\ 0& 0\end{array}\right)\in S$ is the right identity in S.
3. It is proved that J is not the left identity in S.

Step-by-step solution

Prove that S is a ring  

It is given that$S=\left\{\overline{)\left(\begin{array}{cc}a& a\\ b& b\end{array}\right)}a,b\in R\right\}\subseteq M\left(R\right)$ . Assume that $\left(\begin{array}{cc}a& a\\ b& b\end{array}\right),\left(\begin{array}{cc}c& c\\ d& d\end{array}\right)\in S$.

Show the properties as:

1. S is closed under addition.

role="math" localid="1646300861679" $$\begin{gathered} \left(\begin{array}{cc}a& a\\ b& b\end{array}\right)+\left(\begin{array}{cc}c& c\\ d& d\end{array}\right)=\left(\begin{array}{cccccc}a& +& c& a& +& c\\ b& +& d& b& +& d\end{array}\right) \\ \in S \end{gathered}$$

(ii) S is closed under multiplication

$$\begin{gathered} \left(\begin{array}{cc}a& a\\ b& b\end{array}\right)\left(\begin{array}{cc}c& c\\ d& d\end{array}\right)=\left(\begin{array}{cccccc}ac& +& ad& ac& +& ad\\ bc& +& bd& bc& +& bd\end{array}\right) \\ \in S \end{gathered}$$(iii) $0_R\in S$

$\left(\begin{array}{cc}0& 0\\ 0& 0\end{array}\right)\in S$

(iv) If$a\in S$ , then the solution of the equation $a+x=0_R$ lies in S.

role="math" localid="1646303873776" $\begin{array}{rcl}\left(\begin{array}{cc}-a& -a\\ -b& -b\end{array}\right)& \in & S\\ \left(\begin{array}{cc}a& a\\ b& b\end{array}\right)+\left(\begin{array}{cc}-a& -a\\ -b& -b\end{array}\right)& =& \left(\begin{array}{cc}0& 0\\ 0& 0\end{array}\right)\\ & =& 0_{M\left(R\right)}\end{array}$

Since all the properties are satisfied, this implies that S is a subring of$M\left(R\right)$ , that is, S is a ring.

Show that J is right identity 

Assume that $\left(\begin{array}{cc}a& a\\ b& b\end{array}\right)\in S$.

Find right identity as:

$\left(\begin{array}{cc}a& a\\ b& b\end{array}\right)\left(\begin{array}{cc}1& 1\\ 0& 0\end{array}\right)=\left(\begin{array}{cc}a& a\\ b& b\end{array}\right)$

As $\left(\begin{array}{cc}a& a\\ b& b\end{array}\right)\left(\begin{array}{cc}1& 1\\ 0& 0\end{array}\right)=\left(\begin{array}{cc}a& a\\ b& b\end{array}\right)$, then by the definition of right identity $\left(\begin{array}{cc}1& 1\\ 0& 0\end{array}\right)\in S$ is the right identity in S.

Hence, it is proved.

Show that J is right identity

Assume that$\left(\begin{array}{cc}1& 1\\ 1& 1\end{array}\right)\in S$ .

Find right identity as:

$\begin{array}{rcl}\left(\begin{array}{cc}1& 1\\ 0& 0\end{array}\right)\left(\begin{array}{cc}1& 1\\ 1& 1\end{array}\right)& =& \left(\begin{array}{cc}2& 2\\ 0& 0\end{array}\right)\\ & \ne & \left(\begin{array}{cc}1& 1\\ 1& 1\end{array}\right)\end{array}$

As $\left(\begin{array}{cc}1& 1\\ 0& 0\end{array}\right)\left(\begin{array}{cc}1& 1\\ 1& 1\end{array}\right)\ne \left(\begin{array}{cc}1& 1\\ 1& 1\end{array}\right)$, then by the definition of left right identity $\left(\begin{array}{cc}1& 1\\ 0& 0\end{array}\right)\in S$ is not the left identity in S.

Hence, it is proved.