Question

Let $\mathit{S}$be the ring in Exercise 11.

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

b. Prove that the matrix $\left(\begin{array}{cc}x& x\\ y& y\end{array}\right)$is a right identity in $\mathit{S}$if and only if $\mathit{x}\mathbf{+}\mathit{y}\mathbf{=}\mathbf{1}$

.

Short answer

It is proved that $\left(\begin{array}{cc}x& x\\ y& y\end{array}\right)$is a right identity in$S$ if and only if $x+y=1$.

Step-by-step solution

Right identity in a ring R

A ring with right identity is a ring $R$ that contains an element$1_R$ satisfying this axiom:

$a.1_R=a,\forall a\in R$.

Conclusion

Here, $\left(\begin{array}{cc}x& x\\ y& y\end{array}\right)$will be a right identity in $S$

$$\begin{gathered} \iff A.\left(\begin{array}{cc}x& x\\ y& y\end{array}\right)=A,\forall A\in S \\ \iff \left(\begin{array}{cc}a& a\\ b& b\end{array}\right).\left(\begin{array}{cc}x& x\\ y& y\end{array}\right)=\left(\begin{array}{cc}ax+ay& ax+ay\\ bx+by& bx+by\end{array}\right)=\left(\begin{array}{cc}a& a\\ b& b\end{array}\right),\forall a,b\in R \\ \iff \left(\begin{array}{cc}ax+ay& ax+ay\\ bx+by& bx+by\end{array}\right)=\left(\begin{array}{cc}\left(x+y\right)a& \left(x+y\right)a\\ \left(x+y\right)b& \left(x+y\right)b\end{array}\right)=\left(\begin{array}{cc}a& a\\ b& b\end{array}\right),\forall a,b\in R \\ \iff \left(x+y\right)\left(\begin{array}{cc}a& a\\ b& b\end{array}\right)=\left(\begin{array}{cc}a& a\\ b& b\end{array}\right),\forall a,b\in R \\ \iff \left(x+y\right)=1 \end{gathered}$$

Hence, $\left(\begin{array}{cc}x& x\\ y& y\end{array}\right)$is a right identity in$S$ if and only if,$x+y=1$