Question

Show that the set S of matrices of the form $\left(\begin{array}{cc}a& 4b\\ b& a\end{array}\right)$with a and b real numbers is a subring of $\mathit{M}\left(ℝ\right)$.

Short answer

It is proved that S is a subring of$M\left(\mathrm{ℝ}\right)$ .

Step-by-step solution

Theorem

According to theorem 3.6, a non-empty subset, S of a ring, R is closed under subtraction and multiplication or it can be said that, if $a,b\in S$, then$a-b\in S$ and$ab\in S$ , then S is also a subring of R.

Check for closed under subtraction

The given set is S, and $A=\left(\begin{array}{cc}a& 4b\\ b& a\end{array}\right)$and $B=\left(\begin{array}{cc}c& 4d\\ d& c\end{array}\right)$

Find $A-B$

$$\begin{gathered} A-B=\left(\begin{array}{cc}a& 4b\\ b& a\end{array}\right)-\left(\begin{array}{cc}c& 4d\\ d& c\end{array}\right) \\ =\left(\begin{array}{cc}a-c& 4\left(b-d\right)\\ b-d& a-c\end{array}\right) \end{gathered}$$
Hence, subtraction is closed under subtraction.

Check for closed under multiplication 

Find AB

$$\begin{gathered} AB=\left(\begin{array}{cc}a& 4b\\ b& a\end{array}\right)\left(\begin{array}{cc}c& 4d\\ d& c\end{array}\right) \\ =\left(\begin{array}{cc}ac+4bd& 4\left(ab+bc\right)\\ bc+ad& ac+b4d\end{array}\right) \end{gathered}$$

Hence, subtraction is closed under multiplication. So, set Sis a subring of $M\left(\mathrm{ℝ}\right)$