Opposite Ring of $M_2(\mathbb R)$

Detailed Solution

The opposite ring $R^{\mathrm{op}}$ has exactly the same elements and addition as $R$, but its multiplication is reversed: $$ A*B=BA. $$ Consider the transpose map $$ \Phi:R\longrightarrow R^{\mathrm{op}}, \qquad \Phi(A)=A^T. $$ We show that $\Phi$ is a ring isomorphism. For $A,B\in R$, $$ \Phi(A+B) =(A+B)^T =A^T+B^T =\Phi(A)+\Phi(B). $$ Thus $\Phi$ is additive. In the ordinary matrix ring $R$, $$ \Phi(AB)=(AB)^T=B^TA^T. $$ But multiplication in $R^{\mathrm{op}}$ is defined by $$ X*Y=YX. $$ Therefore, $$ \Phi(A)*\Phi(B) =A^T*B^T =B^TA^T. $$ Hence $$ \Phi(AB)=\Phi(A)*\Phi(B). $$ Thus the transpose map is a ring homomorphism $$ R\longrightarrow R^{\mathrm{op}}. $$ Since $$ (A^T)^T=A, $$ the transpose map is its own inverse. Hence $\Phi$ is bijective. Therefore, $$ \boxed{\Phi(A)=A^T} $$ is a ring isomorphism from $M_2(\mathbb R)$ onto its opposite ring. Hence the correct answer is $$ \boxed{\mathrm{B}}. $$ **Option A:** False. The opposite multiplication $$ a*b=ba $$ still satisfies associativity and the ring axioms. Thus $R^{\mathrm{op}}$ is indeed a ring. **Option C:** False in general. Although $R$ and $R^{\mathrm{op}}$ have the same underlying set and additive structure, their multiplications are reversed. Since $M_2(\mathbb R)$ is noncommutative, the two ring structures are not literally the same. For example, if $$ A= \begin{pmatrix} 1&0\\ 0&0 \end{pmatrix}, \qquad B= \begin{pmatrix} 0&1\\ 0&0 \end{pmatrix}, $$ then $$ AB= \begin{pmatrix} 0&1\\ 0&0 \end{pmatrix}, \qquad BA= \begin{pmatrix} 0&0\\ 0&0 \end{pmatrix}. $$ Thus $AB\neq BA$. **Option D:** False. By definition, $R^{\mathrm{op}}$ has **exactly the same additive structure** as $R$. Only the multiplication is reversed. Therefore, $$ \boxed{\text{The transpose map }A\mapsto A^T\text{ gives }R\cong R^{\mathrm{op}}.} $$