[proofplan] We compute directly from the elementary matrix definition $e_{ij}(r)=I_n+rE_{ij}$. The only algebraic input is the matrix-unit multiplication rule $E_{ab}E_{cd}=\delta_{bc}E_{ad}$, with the order of coefficients preserved because $R$ need not be commutative. First we record the inverse formula, then multiply the relevant elementary matrices and use the stated index inequalities to determine which matrix-unit products vanish. [/proofplan] [step:Record the matrix-unit products and the inverse formula] For matrix units in $M_n(R)$, multiplication gives \begin{align*} E_{ab}E_{cd}=\delta_{bc}E_{ad}, \end{align*} where $\delta_{bc}=1$ if $b=c$ and $\delta_{bc}=0$ otherwise. In particular, if $a\ne b$, then \begin{align*} E_{ab}^2=E_{ab}E_{ab}=0. \end{align*} Fix $a\ne b$ and $t\in R$. Since $E_{ab}^2=0$, we have \begin{align*} (I_n+tE_{ab})(I_n-tE_{ab})=I_n. \end{align*} The same computation in the opposite order gives \begin{align*} (I_n-tE_{ab})(I_n+tE_{ab})=I_n. \end{align*} Therefore \begin{align*} e_{ab}(t)^{-1}=e_{ab}(-t). \end{align*} [/step] [step:Multiply elementary matrices with the same off-diagonal position] Let $i\ne j$ and let $r,s\in R$. Using $E_{ij}^2=0$, we compute \begin{align*} e_{ij}(r)e_{ij}(s)=(I_n+rE_{ij})(I_n+sE_{ij})=I_n+(r+s)E_{ij}. \end{align*} By the definition of $e_{ij}(r+s)$, this is \begin{align*} e_{ij}(r)e_{ij}(s)=e_{ij}(r+s). \end{align*} [guided] We are multiplying two matrices whose only possible non-identity entry is in the same off-diagonal position $(i,j)$. Because $i\ne j$, the matrix unit satisfies \begin{align*} E_{ij}^2=E_{ij}E_{ij}=0. \end{align*} Thus the product has no quadratic term: \begin{align*} (I_n+rE_{ij})(I_n+sE_{ij})=I_n+rE_{ij}+sE_{ij}+rsE_{ij}^2. \end{align*} The final term is zero, so the product becomes \begin{align*} I_n+(r+s)E_{ij}. \end{align*} By definition, this matrix is exactly $e_{ij}(r+s)$. Hence \begin{align*} e_{ij}(r)e_{ij}(s)=e_{ij}(r+s). \end{align*} [/guided] [/step] [step:Expand the adjacent commutator and keep the ordered coefficient $rs$] Let $i,j,k$ be pairwise distinct, and let $r,s\in R$. Set \begin{align*} A:=rE_{ij}, \qquad B:=sE_{jk}. \end{align*} Then $A^2=0$, $B^2=0$, and \begin{align*} AB=rsE_{ik}. \end{align*} Also $BA=0$, because $k\ne i$. Therefore \begin{align*} [e_{ij}(r),e_{jk}(s)]=(I_n+A)(I_n+B)(I_n-A)(I_n-B). \end{align*} First, \begin{align*} (I_n+A)(I_n+B)=I_n+A+B+AB. \end{align*} Multiplying by $I_n-A$ on the right gives \begin{align*} (I_n+A+B+AB)(I_n-A)=I_n+B+AB. \end{align*} Indeed, the terms $A$ and $-A$ cancel, while $A^2=0$, $BA=0$, and $ABA=0$ since $BA=0$. Finally, \begin{align*} (I_n+B+AB)(I_n-B)=I_n+AB. \end{align*} Here $B^2=0$ and $ABB=AB^2=0$. Hence \begin{align*} [e_{ij}(r),e_{jk}(s)]=I_n+rsE_{ik}=e_{ik}(rs). \end{align*} [/step] [step:Use the separated index conditions to prove the remaining elementary matrices commute] Let $i,j,k,l$ satisfy $i\ne j$, $k\ne l$, $i\ne l$, and $j\ne k$, and let $r,s\in R$. Define \begin{align*} A:=rE_{ij}, \qquad B:=sE_{kl}. \end{align*} The condition $j\ne k$ gives \begin{align*} AB=rsE_{ij}E_{kl}=0. \end{align*} The condition $l\ne i$ gives \begin{align*} BA=srE_{kl}E_{ij}=0. \end{align*} Therefore \begin{align*} (I_n+A)(I_n+B)=I_n+A+B=(I_n+B)(I_n+A). \end{align*} Thus $e_{ij}(r)$ and $e_{kl}(s)$ commute in $GL_n(R)$. Using the definition of the group commutator, \begin{align*} [e_{ij}(r),e_{kl}(s)]=I_n. \end{align*} This proves all three stated elementary matrix relations. [/step]