Let $R$ be a commutative ring with identity. Suppose the matrix algebra $\operatorname{Mat}_n (R)$ and matrices $$E_{mk}:=(a_{ij})=\begin{cases} a_{ij}=1 \ \text{if} \ i=m \ \text{and} \ j=k \\ a_{ij}=0 \ \text{otherwise} \end{cases}.$$
Now consider algebras homomorphism from the free $R$-algebra $$f\colon R\langle t_{ij}\rangle \to \operatorname{Mat}_n (R), \ t_{ij}\mapsto E_{ij}.$$ Is there a description of generator set for $\operatorname{Ker} f$?
It's easy to establish that $f$ is surjective. Hence, in other words, this question is about some "nice" presentation of the matrix algebra.
Thanks for paying attention
The ideal $\operatorname{Ker} f$ is generated by the differences $t_{ij}t_{k\ell} - \delta_{jk} t_{i\ell}$ (where $\delta$ means the Kronecker delta) and the difference $t_{11}+t_{22}+\cdots+t_{nn}-1$.
Proof idea. These differences are clearly in $\operatorname{Ker} f$ (since $f$ sends them to the matrices $E_{ij}E_{k\ell} - \delta_{jk} E_{i\ell} = 0$ and $E_{11}+E_{22}+\cdots+E_{nn}-I_n = 0$). It remains to show that they generate $\operatorname{Ker} f$. To this purpose, we observe that any word (= noncommutative monomial) in the $t_{ij}$'s can be reduced modulo these differences to a linear combination of single $t_{ij}$'s (use $t_{ij}t_{k\ell} - \delta_{jk} t_{i\ell}$ to reduce the degree of a word of degree $> 1$, and use $t_{11}+t_{22}+\cdots+t_{nn}-1$ to increase the degree of the empty word). But a linear combination of single $t_{ij}$'s never lies in $\operatorname{Ker} f$ unless it is trivial (i.e., all coefficients are zero), since their images $f\left(t_{ij}\right) = E_{ij}$ are linearly independent.