Given any positive integer $n$, does the endofunctor $R \mapsto M_n(R)$ of the category of (unital) rings have a left adjoint?
One possible idea for constructing the left adjoint $L(R)$ at a ring $R$ is to consider the quotient of the free ring on $R \times \{1,2,...,n\} \times \{1,2,...,n\}$ by the two-sided ideal generated by the following elements (to ensure that the map $\eta_{R}:R \to M_n(L(R))$, $r \mapsto [(r,i,j)]_{1 \le i,j \le n}$, is a ring homomorphism):
- $(r,i,j)+(s,i,j)-(r+s,i,j)$ for $r,s \in R$ and $1 \le i,j \le n$ (necessary for $\eta_{R}$ to be an additive group homomorphism)
- $\big(\sum_{k=1}^{n} (r,i,k)(s,k,j)\big)-(rs,i,j)$ for $r,s \in R$ and $1 \le i,j \le n$ (necessary for $\eta_{R}$ to preserve multiplication)
- $(1,i,i)-1$ for $1 \le i \le n$ (one of the necessary conditions for $\eta_{R}(1)$ to be the identity matrix)
- $(1,i,j)$ for $1 \le i,j \le n$ with $i \neq j$ (the other necessary condition for $\eta_{R}(1)$ to be the identity matrix)
But does the above quotient ring idea really give a left adjoint?
A silly answer: it preserves limits and is accessible, so because the category of rings is presentable, it has a left adjoint by the adjoint functor theorem.
Slightly less silly: It can be described as $\hom(\mathbb Z[X_{ij}, 1\leq i,j\leq n], -)$ with a certain ring structure, which is given by universal formulas, in other words, by a co-ring object structure on $U=\mathbb Z[X_{ij}, 1\leq i,j\leq n]$ ( a ring structure on that object in the category $\mathrm{Ring}^{op}$).
In particular, $\hom(S,M_n(R)) \cong \hom(S, \hom(U, R))$ and this is definitely representable by some quotient of the free ring on $S\times \{1,...,n\}^2$. One can then work out the relations to impose on this free ring, the ones you outlined are definitely among them, and I think it should be enough.
The conditions you need to impose are exactly those such that if you start from a ring morphism from this quotient to $R$, then $s\mapsto (f(s,i,j))_{i,j}$ is a ring morphism $S\to M_n(R)$.
Then you can work these out and see that exactly what you wrote comes out.