I know that this exercise is easy, but I have problem to solve it:
Let $A$ be an algebra (associative and unital) and $X$ a free $A$-module generated by $\{x_1,\cdots,x_n\}$. If $C$ is the algebra whose elements are modules homomorphisms from $X$ to $X$, prove that $C^{o}$ (the algebra opposite to $C$) is isomorphic (as algebra) to $M_n(A)$.
My attempt was to define $F$ such that $F(f)=\big(f(x_1)|f(x_2)|\cdots|f(x_n)\big)\in{M_n(A)}$ for every $f\in{C^o}$. Then $F(f\cdot{}g)=F(g\circ{}f)=F(g)F(f)$, the contrary what I need.
Could someone help me?
Assuming we are taking left $A$-modules, and writing homomorphisms on the left, then there is a ring isomorphism $\mathrm{End}_A(A^n)\cong\mathbb M_n(A^{\mathrm{op}})$. Note that the opposite is on the entries of the matrix. This sends an endomorphism $f$ to the matrix $F=(f_{ij})$ such that $f(x_j)=\sum_if_{ij}x_i$.
To check the multiplication, suppose $g\mapsto G=(g_{ij})$. Then $$ gf(x_j) = \sum_{i,k}f_{kj}g_{ik}x_i $$ so $gf$ corresponds to the matrix $g\cdot f$ provided we take the matrix multiplication in $\mathbb M_n(A^{\mathrm{op}})$.
We can now use the transpose to obtain a ring isomorphism $$ \mathbb M_n(A^{\mathrm{op}}) \cong \mathbb M_n(A)^{\mathrm{op}}, \quad M\mapsto M^t. $$
In general, we have a matrix decomposition for direct sums, so that $$ \mathrm{End}_A(X\oplus Y) \cong \begin{pmatrix}\mathrm{Hom}_A(X,X)&\mathrm{Hom}_A(Y,X)\\\mathrm{Hom}_A(X,Y)&\mathrm{Hom}_A(Y,Y)\end{pmatrix}. $$ In your question we have $X\cong A^n$ and $\mathrm{End}_A(A)\cong A^{\mathrm{op}}$ via $f\mapsto f(1)$.