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Use ordinary matrix addition and multiplication on $R=K[A]$. Sums and products of polynomials in $A$ remain such polynomials, and they commute because powers of the same matrix commute. The identity is $A^0=I$. Evaluation $\varphi (\sum c_i{X}^i)=\sum c_i{A}^i$ is a surjective unital algebra homomorphism over $K$. It is unique among maps fixing the scalar matrices and sending $X$ to $A$.

The scalar condition is needed for the word “unique”: as an unrestricted ring-homomorphism assertion over an arbitrary field, prescribing only the image of $X$ does not suffice. For example, over $\mathbb{C}$ with a real scalar matrix $A$, conjugating every coefficient gives another unital ring homomorphism with the same image of $X$. The intended polynomial-evaluation map is the $K$-algebra map just defined.

The kernel of a ring homomorphism is an ideal of the principal ideal domain $K[X]$. It is nonzero because finitely many powers of $A$ must become linearly dependent in the finite-dimensional matrix space. Its unique monic generator is precisely the minimal polynomial $m$: division by $m$ shows that a polynomial killing $A$ has zero remainder, by minimality. Hence $\ker \varphi =(m)$ and $R\cong K[X]/(m)$.

If $\gcd (f,m)=1$, Bézout's identity gives $uf+vm=1$; evaluation gives $u(A)f(A)=I$. Conversely an inverse in $R$ has form $g(A)$, so $fg-1\in (m)$, which forces $\gcd (f,m)=1$. Thus invertibility in this ring is exactly coprimality with the minimal polynomial.