The question has two parts.
Let $m_{A}$ be the minimal polynomial of square matrix A. Show that there exists a unique polynomial $q$, with $deg(q) < deg(m_{A})$, such that $A^{-1} = q(A)$.
Let the n-by-n matrix $A$ be diagonalizable, with $k$ distinct eigenvalues $\{a_{j}\}$. Let $p(t)$ be a degree $k-1$ polynomial that interpolates the function $f(t)=t^{-1}$, at $a_{j}$, namely $p(a_{j}) = a_{j}^{-1}$. Show that the polynomial $q$ of the first part is $p$.
I believe I have answered the first part, and have an equation of the form $I = Aq(A)$. But I'm not sure what to do about the second part.
If $A$ is diagonalizable, it is similar to a diagonal matrix $D = \operatorname{Diag}(c_1, \dots, c_n)$ with diagonal values $c_1, \dots, c_n$ taken from $\{ a_1, \dots, a_k \}$. That is, there exists an invertible $P$ such that $A = PDP^{-1}$ or $D = P^{-1}AP$. Now, for any polynomial $p$, we have
$$ p(A) = p(PDP^{-1}) = Pp(D)P^{-1}. $$
For any diagonal matrix $D$, the result of applying $p$ to $D$ is just applying $p$ to each of the diagonal entries of $D$ and since $p$ interpolates $f$, we have $p(D) = \operatorname{Diag}( c_1^{-1}, \dots, c_n^{-1})$ which implies that $p(D)$ is the inverse of $D$. But then
$$ Ap(A) = APp(D)P^{-1} = PP^{-1}APp(D)P^{-1} = PDp(D)P^{-1} = PP^{-1} = I $$
so $p(A)$ is the inverse of $A$. Since $A$ is diagonalizable with $k$ distinct eigenvalues, the minimal polynomial of $A$ has degree $k$ while the polynomial $p$ has degree $k-1$ so by uniqueness of part (1), it must be $q$.