Finding $A^{-1}=bA+cI$ with the eigenvalues of $A$ 
Let $A_{n\times n}$ (EDIT: $A$ is diagnolizable) and it's eigenvalues are $2,-3$, find an expression for $A^{-1}$ such that $A^{-1}=bA+cI, b,c\in \mathbb R$

My try:
Let's change the expression to a polynomial: $x^{-1}=bx+c\Rightarrow bx^2+cx-1=0$.
Using the given eigenvalues: $(A-2)(A+3)=A^2+A-6=0$.
So we get that $b=c=\frac 1 6$? Is it really ok to express $A^{-1}=bA+cI$ with a polynomial?
 A: Transform to $I=bA^2+cA$, apply eigenvectors $v=vI=(b\lambda^2 +c\lambda)v$ resulting in $b\lambda^2+c\lambda=1$ for $\lambda \in \{2, -3\}$ and solve the set of linear equations $4b+2c=1$ and $9b-3c=1$. This gives you $b=c=\frac 1 6$ even without having to think about matrix polynomials.
Edit: As hinted by @Travis, this gives an easy way to calculate $b,c$, but does not imply on its own that $A^{-1}$ is given by $bA+cI$.
A: The assertion is not true as stated: Indeed, The matrix $$A := \pmatrix{2&1&0\\0&2&0\\0&0&-3}$$ is upper triangular so its eigenvalues are the values $2, -3$ that occur on the diagonal, but computing gives $$A^2 + A - 6 \neq 0; $$ correspondingly there are no $b, c$ for which $A^{-1} = bA + cI$.

In general, the constant term of the characteristic polynomial $p_B(t) := \det(t I - B)$ of the $n \times n$ polynomial $B$ is $p_B(0) = \det(-B) = (-1)^n \det B$, so we can write $$p_B(t) = t q(t) + (-1)^n (\det B)$$ for some polynomial $q(t)$ of degree $n - 1$. The Cayley-Hamilton Theorem states that any square matrix satisfies its own characteristic polynomial, that is, that $p_B(B) = 0$, and so
$$0 = p_B(B) = B q(B) + (-1)^n (\det B) I.$$
So, if $B$ is invertible, or equivalently, $\det B \neq 0$, dividing and rearranging gives
$$B \cdot \color{red}{(-1)^{n + 1} (\det B)^{-1} q(B)} = I.$$
Thus, the quantity in red is $B^{-1}$, which in particular is expressed as a polynomial in $B$.
Note that the only features of $B$ we used in this construction are that (1) $p_B(0) \neq 0$ if $B$ invertible and (2) $p_B(B) = 0$, and both of these features also hold for the minimal polynomial $m_B(t)$ of $B$ (the unique nonzero monic polynomial of minimal degree for which $m_B(B) = 0$), so one can analogously give for $B^{-1}$ a formula polynomial in $B$ of degree $\deg m_B - 1$, and by minimality this is the smallest degree polynomial for which such a formula exists.

Returning to our case, $$p_A(t) = (t - 2)^2 (t - 3) = t^3 - t^2 - 8 t + 12,$$ giving
$$A^{-1} = \tfrac{1}{12}(-A^2 + A + 8 I). $$ Moreover, $m_A(t) = p_A(t)$, and so this is the best one can do, i.e., there is no linear polynomial expression in $A$ for $A^{-1}$.
On the other hand, if a matrix $B$ is diagonalizable, its minimal polynomial is $\prod_k (t - \lambda_k)$, where the $\lambda_k$ are the eigenvalues of $B$, not including multiplicity. So, in our case, if we impose the additional condition that $A$ is diagonalizable, we have $$m_A(t) = (t - 2)(t + 3) = t^2 + t - 6$$ and rearranging $$m_A(A) = 0$$ gives $$A^{-1} = \tfrac{1}{6}(A + I),$$ and so in this case we can take $b = c = \frac{1}{6}$ as originally claimed.
