If $A$ is a square matrix find $(c,d)$ if $6A^{-1}=A^2+cA+dI$ 
Given,$$A=\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & -2 & 4 \end{pmatrix}$$  and $$6A^{-1}=A^2+cA+dI$$ then $(c,d)=?$

I have no other clue than putting the values of $A$ and $A^{-1}$ and explicitly solving $c$ and $d$. Of course thats not too decent. Is there a better way out?  
Thanks!!
 A: The characteristic polynomial must be
$$f(x)=x^3+cx^2+dx-6.$$
By inspection $1$ is an eigenvalue of $A$. Therefore $f(1)=0$ and so
$$c+d=5.$$
The trace of $A$ is $6$, so $c=-6$ and therefore $d=11$.
A: Note that $$\rho_A(x)=(1-x)(x-3)(x-2)$$ so by Cayley-Hamilton $$(I-A)(A-3I)(A-2I)=0=A^3-6A^2+11A-6I$$  and so $$6I=A^3-6A^2+11A$$ so $$6A^{-1}=A^2-6A+11I$$ since $A$ is invertble!
A: Hint : 
Pre multiplying both sides by $A$ we  get $$A^3+cA^2+dA-6I=0$$ 
That means $A$ satisfies the equation $$x^3+cx^2+dx+6=0$$
So by Cayley Hamilton theorem you just need to find the characteristic equation of $A$ and compare it with the cubic above to get $c, d$
A: The characteristic polynomial:
$$p(\lambda)=\det(A-\lambda I_3)=\begin{vmatrix}1-\lambda & 0 & 0\\ 0&1-\lambda & 1\\ 0&-2&4-\lambda\end{vmatrix}=0 \Rightarrow \\
p(\lambda)=-\lambda ^3+6\lambda ^2-11\lambda +6=0 \Rightarrow \\
p(A)=-A^3+6A^2-11A+6I=0 \Rightarrow \\
6I=A^3-6A^2+11A \stackrel{\text{multiply by } A^{-1}}\Rightarrow \\
6A^{-1}=A^2-6A+11I \Rightarrow c=-6, d=11.$$
