Proving $A^{-1}=-\frac{1}{6}A^{2}+\frac{2}{3}A-\frac{1}{6}I$ I have a matrix A 
\begin{bmatrix}
1 & 1 & 2\\ 
 0& 2 &0 \\ 
 2&3  & 1
\end{bmatrix}
I have calculated the $A^{-1}$ 
\begin{bmatrix}
-\frac{1}{2} & -\frac{5}{4} & 2\\ 
 0& \frac{3}{4} &0 \\ 
 1&\frac{1}{4}  & -\frac{1}{2}
\end{bmatrix}
And I want to prove that $A^{-1}=-\frac{1}{6}A^{2}+\frac{2}{3}A-\frac{1}{6}I$
I tried doing all the calculations (not really productive, I know) but the result wasn't correct, or I made some calculation mistake. Is there any more productive way to prove this?
I have also found the characteristic polynomial, and it's $-\lambda ^{3}+4\lambda ^{2}-\lambda-6$
I would prefer a little tip to help me rather than the solution itself
 A: A hunch at what they're going for:
Suppose that an invertible matrix $A$ has characteristic polynomial (say $A$ is $3 \times 3$) $a_3 x^3 + a_2x^2 + a_1 x + a_0$.  By the Cayley Hamilton theorem, we have
$$
a_3 A^3 + a_2 A^2 + a_1 A + a_0 I = 0
$$
That is,
$$
a_3A^3 + a_2 A^2 + a_1 A = -a_0 I
$$
Which we can factor to get
$$
-\frac{1}{a_0}\left(
a_3A^2 + a_2 A + a_1I 
\right)A = I
$$
So that $a_3A^2 + a_2 A + a_1I$ must be the inverse of $A$.
A: Note that $\operatorname{char}_A(\lambda)=-\lambda^3+4\lambda^2-\lambda-6$. The Cayley-Hamilton Theorem then implies 
$$
-A^3+4A^2-A-6I=\mathbf 0
$$
so that
$$
A\left(-A^2+4A-I\right)=6I
$$
That is,
$$
A^{-1}=
-\frac{1}{6}A^2+\frac{2}{3}A-\frac{1}{6}I
$$
as advertised.
A: Since, by the Cayley-Hamilton theorem, $A$ satisfies its own characteristic polynomial, we have $A^3 -4A^2 + A + 6 = 0$.  Re-arranging things a little bit, we see that $A(A^2 -4A + I) = -6I$, or $A(-(1/6)A^2 + (2/3)A - (1/6)) = I$.  Thus $A^{-1}$ exists and we have $A^{-1} = -(1/6)A^2 + (2/3)A - (1/6)I$, the requisite result.  QED!!!
***Note:  the CayleyHamilton theorem is a central result in matrix theory and a wealth of information may easily be found by googling around a bit.  Or see http://en.m.wikipedia.org/wiki/Cayley%E2%80%93Hamilton_theorem.
Hope this helps.  Cheers,
and as always,
Fiat Lux!!!
