Oh yeah, thanks, so
$\displaystyle A^{-1}= -(\det(A))^{-1} (A^{n-1} + \cdot \cdot \cdot)$
for part b, n=2
Let $\displaystyle A= \begin{bmatrix} a & b\\c & d \end{bmatrix}$
$\displaystyle p(A)=A^2-(a+d)A+(ad-bc)=0$
So $\displaystyle A^{-1}= -(\det(A))^{-1} (A^{2-1} +I)$
But this does not give me the correct $\displaystyle A^{-1}$ for a 2x2 matrix.
you are not quite applying the formula correctly.
$\displaystyle p(A) = A^2 - (a+d)A + \det(A)I = 0$
$\displaystyle A^2 - (a+d)A = -\det(A)I$
$\displaystyle \left(\frac{-1}{\det(A)}\right)(A - (a+d)I)A = I$, so
$\displaystyle A^{-1} = -(\det(A))^{-1}(A - (a+d)I)$
does that look any better?