Finding the inverse and power of a matrix

Question

So I was given this question Let A = $$ \left[ \begin{array}{cc|c} -1&2\\ 1&3 \end{array} \right] $$

Find a) $A^{-1}$, b)$A^{3}$, c) $(A^{-1})^3$, and finally d) use answers to (b) and (c) to show that $(A^{-1})^3$ is the inverse of $A^3$

So here is my take on it.

So for part a) i did,

$A^{-1}$ = $$ \left[ \begin{array}{cc|c} d/(ad-bc)&-b/(ad-bc)\\ -c/(ad-bc)&a/(ad-bc) \end{array} \right] $$

= $$ \left[ \begin{array}{cc|c} 3/((-1*3)-(2*1))&-2/((-1*3)(2*1)\\ -1/((-1*3)-(2*1))&-1/((-1*3)-(2*1) \end{array} \right] $$

= $$ \left[ \begin{array}{cc|c} 2/-5&-2/-5\\ -1/-5&-1/-5 \end{array} \right] $$

I'm pretty sure I'm correct but I'm completely confused about the rest and how to go about it.

Answer 1

The matrix $A^3$ is, in fact (a bit obvious), $A \times A \times A$. Do the multiplication. For b), $(A^{-1})^3$ is $A^{-1} \times A^{-1} \times A^{-1} $, do the multiplication. Since $(A^x)^y = A^{xy}$, onde could just state that $A^{-3} = (A^{-1})^{3} = (A^3)^{-1}$, but, since the exercise wants you to show this via the results, just show that $(A^{-1})^{3}$ and $(A^3)^{-1}$ are both equal to $I_{2\times2}$.