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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.

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    $A^3 = AAA$. Just do the multiplication. Likewise for $(A^{-1})^3$. Then to show that $(A^{-1})^3$ is the inverse of $A^3$, you'll need to show that **both** products $(A^{-1})^3 \cdot A^3$ and $A^3 \cdot (A^{-1})^3$ are equal to $\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$. – 2017-02-28
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    Can you explain what u mean by $A^3$ = $AAA$ and how would that work with the inverse – 2017-02-28
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    $AAA$ means $A$ multiplied by $A$ and then the result multiplied by $A$. – 2017-02-28
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    @ZeroInf, see the first two sentences of Dovah-king's answer. It's just exponent notation. Just like how $2^3 = 2 \cdot 2 \cdot 2$, the notation $A^3$ means $A \cdot A \cdot A$ (or just $AAA$ if we want to write even less). – 2017-02-28

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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 $I_{2\times2}$.

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    Re: your last sentence, don't show that $(A^{-1})^3$ and $(A^{3})^{-1}$ are both equal to $I_{2 \times 2}$. They're not and that wouldn't prove the desired result anyway. The goal is to show that $(A^{-1})^3 \cdot A^3$ and $A^3 \cdot (A^{-1})^3$ are both equal to $I_{2 \times 2}$. This would show that $(A^3)^{-1}$ (the inverse of $A^3$) is equal to $(A^{-1})^3$. – 2017-02-28
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    Yes, my bad. Realized it now. Thanks. – 2017-02-28