Help with inverse matrix problem? (Specific problem in description) \begin{equation} 
\text{If}
\begin{vmatrix}A\end{vmatrix}
\text{=}\frac{1}{24}
\text{, solve }
\begin{vmatrix}
\begin{pmatrix}\frac{1}{3}A\end{pmatrix}^{-1} - 120 \text{ }A^*
\end{vmatrix}
\end{equation}
My attempt so far comes from the definition of an inverted matrix, namely
\begin{equation}
A^{-1} = \frac{A^{*}}{\begin{vmatrix}A\end{vmatrix}}
\end{equation}
So substituting gets
\begin{vmatrix}
\begin{pmatrix}\frac{1}{3}\frac{A^{*}}{\begin{vmatrix}A\end{vmatrix}}\end{pmatrix}^{-1} - 120 \text{ }\frac{A^{-1}}{\begin{vmatrix}A\end{vmatrix}}
\end{vmatrix}
Simplifying
\begin{vmatrix}
\begin{pmatrix}\frac{1}{72}A^{*}
\end{pmatrix}^{-1} - 5 \text{ }A^{-1}
\end{vmatrix}
Substituting using the inverse matrix definition again
\begin{vmatrix}
\begin{pmatrix}\frac{1}{72}A^{-1}|A|\end{pmatrix}^{-1} - 5A^{-1}
\end{vmatrix}
Using 
\begin{equation}
(AB)^{-1} = B^{-1}\cdot A^{-1}
\end{equation}
And I get this?
\begin{vmatrix}
\begin{pmatrix}\frac{1}{72}|A|\end{pmatrix}^{-1}
\begin{pmatrix}A^{-1}\end{pmatrix}^{-1} - 5A^{-1}
\end{vmatrix}
I think I went about this the wrong way completely, but I'm not getting any new ideas for now - woud appreciate any tips or hints regarding this. Thanks. 

Use the $A^{-1}$ only in second part or first part of the problem not both 
\begin{equation}
A^{*} = \frac{1}{24}{A^{-1}}
\end{equation}
sub this where $A^*$ is in second part and don't change $A^{-1}$ until fully simplified 
$=|\frac13 A^{-1} - 120 (\frac{1}{24} A^{-1})|$
$=|\frac13 A^{-1} - 5 A^{-1})|$
$=(\frac{-14}{3})^n |A^{-1}|$
$=(\frac{-14}{3})^n (24)$
