\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
$=|\frac13 A^{-1} - 120 (\frac{1}{24} A^{-1})|$
$=|\frac13 A^{-1} - 5 A^{-1})|$
$=(\frac{-14}{3})^n |A^{-1}|$
$=(\frac{-14}{3})^n (24)$