It is given that the matrices $A$ and $B$ are non-singular of order 3 and $|B|=k$, $a$ positive integer, so
$\begin{aligned}
\left|k^{-1} A^{-1}\right| & =\left(k^{-1}\right)^3\left|A^{-1}\right|=\frac{1}{k^3|A|} \quad\left[\because\left|A^{-1}\right|=\frac{1}{|A|}\right] \\
& =\frac{1}{|B|^3|A|} \\
\because\left|\operatorname{adj}\left(A^{-1}\right)\right| & =\left|A^{-1}\right|^{3-1}=\left|A^{-1}\right|^2=\frac{1}{|A|^2}
\end{aligned}$
$\left[\because|\operatorname{adj}(A)|=|A|^{n-1}\right.$, where $n$ is the order of matrix $A$. ]
Now, since it is given that $B A B^{-1}=I$
$\begin{aligned}
\Rightarrow \quad A B^{-1} & =B^{-1} \\
\therefore \quad B A^k B^{-1} & =B A^{k-1}\left(A B^{-1}\right)=B A^{k-1} B^{-1} \\
& =B A^{k-2}\left(A B^{-1}\right)=B A^{k-2} B^{-1} \\
\therefore \quad B A^k B^{-1} & =B A B^{-1}=I \\
\text { and }, B \frac{\operatorname{adj}(B)}{|B|} & =B B^{-1}=I
\end{aligned}$
Therefore, $B A^k B^{-1}=B \frac{\operatorname{adj}(B)}{|B|}$, if $B A B^{-1}=I$
Now, the $\operatorname{adj}\left(\operatorname{adj}\left(A^{-1}\right)\right)=\left|A^{-1}\right|^{3-2}\left(A^{-1}\right)$
$\left[\because \operatorname{adj}(\operatorname{adj} A)=|A|^{n-2} A\right.$ where $n$ is the order of matrix $A]$
Activate Wind
$=\frac{1}{|A|}\left(A^{-1}\right)$
Go to Settings to a