matrix-finding determinant of adj of inverse matirx if A is a $3$x$3$ matrix and let A=$2$,then what will be the value of det(adj(adj(adj($A^{-1}$)))?
1.$\dfrac{1}{512}$
2.$\dfrac{1}{1024}$
3.$\dfrac{1}{128}$
4.$\dfrac{1}{256}$
 A: Assume you meant $det(A)=2$. Then $det(A^{-1})=\frac{1}{2}$. Work the problem outward:
\begin{align*}
adj(A^{-1})&=det(A^{-1})A=\frac{1}{2}A,\\
adj(\frac{1}{2}A)&=(1/2)^2 adj(A)=\frac{1}{4}det(A)A^{-1}=\frac{1}{2}A^{-1},\\
adj(\frac{1}{2}A^{-1})&=(1/2)^2adj(A^{-1})=\frac{1}{4}det(A^{-1})A=\frac{1}{8}A,\\
det(\frac{1}{8}A)&=(1/8)^3det(A)=\frac{1}{512}2=\frac{1}{256}.
\end{align*}
For justifications of the steps, you can wiki "adjugate matrix" and "determinant."
A: To find: $det(adj(adj(adj(A^{-1})))$
Now denote $det(A)$ as $|A|$. 
Note that by adjoint rule we must have: $$A.adj(A)=adj(A).A=|A|I_n.$$ In case, if $A$ be non singular, then deduction comes out as $|adj(A)|=|A|^{n-1}$. Hence $|adj(adj(A))|=|adj(A)|^{n-1}=(|A|^{n-1})^{n-1}=|A|^{(n-1)^2}.$ So $$|adj(adj(adj(A)))|=|adj(adj(A))|^{n-1}=|A|^{(n-1)^3}$$.
Therefore, $$|adj(adj(adj(A^{-1})))|=|A^{-1}|^{(n-1)^3}=|A|^{-(n-1)^3}$$
Since $n=3$ and $|A|=2$ so we have 
 $$|adj(adj(adj(A^{-1})))|=2^{-8}=\frac{1}{256}$$ so that option 4 is correct
