Question

Let $A$ be a $3\times 3$ real matrix. If $det\left(\text{2 Adj(2 Adj(Adj(2A)))}\right)=2^{41},$ then the value of $det\left(A^2\right)$ equals

Answer 1

We know that, if $A$ is a square matrix of order $n,$ then
$|\text{Adj}A|=|A{|}^{n-1}$
$\text{Adj(Adj}A)=|A{|}^{n-2}A$
$|kA|=k^n|A|$ where $k$ is a scalar
$\text{Adj}\left(kA\right)=k^{n-1}\text{Adj}\left(A\right)$

$det\left(\text{2 Adj(2 Adj(Adj(2A)))}\right)=2^{41}$
$\Rightarrow det\left(\text{2 Adj(2 Adj(}{2}^2\text{Adj(A)}\right)\left)\right)=2^{41}$
$\Rightarrow det\left(\text{2Adj(}2\cdot 2^4\text{Adj(Adj(A)}\right)\left)\right)=2^{41}$
$\Rightarrow det\left(\text{2Adj}\right(2^5|A|A\left)\right)=2^{41}$
$\Rightarrow det(2\left(32|A|{)}^2\right(\text{Adj}\left(A\right))=2^{41}$
$\Rightarrow det(2^{11}|A{|}^2\left(\text{Adj}\right(A))=2^{41}$
$\Rightarrow (2^{11}|A{|}^2{)}^3|\text{Adj}A|=2^{41}$
$\Rightarrow 2^{33}|A{|}^6|A{|}^2=2^{41}$
$\Rightarrow |A{|}^8=2^8\Rightarrow |A|=\pm 2$
$\Rightarrow |A{|}^2=4$