Question

Let $A=\left[\begin{array}{ccc}6& 3& 0\\ 2& 5& 2\\ 0& 1& 1\end{array}\right]$ and $B$ be the adjoint of $A.$ Then the value of $det(det\left(A^{-1}\right)(A^{2}B{)}^T)$ is

Answer 1

$A=\left[\begin{array}{ccc}6& 3& 0\\ 2& 5& 2\\ 0& 1& 1\end{array}\right]$

$|A|=18-6=12$

Now, $det(det\left(A^{-1}\right)(A^{2}B{)}^T)$
$=(det\left(A^{-1}\right){)}^3\cdot det((A^{2}B{)}^T)$
$=(det\left(A^{-1}\right){)}^3\cdot det\left(A^{2}B\right)$
$=(det\left(A^{-1}\right){)}^3\cdot (det(A){)}^{2}det(B)$
$=(det\left(A^{-1}\right){)}^3\cdot (det(A\left){)}^2\right(detA{)}^2$
$[\because det\left(adjA\right))=(detA{)}^{n-1},$ where $n$ is the order of $A.]$
$=\frac{1}{(det(A){)}^3}\cdot (det(A){)}^4$
$=detA=12$