Question

Let $A$ be a square matrix of order $3$ whose elements are real numbers and $adj(adj\left(adjA\right))=\left[\begin{array}{ccc}16& 0& -3\\ 0& 4& 0\\ 0& 3& 4\end{array}\right].$ Then the absolute value of $trace\left(A^{-1}\right)$ is

Answer 1

$adj\left(adjA\right)=|A{|}^{n-2}A$
So, $adj\left(adjadjA\right)=|adjA{|}^{n-2}adjA$
$=(|A{|}^{n-1}{)}^{n-2}adjA$
Putting $n=3,$ we get
$adj\left(adjadjA\right)=|A{|}^{2}adjA=\left[\begin{array}{ccc}16& 0& -3\\ 0& 4& 0\\ 0& 3& 4\end{array}\right]\cdots \left(1\right)$
Taking determinant both sides
$|adj\left(adjadjA\right)|=|A{|}^6|adjA|=256$
$\Rightarrow |adj\left(adjadjA\right)|=|A{|}^6|A{|}^2=2^8$
$\Rightarrow |A|=\pm 2$

From $\left(1\right),$
$adjA=\frac{1}{4}\left[\begin{array}{ccc}16& 0& -3\\ 0& 4& 0\\ 0& 3& 4\end{array}\right]=\left[\begin{array}{ccc}4& 0& -\frac{3}{4}\\ 0& 1& 0\\ 0& \frac{3}{4}& 1\end{array}\right]$

$A^{-1}=\frac{adjA}{|A|}=\pm \left[\begin{array}{ccc}2& 0& -\frac{3}{8}\\ 0& \frac{1}{2}& 0\\ 0& \frac{3}{8}& \frac{1}{2}\end{array}\right]\therefore |trace\left(A^{-1}\right)|=3$