Question

Let $M$ be a square matrix of order $3$ whose elements are real numbers and $\text{adj}(\text{adj}M)=\left[\begin{array}{ccc}36& 0& -4\\ 0& 6& 0\\ 0& 3& 6\end{array}\right]$, then the absolute value of $\text{Tr}(M)$ is
[Here, adj $P$ denotes adjoint matrix of $P$ and $\text{Tr}(P)$ denotes trace of matrix $P$ i.e. sum of all principal diagonal elements of matrix $P$ ]

Answer 1

Answer: 8

$\text{adj}(\text{adj}M)=|M{|}^{n-2}M$
$\therefore$ order of $M$ is $3$
$\text{adj}(\text{adj}M)=|M|M=\left[\begin{array}{ccc}36& 0& -4\\ 0& 6& 0\\ 0& 6& 6\end{array}\right]$
Taking determinant on both the sides
$|M{|}^4=\left[\begin{array}{ccc}36& 0& -4\\ 0& 6& 0\\ 0& 3& 6\end{array}\right]$
$|M{|}^4=6^4$
$\Rightarrow |M|=\pm 6$
Hence, $M=\left[\begin{array}{ccc}\pm 6& 0& \pm 2/3\\ 0& \pm 1& 0\\ 0& \pm 1/2& \pm 1\end{array}\right]$
$\left|T_r(M)\right|=8$