Question

Let $M$ be a square matrix of order $3$ whose elements are real numbers and

$adj\left(adjM\right)=\left[\begin{array}{ccc}36& 0& -4\\ 0& 6& 0\\ 0& 3& 6\end{array}\right]$ , then the absolute value of $Tr\left(M\right)$ is

[Here, $adjP$ denotes adjoint matrix of $P$ and $T_r\left(P\right)$ denotes trace of matrix $P$ i.e. sum of all principal diagonal elements of matrix $P$ ]

Answer 1

Answer: 8

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