Let Z and A are two   matrices such that $\mathrm{Z}=\left[\begin{array}{ccc}1& 1& 2\\ 2& 3& 5\\ 4& 1& 0\end{array}\right] \mathrm{and} \mathrm{A}=\left[\begin{array}{ccc}1& 0& 1\\ 2& 1& 0\\ 0& 3& 1\end{array}\right]$ if  $\mathrm{Z}={\mathrm{AB}}^{-1}$where B is a square matrix of order 3 and $\mathrm{t}=\mathrm{trace}\hspace{0.17em}\hspace{0.17em}\mathrm{of}\mathrm{\hspace{0.17em}}\left(\mathrm{A}(\mathrm{adj}\hspace{0.17em}\hspace{0.17em}\mathrm{B})\right)$, then $\frac{28}{\left|\mathrm{t}\right|}$  is:

Given $\mathrm{Z}={\mathrm{AB}}^{-1}\Rightarrow \left|\mathrm{Z}\right|=\left|{\mathrm{AB}}^{-1}\right|\Rightarrow \left|\mathrm{Z}\right|=\left|\mathrm{A}\right|\times {\left|\mathrm{B}\right|}^{-1}$
$\Rightarrow -5=7\left(\left|{\mathrm{B}}^{-1}\right|\right)\Rightarrow \left|{\mathrm{B}}^{-1}\right|=-\frac{5}{7}$
Again
 $$\begin{gathered} \mathrm{Z}={\mathrm{AB}}^{-1}\Rightarrow \mathrm{Z}=\mathrm{A}\frac{1}{\left|\mathrm{B}\right|}\mathrm{adjB} \\ \mathrm{Z}=\mathrm{A}\left(\frac{1}{\left|\mathrm{B}\right|}\mathrm{adjB}\right) \Rightarrow \mathrm{Z}\left|\mathrm{B}\right|=\mathrm{A}\left(\mathrm{adjB}\right) \\ \mathrm{trace} \left(\mathrm{A}\left(\mathrm{adjB}\right)\right) = \mathrm{trace} \left(\mathrm{Z}\left|\mathrm{B}\right|\right)=\left|\mathrm{B}\right|\mathrm{trace}\left(\mathrm{Z}\right)=4(-\frac{7}{5})=-\frac{28}{5} \end{gathered}$$