Let $\mathrm{A}=\left[{\mathrm{a}}_{\mathrm{ij}}\right]=\left[\begin{array}{cc}{\log }_{5}128& {\log }_{4}5\\ {\log }_{5}8& {\log }_{4}25\end{array}\right].$ If $A_{ij}$ is the cofactor of ${\mathrm{a}}_{\mathrm{ij}},{\mathrm{C}}_{\mathrm{ij}}=\sum _{\mathrm{i}=1}^2 {\mathrm{a}}_{\mathrm{ik}}{\mathrm{A}}_{\mathrm{jk}},1\le \mathrm{i}, \mathrm{j}\le 2,$ and $C=\left[C_{ij}\right]$, then $8\left|C\right|$ is equal to

$\begin{array}{l}\left|\mathrm{A}\right|=\frac{11}{2}\\ {\mathrm{C}}_{11}=\sum _{\mathrm{k}=1}^2 {\mathrm{a}}_{1\mathrm{k}}\cdot {\mathrm{A}}_{1\mathrm{k}}={\mathrm{a}}_{11}{\mathrm{A}}_{11}+{\mathrm{a}}_{12}{\mathrm{A}}_{12}=\left|\mathrm{A}\right|=\frac{11}{2}\\ {\mathrm{C}}_{12}=\sum _{\mathrm{k}=1}^2 {\mathrm{a}}_{1\mathrm{k}}\cdot {\mathrm{A}}_{2\mathrm{k}}=0\\ {\mathrm{C}}_{21}=\sum _{\mathrm{k}=1}^2 {\mathrm{a}}_{2\mathrm{k}}\cdot {\mathrm{A}}_{1\mathrm{k}}=0\\ {\mathrm{C}}_{22}=\sum _{\mathrm{k}=1}^2 {\mathrm{a}}_{2\mathrm{k}}\cdot {\mathrm{A}}_{2\mathrm{k}}=\left|\mathrm{A}\right|=\frac{11}{2}\\ \mathrm{C}=\left[\begin{array}{cc}11/2& 0\\ 0& 11/2\end{array}\right]\\ \left|\mathrm{C}\right|=\frac{121}{4}\end{array}$
8|C| = 242