$\text{ Let }\mathrm{A}={\left[{\mathrm{a}}_{\mathrm{ij}}\right]}_{3\times 3}\text{ be a matrix such that }{\mathrm{AA}}^{\mathrm{T}}=4\mathrm{I}\text{ and }{\mathrm{a}}_{\mathrm{ij}}+2{\mathrm{c}}_{\mathrm{ij}}=0\text{, where }{\mathrm{c}}_{\mathrm{ij}}\text{ is the }$$\text{ cofactor of }{\mathrm{a}}_{\mathrm{ij}}\text{ and }\mathrm{I}\text{ is the unit matrix of order 3 }$$\left|\begin{array}{ccc}{\mathrm{a}}_{11}+4& {\mathrm{a}}_{12}& {\mathrm{a}}_{13}\\ {\mathrm{a}}_{21}& {\mathrm{a}}_{22}+4& {\mathrm{a}}_{23}\\ {\mathrm{a}}_{31}& {\mathrm{a}}_{32}& {\mathrm{a}}_{33}+4\end{array}\right|+5\mathrm{\lambda }\left|\begin{array}{ccc}{\mathrm{a}}_{11}+1& {\mathrm{a}}_{12}& {\mathrm{a}}_{13}\\ {\mathrm{a}}_{21}& {\mathrm{a}}_{22}+1& {\mathrm{a}}_{23}\\ {\mathrm{a}}_{31}& {\mathrm{a}}_{32}& {\mathrm{a}}_{33}+1\end{array}\right|=0 \mathrm{then} \mathrm{the} \mathrm{value} \mathrm{of} 10\mathrm{\lambda } \mathrm{is}$ ______.

Given that ${\mathrm{AA}}^{\mathrm{T}}=4\mathrm{I}$
$\begin{array}{l}\Rightarrow |\mathrm{A}{|}^2=4\Rightarrow |\mathrm{A}|=\pm 2\\ {\mathrm{A}}^{\mathrm{T}}=4{\mathrm{A}}^{-1}=4\frac{\text{ adj }\mathrm{A}}{\left|\mathrm{A}\right|}\\ \Rightarrow \left[\begin{array}{lll}{\mathrm{a}}_{11}& {\mathrm{a}}_{21}& {\mathrm{a}}_{31}\\ {\mathrm{a}}_{12}& {\mathrm{a}}_{22}& {\mathrm{a}}_{32}\\ {\mathrm{a}}_{13}& {\mathrm{a}}_{23}& {\mathrm{a}}_{33}\end{array}\right]=\frac{4}{\left|\mathrm{A}\right|}\left[\begin{array}{lll}{\mathrm{c}}_{11}& {\mathrm{c}}_{21}& {\mathrm{c}}_{31}\\ {\mathrm{c}}_{12}& {\mathrm{c}}_{22}& {\mathrm{c}}_{32}\\ {\mathrm{c}}_{13}& {\mathrm{c}}_{23}& {\mathrm{c}}_{33}\end{array}\right]\end{array}$
$\text{ Now }{\mathrm{a}}_{\mathrm{ij}}=\frac{4}{\left|\mathrm{A}\right|}{\mathrm{c}}_{\mathrm{ij}}$
$\begin{array}{l}\Rightarrow -2{\mathrm{c}}_{\mathrm{ij}}=\frac{4}{\left|\mathrm{A}\right|}{\mathrm{c}}_{\mathrm{ij}}\left({\mathrm{asa}}_{\mathrm{ij}}+2{\mathrm{c}}_{\mathrm{ij}}=0\right)\\ \Rightarrow \left|\mathrm{A}\right|=-2\end{array}$
Now $|\mathrm{A}+4\mathrm{I}|=\left|\mathrm{A}+{\mathrm{AA}}^{\mathrm{T}}\right|$
$\begin{array}{l}=\left|\mathrm{A}\right|\left|\mathrm{I}+{\mathrm{A}}^{\mathrm{T}}\right|\\ =-2\left|\mathrm{I}+{\mathrm{A}}^{\mathrm{T}}\right|\\ =-2|\mathrm{I}+\mathrm{A}|\end{array}$
$\Rightarrow |\mathrm{A}+4\mathrm{I}|+2|\mathrm{A}+\mathrm{I}|=0$
So on comparing, we get $5\mathrm{\lambda }=2\Rightarrow \mathrm{\lambda }=\frac{2}{5}$
$\text{ Hence, }10\mathrm{\lambda }=4$