$\because $ $\left[ {{{\left( {A + B} \right)}^T} = {{\left( {2B} \right)}^T}} \right]$

$ \Rightarrow B = \frac{{{A^T} + {B^T}}}{2} = A + \left( {\frac{{{A^T} + {B^T}}}{2}} \right) = 2{B^T}$

                          $2A + {A^T} = 2{B^T}$

$ \Rightarrow A = \frac{{3{B^T} + {A^T}}}{2}$

                          $ = 3A + 2B = {I_{3\,\,\,\,\,\,}}\,\,\,\,\,.....\left( i \right)$

$ \Rightarrow 3\left( {\frac{{3{B^T} + {A^T}}}{2}} \right) + 2\left( {\frac{{{A^T} + {B^T}}}{2}} \right) = {I_3}$

$ \Rightarrow 11{B^T} - {A^T} = 2{I_3}\,\,\,\,\,\,\,\,\,\,\,......\left( {ii} \right)$

Add $(i)$ and $(ii)$

$35B = 7{I_3}$

$ \Rightarrow B = \frac{{{I_3}}}{5}$

         $11\frac{{{I_3}}}{5} - A = 2{I_3}$

$ \Rightarrow 11\frac{{{I_3}}}{5} - 2{I_3} = A$

$ \Rightarrow A = \frac{{{I_3}}}{5}$

$\because $ $5A = 5B = {I_3}$

$ \Rightarrow 10A + 5B = 3{I_3}$