For two $3\times 3$ matrices $A$and $B$, let $A+B=2B^{\mathrm{\prime }}$ and $3A+2B=I_3$, where $B^{\mathrm{\prime }}$ is the transpose of $B$ and $I_3$ is $3\times 3$ identity matrix. Then

$A+B=2B^{\mathrm{\prime }}$

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

Now, $A+\left(\frac{B^{\mathrm{\prime }}+A^{\mathrm{\prime }}}{2}\right)=2B^{\mathrm{\prime }}\left[\because A+B=2B^{\mathrm{\prime }}\right]$

$\begin{array}{l}\Rightarrow 2A+B^{\mathrm{\prime }}+A^{\mathrm{\prime }}=4B^{\mathrm{\prime }}\Rightarrow 2A+A^{\mathrm{\prime }}=3B^{\mathrm{\prime }}\\ \Rightarrow A=\frac{3B^{\mathrm{\prime }}-A^{\mathrm{\prime }}}{2}\end{array}$

Also, $3A+2B=I_3$                                                                          $\dots \left(\mathrm{i}\right)$

$\begin{array}{l}\Rightarrow 3\left(\frac{3B^{\mathrm{\prime }}-A^{\mathrm{\prime }}}{2}\right)+2\left(\frac{A^{\mathrm{\prime }}+B^{\mathrm{\prime }}}{2}\right)=I_3\\ \Rightarrow \left(\frac{9B^{\mathrm{\prime }}+2B^{\mathrm{\prime }}}{2}\right)+\left(\frac{2A^{\mathrm{\prime }}-3A^{\mathrm{\prime }}}{2}\right)=I_3\end{array}$

$\begin{array}{l}\Rightarrow 11B^{\mathrm{\prime }}-A^{\mathrm{\prime }}=2I_3\Rightarrow {\left(11B^{\mathrm{\prime }}-A^{\mathrm{\prime }}\right)}^{\mathrm{\prime }}={\left(2I_3\right)}^{\mathrm{\prime }}\\ \Rightarrow 11B-A=2I_3\end{array}$                                    $\text{ ...(ii) }$

Multiplying $\text{ (ii) }$by 3 and then adding $\text{ (i), }$we get

$35B=7I_3\Rightarrow B=\frac{I_3}{5}$

From $\text{ (ii), }11\frac{I_3}{5}-A=2I_3\Rightarrow 11\frac{I_3}{5}-2I_3=A$        

$\begin{array}{l}\Rightarrow A=\frac{I_3}{5}\\ \therefore 5A=5B=I_3\Rightarrow 10A+5B=3I_3\end{array}$