For two 3× 3 matrices A and B, let A+B=2B^' and 3A+2B=I_3, where B^' is the transpose of B and I_3 is 3× 3 identity matrix. Then :

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

$$\Rightarrow {A}^{T}+{B}^{T}=2B$$⇒AT+BT=2B

$$\Rightarrow B=\frac{{A}^{T}+{B}^{T}}{2}$$⇒B=2AT+BT​

$$=A+(\frac{{B}^{T}+{A}^{T}}{2})=2{B}^{T}$$=A+(2BT+AT​)=2BT

$$2A+{A}^{T}=3{B}^{T}$$2A+AT=3BT

$$\Rightarrow A=\frac{3{B}^{T}-{A}^{T}}{2}$$⇒A=23BT−AT​

$3A+2B={I}_{3}$3A+2B=I3​ ......$(i)$(i)

$$\Rightarrow 3(\frac{3{B}^{T}-{A}^{T}}{2})+2(\frac{{A}^{T}+{B}^{T}}{2})={I}_{3}$$⇒3(23BT−AT​)+2(2AT+BT​)=I3​

$$\Rightarrow (\frac{3{B}^{T}+2{B}^{T}}{2})+2(\frac{{A}^{T}+{B}^{T}}{2})={I}_{3}$$⇒(23BT+2BT​)+2(2AT+BT​)=I3​

$\Rightarrow 11{B}^{T}-{A}^{T}=2{I}_{3}$⇒11BT−AT=2I3​ .......$(ii)$(ii)

Equation $(i)+(ii)$(i)+(ii)

$$35B=7{I}_{3}$$35B=7I3​

$$\Rightarrow B=\frac{{I}_{3}}{5}$$⇒B=5I3​​

$$11 \frac{{I}_{3}}{5}-A=2{I}_{3}$$115I3​​−A=2I3​

$$\Rightarrow 11\frac{{I}_{3}}{5}-2{I}_{3}=A$$⇒115I3​​−2I3​=A

$$\Rightarrow A=\frac{{I}_{3}}{5}$$⇒A=5I3​​

$$\because 5A=5B={I}_{3}$$∵5A=5B=I3​

$$\Rightarrow 10A+5B=3{I}_{3}$$⇒10A+5B=3I3​