A) \[10A+5B=3{{I}_{3}}\]
B) \[3A+6B=2{{I}_{3}}\]
C) \[5A+10B-2{{I}_{3}}\]
D) \[B+2A={{I}_{3}}\]
Correct Answer: A
Solution :
\[{{A}^{T}}+{{B}^{T}}\,=2B\] \[\Rightarrow \,\,B=\frac{{{A}^{T}}+{{B}^{T}}}{2}\] \[=A\left( \frac{{{B}^{T}}+{{A}^{T}}}{2} \right)=2{{B}^{T}}\] \[=2A+{{A}^{T}}=2{{B}^{T}}\] \[\Rightarrow \,A=\frac{3{{B}^{T}}-{{A}^{T}}}{2}\] \[3A+2B={{I}_{3}}\] ?(i) \[\Rightarrow \,\,3\left( \frac{3{{B}^{T}}-{{A}^{T}}}{2} \right)+2\left( \frac{{{A}^{T}}+{{B}^{T}}}{2} \right)={{I}_{3}}\] \[\Rightarrow \,\left( \frac{3{{B}^{T}}+2{{B}^{T}}}{2} \right)+\left( \frac{2{{A}^{T}}-3{{A}^{T}}}{2} \right)={{I}_{3}}\] \[\Rightarrow \,11{{B}^{T}}\,-{{A}^{T}}=2{{I}_{3}}\] ?(ii) Equation (i) + (ii) \[35B=7{{I}_{3}}\] \[\Rightarrow \,B=\frac{{{I}_{3}}}{5}\] \[11\frac{{{I}_{3}}}{5}-A=2{{I}_{3}}\] \[\Rightarrow \,11\frac{{{I}_{3}}}{5}\,-2{{I}_{3}}=A\] \[\Rightarrow \,A=\frac{{{I}_{3}}}{5}\] \[\because \,\,5A=5B={{I}_{3}}\] \[\Rightarrow \,10A+5B=3{{I}_{3}}\]
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