Question

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

• A. $5A + 10B = 2I_3$
• B. $10A + 5B= 3I_3$
• C. $B + 2A = I_3$
• D. $3A + 6B = 2I_3$

Answer 1

Answer: (B)

$A^{T}+B^{T}=2 B$
$\Rightarrow A+\left(\frac{B^{T}+A^{T}}{2}\right)=2 B^{T}$
$2 A+A^{T}=2 B^{T}$
$\Rightarrow A=\frac{3 B^{T}-A^{T}}{2}$
$3 A+2 B=I_{3} \,\,\,\,\,\,\ldots(i)$
$\Rightarrow 3\left(\frac{3 B ^{ T }- A ^{ T }}{2}\right)+2\left(\frac{ A ^{ T }+ B ^{ T }}{2}\right)= I _{3}$
$\Rightarrow \left(\frac{3 B^{T}+2 B^{T}}{2}\right)+\left(\frac{2 A^{T}-3 A^{T}}{2}\right)=I_{3}$
$\Rightarrow 11 B^{T}-A^{T}=2 I_{3}\,\,\,\,\,\, \ldots(ii)$
Equation $(i)+(ii)$
$35 B=7 I_{3}$
$\Rightarrow B=\frac{I_{3}}{5}$
$11 \frac{I_{3}}{5} A=2 I_{3}$
$\Rightarrow 11 \frac{I_{3}}{5}-2 I_{3}=A$
$\Rightarrow A=\frac{I_{3}}{4}$
$\because 5 A=5 B=I_{3}$
$\Rightarrow 10 A+5 B=3 I_{3}$