Question

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

• A. $5A+10B=2I_3$
• B. $3A+2B=2I_3$
• C. $10A+5B={3I}_3$
• D. $3A+6B=2I_3$

Answer 1

The correct option is C $10A+5B={3I}_3$

$A+B=2B^{\prime }\cdots \left(i\right)$
Taking transpose on both sides, we have
$A^{\prime }+B^{\prime }=2B\cdots \left(ii\right)3A+2B=I_3\cdots \left(iii\right)$
Taking transpose on both sides, we get
$3A^{\prime }+2B^{\prime }=I_3\cdots \left(iv\right)$
Substituting equation $\left(ii\right)$ in (iv), we have
$3(2B-B^{\prime })+2B^{\prime }=I_3$
i.e. $6B-B^{\prime }=I_3$
Writing $B^{\prime }=\frac{A+B}{2}$, we get
$6B-\frac{A+B}{2}=I_3\therefore 12B-A-B=2I_3$
i.e. $11B-A=2I_3$
$\Rightarrow 11B-A=6A+4B\therefore 7B=7A\Rightarrow A=B$
Put this in equation $\left(iii\right)$
$\Rightarrow I_3=3A+2A=5A=5B$
On verifying options $\Rightarrow 10A+5B=10A+5A=15A=3I_3$