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 $3\times 3$ identify matrix. Then :

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

Answer 1

The correct option is C $10A+5B=3I_3$

$A+B=2B^{\prime }..\left(1\right)$

Taking transpose on both sides, we have $A^{\prime }+B^{\prime }=2B..\left(2\right)$

$3A+2B=I_3..\left(3\right)$

Taking transpose on both sides, we get $3A^{\prime }+2B^{\prime }=I_3..\left(4\right)$

Substituting equation (2) in (4), 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 (3)

$\Rightarrow I_3=3A+2A=5A=5B$

$\Rightarrow 10A+5B=10A+5A=15A=3I_3$

Hence, option $C$ is correct.