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

Answer 1

Answer: (B)

$10A+5B=3I_3$

Given, $A+B=2B^{\prime }..eq.13A+2B=I_3..eq.2$

Let A = $[a_t{]}_{3\times 3},$ B = $[b_t{]}_{3\times 3}$ where , i, j = 1, 2, 3

From eq.1

$\left[a_{ij}\right]+\left[b_{ij}\right]=2\left[b_{ji}\right]$

for i = j

$a_{jj}+b_{jj}=2b_{jj}\Rightarrow a_{jj}=b_{jj}..eq.3$

For $i\ne j$

$\left[a_{ij}\right]+\left[b_{ij}\right]=2\left[b_{ji}\right]..eq.4$

From eq.2

$3a_{ij}+2_{ij}=I_3$

for I = j

$3a_{jj}+2b_{jj}=1$

using eq.3,

$\Rightarrow 3a_{jj}+2a_{jj}=1$

$\Rightarrow 5a_{jj}=1$

$\Rightarrow a_{jj}=\frac{1}{5}$

For $i\ne j$

$3a_{ij}+2_{ij}=0\Rightarrow 3a_{ij}=-2b_{ij}{a}_{ij}=\frac{-2}{3}{b}_{ij}$

using eq.4

$a_{ij}+b_{ij}=2b_{ji}\Rightarrow \frac{-2}{3}{b}_{ij}+b_{ij}=2b_{ji}\Rightarrow b_{ij}=6b_{ji}$

Hence $10A+5B=3I_3$