Question

Let $A$ be a $3\times 3$ matrix such that $det\left(A\right)=a=3$ and $B=adj\left(A\right)$ such that $det\left(B\right)=b.$ Then the value of $(3b^2+9b+1)S,$ where $\frac{1}{2}S=\frac{a}{b}+\frac{a^2}{b^3}+\frac{a^3}{b^5}+\cdots \text{upto}\infty ,$ is

• A. $175$
• B. $224$
• C. $225$
• D. $325$

Answer 1

The correct option is C $225$

We know that
$A\cdot adj\left(A\right)=|A|I$
$\Rightarrow AB=3I$
$\Rightarrow |A||B|=3^3$
$\Rightarrow |B|=b=9$

Now, $S=2(\frac{a}{b}+\frac{a^2}{b^3}+\frac{a^3}{b^5}+\cdots \text{upto}\infty )$
$\Rightarrow S=2\left(\frac{a/b}{1-a/b^2}\right)(\because r=\frac{a}{b^2}<1)$
$\Rightarrow S=\frac{9}{13}$

$\therefore (3b^2+9b+1)S$
$=(3\cdot 81+9\cdot 9+1)\frac{9}{13}$
$=\left(325\right)\frac{9}{13}=25\times 9=225$