Question

If $A=\left(\begin{array}{cc}3& 1\\ -9& -3\end{array}\right)$ then ${(1+2A+3A^2+....\infty )}^{-1}$ equals

• A. $\left(\begin{array}{cc}-5& -2\\ 18& 7\end{array}\right)$
• B. $\left(\begin{array}{cc}-5& 18\\ -2& 7\end{array}\right)$
• C. $\left(\begin{array}{cc}7& -2\\ 18& -5\end{array}\right)$
• D. None of these

Answer 1

Answer: (A)

$\left(\begin{array}{cc}-5& -2\\ 18& 7\end{array}\right)$

$A=\left(\begin{array}{cc}3& 1\\ -9& -3\end{array}\right)$
$\therefore A^2=\left(\begin{array}{cc}3& 1\\ -9& -3\end{array}\right)\left(\begin{array}{cc}3& 1\\ -9& -3\end{array}\right)=\left(\begin{array}{cc}0& 0\\ 0& 0\end{array}\right)$

$\therefore 1+2A+3A^2+....\infty =1+2A$
$=\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)+\left(\begin{array}{cc}6& 2\\ -18& -6\end{array}\right)=\left(\begin{array}{cc}7& 2\\ -18& -5\end{array}\right)$

$\therefore (1+2A+3A^2+....\infty {)}^{-1}=(1+2A{)}^{-1}$
$={\left(\begin{array}{cc}7& 2\\ -18& -5\end{array}\right)}^{-1}=\left(\begin{array}{cc}-5& -2\\ 18& 7\end{array}\right)$