Question

If $M=\left[\begin{array}{cc}1& -3\\ -1& 1\end{array}\right],$ then the value of $M-\frac{1}{3}{M}^2+\frac{1}{9}{M}^3-\frac{1}{27}{M}^4+....\infty$ is

• A. $\frac{1}{13}\left[\begin{array}{cc}-1& 9\\ 3& -1\end{array}\right]$
• B. $\frac{3}{13}\left[\begin{array}{cc}-1& 9\\ 3& -1\end{array}\right]$
• C. $\frac{1}{13}\left[\begin{array}{cc}1& -9\\ -3& 1\end{array}\right]$
• D. $\frac{3}{13}\left[\begin{array}{cc}1& -9\\ -3& 1\end{array}\right]$

Answer 1

The correct option is D $\frac{3}{13}\left[\begin{array}{cc}1& -9\\ -3& 1\end{array}\right]$

Let $N=M-\frac{M^2}{3}+\frac{M^3}{9}-\frac{M^4}{27}+....\infty ...\left(1\right)$
$MN=M^2-\frac{M^3}{3}+\frac{M^4}{9}-\frac{M^5}{27}+....\infty$
$\frac{MN}{3}=\frac{M^2}{3}-\frac{M^3}{9}+\frac{M^4}{27}-....\infty ...\left(2\right)$
Adding $\left(1\right)$ and $\left(2\right),$ we get
$N+\frac{MN}{3}=M$
$\Rightarrow N(I+\frac{M}{3})=M$
$\Rightarrow N=\left[\begin{array}{cc}3& -9\\ -3& 3\end{array}\right]{\left[\begin{array}{cc}4& -3\\ -1& 4\end{array}\right]}^{-1}=\frac{1}{13}\left[\begin{array}{cc}3& -27\\ -9& 3\end{array}\right]=\frac{3}{13}\left[\begin{array}{cc}1& -9\\ -3& 1\end{array}\right]$