M (1)

Given that $ \displaystyle A=\left( {\begin{array}{*{20}{c}} {\frac{1}{2}} & 1 \\ 0 & {\frac{1}{3}} \end{array}} \right)\ \ $ , find $ \displaystyle {{A}^{n}}$ , where $ \displaystyle n$ is a natural number. If $ \displaystyle \ \ B=A+{{A}^{2}}+{{A}^{3}}+{{A}^{4}}+...\ \ $ , then find $ \displaystyle \ {{B}^{{-1}}}$ .

Solution

$ \displaystyle {{A}^{n}}=\left( {\begin{array}{*{20}{c}} {\frac{1}{2}} & 1 \\ 0 & {\frac{1}{3}} \end{array}} \right)\ =\left( {\begin{array}{*{20}{c}} {\frac{1}{2}} & {6(\frac{1}{2}-\frac{1}{3})} \\ 0 & {\frac{1}{3}} \end{array}} \right)\ $

$ \displaystyle {{A}^{2}}=\ \left( {\begin{array}{*{20}{c}} {\frac{1}{2}} & 1 \\ 0 & {\frac{1}{3}} \end{array}} \right)\left( {\begin{array}{*{20}{c}} {\frac{1}{2}} & 1 \\ 0 & {\frac{1}{3}} \end{array}} \right)=\left( {\begin{array}{*{20}{c}} {\frac{1}{4}} & {\frac{5}{6}} \\ 0 & {\frac{1}{9}} \end{array}} \right)\ =\left( {\begin{array}{*{20}{c}} {\frac{1}{{{{2}^{2}}}}} & {6(\frac{1}{{{{2}^{2}}}}-\frac{1}{{{{3}^{2}}}})} \\ 0 & {\frac{1}{{{{3}^{2}}}}} \end{array}} \right)\ $

$ \displaystyle {{A}^{3}}=\left( {\begin{array}{*{20}{c}} {\frac{1}{4}} & {\frac{5}{6}} \\ 0 & {\frac{1}{9}} \end{array}} \right)\ \left( {\begin{array}{*{20}{c}} {\frac{1}{2}} & 1 \\ 0 & {\frac{1}{3}} \end{array}} \right)=\left( {\begin{array}{*{20}{c}} {\frac{1}{8}} & {\frac{{19}}{{36}}} \\ 0 & {\frac{1}{{27}}} \end{array}} \right)\ =\left( {\begin{array}{*{20}{c}} {\frac{1}{{{{2}^{3}}}}} & {6(\frac{1}{{{{2}^{3}}}}-\frac{1}{{{{3}^{3}}}})} \\ 0 & {\frac{1}{{{{3}^{3}}}}} \end{array}} \right)\ $

$ \displaystyle {{A}^{4}}=\ \left( {\begin{array}{*{20}{c}} {\frac{1}{8}} & {\frac{{19}}{{36}}} \\ 0 & {\frac{1}{{27}}} \end{array}} \right)\ \left( {\begin{array}{*{20}{c}} {\frac{1}{2}} & 1 \\ 0 & {\frac{1}{3}} \end{array}} \right)=\left( {\begin{array}{*{20}{c}} {\frac{1}{{16}}} & {\frac{{65}}{{216}}} \\ 0 & {\frac{1}{{81}}} \end{array}} \right)=\left( {\begin{array}{*{20}{c}} {\frac{1}{{{{2}^{4}}}}} & {6(\frac{1}{{{{2}^{4}}}}-\frac{1}{{{{3}^{4}}}})} \\ 0 & {\frac{1}{{{{3}^{4}}}}} \end{array}} \right)$
.
.
.
$ \displaystyle {{A}^{n}}=\left( {\begin{array}{*{20}{c}} {\frac{1}{{{{2}^{n}}}}} & {6(\frac{1}{{{{2}^{n}}}}-\frac{1}{{{{3}^{n}}}})} \\ 0 & {\frac{1}{{{{3}^{n}}}}} \end{array}} \right)\ $

$ \displaystyle B=A+{{A}^{2}}+{{A}^{3}}+{{A}^{4}}+...\ $

$ \displaystyle B=\left( {\begin{array}{*{20}{c}} {\frac{1}{2}+\frac{1}{{{{2}^{2}}}}+\frac{1}{{{{2}^{3}}}}+\frac{1}{{{{2}^{4}}}}+...} & {6((\frac{1}{2}+\frac{1}{{{{2}^{2}}}}+\frac{1}{{{{2}^{3}}}}+\frac{1}{{{{2}^{4}}}}+...)-(\frac{1}{3}+\frac{1}{{{{3}^{2}}}}+\frac{1}{{{{3}^{3}}}}+\frac{1}{{{{3}^{4}}}}+...))} \\ 0 & {\frac{1}{3}+\frac{1}{{{{3}^{2}}}}+\frac{1}{{{{3}^{3}}}}+\frac{1}{{{{3}^{4}}}}+...} \end{array}} \right)\ $

$ \displaystyle B=\left( {\begin{array}{*{20}{c}} 1 & {6(1-\frac{1}{2})} \\ 0 & {\frac{1}{2}} \end{array}} \right)\ $

$ \displaystyle \therefore B=\ \left( {\begin{array}{*{20}{c}} 1 & 3 \\ 0 & {\frac{1}{2}} \end{array}} \right)\ $

$ \displaystyle \det B=\frac{1}{2}\ -0=\frac{1}{2}\ne 0,\ {{B}^{{-1}}}\ \text{exists }\text{.}$

$ \displaystyle \therefore \ {{B}^{{-1}}}=2\left( {\begin{array}{*{20}{c}} {\frac{1}{2}} & {-3} \\ 0 & 1 \end{array}} \right)\ =\left( {\begin{array}{*{20}{c}} 1 & {-6} \\ 0 & 2 \end{array}} \right)$

Note
$ \displaystyle *\ \ \ \ S=\frac{a}{{1-r}}$