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Let three matrices $A =$$\left[ {\begin{array}{*{20}{c}}2&1\\4&1\end{array}} \right]$ ; $B =$$\left[ {\begin{array}{*{20}{c}}3&4\\2&3\end{array}} \right]$ and $C =$$\left[ {\begin{array}{*{20}{c}}3&{ - 4}\\{ - 2}&3\end{array}} \right]$ then $T_r(A) + t_r$ $\left( {\frac{{ABC}}{2}} \right)$ $+$ $t_r$$\left( {\frac{{A{{(BC)}^2}}}{4}} \right)$ $+$ $t_r$ $\left( {\frac{{A{{(BC)}^3}}}{8}} \right)$ $+$ ....... $+ \infty =$
$6$
$9$
$12$
none
Solution
$BC =$ $\left[ {\begin{array}{*{20}{c}}3&4\\2&3\end{array}} \right]$ $\left[ {\begin{array}{*{20}{c}}3&{ – 4}\\{ – 2}&3\end{array}} \right]$ ==> $BC =$ $\left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right]$ $= I$
$t_r(A) + t_r$ $\left( {\frac{A}{2}} \right)$ $+$ $t_r$ $\left( {\frac{A}{{{2^2}}}} \right)$ $+$ …….
$= t_r(A) +$ $\frac{1}{2}$ $t_r(A)$ $+$ $\frac{1}{{{2^2}}}$ $t_r(A) +$ …….
$=$ $\frac{{{t_r}(A)}}{{1 – \left( {\frac{1}{2}} \right)}}$ $= 2 t_r(A)$ $= 2(2 + 1) = 6$