Let $A = [{a_{ij}}]$ be a square matrix of order 3 such that ${a_{ij}} = {2^{j - i}}$, for all i, j = 1, 2, 3. Then, the matrix A² + A³ + ...... + A¹⁰ is equal to :

<p>Given, ${a_{ij}} = {2^{j - i}}$</p> <p>Now, $$A = \left[ {\matrix{ {{2^0}} & {{2^1}} & {{2^2}} \cr {{2^{ - 1}}} & {{2^0}} & {{2^1}} \cr {{2^{ - 2}}} & {{2^{ - 1}}} & {{2^0}} \cr } } \right]$$</p> <p>$$ = \left[ {\matrix{ 1 & 2 & 4 \cr {{1 \over 2}} & 1 & 2 \cr {{1 \over 4}} & {{1 \over 2}} & 1 \cr } } \right]$$</p> <p>$${A^2} = \left[ {\matrix{ 1 & 2 & 4 \cr {{1 \over 2}} & 1 & 2 \cr {{1 \over 4}} & {{1 \over 2}} & 1 \cr } } \right]\left[ {\matrix{ 1 & 2 & 4 \cr {{1 \over 2}} & 1 & 2 \cr {{1 \over 4}} & {{1 \over 2}} & 1 \cr } } \right]$$</p> <p>$$ = \left[ {\matrix{ {1 + 1 + 1} & {2 + 2 + 2} & {4 + 4 + 4} \cr {{1 \over 2} + {1 \over 2} + {1 \over 2}} & {1 + 1 + 1} & {2 + 2 + 2} \cr {{1 \over 4} + {1 \over 4} + {1 \over 4}} & {{1 \over 2} + {1 \over 2} + {1 \over 2}} & {1 + 1 + 1} \cr } } \right]$$</p> <p>$$ = \left[ {\matrix{ 3 & 6 & {12} \cr {{3 \over 2}} & 3 & 6 \cr {{3 \over 4}} & {{3 \over 2}} & 3 \cr } } \right]$$</p> <p>$$ = 3\left[ {\matrix{ 1 & 2 & 4 \cr {{1 \over 2}} & 1 & 2 \cr {{1 \over 4}} & {{1 \over 2}} & 1 \cr } } \right]$$</p> <p>$= 3A$</p> <p>Similarly, ${A^3} = {3^2}A$</p> <p>${A^4} = {3^3}A$</p> <p>$\therefore$ ${A^2} + {A^3} + \,\,......\,\, + \,\,{A^{10}}$</p> <p>$= 3A + {3^2}A + {3^3}A + \,\,......\,\, + \,\,{3^9}A$</p> <p>$= A(3 + {3^2} + {3^3} + \,\,......\,\, + \,\,{3^9})$</p> <p>$= A\left( {{{3({3^9} - 1)} \over {3 - 1}}} \right) = {{3({3^9} - 1)} \over 2}A$ = $\left( {{{{3^{10}} - 3} \over 2}} \right)A$</p>