दिया गया, $${a_{ij}} = {2^{j - i}}$$

अब, $$A = \left[ {\matrix{ {{2^0}} & {{2^1}} & {{2^2}} \cr {{2^{ - 1}}} & {{2^0}} & {{2^1}} \cr {{2^{ - 2}}} & {{2^{ - 1}}} & {{2^0}} \cr } } \right]$$

$$ = \left[ {\matrix{ 1 & 2 & 4 \cr {{1 \over 2}} & 1 & 2 \cr {{1 \over 4}} & {{1 \over 2}} & 1 \cr } } \right]$$

$${A^2} = \left[ {\matrix{ 1 & 2 & 4 \cr {{1 \over 2}} & 1 & 2 \cr {{1 \over 4}} & {{1 \over 2}} & 1 \cr } } \right]\left[ {\matrix{ 1 & 2 & 4 \cr {{1 \over 2}} & 1 & 2 \cr {{1 \over 4}} & {{1 \over 2}} & 1 \cr } } \right]$$

$$ = \left[ {\matrix{ {1 + 1 + 1} & {2 + 2 + 2} & {4 + 4 + 4} \cr {{1 \over 2} + {1 \over 2} + {1 \over 2}} & {1 + 1 + 1} & {2 + 2 + 2} \cr {{1 \over 4} + {1 \over 4} + {1 \over 4}} & {{1 \over 2} + {1 \over 2} + {1 \over 2}} & {1 + 1 + 1} \cr } } \right]$$

$$ = \left[ {\matrix{ 3 & 6 & {12} \cr {{3 \over 2}} & 3 & 6 \cr {{3 \over 4}} & {{3 \over 2}} & 3 \cr } } \right]$$

$$ = 3\left[ {\matrix{ 1 & 2 & 4 \cr {{1 \over 2}} & 1 & 2 \cr {{1 \over 4}} & {{1 \over 2}} & 1 \cr } } \right]$$

$$ = 3A$$

इसी तरह, $${A^3} = {3^2}A$$

$${A^4} = {3^3}A$$

$$\therefore$$ $${A^2} + {A^3} + \,\,......\,\, + \,\,{A^{10}}$$

$$ = 3A + {3^2}A + {3^3}A + \,\,......\,\, + \,\,{3^9}A$$

$$ = A(3 + {3^2} + {3^3} + \,\,......\,\, + \,\,{3^9})$$

$$ = A\left( {{{3({3^9} - 1)} \over {3 - 1}}} \right) = {{3({3^9} - 1)} \over 2}A$$ = $$\left( {{{{3^{10}} - 3} \over 2}} \right)A$$