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Let $A [a_{i j}]$ be a square matrix of order 3 such that $a_{i j} = 2^{j - i}$ , for all $i, j = 1, 2, 3$ .then the matrix $A^{2} + A^{3} + ... + A^{10}$ is equal to .

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$[\frac{3^{10} - 1}{2}] A$

$[\frac{3^{10} - 3}{2}] A$

$[\frac{3^{10} + 3}{2}] A$

$[\frac{3^{10} + 1}{2}] A$

answer is A.

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Detailed Solution

$A = (\begin{matrix} 1 & 2 & 2^{2} \\ 1 / 2 & 1 & 2 \\ 1 / 2^{2} & 1 / 2 & 1 \end{matrix})$

$A^{2} = 3 A, A^{3} = 3^{2} A, .........$

$A^{2} + A^{3} + A^{10} = 3 A + 3^{2} A + {...3}^{9} A = \frac{3 (3^{9} - 1)}{3 - 1} A = \frac{3^{10} - 3}{2} A$

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