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 $A^{2} + A^{3} + \dots + A^{10}$ is equal to:

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Solution

$A = ⎡ ⎢ ⎣ \begin{matrix} a_{11} & a_{12} & a_{13} a_{21} & a_{22} & a_{23} a_{31} & a_{32} & a_{33} \end{matrix} ⎤ ⎥ ⎦ = ⎡ ⎢ ⎣ \begin{matrix} 2^{0} & 2^{1} & 2^{2} 2^{- 1} & 2^{0} & 2^{1} 2^{- 2} & 2^{- 1} & 2^{0} \end{matrix} ⎤ ⎥ ⎦$
$A^{2} = ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ \begin{matrix} 1 & 2 & 4 \frac{1}{2} & 1 & 2 \frac{1}{4} & \frac{1}{2} & 1 \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ \begin{matrix} 1 & 2 & 4 \frac{1}{2} & 1 & 2 \frac{1}{4} & \frac{1}{2} & 1 \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ = ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ \begin{matrix} 3 & 6 & 12 \frac{3}{2} & 3 & 6 \frac{3}{4} & \frac{3}{2} & 3 \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ = 3 A$
$A^{2} = 3 A$
$A^{3} = A \cdot A^{2} = A (3 A) = 3 A^{2} = 3^{2} A$
$A^{4} = 3^{3} A$
Now
$A^{2} + A^{3} + \dots + A^{10}$
$A [3^{1} + 3^{2} + 3^{3} + \dots + 3^{9}]$
$= \frac{3 [3^{9} - 1]}{3 - 1} A$
$= \frac{(3^{10} - 3)}{2} A$

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