Let A = [ a _ ij ] be a square matrix of order 3 such that a _ ij =...

Given, $A = {\left[{a}_{ij}\right]}_{3 \times 3}$ where ${a}_{ij} = {2}^{j - i}$ we get,

A = \left[\begin{cases}1 & 2 & 4 \\ \dfrac{1}{2} & 1 & 2 \\ \dfrac{1}{4} & \dfrac{1}{2} & 1\end{cases}\right]

{A}^{2} = \left[\begin{cases}1 & 2 & 4 \\ \dfrac{1}{2} & 1 & 2 \\ \dfrac{1}{4} & \dfrac{1}{2} & 1\end{cases}\right]\left[\begin{cases}1 & 2 & 4 \\ \dfrac{1}{2} & 1 & 2 \\ \dfrac{1}{4} & \dfrac{1}{2} & 1\end{cases}\right]

⇒ {A}^{2} = \left[\begin{cases}3 & 6 & 12 \\ \dfrac{3}{2} & 3 & 6 \\ \dfrac{3}{4} & \dfrac{3}{2} & 3\end{cases}\right] = 3A $⇒ {A}^{3} = 3{A}^{2} = 9A$

Now, ${A}^{2} + {A}^{3} + {A}^{4} + ⋯ {A}^{10}$

$= 3A + {3}^{2}A + {3}^{3}A + ⋯ {3}^{9}A$

$= 3A\left[\dfrac{{3}^{9} - 1}{3 - 1}\right] = A\left[\dfrac{{3}^{10} - 3}{2}\right]$