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 A^2+A^3+… +A^10 is equal to

Given, $A={[{a}_{ij}]}_{3\times 3}$A=[aij​]3×3​ where ${a}_{ij}={2}^{j-i}$aij​=2j−i we get,

$$A=[\begin{matrix}1 & 2 & 4 \\ \frac{1}{2} & 1 & 2 \\ \frac{1}{4} & \frac{1}{2} & 1\end{matrix}]$$A=[121​41​​2121​​421​]

$${A}^{2}=[\begin{matrix}1 & 2 & 4 \\ \frac{1}{2} & 1 & 2 \\ \frac{1}{4} & \frac{1}{2} & 1\end{matrix}][\begin{matrix}1 & 2 & 4 \\ \frac{1}{2} & 1 & 2 \\ \frac{1}{4} & \frac{1}{2} & 1\end{matrix}]$$A2=[121​41​​2121​​421​][121​41​​2121​​421​]

$\Rightarrow {A}^{2}=[\begin{matrix}3 & 6 & 12 \\ \frac{3}{2} & 3 & 6 \\ \frac{3}{4} & \frac{3}{2} & 3\end{matrix}]=3A$⇒A2=[323​43​​6323​​1263​]=3A $\Rightarrow {A}^{3}=3{A}^{2}=9A$⇒A3=3A2=9A

Now, ${A}^{2}+{A}^{3}+{A}^{4}+\cdots {A}^{10}$A2+A3+A4+⋯A10

$$=3A+{3}^{2}A+{3}^{3}A+\cdots {3}^{9}A$$=3A+32A+33A+⋯39A

$$=3A[\frac{{3}^{9}-1}{3-1}]=A[\frac{{3}^{10}-3}{2}]$$=3A[3−139−1​]=A[2310−3​]