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(A) Given $A = [a_{ij}]$ of order $3 	imes 3$ with $a_{ij} = 2^{j-i}$.$A = \begin{bmatrix} 2^{1-1} & 2^{2-1} & 2^{3-1} \ 2^{1-2} & 2^{2-2} & 2^{3-2}

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$\left(\frac{3^{10}-3}{2}\right) A$

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(A) Given $A = [a_{ij}]$ of order $3 	imes 3$ with $a_{ij} = 2^{j-i}$.$A = \begin{bmatrix} 2^{1-1} & 2^{2-1} & 2^{3-1} \ 2^{1-2} & 2^{2-2} & 2^{3-2} \ 2^{1-3} & 2^{2-3} & 2^{3-3} \end{bmatrix} = \begin{bmatrix} 1 & 2 & 4 \ 1/2 & 1 & 2 \ 1/4 & 1/2 & 1 \end{bmatrix}$.Calculate $A^2$:$A^2 = \begin{bmatrix} 1 & 2 & 4 \ 1/2 & 1 & 2 \ 1/4 & 1/2 & 1 \end{bmatrix} \begin{bmatrix} 1 & 2 & 4 \ 1/2 & 1 & 2 \ 1/4 & 1/2 & 1 \end{bmatrix} = \begin{bmatrix} 1+1+1 & 2+2+2 & 4+4+4 \ 1/2+1/2+1/2 & 1+1+1 & 2+2+2 \ 1/4+1/4+1/4 & 1/2+1/2+1/2 & 1+1+1 \end{bmatrix} = \begin{bmatrix} 3 & 6 & 12 \ 3/2 & 3 & 6 \ 3/4 & 3/2 & 3 \end{bmatrix} = 3A$.Since $A^2 = 3A$,we have $A^3 = A^2 \cdot A = (3A) \cdot A = 3A^2 = 3(3A) = 3^2 A$.By induction,$A^n = 3^{n-1} A$.We need to find $S = A^2 + A^3 + \ldots + A^{10}$.$S = 3A + 3^2 A + \ldots + 3^9 A = (3 + 3^2 + \ldots + 3^9) A$.This is a geometric progression with $n=9$ terms,first term $a=3$,and common ratio $r=3$.Sum $= \frac{a(r^n - 1)}{r - 1} = \frac{3(3^9 - 1)}{3 - 1} = \frac{3^{10} - 3}{2}$.Thus,$S = \left(\frac{3^{10} - 3}{2}ight) A$.

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

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