This problem requires us to first understand the structure of a given matrix A, then discover a pattern in its powers, and finally use the properties of geometric progressions to sum a series of these matrix powers.

1. Understanding and Constructing the Matrix A

The matrix A is defined as a 3x3 square matrix $A = [a_{ij}]$ where each element $a_{ij}$ is given by the formula $a_{ij} = 2^{j-i}$ for $i, j = 1, 2, 3$. Our first step is to explicitly write out this matrix. This helps us visualize its structure and prepare for subsequent calculations.

Let's calculate each element:

• For the first row ($i=1$):
  • $a_{11} = 2^{1-1} = 2^0 = 1$
  • $a_{12} = 2^{2-1} = 2^1 = 2$
  • $a_{13} = 2^{3-1} = 2^2 = 4$
• For the second row ($i=2$):
  • $a_{21} = 2^{1-2} = 2^{-1} = \frac{1}{2}$
  • $a_{22} = 2^{2-2} = 2^0 = 1$
  • $a_{23} = 2^{3-2} = 2^1 = 2$
• For the third row ($i=3$):
  • $a_{31} = 2^{1-3} = 2^{-2} = \frac{1}{4}$
  • $a_{32} = 2^{2-3} = 2^{-1} = \frac{1}{2}$
  • $a_{33} = 2^{3-3} = 2^0 = 1$

Thus, the matrix A is: $A = \begin{bmatrix} 1 & 2 & 4 \\ \frac{1}{2} & 1 & 2 \\ \frac{1}{4} & \frac{1}{2} & 1 \end{bmatrix}$ Notice the interesting structure: elements along the main diagonal are $2^0=1$, elements along the first upper diagonal are $2^1=2$, and so on. This structure often hints at simple behavior under matrix multiplication.

2. Discovering the Pattern: Calculating $A^2$

To evaluate the sum $A^2 + A^3 + \dots + A^{10}$, we need to understand how powers of A behave. The most effective way to do this is by calculating the first few powers, starting with $A^2 = A \times A$. This step is critical as it often reveals a simplifying pattern for matrices in competitive exams.

Let's compute $A^2$: $A^2 = \begin{bmatrix} 1 & 2 & 4 \\ \frac{1}{2} & 1 & 2 \\ \frac{1}{4} & \frac{1}{2} & 1 \end{bmatrix} \begin{bmatrix} 1 & 2 & 4 \\ \frac{1}{2} & 1 & 2 \\ \frac{1}{4} & \frac{1}{2} & 1 \end{bmatrix}$

We perform matrix multiplication, where each element $(A^2)_{ik}$ is the dot product of the $i$-th row of A and the $k$-th column of A.

• $(A^2)_{11} = (1)(1) + (2)(\frac{1}{2}) + (4)(\frac{1}{4}) = 1 + 1 + 1 = 3$
• $(A^2)_{12} = (1)(2) + (2)(1) + (4)(\frac{1}{2}) = 2 + 2 + 2 = 6$
• $(A^2)_{13} = (1)(4) + (2)(2) + (4)(1) = 4 + 4 + 4 = 12$
• $(A^2)_{21} = (\frac{1}{2})(1) + (1)(\frac{1}{2}) + (2)(\frac{1}{4}) = \frac{1}{2} + \frac{1}{2} + \frac{1}{2} = \frac{3}{2}$
• $(A^2)_{22} = (\frac{1}{2})(2) + (1)(1) + (2)(\frac{1}{2}) = 1 + 1 + 1 = 3$
• $(A^2)_{23} = (\frac{1}{2})(4) + (1)(2) + (2)(1) = 2 + 2 + 2 = 6$
• $(A^2)_{31} = (\frac{1}{4})(1) + (\frac{1}{2})(\frac{1}{2}) + (1)(\frac{1}{4}) = \frac{1}{4} + \frac{1}{4} + \frac{1}{4} = \frac{3}{4}$
• $(A^2)_{32} = (\frac{1}{4})(2) + (\frac{1}{2})(1) + (1)(\frac{1}{2}) = \frac{1}{2} + \frac{1}{2} + \frac{1}{2} = \frac{3}{2}$
• $(A^2)_{33} = (\frac{1}{4})(4) + (\frac{1}{2})(2) + (1)(1) = 1 + 1 + 1 = 3$

So, the matrix $A^2$ is: $A^2 = \begin{bmatrix} 3 & 6 & 12 \\ \frac{3}{2} & 3 & 6 \\ \frac{3}{4} & \frac{3}{2} & 3 \end{bmatrix}$ By comparing $A^2$ with the original matrix A, we can see a clear relationship: every element in $A^2$ is exactly 3 times the corresponding element in A. $A^2 = 3 \begin{bmatrix} 1 & 2 & 4 \\ \frac{1}{2} & 1 & 2 \\ \frac{1}{4} & \frac{1}{2} & 1 \end{bmatrix} = 3A$ This discovery, $A^2 = 3A$, is the key to solving the problem efficiently.

3. Generalizing Matrix Powers: $A^k$

Now that we know $A^2 = 3A$, we can easily find a general formula for any positive integer power of A. This generalization allows us to express each term in the sum in a simpler form.

• For $A^3$: $A^3 = A \cdot A^2 = A \cdot (3A)$ Since scalar multiplication commutes with matrix multiplication: $A^3 = 3(A \cdot A) = 3A^2$ Substitute $A^2 = 3A$ again: $A^3 = 3(3A) = 3^2A$

• For $A^4$: $A^4 = A \cdot A^3 = A \cdot (3^2A) = 3^2(A \cdot A) = 3^2A^2 = 3^2(3A) = 3^3A$

Following this pattern, we can generalize that for any integer $k \ge 1$: $A^k = 3^{k-1}A$ Let's quickly check this for $k=1$: $A^1 = 3^{1-1}A = 3^0A = 1A = A$, which is correct.

4. Summing the Series of Matrices

We need to calculate the sum $S = A^2 + A^3 + \dots + A^{10}$. Using our generalized formula $A^k = 3^{k-1}A$, we can substitute each term:

• $A^2 = 3^{2-1}A = 3^1A$
• $A^3 = 3^{3-1}A = 3^2A$
• ...
• $A^{10} = 3^{10-1}A = 3^9A$

Now, substitute these into the sum: $S = 3A + 3^2A + 3^3A + \dots + 3^9A$ Since scalar multiplication distributes over matrix addition, we can factor out the matrix A: $S = A(3 + 3^2 + 3^3 + \dots + 3^9)$ This transforms the problem into finding the sum of a scalar series and then multiplying it by A.

5. Evaluating the Geometric Progression (GP)

The expression inside the parenthesis, $3 + 3^2 + 3^3 + \dots + 3^9$, is a finite geometric progression. To find its sum, we need to identify its components:

• First term ($a$): $3$
• Common ratio ($r$): $\frac{3^2}{3} = 3$
• Number of terms ($n$): The powers of 3 range from 1 to 9, so there are 9 terms.

The formula for the sum of the first $n$ terms of a geometric progression is $S_n = \frac{a(r^n - 1)}{r - 1}$. Substituting our values ($a=3, r=3, n=9$): $S_9 = \frac{3(3^9 - 1)}{3 - 1}$ $S_9 = \frac{3(3^9 - 1)}{2}$ $S_9 = \frac{3^{10} - 3}{2}$

6. Final Solution and Conclusion

Now, we substitute this sum back into our expression for S: $S = A \left( \frac{3^{10} - 3}{2} \right)$ $S = \left( \frac{3^{10} - 3}{2} \right)A$

Comparing this result with the given options: (A) $\left( {{{{3^{10}} - 3} \over 2}} \right)A$ (B) $\left( {{{{3^{10}} - 1} \over 2}} \right)A$ (C) $\left( {{{{3^{10}} + 1} \over 2}} \right)A$ (D) $\left( {{{{3^{10}} + 3} \over 2}} \right)A$

Our calculated sum matches option (A).

The final answer is $\boxed{\text{(A)}}$.

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Tips for JEE Aspirants & Key Takeaway:

1. Pattern Recognition is Crucial: For problems involving powers or sums of powers of matrices, always calculate the first few powers ($A^2, A^3$) to identify a pattern. Often, matrices simplify to scalar multiples of themselves, the identity matrix, or follow a cyclic pattern. This saves immense calculation time.
2. Careful Matrix Multiplication: Matrix multiplication is a common source of errors. Be meticulous with your calculations, especially during the initial step of finding $A^2$. One mistake here can propagate through the entire solution.
3. Geometric Progression (GP) Formula: Ensure you are comfortable with the sum formula for a GP: $S_n = \frac{a(r^n - 1)}{r - 1}$. Pay close attention to identifying the correct first term ($a$), common ratio ($r$), and the number of terms ($n$) in the series.
4. Factorization: Remember that scalar multiplication distributes over matrix addition, allowing you to factor out the matrix A when summing scalar multiples of A.