JEE 2022 — Mathematics PYQ

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Question

JEE 2022 — Mathematics PYQ

JEE | Mathematics | 2022

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

Choose the correct answer:

Explanation

Solution

Step 1: Find $A^2$ .

$A = \begin{bmatrix} 2^0 & 2^1 & 2^2 \\ 2^{-1} & 2^0 & 2^1 \\ 2^{-2} & 2^{-1} & 2^0 \end{bmatrix}$ . Calculating $A^2$ , we find that $A^2 = 3A$ .

Step 2: Generalize $A^n$ .

By induction, $A^n = 3^{n-1}A$ .

Step 3: Sum the series.

S = A^2 + A^3 + \dots + A^{10} = 3A + 3^2A + \dots + 3^9A

S = A(3 + 3^2 + \dots + 3^9)

This is a Geometric Progression sum: $a=3, r=3, n=9$ .

Sum = 3 \left( \frac{3^9 - 1}{3 - 1} \right) = 3 \left( \frac{3^9 - 1}{2} \right) = \frac{3^{10} - 3}{2}

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

Correct Option: (A)