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If matrix Ap=[pp+1p−1p]p∈NA_p = \begin{bmatrix} p & p+1 \\ p-1 & p \end{bmatrix}_{p \in N} (where N is the set of natural numbers), then the value of ∣A1∣+∣A2∣+∣A3∣+⋯+∣A2025∣|A_1| + |A_2| + |A_3| + \dots + |A_{2025}| is:

Solution

✅ Correct Option: 1

The matrix is a 2×2 matrix:

Ap=[pp+1p−1p]A_p = \begin{bmatrix} p & p+1 \\ p-1 & p \end{bmatrix}

Step 1: Find |ApA_p|

∣Ap∣=p⋅p−(p+1)(p−1)=p2−(p2−1)=1|A_p| = p \cdot p - (p+1)(p-1) = p^2 - (p^2 - 1) = 1

So every determinant equals 1, no matter what p is.

Step 2: Add them up

∣A1∣+∣A2∣+⋯+∣A2025∣=1+1+⋯+1⏟2025 terms=2025|A_1| + |A_2| + \dots + |A_{2025}| = \underbrace{1 + 1 + \dots + 1}_{2025 \text{ terms}} = 2025

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