NDA Mathematics Question (2026) — Solution

Step-by-step solution

Given M_k = bmatrix k & k-1 \\ k-1 & k bmatrix The determinant of M_k is given by: |M_k| = k^2 - (k-1)^2 |M_k| = k^2 - (k^2 - 2k + 1) = 2k - 1 The required sum is: |M_1| + |M_2| + |M_3| + + |M_ 50 | = _ k=1 ^ 50 (2k - 1) This represents the sum of the first 50 odd natural numbers, which forms an arithmetic progression: 1 + 3 + 5 + + 99 The sum of an arithmetic progression is given by n 2 (a + l), where n = 50, a = 1, and l = 99: Sum = 50 2 (1 + 99) = 25 100 = 2500 Answer: 2500