Let A=[1, 2; 0, 1] and B=I+adj(A)+(adj A)²+…+(adj A)¹⁰. Then, the sum of all the elements of the matrix B is:

Solution

Maths sir · 4 steps

Find adj A first, then notice its powers follow a clean pattern: only the top-right entry grows.

1. Find adj A. For a 2 × 2 matrix, swap the diagonal entries and change the sign of the other two. So adj A = [1, -2; 0, 1].
2. Spot the pattern. Squaring gives [1, -4; 0, 1], and each extra factor adds another -2 to the top-right entry. In general (adj A)ⁿ = [1, -2n; 0, 1], and n = 0 gives I.

   B = Σₙ₌₀¹⁰ [1, -2n; 0, 1] = [11, -2(0+1+…+10); 0, 11] = [11, -110; 0, 11]

3. Count the terms. There are 11 matrices in the sum, I plus ten powers, so each diagonal entry is 11. The top-right entry is -2 × 55 = -110.
4. So. The sum of all elements of B is 11 - 110 + 0 + 11 = -88, which is option (d).

The idea to keep. For an upper triangular matrix with 1s on the diagonal, [1, k; 0, 1]ⁿ = [1, nk; 0, 1]. Find the pattern and add entry by entry.

Answer from the NTA final answer key. Solution by PrepWiser.