Let A = ( 1, 0, 0; 0, 4, - 1; 0, 12, - 3 ). Then the sum of the diagonal elements of the matrix (A + I)¹¹ is equal to :

Solution

Maths sir · 4 steps

Square A first. It comes back as A, and that turns the 11th power into a one-line binomial expansion.

1. Check A squared. The first row and column only hold the 1, so look at the block (4, -1; 12, -3). Its square is (16-12, -4+3; 48-36, -12+9) = (4, -1; 12, -3), and 1² = 1. So A² = A, and every higher power Aᵏ is also A.
2. Expand binomially. A and I commute, so the binomial theorem applies. Every Aᵏ with k ≥ 1 becomes A, and the coefficients add up to C(11, 1) + … + C(11, 11) = 2¹¹ - 1 = 2047.

   (A+I)¹¹ = I + (2¹¹-1)A = I + 2047A

3. Take the trace. tr(I) = 3 and tr(A) = 1 + 4 - 3 = 2. So the trace is 3 + 2047 × 2 = 3 + 4094.
4. So. The sum of the diagonal elements is 4097, which is option (d).

The idea to keep. Before raising a matrix to a big power, check whether A² = A. If it is, (A+I)ⁿ = I + (2ⁿ - 1)A, and the trace is linear, so add the pieces separately.

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