If A = [ 1, 1, 1; 0, 1, 1; 0, 0, 1 ] and M = A + A² + A³ + ....... + A²⁰, then the sum of all the elements of the matrix M is equal to _____________.

Solution

Maths sir · 5 steps

Write A as I plus a nilpotent N. Since N cubed is zero, every power of A has a neat closed form.

1. Split A. Write A = I + N, where N has 1s at positions (1,2), (1,3), (2,3) and 0 elsewhere. Then N² has a single 1 at position (1,3), and N³ = O.
2. Binomial expansion. I commutes with N, so the binomial theorem applies. Every term with N³ or higher vanishes, which gives Aⁿ = I + nN + n(n-1)/2N².

   Aⁿ = (1, n, n(n+1)/2; 0, 1, n; 0, 0, 1)

3. Entries of one power. The (1,3) entry is n + n(n-1)/2 = n(n+1)/2. So the sum of all entries of Aⁿ is 3 + 2n + n(n+1)/2. Check with n = 2: A² has row 1 equal to 1, 2, 3, which matches.
4. Sum over powers. Add these sums for n = 1 to 20. The 3s give 3 × 20 = 60. The 2n terms give 2 × 210 = 420. The triangular numbers give Σₙ₌₁²⁰ n(n+1)/2 = (20 × 21 × 22)/6 = 1540.
5. So. The sum of all elements of M is 60 + 420 + 1540 = 2020.

The common slip: 690. Dropping the n(n-1)/2N² term makes the corner entry n instead of n(n+1)/2, and the total comes out as 690.

The idea to keep. For an upper triangular matrix with 1s on the diagonal, write it as I + N with N nilpotent. The binomial expansion then stops after a few terms and gives Aⁿ for every n.

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