Let the matrix A=[1, 0, 0; 1, 0, 1; 0, 1, 0] satisfy Aⁿ=Aⁿ⁻²+A²-I for n ≥ 3. Then the sum of all the elements of A⁵⁰ is :

Solution

Maths sir · 4 steps

Read the given relation as a jump of two powers at a time, and A⁵⁰ falls straight down to A².

1. Rewrite the relation. Aⁿ - Aⁿ⁻² = A² - I for every n ≥ 3. So each time the power goes up by 2, you add the same matrix A² - I.
2. Step down to A². From A⁵⁰ down to A² is 48 ÷ 2 = 24 jumps, and each jump removes one copy of A² - I. So A⁵⁰ = A² + 24(A² - I).

   A⁵⁰ = 25A² - 24I

3. Find A². Multiplying gives A² = (1, 0, 0; 1, 1, 0; 1, 0, 1). Its elements add up to 5, and the elements of I add up to 3.
4. So. The sum of all elements of A⁵⁰ is 25 × 5 - 24 × 3 = 125 - 72 = 53. Writing it out, A⁵⁰ = (1, 0, 0; 25, 1, 0; 25, 0, 1), which confirms it. The answer is (d).

The idea to keep. When a matrix relation links Aⁿ to Aⁿ⁻², treat it like an arithmetic progression in steps of 2. Telescope down to a small power you can multiply out by hand.

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