Let A = ( 1 + i, 1; - i, 0 ) where i = √(- 1). Then, the number of elements in the set n ∈ 1, 2, ......, 100 : Aⁿ = A is ____________.

Solution

Maths sir · 4 steps

Keep multiplying A by itself until you hit the identity; after that, the powers just repeat.

1. Square it. Multiply row by column, using i² = -1. For example, the top-left entry is (1+i)² + 1 · (-i) = 2i - i = i. This gives A² = (i, 1+i; 1-i, -i), which is not A.
2. Next two powers. Next, A³ = A² A = (0, i; 1, 1-i), which is still not A or I. One more multiplication gives the identity matrix.

   A⁴ = A³ A = (0, i; 1, 1-i)(1+i, 1; -i, 0) = (1, 0; 0, 1) = I

3. When is it A. Since A⁴ = I, the powers repeat in a cycle of length 4: A, A², A³, I, A, …. These four matrices are all different, so Aⁿ = A exactly when n leaves remainder 1 on division by 4.
4. So. The valid values are n = 1, 5, 9, …, 97. That is (97 - 1)/4 + 1 values, so the set has 25 elements.

The common slip: 24. Starting the list at n = 5 and forgetting n = 1 gives 24, but A¹ = A also counts.

The idea to keep. When a question asks for many powers of a small matrix, compute a few powers until one equals I. After that, every power depends only on the remainder of n when divided by that cycle length.

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