Explanation:

Let’s break it down:

- The matrix A is 10×10, and each entry a_ij = w^i+j, where w is a cube root of unity (so w³ = 1).

- The trace of a matrix, tr(A), is just the sum of the diagonal elements: tr(A) = ?a_ii.

- Each diagonal element a_ii = wⁱ⁺ⁱ = w²ⁱ.

Now, w cycles through the cube roots, so w° = 1, w¹ = w, w² = w², w³ = 1 again, and so on.

So tr(A) = w² + w4 + w6 + ... + w²° (for i = 1 to 10).

But since w³ = 1, w4 = w, w5 = w², w6 = 1, and so forth—it cycles every 3. If you actually add up w²ⁱ for i = 1 to 10, these cycle through the same three values over ten terms.

If you sum any whole number of cycles (any multiple of 3), the sum of the cube roots over a cycle is always zero (that’s a fundamental property). For ten terms (three full cycles and one extra), the sum is 0 + w² (last term, since 2×10 = 20, and 20 mod 3 = 2), so your sum is w².

But the cube roots of unity sum to zero. Specifically, 1 + w + w² = 0, so w² = –(1 + w). That’s not zero or one or three.

Let’s check the options:

1. 0 — Not generally true for 10 terms, because of the leftover.

2. 1 — No, this doesn’t match the sum.

3. 3 — Definitely not, the sum can’t jump to three.

4. None — This is the only sensible choice here.

So, Option 4: none is correct.