To solve this, we need to carefully analyze the situation step by step.

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Step 1: Matrix $A$

• $A$ is a square matrix of size $n \times n$.
• All off-diagonal elements are $0$, and only the diagonal elements are non-zero.
• These diagonal elements correspond to numbers $1$ to $n$, arranged in alphabetical order based on their English spelling.

Example for $n = 3$:

The numbers $1, 2, 3$ are spelled as:

1. One
2. Two
3. Three

Arranged alphabetically:

1. One (1)
2. Three (3)
3. Two (2)

So, the diagonal of $A$ is: $\text{diag}(A) = [1, 3, 2]$

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Step 2: $A^2$

Since $A$ is a diagonal matrix, squaring it results in squaring each diagonal entry: $A^2 = \text{diag}(a_1^2, a_2^2, \dots, a_n^2),$ where $a_1, a_2, \dots, a_n$ are the diagonal elements of $A$ in alphabetical order.

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Step 3: Trace of $A^2$

The trace of $A^2$ is the sum of the diagonal elements of $A^2$, which is: $\text{Trace}(A^2) = a_1^2 + a_2^2 + \cdots + a_n^2$

But the diagonal elements $a_1, a_2, \dots, a_n$ are just a permutation of the numbers $1, 2, \dots, n$. Therefore, regardless of the alphabetical order, the sum of their squares remains the same: $\text{Trace}(A^2) = 1^2 + 2^2 + \cdots + n^2$

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Step 4: Formula for Sum of Squares

The formula for the sum of squares of the first $n$ natural numbers is: $\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}$

So, the trace of $A^2$ is: $\text{Trace}(A^2) = \frac{n(n+1)(2n+1)}{6}$

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Final Answer:

The correct option is: C) $\frac{n(n+1)(2n+1)}{6}$

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Would you like further details or clarifications? Here are five related questions to explore:

1. What is the trace of $A$ itself in this example?
2. How does the sum of squares formula generalize to other sequences?
3. What changes if $A$ is not diagonal but symmetric?
4. Can you calculate $A^k$ for $k > 2$ in this scenario?
5. How does this problem relate to permutations of natural numbers?

Tip: When dealing with diagonal matrices, operations like squaring or finding traces are simplified because only the diagonal elements matter!