Question:

Step 1: Identify the elements of the matrix and define the trace. The matrix $A = [a_{ij}]_{n \times n}$ is given, where its elements are defined by the formula: $$a_{ij} = (i^2 + j^2 - ij)(j - i)$$ The trace of a square matrix $A$, denoted as $tr(A)$, is the sum of its diagonal elements. For an $n \times n$ matrix, this is: $$tr(A) = \sum_{k=1}^{n} a_{kk}$$ Step 2: Calculate the diagonal elements $a_{kk}$. To find the diagonal elements, we set the row index $i$ equal to the column index $j$. Let's use $k$ for both indices, so $i=k$ and $j=k$. Substituting these into the formula for $a_{ij}$: $$a_{kk} = (k^2 + k^2 - k \cdot k)(k - k)$$ $$a_{kk} = (k^2 + k^2 - k^2)(0)$$ $$a_{kk} = (k^2)(0)$$ $$a_{kk} = 0$$ This shows that every diagonal element of the matrix $A$ is $0$. Step 3: Calculate the trace of the matrix $A$. Since all diagonal elements $a_{kk}$ are $0$, the trace of matrix $A$ is the sum of these zero elements: $$tr(A) = a_{11} + a_{22} + \dots + a_{nn}$$ $$tr(A) = 0 + 0 + \dots + 0$$ $$tr(A) = 0$$ Step 4: Verify if the matrix is skew-symmetric (alternative method). A matrix $A$ is skew-symmetric if its transpose is equal to its negative, i.e., $A^T = -A$. This implies that its elements satisfy $a_{ji} = -a_{ij}$ for all $i, j$. Let's find the expression for $a_{ji}$: $$a_{ji} = (j^2 + i^2 - ji)(i - j)$$ We can rearrange the terms in the first parenthesis: $j^2 + i^2 - ji = i^2 + j^2 - ij$. We can also factor out $-1$ from the second parenthesis: $i - j = -(j - i)$. Substituting these into the expression for $a_{ji}$: $$a_{ji} = (i^2 + j^2 - ij)(-(j - i))$$ $$a_{ji} = -(i^2 + j^2 - ij)(j - i)$$ By definition, $a_{ij} = (i^2 + j^2 - ij)(j - i)$. Therefore, we have: $$a_{ji} = -a_{ij}$$ This confirms that the matrix $A$ is skew-symmetric. For any skew-symmetric matrix, its diagonal elements must be zero ($a_{ii} = -a_{ii} \implies 2a_{ii} = 0 \implies a_{ii} = 0$). Consequently, the trace of any skew-symmetric matrix is always zero, regardless of its dimension $n$. Step 5: Conclude the final answer. Based on the calculations, all diagonal elements of matrix $A$ are $0$. Therefore, the trace of matrix $A$ is $0$. The final answer is $\boxed{0}$.