Question:

Step 1: Identify the constraint from the trace condition. The trace of matrix $A$ is the sum of its diagonal elements: $$\text{Trace}(A) = a + b + c = 10$$ We need to find all possible combinations of three distinct values from $\{1, 2, 3, 4, 5, 6\}$ that sum to 10. Step 2: Find all valid sets for the diagonal elements. We systematically check which three distinct numbers from $\{1, 2, 3, 4, 5, 6\}$ sum to 10: - $\{1, 3, 6\}$: $1 + 3 + 6 = 10$ ✓ - $\{1, 4, 5\}$: $1 + 4 + 5 = 10$ ✓ - $\{2, 3, 5\}$: $2 + 3 + 5 = 10$ ✓ There are exactly **3 valid sets** for the diagonal elements. Step 3: Count arrangements of diagonal elements. For each valid set of three values, we can arrange them as $(a, b, c)$ in the diagonal positions. The number of ways to arrange 3 distinct elements is: $$3! = 6 \text{ ways}$$ Step 4: Count arrangements of off-diagonal elements. Once we choose 3 values for the diagonal, the remaining 3 values from $\{1, 2, 3, 4, 5, 6\}$ must be assigned to the off-diagonal positions $x, y, z$. These 3 remaining values can be arranged in: $$3! = 6 \text{ ways}$$ Step 5: Calculate the total number of matrices. Using the multiplication principle, the total number of matrices is: $$\text{Total} = (\text{Number of valid sets}) \times (3! \text{ arrangements of diagonal}) \times (3! \text{ arrangements of off-diagonal})$$ $$\text{Total} = 3 \times 3! \times 3! = 3(3!)^2$$ The answer is **Option 1: $3(3!)^2$**