Let $M$ denote the set of all real matrices of order $3 \times 3$ and let $S = \{-3, -2, -1, 1, 2\}$. Let $S_1 = \{A = [a_{ij}] \in M: A = A^T \text{ and } a_{ij} \in S, \forall i, j\}$, $S_2 = \{A = [a_{ij}] \in M: A = -A^T \text{ and } a_{ij} \in S, \forall i, j\}$, $S_3 = \{A = [a_{ij}] \in M: a_{11} + a_{22} + a_{33} = 0 \text{ and } a_{ij} \in S, \forall i, j\}$. If $n(S_1 \cup S_2 \cup S_3) = 125 \alpha$, then $\alpha$ equals

Step 1: Determine the number of elements in S

The set $S$ contains 5 distinct real numbers: $\{-3, -2, -1, 1, 2\}$. This means that each entry $a_{ij}$ of a matrix $A$ can take any of these 5 values.

Step 2: Calculate the number of matrices in S1 (Symmetric Matrices)

For a $3 \times 3$ symmetric matrix, the elements $a_{ij}$ must be equal to $a_{ji}$. This means we only need to choose the values for the diagonal elements ($a_{11}, a_{22}, a_{33}$) and the upper triangular elements ($a_{12}, a_{13}, a_{23}$). The remaining elements are determined by these choices. There are 3 diagonal elements and 3 upper triangular elements, making a total of $3+3=6$ independent choices. Since each choice can be any of the 5 elements in $S$, the number of such matrices is $5^6$.

Step 3: Calculate the number of matrices in S2 (Skew-Symmetric Matrices)

For a $3 \times 3$ skew-symmetric matrix, $a_{ij} = -a_{ji}$. This implies that the diagonal elements must be zero ($a_{ii} = -a_{ii} \implies 2a_{ii} = 0 \implies a_{ii} = 0$). However, $0 \notin S$. Therefore, no $3 \times 3$ skew-symmetric matrix can be formed using elements only from $S$. Thus, $n(S_2) = 0$.

Step 4: Calculate the number of matrices in S3 (Matrices with trace zero)

For matrices in $S_3$, the sum of the diagonal elements must be zero. There are 9 elements in a $3 \times 3$ matrix. The values of $a_{11}$ and $a_{22}$ can be chosen independently in $5 \times 5 = 25$ ways. For each pair of $a_{11}$ and $a_{22}$, $a_{33}$ is uniquely determined as $a_{33} = -(a_{11} + a_{22})$. We need to check if this $a_{33}$ is in $S$. The possible sums $a_{11} + a_{22}$ range from $-3 + (-3) = -6$ to $2 + 2 = 4$. The possible values for $a_{33}$ would then range from $-4$ to $6$. The values in $S$ are $\{-3, -2, -1, 1, 2\}$. We need to count the pairs $(a_{11}, a_{22})$ such that $-(a_{11} + a_{22}) \in S$. There are 13 such pairs. The remaining $9-3=6$ off-diagonal elements can be chosen independently in $5^6$ ways. So, $n(S_3) = 13 \times 5^6$.

Step 5: Calculate the number of matrices in S1 intersection S3

For matrices in $S_1 \cap S_3$, the matrix must be symmetric and its trace must be zero. We have 6 independent elements for a symmetric matrix ($a_{11}, a_{22}, a_{33}, a_{12}, a_{13}, a_{23}$). The condition $a_{11} + a_{22} + a_{33} = 0$ means that $a_{11}$ and $a_{22}$ can be chosen in $5 \times 5 = 25$ ways, and $a_{33}$ is determined. As before, there are 13 pairs $(a_{11}, a_{22})$ such that $a_{33} \in S$. The remaining 3 independent off-diagonal elements ($a_{12}, a_{13}, a_{23}$) can be chosen in $5^3$ ways. So, $n(S_1 \cap S_3) = 13 \times 5^3$.

Step 6: Calculate n(S1 union S2 union S3)

We use the Principle of Inclusion-Exclusion. Since $n(S_2) = 0$, all terms involving $S_2$ become zero. Therefore, $n(S_1 \cup S_2 \cup S_3) = n(S_1) + n(S_3) - n(S_1 \cap S_3)$. $n(S_1 \cup S_2 \cup S_3) = 5^6 + 13 \times 5^6 - 13 \times 5^3 = 14 \times 5^6 - 13 \times 5^3 = 14 \times 15625 - 13 \times 125 = 218750 - 1625 = 217125$.

Step 7: Find the value of alpha

Given that $n(S_1 \cup S_2 \cup S_3) = 125 \alpha$, we can substitute the calculated value and solve for $\alpha$. Dividing 217125 by 125 gives $\alpha = 1737$.

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