Question

Let \( M \) denote the set of all real matrices of order 3 x 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 \(n(S_1 \cup S_2 \cup S_3) = 125\), then \( \alpha \) equals:

Hint:

For problems involving sets of matrices, carefully consider symmetry properties and constraints like the trace condition to determine the set size.

Answer 1

Correct Answer: 125

Step 1: Analyze each set.

\( S_1 \) comprises symmetric matrices. The elements above the main diagonal dictate the matrix's structure. Given 5 options for each of the 6 such positions, the cardinality is: \[ |S_1| = 5^6 \] \( S_2 \) consists of skew-symmetric matrices. Non-diagonal elements are independently chosen, while diagonal elements must be 0. As \( S \) does not contain diagonal elements, \( S_2 \) is rendered invalid. Therefore: \[ |S_2| = 0 \] \( S_3 \) requires the trace to sum to zero. Free selection of two elements determines the third element: \[ |S_3| = 5^2 \times (\text{number of valid third elements}) \]

Step 2: Calculate the union of sets.

Apply the inclusion-exclusion principle to determine \( n(S_1 \cup S_2 \cup S_3) \): \[ n(S_1 \cup S_2 \cup S_3) = |S_1| + |S_2| + |S_3| - (\text{intersections}) = 125 \]