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

Step 1: Analyze each set.

\( S_1 \) includes symmetric matrices, so elements above the diagonal determine the matrix. With 5 choices for each, and 6 such positions:

\[ |S_1| = 5^6 \]

\( S_2 \) includes skew-symmetric matrices, non-diagonal elements are independent, and diagonal elements must be 0 (not in \( S \)), invalidating \( S_2 \).

\[ |S_2| = 0 \]

\( S_3 \) must balance the trace to be zero. Choosing two elements freely allows the third to be determined:

\[ |S_3| = 5^2 \times (\text{number of valid third elements}) \]

Step 2: Calculate the union of sets.

Using the inclusion-exclusion principle, find \(n(S_1 \cup S_2 \cup S_3)\):

\[ n(S_1 \cup S_2 \cup S_3) = |S_1| + |S_2| + |S_3| - (\text{intersections}) = 125 \]