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:

Answer 1

Correct Answer: 1613

\( \begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix} \)

No. of elements in \( S_1 \): \( A = A^T \Rightarrow 5^3 \times 5^3 \)

No. of elements in \( S_2 \): \( A = -A^T \Rightarrow 0 \) since no zero in \( S_2 \)

No. of elements in \( S_3 \):

\( a_{11} + a_{22} + a_{33} = 0 \Rightarrow (1, 2, -3) \Rightarrow 31 \)

or \( (1,1, -2) \Rightarrow 3 \)

or \( (-1,-1,2) \Rightarrow 3 \)

So, it simplifies to \( 12 \times 5^6 \)

\( n(S_1 \cap S_3) = 12 \times 5^3 \)

\( n(S_1 \cup S_2 \cup S_3) = 5^6(1 + 12 - 12) \Rightarrow 5^3 \times [13 \times 5^3 - 12] = 125\alpha \)

Thus, \( \alpha = 1613 \)