Matrices & Determinants
Counting Matrix Types
nta_pyq_2025_apr
Grade 12
Question:
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-by-Step Solution
Key Concept: Count elements in $S_1$ (symmetric: 6 independent entries, each from $S$), $S_2$ (skew-symmetric: diagonal must be 0, but 0 $\notin$ $S$, so $S_2 = \emptyset$), and $S_3$ (trace-zero constraint). Apply inclusion-exclusion.
$|S_1| = 5^6$ (6 free entries). $|S_2| = 0$. For $S_3$: diagonal triples $(a_{11},a_{22},a_{33})$ with sum 0 from $S^3$ — found to give $12 \times 5^6$ total matrices. After inclusion-exclusion: $n(S_1 \cup S_2 \cup S_3) = 5^3[13 \times 5^3 - 12] = 125\alpha$, so $\alpha = 1613$.
Correct Answer: 1613