JEE Main 2025 Jan 28 Shift 1, Mathematics Q21: Algebra Of Matrices

Note that $0\notin S$, so any skew-symmetric matrix (which forces diagonal entries to be 0) cannot lie in $S_2$. Hence $S_2=\varnothing$, and $n(S_1\cup S_2\cup S_3)=n(S_1\cup S_3)=n(S_1)+n(S_3)-n(S_1\cap S_3)$.

Count $n(S_1)$ (symmetric matrices over $S$). A $3\times 3$ symmetric matrix is determined by 3 diagonal entries and 3 above-diagonal entries (6 independent slots), each chosen from $S$ (5 values): $n(S_1)=5^6$.

Count $n(S_3)$ (trace zero). The diagonal triples $(a_{11},a_{22},a_{33})$ from $S^3$ summing to zero, with $S=\{-3,-2,-1,1,2\}$. Unordered triples are $\{1,2,-3\}$ (giving $3!=6$ ordered triples), $\{1,1,-2\}$ (giving 3 ordered triples), and $\{-1,-1,2\}$ (giving 3 ordered triples). Total: $6+3+3=12$ choices of diagonal. The remaining 6 off-diagonal slots are free over $S$, contributing $5^6$ each. So $n(S_3)=12\cdot 5^6$.

Count $n(S_1\cap S_3)$. Symmetric with trace zero: 12 diagonal choices, 3 above-diagonal slots each free over $S$, giving $12\cdot 5^3$.

$n(S_1\cup S_3)=5^6+12\cdot 5^6-12\cdot 5^3=13\cdot 5^6-12\cdot 5^3=5^3(13\cdot 5^3-12)=125(1625-12)=125\cdot 1613$.

$\therefore \alpha =1613$.