JEE Main · 2025 · 28 Jan 2025 · Shift 1 · Matrices

Each matrix has entries drawn from $S=\{-3,-2,-1,1,2\}$ (5 values), and $0\notin S$.

**$S_1$ (symmetric):** A symmetric $3\times 3$ matrix is fixed by its 3 diagonal and 3 upper off-diagonal entries — 6 free entries, each with 5 choices: $n(S_1)=5^6$.

**$S_2$ (skew-symmetric):** $A=-A^T$ forces every diagonal entry to satisfy $a_{ii}=-a_{ii}$, i.e. $a_{ii}=0$. Since $0\notin S$, no such matrix exists: $n(S_2)=0$.

**$S_3$ (trace zero):** Need ordered triples $(a_{11},a_{22},a_{33})$ from $S$ with sum 0. Checking all combinations, the valid unordered sets are $\{-3,1,2\}$ (all distinct, giving $3!=6$ orderings), $\{-2,1,1\}$ (giving $3$ orderings) and $\{-1,-1,2\}$ (giving $3$ orderings) — 12 ordered diagonal triples in all. The remaining 6 off-diagonal entries are free: $n(S_3)=12\times 5^6$.

Union: Since $S_2=\varnothing$,

$n(S_1\cup S_2\cup S_3)=n(S_1)+n(S_3)-n(S_1\cap S_3)$

For $S_1\cap S_3$ (symmetric with trace 0), the diagonal is again restricted to the 12 valid triples, while the 3 upper off-diagonal entries remain free: $n(S_1\cap S_3)=12\times 5^3$.

$n(S_1\cup S_2\cup S_3)=5^6+12\times 5^6-12\times 5^3=5^3\left(13\times 5^3-12\right)=125(1625-12)=125\times 1613$

So $\alpha =1613$