Matrices & Determinants
Matrices
nta_pyq_2025_jan
Grade 12
Question:
Let $M$ denote the set of all real matrices of order $3\times 3$ and let $S=\{-3,-2,-1,1,2\}$. $S_{1}=\{A\in M\,:\,A=A^{T},\ a_{ij}\in S\}$, $S_{2}=\{A\in M\,:\,A=-A^{T},\ a_{ij}\in S\}$, $S_{3}=\{A=[a_{ij}]\in M\,:\,a_{11}+a_{22}+a_{33}=0,\ a_{ij}\in S\}$. If $n(S_{1}\cup S_{2}\cup S_{3})=125\alpha$, then $\alpha$ equals \rule{2cm}{0.4pt}.
Step-by-Step Solution
Key Concept: $|S_{1}|$: $6$ independent entries (3 diagonal + 3 above), $|S|^{6}.$ $|S_{2}|=0$ because skew-symmetric matrices have zero diagonal but $0\notin S.$ $|S_{3}|$: pick diagonal triples summing to $0$ (combinatorial), off-diagonals free.
$|S_{1}|=|S|^{6}=5^{6}=15625.$
$|S_{2}|=0$ (skew-symmetric requires $a_{ii}=0\notin S$).
For $|S_{3}|$: count ordered diagonals $(a_{11},a_{22},a_{33})$ with $a_{ii}\in S$ and sum $0.$
Unordered triples: $\{-3,1,2\}\to 6$ permutations; $\{-2,1,1\}\to 3$; $\{-1,-1,2\}\to 3.$ Total $=12$ ordered triples.
Off-diagonals free: $|S|^{6}=15625.$ So $|S_{3}|=12\cdot 5^{6}=187500.$
$|S_{1}\cap S_{3}|$: symmetric with trace $0$. Diagonal triples: $12$; off-diagonals (only the upper triangle, $3$ entries): $|S|^{3}=125.$ So $|S_{1}\cap S_{3}|=12\cdot 125=1500.$
$|S_{2}\cap\bullet|=0$ as before. By inclusion-exclusion:
$|S_{1}\cup S_{2}\cup S_{3}|=15625+0+187500-0-1500-0+0=201625=125\cdot 1613.$
Hence $\alpha=1613.$
Correct Answer: 1613