Let M denote the set of all real 33 matrices and S=-3,-2,-1,1,2.

Question

Let $M$ denote the set of all real $3\times3$ matrices and $S=\{-3,-2,-1,1,2\}$. Let
$$S_1=\{A=[a_{ij}]\in M: A=A^T,\ a_{ij}\in S\},$$
$$S_2=\{A=[a_{ij}]\in M: A=-A^T,\ a_{ij}\in S\},$$

$$S_3=\{A=[a_{ij}]\in M: a_{11}+a_{22}+a_{33}=0,\ a_{ij}\in S\}.$$
If $n(S_1\cup S_2\cup S_3)=125\alpha$, then $\alpha$ equals \_\_\_\_\_\_\_\_\_\_.

Answer: 1613

Accepted Answer

$S_1$ (symmetric): determined by the 6 independent entries $a_{11},a_{22},a_{33},a_{12},a_{13},a_{23}$, each from $S$ (5 choices), giving $n(S_1)=5^6$.

$S_2$ (skew-symmetric): all diagonal entries must be $0$, but $0\notin S$, so $n(S_2)=0$.

$S_3$: the diagonal entries $(a_{11},a_{22},a_{33})$ must sum to zero using values from $S=\{-3,-2,-1,1,2\}$. The valid ordered triples are permutations of $(1,2,-3)$ --- giving $3!=6$ arrangements --- permutations of $(1,1,-2)$ --- giving $3$ arrangements --- and permutations of $(-1,-1,2)$ --- giving $3$ arrangements --- totaling $12$ diagonal patterns; the off-diagonal 6 entries are free with 5 choices each, giving $n(S_3)=12\times5^6$.

$S_1\cap S_3$: symmetric matrices with trace $0$: same $12$ diagonal patterns, with the 3 independent off-diagonal entries free: $n(S_1\cap S_3)=12\times5^3$. ($S_2\cap S_3=S_2\cap S_1=\varnothing$ effectively since $n(S_2)=0$.)

$$n(S_1\cup S_2\cup S_3)=5^6(1+12)-12\times5^3=5^3\big[13\times5^3-12\big]=125\times1613.$$

So $\alpha=1613$.

Solutions · 1