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(A) For a $3 	imes 3$ matrix $A$,the number of elements in $S_1$ (symmetric matrices) is determined by the independent entries $a_{11}, a_{22}, a_{33

Vedclass · Accepted Answer

$1613$

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(A) For a $3 	imes 3$ matrix $A$,the number of elements in $S_1$ (symmetric matrices) is determined by the independent entries $a_{11}, a_{22}, a_{33}, a_{12}, a_{13}, a_{23}$. Since each can take $5$ values,$n(S_1) = 5^6 = 15625$. For $S_2$ (skew-symmetric matrices),$a_{ii}=0$ for all $i$. Since $0 
otin S$,$n(S_2) = 0$. For $S_3$,the condition is $a_{11}+a_{22}+a_{33}=0$. The number of ways to choose $(a_{11}, a_{22}, a_{33})$ from $S$ such that their sum is $0$ is: $(1, 2, -3)$ in $3! = 6$ permutations,$(1, -1, 0)$ is not possible,$(2, 1, -3)$ is same as first,$(1, 1, -2)$ in $3$ permutations,$(-1, -1, 2)$ in $3$ permutations. Total ways = $6+3+3 = 12$. The other $6$ entries can be any of the $5$ values,so $n(S_3) = 12 	imes 5^6$. $n(S_1 \cap S_3)$ requires $A=A^T$ and $a_{11}+a_{22}+a_{33}=0$. The diagonal entries must satisfy the sum condition ($12$ ways) and the off-diagonal entries $a_{12}, a_{13}, a_{23}$ can be any of the $5$ values. Thus $n(S_1 \cap S_3) = 12 	imes 5^3$. $n(S_1 \cup S_2 \cup S_3) = n(S_1) + n(S_2) + n(S_3) - n(S_1 \cap S_2) - n(S_2 \cap S_3) - n(S_1 \cap S_3) + n(S_1 \cap S_2 \cap S_3)$. Since $S_2$ is empty,all intersections involving $S_2$ are $0$. $n(S_1 \cup S_2 \cup S_3) = 5^6 + 0 + 12 	imes 5^6 - 0 - 0 - 12 	imes 5^3 + 0 = 13 	imes 5^6 - 12 	imes 5^3 = 5^3(13 	imes 5^3 - 12) = 125(1625 - 12) = 125(1613)$. Therefore,$\alpha = 1613$.

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