Using elements $- 3, - 2, - 1, 0, 1, 2, 3$ as entries

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The number of $3 \times 3$ skew symmetric matrices is $7^{3}$

The number of $3 \times 3$ matrices is $7^{9}$

The number of symmetric matrices is $7^{6}$

The number of $3 \times 3$ matrices having trace 0 is $37 (7^{6})$

answer is A, B, C, D.

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Detailed Solution

$a_{11} + a_{22} + a_{33} = 0$ remaining 6 elements can be filled in $7^{6}$ ways
$\begin{array}{l} \begin{array}{l} (- 3, 0, 3), (- 2, 0, 2), (- 1, 0, 1) \\ (- 3, 1, 2), (3, - 1, - 2) \end{array}}, 3! (5) = 30 \\ (- 2, 11), (2, - 1, 1) \to 3.2 = 6 \end{array}$
$(0, 0, 0) \to 1$
Total no of matrices with trace 0 = 37
B. Each of 9 elements can be filled in 7ways so $7^{9}$
C. $[\begin{array}{l} 0 0 0 \\ 0 0 \\ 0 \end{array}]$ all diagonals elements are zero and 3 places above diagonal can be filled in $7^{3}$ ways
D. Diagonal places and 3 places either above or below. The diagonal places can be filled in $7^{6} .$