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(D) For a skew-symmetric matrix $A$, we have $A = -A^T$. This implies that the diagonal elements must be zero, so $a = e = i = 0$. Thus, the matrix is

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$\frac{-dh}{fg}$

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(D) For a skew-symmetric matrix $A$, we have $A = -A^T$. This implies that the diagonal elements must be zero, so $a = e = i = 0$. Thus, the matrix is $A = \begin{bmatrix} 0 & b & c \ d & 0 & f \ g & h & 0 \end{bmatrix}$. Since $A$ is a skew-symmetric matrix, we have $d = -b$, $g = -c$, and $h = -f$. Substituting these into the expression $\frac{b}{c}$: We know that for any skew-symmetric matrix of odd order, the determinant is $0$. $|A| = 0 \cdot (0 - fh) - b(0 - gf) + c(dh - 0) = 0$. $bgf + cdh = 0$. $bgf = -cdh$. Dividing by $cf$, we get $\frac{b}{c} = \frac{-dh}{fg}$.

If $\begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end{bmatrix}$ is a skew-symmetric matrix and $b, c, f$ are non-zero real numbers, then $\frac{b}{c} = $

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