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(D) Let $S_A = A - A^T$. Since $S_A^T = (A - A^T)^T = A^T - A = -(A - A^T) = -S_A$,$S_A$ is a skew-symmetric matrix of order $3$.Similarly,let $S_B =

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(D) Let $S_A = A - A^T$. Since $S_A^T = (A - A^T)^T = A^T - A = -(A - A^T) = -S_A$,$S_A$ is a skew-symmetric matrix of order $3$.Similarly,let $S_B = B - B^T$. Since $S_B^T = (B - B^T)^T = B^T - B = -(B - B^T) = -S_B$,$S_B$ is a skew-symmetric matrix of order $3$.Let $M = S_A + S_B$. Then $M^T = (S_A + S_B)^T = S_A^T + S_B^T = -S_A - S_B = -(S_A + S_B) = -M$.Thus,$M$ is a skew-symmetric matrix of order $3$.The determinant of a skew-symmetric matrix of odd order $n$ is always $0$,because $|M| = |M^T| = |-M| = (-1)^n |M| = -|M|$ for odd $n$,which implies $2|M| = 0$,so $|M| = 0$.Since $n = 3$ is odd,$|M| = |(A-A^T)+(B-B^T)| = 0$.

If $A$ and $B$ are square matrices of order $3$,then $|(A-A^T)+(B-B^T)|=$

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