Let $X$ and $Y$ be two arbitrary, $3 \times 3$, non-zero, skew-symmetric matrices and $Z$ be an arbitrary $3 \times 3$, non zero, symmetric matrix. Then which of the following matrices is (are) skew symmetric?

Let X and Y be two arbitrary, 3 × 3, non-zero, skew-symmetric matrices and Z be an arbitrary 3 × 3, non zero, symmetric matrix. Then which of the following matrices is (are) skew symmetric? $X = -X^{T}, \qquad Y = -Y^{T} \quad (\text{skew-symmetric})$ $Z^{T} = Z \quad (\text{symmetric})$ $(1)\ Y^{3}Z^{4} - Z^{4}Y^{3} \qquad (2)\ X^{44} + Y^{44}$ $(3)\ X^{4}Z^{3} - Z^{3}X^{4} \qquad (4)\ X^{23} + Y^{23}$ $(AB)^{T} = B^{T}A^{T}, \qquad (A^{n})^{T} = (A^{T})^{n}$ $(X^{n})^{T} = (-1)^{n}X^{n} \Rightarrow \text{odd } n \text{ skew, even } n \text{ symmetric}$ $(Z^{n})^{T} = Z^{n}, \quad \text{skew}+\text{skew}=\text{skew}, \quad \text{sym}+\text{sym}=\text{sym}$ $(AB-BA)^{T} = B^{T}A^{T} - A^{T}B^{T}$ $(1)\ A=Y^{3}\ (\text{skew}),\ B=Z^{4}\ (\text{sym})$ $M_1^{T} = B^{T}A^{T}-A^{T}B^{T} = -BA+AB = M_1$ $\Rightarrow M_1 \text{ is symmetric (not skew)}$ $(2)\ X^{44},\ Y^{44} \text{ symmetric (even powers)}$ $M_2^{T} = X^{44}+Y^{44} = M_2 \Rightarrow \text{symmetric}$ $(3)\ A=X^{4},\ B=Z^{3}\ (\text{both symmetric})$ $M_3^{T} = BA-AB = -(AB-BA) = -M_3$ $\Rightarrow M_3 \text{ is skew-symmetric}$ $(4)\ X^{23},\ Y^{23} \text{ skew (odd powers)}$ $M_4^{T} = -X^{23}-Y^{23} = -M_4 \Rightarrow \text{skew}$ $\text{Skew-symmetric: options } (3) \text{ and } (4)$ $\boxed{\text{Answer: (3), (4)}}$