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

$\begin{array}{l}{\text{ Given X}}^T=-X, Y^T=-Y, Z^T=Z\\ {\left(Y^3{Z}^4-Z^4{Y}^3\right)}^T={\left(Y^3{Z}^4\right)}^T-{\left(Z^4{Y}^3\right)}^T\\ ={\left(Z^T\right)}^4{\left(Y^T\right)}^3-{\left(Y^T\right)}^3{\left(Z^T\right)}^4\\ =Z^4(-Y{)}^3-(-Y{)}^3(Z{)}^4\\ =-Z^4{Y}^3+Y^3{Z}^4=-\left(Z^4{Y}^3-Y^3{Z}^4\right)\\ \Rightarrow \mathrm{Not}\mathrm{skew}-\mathrm{symmetric}\end{array}$

$\begin{array}{l}{\left(X^{44}+Y^{44}\right)}^T={\left(Y^T\right)}^4+{\left(Y^T\right)}^4=(-X{)}^{44}+(-Y{)}^{44}=X^{44}+Y^{44}\\ \Rightarrow \mathrm{Symmetric}\end{array}$

$\begin{array}{l}{\left(X^4{Z}^3-Z^3{X}^4\right)}^T={\left(Z^T\right)}^3{\left(X^T\right)}^4-{\left(X^T\right)}^4{\left(Z^T\right)}^3\\ =\left(Z^3\right)(-X{)}^4-(-X{)}^4(Z{)}^3\\ =Z^3{X}^4-X^4{Z}^3\\ =-\left(X^4{Z}^3-Z^3{X}^4\right)\\ \Rightarrow \mathrm{Skew}symmetric\end{array}$

$\begin{array}{l}{\left(X^{23}+Y^{23}\right)}^T={\left(X^T\right)}^{23}+{\left(Y^T\right)}^{23}=(-X{)}^{23}+(-Y{)}^{23}\\ =-X^{23}-Y^{23}=-\left(X^{23}+Y^{23}\right)\\ \Rightarrow \mathrm{Skew}-\mathrm{symmetric}\end{array}$