Solve this

Given : $A=\left[\begin{array}{ll}2 & 3 \\ 4 & 5\end{array}\right]$

$A^{T}=\left[\begin{array}{ll}2 & 4 \\ 3 & 5\end{array}\right]$

Now,

$\left(A-A^{T}\right)=\left[\begin{array}{ll}2 & 3 \\ 4 & 5\end{array}\right]-\left[\begin{array}{ll}2 & 4 \\ 3 & 5\end{array}\right]$

$\Rightarrow\left(A-A^{T}\right)=\left|\begin{array}{ll}2-2 & 3-4 \\ 4-3 & 5-5\end{array}\right|$

$\Rightarrow\left(A-A^{T}\right)=\left[\begin{array}{cc}0 & -1 \\ 1 & 0\end{array}\right]$          ....(1)

$\left(A-A^{T}\right)^{T}=\left[\begin{array}{cc}0 & -1 \\ 1 & 0\end{array}\right]^{1}$

$\Rightarrow\left(A-A^{T}\right)^{T}=\left[\begin{array}{cc}0 & 1 \\ -1 & 0\end{array}\right]$

$\Rightarrow\left(A-A^{T}\right)^{T}=-\left[\begin{array}{cc}0 & -1 \\ 1 & 0\end{array}\right]$

$\Rightarrow\left(A-A^{T}\right)=-\left(A-A^{T}\right)^{T} \quad[$ Using eq. (1) $]$

Thus, $\left(A-A^{T}\right)$ is a skew-symmetric matrix.