If $\mathrm{A}=\left[\begin{array}{ll}2 & 3 \\ 4 & 5\end{array}\right]$, and show that $\left(\mathrm{A}-\mathrm{A}^{\prime}\right)$ is skew-symmetric

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If $\mathrm{A}=\left[\begin{array}{ll}2 & 3 \\ 4 & 5\end{array}\right]$ , and show that $\left(\mathrm{A}-\mathrm{A}^{\prime}\right)$ is skew-symmetric

If $\mathrm{A}=\left[\begin{array}{ll}2 & 3 \\ 4 & 5\end{array}\right]$ , and show that $\left(\mathrm{A}-\mathrm{A}^{\prime}\right)$ is skew-symmetric

Solution:

We have $\left(\begin{array}{ll}2 & 3 \\ 4 & 5\end{array}\right)$ .
The transpose of the matrix is an operation of making interchange of elements by the rule on positioned element $a_{j i}$ shifted to new position $a_{j i}$ .
The symmetric matrix is defined as similarity of transpose of matrix with it self. i.e, $A^{T}=A$ .

The skew-symmetric matrix is defined as similarity of transpose of matrix with it self. i.e, $A^{T}=-A$ .
$A^{T}=\left(\begin{array}{ll} 2 & 3 \\ 4 & 5 \end{array}\right)^{T}=\left(\begin{array}{ll} 2 & 4 \\ 3 & 5 \end{array}\right)$
Therefore finally we have
$A-A^{T}=\left(\begin{array}{ll} 2 & 3 \\ 4 & 5 \end{array}\right)-\left(\begin{array}{cc} 2 & 4 \\ 3 & 5 \end{array}\right)=\left(\begin{array}{cc} 0 & -1 \\ 1 & 0 \end{array}\right)$
Here $\left(A-A^{T}\right)^{T}=\left(\begin{array}{cc}0 & -1 \\ 1 & 0\end{array}\right)^{T}=\left(\begin{array}{cc}0 & 1 \\ -1 & 0\end{array}\right)=-\left(\begin{array}{cc}0 & -1 \\ 1 & 0\end{array}\right)=A-$ $A^{T} .$
Therefore $A+A^{T}$ is skew symmetric.