If $A$ is a square matrix then $A-{{A}^{'}}$ is a
A. diagonal matrix \[\]
B. skew symmetric matrix \[\]
C. Symmetric matrix\[\]
D. None of these \[\]

Hint: We recall transpose of a matrix, diagonal matrix, skew symmetric matrix, symmetric matrix. We take the transpose of the given matrix $A-{{A}^{'}}$ and see if it satisfies the property of diagonal matrix or skew symmetric matrix or symmetric matrix.

Complete answer:
We know that square matrices have the same number of rows and columns. We also know that transpose of a matrix $A$ is denoted as ${{A}^{'}}$ or ${{A}^{T}}$ and obtained by rows of $A$ as columns in ${{A}^{'}}$ or reflecting $A$ over its main diagonal.
We know from property of involution of transpose that
\[{{\left( {{A}^{'}} \right)}^{'}}=A\]
We know from property of addition of transpose that
\[{{A}^{'}}+{{B}^{'}}={{\left( A+B \right)}^{'}}\]
We know that a diagonal matrix is a square matrix with entries except on the main diagonal zero. A skew symmetric matrix is a square matrix whose transpose is equal to its negative which means${{A}^{'}}=-A$. A symmetric matrix is a square matrix whose transpose is equal to itself which means${{A}^{'}}=A$.
We are given that $A$ is a square matrix and are asked to find what type of matrix $A-{{A}^{'}}$ is . Let us take the transpose of the matrix $A-{{A}^{'}}$ and use property of addition of transpose to have;
\[\begin{align}
  & \Rightarrow {{\left( A-{{A}^{'}} \right)}^{'}} \\
 & \Rightarrow {{A}^{'}}-{{\left( {{A}^{'}} \right)}^{'}} \\
\end{align}\]
We use the property of involution of transpose to have;
\[\Rightarrow {{A}^{'}}-A=-\left( A-{{A}^{'}} \right)\]
So we have;
\[{{\left( A-{{A}^{'}} \right)}^{'}}=-\left( {{A}^{'}}-A \right)\]
So the transpose of the matrix $A-{{A}^{'}}$ results in a negative of the matrix $A-{{A}^{'}}$. Hence $A-{{A}^{'}}$ is a skew symmetric matrix.

So the correct choice is B.

Note:
We note that if we can represent the elements of square matrix $A$ with the general entry ${{a}_{ij}}$, the entries of ${{A}^{'}}$ will be ${{a}_{ji}}$. Then the entries of $A-{{A}^{'}}$ will be ${{a}_{ij}}-{{a}_{ji}}$. We see in a diagonal matrix ${{a}_{ij}}-{{a}_{ji}}=0$ except when $i=j$. Since it not necessary that ${{a}_{ij}}={{a}_{ji}}$ for all square matrices then ${{a}_{ij}}-{{a}_{ji}}=0$ then $A-{{A}^{'}}$ is not diagonal matrix. We note that for symmetric matrix ${{a}_{ji}}={{a}_{ji}}$ and for skew symmetric matrix ${{a}_{ji}}=-{{a}_{ij}}$.