If A and B are matrices of same order, then (AB’ – BA’) is a

The correct option is skew symmetric matrix.

To determine the nature of the matrix $C=A{B}^{\prime }-B{A}^{\prime }$, we need to find its transpose, denoted by $C^{\prime }$ (or $C^T$).

Recall the fundamental properties of matrix transposes:
1. The transpose of a difference of two matrices is the difference of their transposes:
${(X-Y)}^{\prime }=X^{\prime }-Y^{\prime }$
2. The transpose of a product of two matrices reverses the order of multiplication:
${\left(XY\right)}^{\prime }=Y^{\prime }{X}^{\prime }$
3. The transpose of the transpose of a matrix returns the original matrix:
${\left(X^{\prime }\right)}^{\prime }=X$

Now, let us apply these properties to our matrix:
${(A{B}^{\prime }-B{A}^{\prime })}^{\prime }={\left(A{B}^{\prime }\right)}^{\prime }-{\left(B{A}^{\prime }\right)}^{\prime }$

Applying the product rule for transposes to each term, we get:
${\left(A{B}^{\prime }\right)}^{\prime }={\left(B^{\prime }\right)}^{\prime }{A}^{\prime }=B{A}^{\prime }$
and
${\left(B{A}^{\prime }\right)}^{\prime }={\left(A^{\prime }\right)}^{\prime }{B}^{\prime }=A{B}^{\prime }$

Substituting these back into our expression, we obtain:
${(A{B}^{\prime }-B{A}^{\prime })}^{\prime }=B{A}^{\prime }-A{B}^{\prime }$

Factoring out a negative sign ($-1$) from the right-hand side:
${(A{B}^{\prime }-B{A}^{\prime })}^{\prime }=-(A{B}^{\prime }-B{A}^{\prime })$

Since the transpose of the matrix is equal to its negative, i.e., $C^{\prime }=-C$, the matrix $(A{B}^{\prime }-B{A}^{\prime })$ is by definition a skew symmetric matrix.