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# If A, B are symmetric matrices of the same order then AB-BA is a : $\begin{array} ((A)\: skew\: symmetric \: matrix \\ (B) \: symmetric\: matrix \\ (C)\: Zero \: matrix \\ (D)\: Identity \: matrix \end{array}$

Toolbox:
• A square matrix A=[a$_{ij}$] is said to be symmetric if A'=A that is $[a_{ij}]=[a_{ji}]$ for all possible value of i and j.
• A square matrix A=[a$_{ij}$] is said to be skew symmetric if A'=-A that is $[a_{ij}]= -[a_{ji}]$ for all possible value of i and j.
Step1:
Given:
A & B $\rightarrow$ Symmetric matrices.
$\Rightarrow$ A'=A
$\quad B'=B$
Step2:
$(AB-BA)'=(AB)'-(BA)'$
From the property of transpose of a matrix we have
(AB)' = B'A'
$\qquad\qquad\;\;=B'A'-A'B'$
$\qquad\qquad\;\;=BA-AB$ $\quad[B'=B,A'=A]$
$\qquad\qquad\;\;=-(AB-BA)$
=>AB-BA is a skew symmetric matrix.
part A is correct