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# If $A, B$ are symmetric matrices of the same order, then $AB - BA$ is

$$\begin{array}{l l} (A) \quad \text{Skew symmetric matrix} & (B) \quad \text{Symmetric Matrix} \\ (C) \quad \text{Zero Matrix} & (D) \quad \text{Identity Matrix} \end{array}$$

Toolbox:
• If A_{i,j} be a matrix m*n matrix , then the matrix obtained by interchanging the rows and column of A is called as transpose of A.
• 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.
• 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.
Given:
A and B are symmetric matrices
A=A'$\rightarrow$ transpose of A
B=B'$\rightarrow$ transpose of B
Now consider AB-BA
Taking transpose of it we get
(AB-BA)'=(AB)'-(BA)'
(AB)'-(BA)'= B'A'-A'B'
From the property of transpose of a matrix we have (AB)'=B'A'.
(AB-BA)'=B'A'-A'B'
Replacing A'=A and B'=B(Since A and B are symmetric matrix)
=BA-AB=-(AB-BA)( By taking out -1)
AB-BA is a skew symmetric matrix.
so (A) is the right option.

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