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(B) Given that $A$ and $B$ are symmetric matrices,we have $A' = A$ and $B' = B$.Consider the transpose of the matrix $(AB - BA)$:$(AB - BA)' = (AB)' -

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Skew-symmetric matrix

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(B) Given that $A$ and $B$ are symmetric matrices,we have $A' = A$ and $B' = B$.Consider the transpose of the matrix $(AB - BA)$:$(AB - BA)' = (AB)' - (BA)'$Using the property $(XY)' = Y'X'$,we get:$(AB - BA)' = B'A' - A'B'$Since $A' = A$ and $B' = B$,we substitute these into the expression:$(AB - BA)' = BA - AB$$(AB - BA)' = -(AB - BA)$Since the transpose of the matrix $(AB - BA)$ is equal to its negative,$(AB - BA)$ is a skew-symmetric matrix.

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

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