Let A and B be two symmetric matrices of the | vedclass

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

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a skew-symmetric matrix

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(B) Given that $A$ and $B$ are symmetric matrices,we have $A^{\prime} = A$ and $B^{\prime} = B$. Consider the transpose of the matrix $(AB - BA)$: $(AB - BA)^{\prime} = (AB)^{\prime} - (BA)^{\prime}$ Using the property $(XY)^{\prime} = Y^{\prime}X^{\prime}$,we get: $(AB - BA)^{\prime} = B^{\prime}A^{\prime} - A^{\prime}B^{\prime}$ Since $A^{\prime} = A$ and $B^{\prime} = B$,we substitute these values: $(AB - BA)^{\prime} = BA - AB$ $(AB - BA)^{\prime} = -(AB - BA)$ Since the transpose of the matrix $(AB - BA)$ is equal to the negative of the matrix itself,$AB - BA$ is a skew-symmetric matrix.

Let $A$ and $B$ be two symmetric matrices of the same order. Then,the matrix $AB - BA$ is

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