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(B) Since $A$ and $B$ are symmetric matrices, we have $A^{\prime} = A$ and $B^{\prime} = B$. Let $P = AB - BA$. Taking the transpose of $P$: $P^{\prim

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a skew-symmetric matrix

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(B) Since $A$ and $B$ are symmetric matrices, we have $A^{\prime} = A$ and $B^{\prime} = B$. Let $P = AB - BA$. Taking the transpose of $P$: $P^{\prime} = (AB - BA)^{\prime} = (AB)^{\prime} - (BA)^{\prime}$ Using the property $(XY)^{\prime} = Y^{\prime}X^{\prime}$, we get: $P^{\prime} = B^{\prime}A^{\prime} - A^{\prime}B^{\prime}$ Substituting $A^{\prime} = A$ and $B^{\prime} = B$: $P^{\prime} = BA - AB = -(AB - BA) = -P$. Since $P^{\prime} = -P$, the matrix $AB - BA$ is a skew-symmetric matrix.

Suppose $A$ and $B$ are two square matrices of the same order. If $A$ and $B$ are symmetric matrices, then $AB - BA$ is

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