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(D) Let $P = A^2 B^2 - B^2 A^2$. Taking the transpose of $P$: $P^T = (A^2 B^2 - B^2 A^2)^T = (A^2 B^2)^T - (B^2 A^2)^T$. Using the property $(XY)^T =

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infinitely many solutions

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(D) Let $P = A^2 B^2 - B^2 A^2$. Taking the transpose of $P$: $P^T = (A^2 B^2 - B^2 A^2)^T = (A^2 B^2)^T - (B^2 A^2)^T$. Using the property $(XY)^T = Y^T X^T$: $P^T = (B^2)^T (A^2)^T - (A^2)^T (B^2)^T$. Since $A$ is symmetric $(A^T = A)$ and $B$ is skew-symmetric $(B^T = -B)$,we have $(A^2)^T = (A^T)^2 = A^2$ and $(B^2)^T = (B^T)^2 = (-B)^2 = B^2$. Substituting these: $P^T = B^2 A^2 - A^2 B^2 = -(A^2 B^2 - B^2 A^2) = -P$. Since $P^T = -P$,$P$ is a skew-symmetric matrix. For any skew-symmetric matrix $P$ of odd order $n$ (here $n=3$),the determinant $\det(P) = 0$. Since $\det(P) = 0$,the system $PX = 0$ has a non-trivial solution,implying it has infinitely many solutions.

Let $A$ and $B$ be $3 \times 3$ real matrices such that $A$ is a symmetric matrix and $B$ is a skew-symmetric matrix. Then the system of linear equations $(A^2 B^2 - B^2 A^2) X = 0$,where $X$ is a $3 \times 1$ column matrix of unknown variables and $0$ is a $3 \times 1$ null matrix,has:

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