Let A and B be 3 times 3 real matrices such t | vedclass

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(C) Given that $A$ is a symmetric matrix,so $A^{T} = A$. Given that $B$ is a skew-symmetric matrix,so $B^{T} = -B$. Let $C = A^{2}B^{2} - B^{2}A^{2}$.

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(C) Given that $A$ is a symmetric matrix,so $A^{T} = A$. Given that $B$ is a skew-symmetric matrix,so $B^{T} = -B$. Let $C = A^{2}B^{2} - B^{2}A^{2}$. Now,consider the transpose of $C$: $C^{T} = (A^{2}B^{2} - B^{2}A^{2})^{T} = (A^{2}B^{2})^{T} - (B^{2}A^{2})^{T}$. Using the property $(PQ)^{T} = Q^{T}P^{T}$,we get: $C^{T} = (B^{2})^{T}(A^{2})^{T} - (A^{2})^{T}(B^{2})^{T}$. Since $(A^{2})^{T} = (A^{T})^{2} = A^{2}$ and $(B^{2})^{T} = (B^{T})^{2} = (-B)^{2} = B^{2}$,we have: $C^{T} = B^{2}A^{2} - A^{2}B^{2} = -(A^{2}B^{2} - B^{2}A^{2}) = -C$. Thus,$C$ is a skew-symmetric matrix. For any skew-symmetric matrix $C$ of odd order $n$ (here $n = 3$),the determinant is always zero,i.e.,$\det(C) = 0$. Since the system is $(C)X = O$ and $\det(C) = 0$,the system has infinitely many solutions.

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