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 $$(A^2B^2 - B^2A^2)X = O$$, where $$X$$ is a $$3 \times 1$$ column matrix of unknown variables and $$O$$ is a $$3 \times 1$$ null matrix, has:

We are given that $$A$$ is a $$3 \times 3$$ symmetric matrix (so $$A^T = A$$) and $$B$$ is a $$3 \times 3$$ skew-symmetric matrix (so $$B^T = -B$$).

Let $$M = A^2B^2 - B^2A^2$$. We compute the transpose of $$M$$: $$M^T = (A^2B^2)^T - (B^2A^2)^T = (B^2)^T(A^2)^T - (A^2)^T(B^2)^T$$.

Since $$A^T = A$$, we have $$(A^2)^T = (A^T)^2 = A^2$$. Since $$B^T = -B$$, we have $$(B^2)^T = (B^T)^2 = (-B)^2 = B^2$$.

Therefore, $$M^T = B^2 \cdot A^2 - A^2 \cdot B^2 = -(A^2B^2 - B^2A^2) = -M$$. This shows that $$M$$ is a skew-symmetric matrix.

For any $$n \times n$$ skew-symmetric matrix with $$n$$ odd, we have $$\det(M) = \det(M^T) = \det(-M) = (-1)^n \det(M)$$. Since $$n = 3$$ is odd, $$\det(M) = -\det(M)$$, which gives $$\det(M) = 0$$.

Since $$\det(M) = 0$$, the matrix $$M$$ is singular. The homogeneous system $$MX = O$$ always has the trivial solution $$X = O$$, and because the determinant is zero, there also exist non-trivial solutions. Hence the system has infinitely many solutions.

Therefore, the system $$(A^2B^2 - B^2A^2)X = O$$ has infinitely many solutions.