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Step-by-Step Solution
Step 1: Identify the matrices and the system
We are given two 3×3 real matrices A and B with the properties:
A is symmetric, i.e., $A^T = A$.
B is skew-symmetric, i.e., $B^T = -B$.
We need to analyze the system of linear equations
$$(A^2 B^2 - B^2 A^2)\, X = \mathbf{0},$$
where $X$ is a 3×1 column matrix of unknowns, and $\mathbf{0}$ is the 3×1 null (zero) matrix.
Step 2: Define the matrix P and show it is skew-symmetric
Let
$$P = A^2 B^2 - B^2 A^2.$$
We want to determine the nature of $P$. Compute its transpose:
\[
P^T
= (A^2 B^2 - B^2 A^2)^T
= (A^2 B^2)^T - (B^2 A^2)^T.
\]
Since $(MN)^T = N^T M^T$, we use the fact that $A^T = A$ and $B^T = -B$ to get:
\[
(A^2 B^2)^T = (B^2)^T (A^2)^T = (B^T)^2 (A^T)^2 = (-B)^2 A^2 = B^2 A^2,
\]
and similarly,
\[
(B^2 A^2)^T = (A^2)^T (B^2)^T = A^2 B^2.
\]
Thus,
\[
P^T = B^2 A^2 - A^2 B^2 = -(A^2 B^2 - B^2 A^2) = -P.
\]
Hence, $P$ is skew-symmetric.
Step 3: General property of a skew-symmetric matrix in 3×3
A real skew-symmetric matrix of order 3 has the form
\[
\begin{bmatrix}
0 & a & b \\
-\,a & 0 & c \\
-\,b & -\,c & 0
\end{bmatrix}.
\]
Such matrices have all diagonal entries equal to zero. When considering the homogeneous system $P\,X = \mathbf{0}$, the determinant of a skew-symmetric matrix of odd order is always zero. This implies that $P$ is not invertible, and hence the system
\[
P\,X = \mathbf{0}
\]
must have nontrivial solutions.
Step 4: Conclude the number of solutions
Because $P$ is skew-symmetric (and in particular its determinant is zero), the system does not have a unique solution. Instead, there is at least one free variable leading to an infinite family of solutions. In other words, the solution space has dimension greater than zero, indicating
infinitely many solutions.
Answer: Infinitely many solutions.
