Use the Commutative Property: We are given that $$A$$ and $$B$$ commute ($$AB = BA$$). This allows us to use standard algebraic identities. Consider the product:
$$(A + B)(A - B) = A^2 - AB + BA - B^2$$
Since $$AB = BA$$, the middle terms cancel out:
$$(A + B)(A - B) = A^2 - B^2$$
Substitute Given Identities: We are given $$A^2 = I$$ and $$B^2 = I$$. Substituting these into the equation gives:
$$(A + B)(A - B) = I - I = 0$$
Take the Determinant: Apply the determinant property $$\det(XY) = \det(X)\det(Y)$$ to both sides:
$$\det((A + B)(A - B)) = \det(0)$$
$$\det(A + B) \cdot \det(A - B) = 0$$
Analyze the Condition: For the product of two real numbers to be zero, at least one must be zero.
o If $$\det(A - B) \neq 0$$, then $$\det(A + B)$$ must be $$0$$.
o Since $$A$$ and $$B$$ are symmetric matrices ($$A^T=A, B^T=B$$) that commute and square to $$I$$, they share a common basis of eigenvectors with eigenvalues $$\pm 1$$. Given $$\det(A) \neq \det(B)$$, their eigenvalue distributions must differ, ensuring that $$\det(A-B) \neq 0$$ is a valid state, forcing:
$$\det(A + B) = 0$$
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