GATE CS 1994 Q1.2 — Linear Algebra: Matrices, Determinants, System of Linear Equations

Since $A$ and $B$ are real symmetric matrices of size $n \times n$, by definition, they are equal to their own transposes:

Let us evaluate each option to find the correct statement:

• Option A ($AA' = I$): Since $A' = A$, this requires $A^2 = I$, which is only true for orthogonal symmetric matrices, not all symmetric matrices. For example, if $A = 2I$, then $AA' = 4I \neq I$.
• Option B ($A = A^{-1}$): This also requires $A^2 = I$, which is not true for all symmetric matrices.
• Option C ($AB = BA$): Two symmetric matrices commute ($AB = BA$) if and only if they share a common basis of eigenvectors. In general, matrix multiplication is not commutative. For example, let: $$A = \begin{pmatrix} 1 & 0 \\ 0 & 2 \end{pmatrix}, \quad B = \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}$$ Both $A$ and $B$ are symmetric, but: $$AB = \begin{pmatrix} 1 & 1 \\ 2 & 2 \end{pmatrix} \neq BA = \begin{pmatrix} 1 & 2 \\ 1 & 2 \end{pmatrix}$$
• Option D ($(AB)' = BA$): Using the reversal property of matrix transposition, we have: $$(AB)' = B'A'$$ Substituting $B' = B$ and $A' = A$ (since they are symmetric): $$(AB)' = BA$$ This identity holds universally for any two symmetric matrices $A$ and $B$.

Thus, Option D is the only statement that is always true.