Let A and B be real symmetric matrices of size n x n. Then which one of the following is true? A. A*A = A^(-1) B. A*B = B*A C. A*B - B*A is skew symmetric D. (A*B)^T = -B*A

GATE 1994 · Engineering Mathematics · Matrix · medium

Answer: D. (A*B)^T = -B*A is the stated answer per the official key. Note: (A*B)^T = B*A for symmetric matrices; option C (AB-BA is skew-symmetric) is actually true.

1. Evaluate option A: A*A = A^(-1): A*A = A^2 which equals A^(-1) only for very special matrices (like involutions where A^3 = I). Not generally true for all symmetric matrices.
2. Evaluate option B: A*B = B*A: Symmetric matrices do NOT generally commute. Example: A=[[1,1],[1,0]], B=[[0,1],[1,1]]. AB != BA in general.
3. Evaluate option C: A*B - B*A is skew-symmetric: (A*B - B*A)^T = B^T*A^T - A^T*B^T = B*A - A*B = -(A*B - B*A). So A*B - B*A IS skew-symmetric!
4. Evaluate option D: (A*B)^T = -B*A: (A*B)^T = B^T * A^T = B * A (since A and B are symmetric). This equals B*A, NOT -B*A. So option D states (A*B)^T = -B*A which is false in general.