Let \( A \) and \( B \) be two symmetric matrices of same order, then
Match List-I with List-II
| List-I | List-II |
| (A). \( AB-BA \) | (I). Neither symmetric nor Skew symmetric |
| (B). \( BA-2AB \) | (II). Skew symmetric |
| (C). \( AB+BA \) | (III). Null matrix |
| (D). \( (AB)^T-BA \) | (IV). Symmetric |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A) - (II), (B) - (I), (C) - (IV), (D) - (III)
Since \( A^T=A \) and \( B^T=B \), the reversal law gives \( (AB)^T=B^TA^T=BA \) and \( (BA)^T=AB \).
(A) \( AB-BA \)
This matches (II) Skew symmetric.
\( \displaystyle (AB-BA)^T=BA-AB=-(AB-BA) \)
(B) \( BA-2AB \)
This matches (I) Neither symmetric nor Skew symmetric.
\( \displaystyle (BA-2AB)^T=AB-2BA \)
This is neither BA − 2AB nor its negative in general.
(C) \( AB+BA \)
This matches (IV) Symmetric.
\( \displaystyle (AB+BA)^T=BA+AB=AB+BA \)
(D) \( (AB)^T-BA \)
This matches (III) Null matrix.
\( \displaystyle (AB)^T-BA=BA-BA=O \)
So the matching is (A) - (II), (B) - (I), (C) - (IV), (D) - (III).