Let A be a symmetric matrix and B be a skew-symmetric matrix of the same order. Which of the following is always a skew-symmetric matrix?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(ABA\)
Here \(A' = A\) and \(B' = -B\). The transpose of a product is the product of the transposes in the reverse order, \((PQR)' = R'Q'P'\):
\((ABA)' = A'B'A'\)
\((ABA)' = A(-B)A = -ABA\)
So ABA is always skew-symmetric.
Why the other options are incorrect
Option (2) \(BAB\) — Incorrect. \((BAB)' = B'A'B' = (-B)A(-B) = BAB\), so it is symmetric.
Option (3) \(A^2 + B^2\) — Incorrect. \((A^2)' = A^2\) and \((B^2)' = (-B)(-B) = B^2\), so it is symmetric.
Option (4) \(AB\) — Incorrect. \((AB)' = B'A' = -BA\), which equals \(-AB\) only when \(AB = BA\). For \(A = \begin{bmatrix}1&0\\0&0\end{bmatrix}\) and \(B = \begin{bmatrix}0&1\\-1&0\end{bmatrix}\), \(AB = \begin{bmatrix}0&1\\0&0\end{bmatrix}\), which is not skew-symmetric.