If A and B are square matrices of the same order, then the matrix \(AB' - BA'\) is always
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → a skew-symmetric matrix
Let \(P = AB' - BA'\). Using the transpose of a product, \((XY)' = Y'X'\), and \((X')' = X\):
\((AB')' = (B')'A' = BA'\)
\((BA')' = (A')'B' = AB'\)
So
\(P' = BA' - AB'\)
\(P' = -(AB' - BA') = -P\)
Hence \(AB' - BA'\) is skew-symmetric.
Why the other options are incorrect
Option (1) a symmetric matrix — Incorrect. A matrix with \(P' = -P\) is symmetric only when it is O; for \(A = \begin{bmatrix}1&0\\0&0\end{bmatrix}\), \(B = \begin{bmatrix}0&0\\1&0\end{bmatrix}\), \(AB' - BA' = \begin{bmatrix}0&1\\-1&0\end{bmatrix}\).
Option (3) the zero matrix — Incorrect. The same example gives a non-zero matrix.
Option (4) the identity matrix — Incorrect. The diagonal elements of a skew-symmetric matrix are all 0, so it is never I.