If A is a symmetric matrix and B is a skew-symmetric matrix of the same order, then the matrix \(AB - BA\) is always
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → a symmetric matrix
We are given
\(A' = A\)
\(B' = -B\)
Using the transpose of a product, \((PQ)' = Q'P'\):
\((AB - BA)' = B'A' - A'B'\)
\(= (-B)A - A(-B)\)
\(= AB - BA\)
So \(AB - BA\) equals its own transpose: it is a symmetric matrix.
For example, with \(A = \begin{bmatrix}1&0\\0&2\end{bmatrix}\) and \(B = \begin{bmatrix}0&1\\-1&0\end{bmatrix}\), \(AB - BA = \begin{bmatrix}0&-1\\-1&0\end{bmatrix}\), which is symmetric but neither skew-symmetric, nor zero, nor the identity.