Let A and B be square matrices of the same order. Consider the following statements:
(A) \(A - A'\) is a skew-symmetric matrix.
(B) All the diagonal elements of a skew-symmetric matrix are zero.
(C) If A and B are symmetric matrices, then AB is always a symmetric matrix.
(D) If A is a symmetric matrix, then \(A^2\) is a symmetric matrix.
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A), (B) and (D) only
A matrix M is symmetric if \(M' = M\) and skew-symmetric if \(M' = -M\). Using \((A')' = A\) and \((AB)' = B'A'\):
\((A - A')' = A' - A = -(A - A')\)
\((AB)' = B'A' = BA\), when \(A' = A\) and \(B' = B\)
\((A^2)' = A'A' = AA = A^2\), when \(A' = A\)
For a skew-symmetric matrix, \(a_{ii} = -a_{ii}\), so \(a_{ii} = 0\).
So \(A - A'\) is skew-symmetric, every diagonal element of a skew-symmetric matrix is 0, \((AB)' = BA\) for symmetric A and B, and \(A^2\) is symmetric when A is.
Checking the statements
(A) \(A - A'\) is a skew-symmetric matrix. — Correct. Its transpose is \(-(A - A')\).
(B) All the diagonal elements of a skew-symmetric matrix are zero. — Correct. \(a_{ii} = -a_{ii}\) gives \(a_{ii} = 0\).
(C) If A and B are symmetric matrices, then AB is always a symmetric matrix. — Incorrect. \((AB)' = BA\), which equals AB only when AB = BA; for \(A = \begin{bmatrix}1&0\\0&2\end{bmatrix}\), \(B = \begin{bmatrix}0&1\\1&0\end{bmatrix}\), \(AB = \begin{bmatrix}0&1\\2&0\end{bmatrix}\) is not symmetric.
(D) If A is a symmetric matrix, then \(A^2\) is a symmetric matrix. — Correct. \((A^2)' = A'A' = A^2\).