Let A and B be square matrices of order 3, then det \([(A - A^T) + (B - B^T)]\) is equal to
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 0
Let \(C = (A - A^T) + (B - B^T)\). Taking the transpose:
\(C^T = (A^T - A) + (B^T - B) = -C\)
So C is skew-symmetric of order 3. For any square matrix of order n, \(\det(-C) = (-1)^n\det C\), and \(\det C^T = \det C\):
\(\det C = \det C^T = \det(-C)\)
\(\det C = (-1)^3\det C = -\det C\)
\(2\det C = 0\)
\(\det C = 0\)
A skew-symmetric matrix of odd order always has determinant 0.