Let A and B be 3 × 3 matrices such that \(A \ne B\). If \(A^3 = B^3\) and \(A^2B = B^2A\), then the determinant of \(A^2 + B^2\) is:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 0
Multiplying \(A^2 + B^2\) on the right by \(A - B\):
\((A^2 + B^2)(A - B) = A^3 - A^2B + B^2A - B^3\)
Since \(A^3 = B^3\) and \(A^2B = B^2A\), the right side is the zero matrix:
\((A^2 + B^2)(A - B) = O\)
Suppose \(|A^2 + B^2| \ne 0\). Then \(A^2 + B^2\) is invertible, and multiplying on the left by its inverse gives
\(A - B = O\), that is, \(A = B\)
This contradicts \(A \ne B\). So
\(|A^2 + B^2| = 0\)