If A and B are invertible matrices of the same order, then \((AB)^{-1}\) is equal to
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(B^{-1}A^{-1}\)
By the reversal law for inverses, \((AB)^{-1} = B^{-1}A^{-1}\). To see why, multiply AB by \(B^{-1}A^{-1}\):
\((AB)(B^{-1}A^{-1}) = A(BB^{-1})A^{-1}\)
\(= AIA^{-1} = AA^{-1} = I\)
In the same way \((B^{-1}A^{-1})(AB) = I\), so \(B^{-1}A^{-1}\) is the inverse of AB.
Why the other options are incorrect
Option (1) \(A^{-1} B^{-1}\) — Incorrect. The order is not reversed; \((AB)(A^{-1}B^{-1})\) does not simplify to I, since matrix multiplication is not commutative.
Option (2) \(A^{-1}B\) — Incorrect. B is not inverted at all.
Option (4) \(AB^{-1}\) — Incorrect. A is not inverted at all.
NCERT (2026-27 edition):
Invertible Matrices
