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If A and B are invertible matrices of the same order, then (AB)−1(AB)^{-1} is equal to

Solution

✅ Correct Option: 3

For invertible matrices AA and BB of the same order, (AB)−1=B−1A−1(AB)^{-1} = B^{-1}A^{-1}

The inverse operation reverses the order of the product.


To verify this is the correct inverse, multiply (AB)(AB) with (B−1A−1)(B^{-1}A^{-1}):

(AB)×(B−1A−1)(AB) \times (B^{-1}A^{-1})

=A(BB−1)A−1= A(BB^{-1})A^{-1}

=A(I)A−1= A(I)A^{-1}

=AA−1= AA^{-1}

=I= I

Since the product equals the identity matrix II, (AB)−1=B−1A−1(AB)^{-1} = B^{-1}A^{-1}


This result extends to products of multiple matrices:

(ABC)−1=C−1B−1A−1(ABC)^{-1} = C^{-1}B^{-1}A^{-1}

The order is always reversed when taking the inverse of a product.


Therefore, (AB)−1=B−1A−1(AB)^{-1} = B^{-1}A^{-1}

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