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NCERT Exemplar · Q96

Q.(AB)−1=A−1⋅B−1(AB)^{-1} = A^{-1} \cdot B^{-1}, where AA and BB are invertible matrices satisfying commutative property with respect to multiplication.

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Under the given condition that AA and BB commute (AB=BAAB = BA), the statement (AB)−1=A−1B−1(AB)^{-1} = A^{-1}B^{-1} is true.

The general rule first

For any two invertible matrices, the inverse of a product reverses the order:

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

This is easy to verify: (AB)(B−1A−1)=A(BB−1)A−1=AIA−1=AA−1=I(AB)(B^{-1}A^{-1}) = A(BB^{-1})A^{-1} = AIA^{-1} = AA^{-1} = I, and similarly (B−1A−1)(AB)=I(B^{-1}A^{-1})(AB) = I. So B−1A−1B^{-1}A^{-1} is indeed the inverse of ABAB.

Using the extra condition

The question adds that AA and BB commute, i.e. AB=BAAB = BA. When two invertible matrices commute, their inverses commute too. To see this, take inverses of both sides of AB=BAAB = BA:

(AB)−1=(BA)−1  ⇒  B−1A−1=A−1B−1.(AB)^{-1} = (BA)^{-1} \;\Rightarrow\; B^{-1}A^{-1} = A^{-1}B^{-1}.

So under the commuting condition, B−1A−1B^{-1}A^{-1} and A−1B−1A^{-1}B^{-1} are the same matrix.

Conclusion …

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